<?xml version="1.0" encoding="UTF-8"?>
<article xmlns="http://specifications.silverchair.com/xsd/1/21/SCJATS-journalpublishing.xsd" xml:lang="en" article-type="research-article" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://specifications.silverchair.com/xsd/1/21/SCJATS-journalpublishing.xsd 1/21/SCJATS-journalpublishing.xsd" xmlns:xlink="http://www.w3.org/1999/xlink">
<front>
<journal-meta>
<journal-id journal-id-type="publisher-id">ptep</journal-id>
<journal-title-group>
<journal-title>Progress of Theoretical and Experimental Physics</journal-title>
</journal-title-group>
<issn pub-type="epub">2050-3911</issn>
<publisher>
<publisher-name>Oxford University Press</publisher-name>
</publisher>
</journal-meta>
<article-meta>
<article-id pub-id-type="doi">10.1093/ptep/ptz160</article-id>
<article-id pub-id-type="publisher-id">ptz160</article-id>
<article-id pub-id-type="arxiv">arXiv:1811.07760</article-id>
<article-categories>
<subj-group subj-group-type="category-toc-heading">
<subject>Papers</subject>
<subj-group subj-group-type="category-toc-heading">
<subject>Theoretical Particle Physics</subject>
</subj-group>
</subj-group>
<subj-group subj-group-type="category-journal-collection">
<subject>JEL/B30</subject>
</subj-group>
</article-categories>
<title-group>
<article-title>A novel approach to the computation of one-loop three- and four-point functions. III. The infrared divergent case</article-title>
</title-group>
<contrib-group>
<contrib contrib-type="author" corresp="yes">
<name><surname>Guillet</surname> <given-names>J Ph.</given-names></name>
<xref ref-type="aff" rid="AFF1"/>
<xref ref-type="corresp" rid="ptz160-cor1"/>
<email xlink:type="simple">guillet@lapth.cnrs.fr</email>
</contrib>
<contrib contrib-type="author" corresp="yes">
<name><surname>Pilon</surname> <given-names>E</given-names></name>
<xref ref-type="aff" rid="AFF1"/>
<xref ref-type="corresp" rid="ptz160-cor1"/>
<email xlink:type="simple">pilon@lapth.cnrs.fr</email>
</contrib>
<contrib contrib-type="author">
<name><surname>Shimizu</surname> <given-names>Y</given-names></name>
<xref ref-type="fn" rid="FM1"/>
<xref ref-type="aff" rid="AFF2"/>
</contrib>
<contrib contrib-type="author" corresp="yes">
<name><surname>Zidi</surname> <given-names>M S</given-names></name>
<xref ref-type="aff" rid="AFF3"/>
<xref ref-type="corresp" rid="ptz160-cor1"/>
<email xlink:type="simple">zidi@lapth.cnrs.fr</email>
</contrib>
</contrib-group>
<aff id="AFF1"><institution>Universit&#x00E9; Grenoble Alpes, Universit&#x00E9; Savoie Mont Blanc</institution>, CNRS, LAPTH, F-74000 Annecy, <country country="FR">France</country></aff>
<aff id="AFF2"><institution>KEK, Oho 1-1, Tsukuba</institution>, Ibaraki 305-0801, <country country="JP">Japan</country></aff>
<aff id="AFF3"><institution>LPTh, Universit&#x00E9; de Jijel</institution>, B.P. 98 Ouled-Aissa, 18000 Jijel, Alg&#x00E9;rie</aff>
<author-notes>
<corresp id="ptz160-cor1">E-mail: <email>guillet@lapth.cnrs.fr</email>; <email>pilon@lapth.cnrs.fr</email>; <email>zidi@lapth.cnrs.fr</email></corresp>
<fn id="FM1"><p>Y. Shimizu passed away during the completion of this series of articles</p></fn>
</author-notes>
<pub-date pub-type="cover">
<month>02</month>
<year>2020</year>
</pub-date>
<pub-date pub-type="collection">
<day>01</day>
<month>02</month>
<year>2020</year>
</pub-date>
<pub-date pub-type="epub" iso-8601-date="2020-02-22">
<day>22</day>
<month>02</month>
<year>2020</year>
</pub-date>
<volume>2020</volume>
<issue>2</issue>
<elocation-id>023B05</elocation-id>
<history>
<date date-type="received">
<day>05</day>
<month>06</month>
<year>2019</year>
</date>
<date date-type="rev-recd">
<day>19</day>
<month>09</month>
<year>2019</year>
</date>
<date date-type="accepted">
<day>23</day>
<month>09</month>
<year>2019</year>
</date>
</history>
<permissions>
<copyright-statement>&#x00A9; The Author(s) 2020. Published by Oxford University Press on behalf of the Physical Society of Japan.</copyright-statement>
<copyright-year>2020</copyright-year>
<license license-type="cc-by" xlink:href="http://creativecommons.org/licenses/by/4.0/">
<license-p>This is an Open Access article distributed under the terms of the Creative Commons Attribution License (<ext-link xmlns:xlink="http://creativecommons.org/licenses/by/4.0/">http://creativecommons.org/licenses/by/4.0/</ext-link>), which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.</license-p>
<license-p>Funded by SCOAP<sup>3</sup></license-p>
</license>
</permissions>
<self-uri xlink:href="ptz160.pdf"/>
<abstract abstract-type="abstract">
<title>Abstract</title>
<p>This article is the third and last of a series presenting an alternative method for computing the one-loop scalar integrals. It extends the results of the first two articles to the infrared divergent case. This novel method enjoys a couple of interesting features as compared with the methods found in the literature. It directly proceeds in terms of the quantities driving algebraic reduction methods. It yields a simple decision tree based on the vanishing of internal masses and one-pinched kinematic matrices, which avoids a profusion of cases. Lastly, it extends to kinematics more general than the physical, e.g. collider processes, relevant at one loop. This last feature may be useful when considering the application of this method beyond one loop using generalized one-loop integrals as building blocks.</p>
</abstract>
<kwd-group kwd-group-type="jel">
<kwd>B30</kwd>
<kwd>B57</kwd>
</kwd-group>

<counts>
<page-count count="63"/>
</counts>
</article-meta>
</front>
<body>
<sec id="SEC1"><title>1. Introduction</title>
<p>This article is the third of a triptych. The first one [<xref ref-type="bibr" rid="B1">1</xref>] presented a method exploiting a Stokes-type identity to compute &#x201C;generalized&#x201D; (in the sense of the underlying kinematics) one-loop three- and four-point scalar integrals for the real mass case. The second article [<xref ref-type="bibr" rid="B2">2</xref>] extended the results of the first paper to the case of general complex masses. The present article widens the results of Refs. [<xref ref-type="bibr" rid="B1">1</xref>] and [<xref ref-type="bibr" rid="B2">2</xref>] to the case where some internal masses are vanishing, leading to infrared divergences. We refer the reader to Ref. [<xref ref-type="bibr" rid="B3">3</xref>] for more details on the motivation of this work.</p>
<p>The scalar Feynman integrals for one-loop three- and four-point functions are all known and have been compiled in a useful article [<xref ref-type="bibr" rid="B4">4</xref>]. This article relies mainly on the results of other publications, especially the important work of Beenakker and Denner [<xref ref-type="bibr" rid="B5">5</xref>]. We should also mention Ref. [<xref ref-type="bibr" rid="B6">6</xref>], which provides a complete set of results for soft and/or collinear divergent four-point functions using different kinds of IR regulators. The purpose of the present article is to extend these results for more general kinematics beyond those relevant for collider processes at the one-loop order. Note that despite the fact that some internal masses may vanish, the others can be real or complex and we treat both cases in this article. The soft and collinear divergences are dealt with using dimensional regularization, <inline-formula><tex-math notation="LaTeX" id="ImEquation1"><![CDATA[$n = 4 - 2 \, \varepsilon$]]></tex-math></inline-formula>, and doing an <inline-formula><tex-math notation="LaTeX" id="ImEquation2"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> expansion.</p>
<p>The outline of this article follows closely that of the preceding articles in Refs. [<xref ref-type="bibr" rid="B1">1</xref>] and [<xref ref-type="bibr" rid="B2">2</xref>]. We start by considering the three-point function <inline-formula><tex-math notation="LaTeX" id="ImEquation3"><![CDATA[$I_{3}^{n}$]]></tex-math></inline-formula> in a spacetime dimension shifted by a small amount from <inline-formula><tex-math notation="LaTeX" id="ImEquation4"><![CDATA[$4$]]></tex-math></inline-formula> to <inline-formula><tex-math notation="LaTeX" id="ImEquation5"><![CDATA[$n$]]></tex-math></inline-formula>. The kinematics leading to infrared divergences are discussed as a warm-up for Sect. <xref ref-type="sec" rid="SEC3">3</xref>. We successively present two variants of the method. The simplest variant, labelled the &#x201C;direct way,&#x201D; is presented in Sect. <xref ref-type="sec" rid="SEC2.1">2.1</xref>. It is well suited for the three-point function, but cannot be extended to the case of the four-point function. Then, in Sect. <xref ref-type="sec" rid="SEC2.2">2.2</xref>, practical implementation of the results of the preceding subsection is discussed and some explicit examples are computed and compared to Ref. [<xref ref-type="bibr" rid="B4">4</xref>]. In Sect. <xref ref-type="sec" rid="SEC2.3">2.3</xref> we present an alternative coined the &#x201C;indirect way,&#x201D; easily applicable to the four-point case which is the subject of Sect. <xref ref-type="sec" rid="SEC3">3</xref>.</p>
<p>We first explain, in Sect. <xref ref-type="sec" rid="SEC3.1">3.1</xref>, how to extend the calculation of <inline-formula><tex-math notation="LaTeX" id="ImEquation6"><![CDATA[$I_4^4$]]></tex-math></inline-formula> developed in Ref. [<xref ref-type="bibr" rid="B1">1</xref>] to the case where the infrared divergences are regulated in <inline-formula><tex-math notation="LaTeX" id="ImEquation7"><![CDATA[$n$]]></tex-math></inline-formula> dimensions. The net result is that the four-point scalar integral can be decomposed on sectors labelled by three indices, and a three-dimensional integral over the first octant of <inline-formula><tex-math notation="LaTeX" id="ImEquation8"><![CDATA[$\mathbb{R}^3$]]></tex-math></inline-formula> is associated to each sector. Then, two cases are distinguished depending on the sectors. In the first case, presented in Sect. <xref ref-type="sec" rid="SEC3.2">3.2</xref>, the determinant of the one-pinched kinematical matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation9"><![CDATA[${\cal S}$]]></tex-math></inline-formula> vanishes but not the internal mass associated with this sector: this case is met when a soft divergence appears. In the second case, presented in Sect. <xref ref-type="sec" rid="SEC3.3">3.3</xref>, both the determinant of the one-pinched <inline-formula><tex-math notation="LaTeX" id="ImEquation10"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix and the internal mass associated with this sector vanish: this case is met when a collinear or a soft and collinear divergence shows up. In Sect. <xref ref-type="sec" rid="SEC3.4">3.4</xref>, the infrared divergent part of the scalar four-point integral is shown to be proportional to a three-point scalar integral, as it should be. Some explicit examples are given and compared to the results found in the literature. We then conclude.</p>
<p>Various appendices gather a number of utilities removed from the main text to facilitate its reading. Accordingly, in Appendix <xref ref-type="sec" rid="SEC6">A</xref> we complete Appendix D of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] and Appendix A of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] by giving a missing case where the power of the integration variable is not an integer as required by dimensional regularization. Appendix <xref ref-type="sec" rid="SEC7">B</xref> shows how to compute an integral appearing in the three-point case in closed form. Appendix <xref ref-type="sec" rid="SEC8">C</xref> collects a bunch of integrals required to compute the three- and four-point functions having soft and/or collinear divergences in the case of general complex masses. Then, Appendix <xref ref-type="sec" rid="SEC9">D</xref> goes through the examples given in Sect. <xref ref-type="sec" rid="SEC2.2">2.2</xref> and explains in detail how the results obtained in the latter subsection can be found again from those derived in the &#x201C;indirect way&#x201D; case. Appendix <xref ref-type="sec" rid="SEC10">E</xref> provides the way to compute the last integration in closed form in terms of dilogarithms for the case of infrared divergent integrals. It complements Appendix E of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] and Appendix B of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]. Lastly, Appendix <xref ref-type="sec" rid="SEC11">F</xref> proves a tricky point used in Sect. <xref ref-type="sec" rid="SEC3">3</xref>: for the real mass case, the sign of the vanishing imaginary part of the denominator in the last integral can be safely changed.</p>
</sec>
<sec id="SEC2"><title>2. Three-point function with infrared divergences</title>
<p>When some internal masses vanish, divergences of collinear or soft origin appear and the approach will be revisited. We regularize these divergences using dimensional regularization, shifting the dimension of the spacetime by a small positive amount from 4 to <inline-formula><tex-math notation="LaTeX" id="ImEquation11"><![CDATA[$n = 4 - 2 \, \varepsilon$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation12"><![CDATA[$\varepsilon < 0$]]></tex-math></inline-formula>. After performing the loop momentum integral, instead of Eq. (2.3) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] we get<sup><xref ref-type="fn" rid="FN1">1</xref></sup>
<disp-formula id="ptz160M2-1"><label>(2.1)</label><tex-math notation="LaTeX" id="Equation1"><![CDATA[$$\begin{equation}
I_3^n
=
- \Gamma(1+\varepsilon) \,
\int_{0}^{1} \prod_{i=1}^3 \, dz_i \,
\delta\bigg(1-\sum_{i=1}^3 z_i\bigg)
\left(
- \, \frac{1}{2} \, z^{\rm T} \cdot {\cal S} \cdot z - i \, \lambda
\right) ^{-1-\varepsilon} .
\label{eqdefi3n}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>To appropriately shift the power of the denominator in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-1">2.1</xref>) so as to apply the Stokes identity in Eq. (1.2) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] as we did in the massive case, we use the following modified integral representation instead of the identity in Eq. (2.9) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] (cf. Appendix B of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]):
<disp-formula id="ptz160UM1"><tex-math notation="LaTeX" id="Equation2"><![CDATA[$$\frac{1}{D^{1+\varepsilon}}
=
\frac{\nu}{B(2-1/\nu,1/\nu)} \;
\int^{+\infty}_{0} \, \frac{d \xi}{(D+\xi^{\nu})^2} ,$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation13"><![CDATA[$\nu = 1/(1-\varepsilon)$]]></tex-math></inline-formula>. Instead of Eq. (2.10) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] we now get
<disp-formula id="ptz160M2-2"><label>(2.2)</label><tex-math notation="LaTeX" id="Equation3"><![CDATA[$$\begin{align}
I_3^n
&= - \, 2^{1+\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{1-\varepsilon} \,
\frac{1}{B(1+\varepsilon,1-\varepsilon)} \,
\int^{+\infty}_0 d \xi \, \int_{\Sigma_{bc}}
\frac{dx_b \, dx_c}{(D^{(a)}(x_b,x_c) + \xi^{\nu} - i \, \lambda)^2} .
\label{eqdefi3n1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We otherwise proceed as in Sect. <xref ref-type="sec" rid="SEC2.1">2.1</xref> of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]. The counterpart of Eq. (2.22) in Ref. [<xref ref-type="bibr" rid="B1">1</xref>] now reads
<disp-formula id="ptz160M2-3"><label>(2.3)</label><tex-math notation="LaTeX" id="Equation4"><![CDATA[$$\begin{align}
I_3^n
&= 2^{\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{1-\varepsilon} \,
\frac{1}{B(1+\varepsilon,1-\varepsilon)} \,
\notag\\
&\quad {} \times
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G)} \,
\int^{+\infty}_0 \frac{d \xi}{\Delta_2 - \xi^{\nu}+ i \, \lambda} \,
\int^1_0 \, \frac{dx}{D^{\{i\}(j)}(x)+ \xi^{\nu} - i \, \lambda} ,
\label{eqdefi3n2}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation14"><![CDATA[$j \in S_3 \setminus \{i\}$]]></tex-math></inline-formula> (<inline-formula><tex-math notation="LaTeX" id="ImEquation15"><![CDATA[$S_3 = \{1,2,3\}$]]></tex-math></inline-formula>). More precisely, we assume that <inline-formula><tex-math notation="LaTeX" id="ImEquation16"><![CDATA[$j$]]></tex-math></inline-formula> is chosen to be <inline-formula><tex-math notation="LaTeX" id="ImEquation17"><![CDATA[$1 + (i \; \mbox{modulo} \; 3)$]]></tex-math></inline-formula>. Similarly to what we did for the three-point function in the massive case, one can also consider both a &#x201C;direct way&#x201D; and an &#x201C;indirect way&#x201D; in the IR case. We first focus on the &#x201C;direct way,&#x201D; which provides a more straightforward and synthetic discussion of the various cases at hand. We then illustrate how these cases are involved in a few examples. The &#x201C;indirect way&#x201D; instead leads to a cumbersome split-up discussion. Notwithstanding this, the calculation of the four-point one-loop integral relying on the approach described in this article proceeds along the &#x201C;indirect way&#x201D; as we found no extension of the &#x201C;direct way&#x201D; approach in this case. In Refs. [<xref ref-type="bibr" rid="B7">7</xref>,<xref ref-type="bibr" rid="B8">8</xref>] it was shown on general grounds using the decomposition<sup><xref ref-type="fn" rid="FN2">2</xref></sup>
<disp-formula id="ptz160M2-4"><label>(2.4)</label><tex-math notation="LaTeX" id="Equation5"><![CDATA[$$\begin{equation}\label{decomp-golem}
\det{({\cal S})} \, I_4^n({\cal S})
=
\sum_{i=1}^{4} \overline{b}_{i} \, I_3^n({\cal S}^{\{i\}}) - \det{(G)} \, (1 - 2 \, \varepsilon) \, I_4^{n+2}({\cal S})
\end{equation}$$]]></tex-math></disp-formula>
that the infrared structure of any IR divergent four-point one-loop integral is carried by IR-divergent three-point one-loop functions resulting from appropriate iterated pinchings. Therefore the comparison of the IR structures on both sides of Eq. (<xref ref-type="disp-formula" rid="ptz160M2-4">2.4</xref>) proceeds most conveniently via a term-by-term comparison using the three-point one-loop functions decomposed according to the &#x201C;indirect way&#x201D; as well. In anticipation, we hereby give the key ingredients to perform this comparison, as well as the general recombination of these &#x201C;indirect way&#x201D; ingredients into the more compact expression obtained from the &#x201C;direct way,&#x201D; thereby checking their equivalence. The extensive collection of expressions computed in closed form which enable us to perform detailed case-by-case comparisons is gathered in Appendix <xref ref-type="sec" rid="SEC9">D</xref> to lighten the presentation.</p>
<sec id="SEC2.1"><title>2.1. Direct way</title>
<p>Soft and/or collinear divergences are caused by some vanishing masses which make <inline-formula><tex-math notation="LaTeX" id="ImEquation18"><![CDATA[$\det{({\cal S})}$]]></tex-math></inline-formula> vanish so that <inline-formula><tex-math notation="LaTeX" id="ImEquation19"><![CDATA[$\Delta_2 = 0$]]></tex-math></inline-formula>, whereas the other internal masses may or may not vanish as well, and may even be complex. We will keep the <inline-formula><tex-math notation="LaTeX" id="ImEquation20"><![CDATA[$- \, i \, \lambda$]]></tex-math></inline-formula> prescription, bearing in mind that it is ineffective in the case of complex masses.</p>
<p>Starting from Eq. (<xref ref-type="disp-formula" rid="ptz160M2-3">2.3</xref>) and performing the <inline-formula><tex-math notation="LaTeX" id="ImEquation21"><![CDATA[$\xi$]]></tex-math></inline-formula> integration using Eq. (<xref ref-type="disp-formula" rid="ptz160M6-4">A.4</xref>), we end up with
<disp-formula id="ptz160M2-5"><label>(2.5)</label><tex-math notation="LaTeX" id="Equation6"><![CDATA[$$\begin{align}
I_3^n
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G)} \,
\int^1_0 dx \,
\big( D^{\{i\}(j)}(x) - i \, \lambda \big)^{-1-\varepsilon} .
\label{eqdirei3n1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the general case <inline-formula><tex-math notation="LaTeX" id="ImEquation22"><![CDATA[$D^{\{i\}(j)}(x)$]]></tex-math></inline-formula> depends on two internal masses squared, <inline-formula><tex-math notation="LaTeX" id="ImEquation23"><![CDATA[$m_j^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation24"><![CDATA[$m_k^2$]]></tex-math></inline-formula>, such that <inline-formula><tex-math notation="LaTeX" id="ImEquation25"><![CDATA[$m_j^2 = D^{\{i\}(j)}(0)/2 = \widetilde{D}_{ik}/2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation26"><![CDATA[$m_k^2 = D^{\{i\}(j)}(1)/2 = \widetilde{D}_{ij}/2$]]></tex-math></inline-formula>, cf. Sect. 2 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]. We introduced the label <inline-formula><tex-math notation="LaTeX" id="ImEquation27"><![CDATA[$k$]]></tex-math></inline-formula>, which is the only element of the complement of <inline-formula><tex-math notation="LaTeX" id="ImEquation28"><![CDATA[$\{i,j\}$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation29"><![CDATA[$S_3$]]></tex-math></inline-formula>. With our assumption on <inline-formula><tex-math notation="LaTeX" id="ImEquation30"><![CDATA[$j$]]></tex-math></inline-formula>, this implies that <inline-formula><tex-math notation="LaTeX" id="ImEquation31"><![CDATA[$k \equiv 1 + ((i+1) \; \mbox {modulo} \;3)$]]></tex-math></inline-formula>. Let us focus on the function <inline-formula><tex-math notation="LaTeX" id="ImEquation32"><![CDATA[$W$]]></tex-math></inline-formula> given by
<disp-formula id="ptz160M2-6"><label>(2.6)</label><tex-math notation="LaTeX" id="Equation7"><![CDATA[$$\begin{equation}
W\big(\det{(G^{\{i\}})},\widetilde{D}_{ij}, \widetilde{D}_{ik}\big) = \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \, \int^1_0 dx \,
\big( D^{\{i\}(j)}(x) - i \, \lambda \big)^{-1-\varepsilon} .
\label{eqdefwi0}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>Remember that (cf. Eqs. (2.16), (2.17), and (2.18) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>])
<disp-formula id="ptz160M2-7"><label>(2.7)</label><tex-math notation="LaTeX" id="Equation8"><![CDATA[$$\begin{equation}
D^{\{i\}(j)}(x)
=
G^{\{i\}(j)} \, x^2 - 2 \, V^{\{i\}(j)} \, x - C^{\{i\}(j)}
\label{eqremd1}
\end{equation}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M2-8"><label>(2.8)</label><tex-math notation="LaTeX" id="Equation9"><![CDATA[$$\begin{align}
G^{\{i\}(j)} &= - {\cal S}_{kk} + 2 \, {\cal S}_{kj} - {\cal S}_{jj} = \det{(G^{\{i\}})} , \notag \\
V^{\{i\}(j)} &= {\cal S}_{kj} - {\cal S}_{jj} = \frac{1}{2} \big[ \det{(G^{\{i\}})} - \widetilde{D}_{ij} + \widetilde{D}_{ik} \big] , \label{eqremd2}\\
C^{\{i\}(j)} &= {\cal S}_{jj} = - \widetilde{D}_{ik} . \notag
\end{align}$$]]></tex-math></disp-formula></p>
<p>The Gram matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation33"><![CDATA[$G^{\{i\}(j)}$]]></tex-math></inline-formula> is built from the one-pinched <inline-formula><tex-math notation="LaTeX" id="ImEquation34"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation35"><![CDATA[${\cal S}^{\{i\}}$]]></tex-math></inline-formula>, and is a real matrix which depends only on a squared external momentum in the three-point case. Notice that in this case <inline-formula><tex-math notation="LaTeX" id="ImEquation36"><![CDATA[$G^{\{i\}(j)}$]]></tex-math></inline-formula> is a <inline-formula><tex-math notation="LaTeX" id="ImEquation37"><![CDATA[$1 \times 1$]]></tex-math></inline-formula> matrix and <inline-formula><tex-math notation="LaTeX" id="ImEquation38"><![CDATA[$V^{\{i\}(j)}$]]></tex-math></inline-formula> a one-dimensional vector this explains the notation<sup><xref ref-type="fn" rid="FN3">3</xref></sup> used in Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-7">2.7</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-8">2.8</xref>). Knowledge of <inline-formula><tex-math notation="LaTeX" id="ImEquation39"><![CDATA[$\det{(G^{\{i\}})}$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation40"><![CDATA[$\widetilde{D}_{ij}$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation41"><![CDATA[$\widetilde{D}_{ik}$]]></tex-math></inline-formula> fully determines the polynomial <inline-formula><tex-math notation="LaTeX" id="ImEquation42"><![CDATA[$D^{\{i\}(j)}(x)$]]></tex-math></inline-formula>. These two internal masses may or may not vanish, and hence three cases must be considered.</p>
<sec id="SEC2.1.1"><title>(a) Neither <inline-formula><tex-math notation="LaTeX" id="ImEquation43"><![CDATA[$m_j^2$]]></tex-math></inline-formula> nor <inline-formula><tex-math notation="LaTeX" id="ImEquation44"><![CDATA[$m_k^2$]]></tex-math></inline-formula> vanishes</title>
<p>We perform a Taylor expansion<sup><xref ref-type="fn" rid="FN4">4</xref></sup> of <inline-formula><tex-math notation="LaTeX" id="ImEquation45"><![CDATA[$W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, \widetilde{D}_{ik})$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation46"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>:
<disp-formula id="ptz160M2-9"><label>(2.9)</label><tex-math notation="LaTeX" id="Equation10"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, \widetilde{D}_{ik}) \notag \\
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \, \left[ \int^1_0 \frac{dx}{D^{\{i\}(j)}(x) - i \, \lambda}
-
\varepsilon
\int^1_0 dx
\frac{\ln \left( D^{\{i\}(j)}(x) - i \, \lambda \right)}
{D^{\{i\}(j)}(x) - i \, \lambda} \right]\!.
\label{eqdirei3n2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Note that <inline-formula><tex-math notation="LaTeX" id="ImEquation47"><![CDATA[$x_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation48"><![CDATA[$x_2$]]></tex-math></inline-formula>, the two roots of <inline-formula><tex-math notation="LaTeX" id="ImEquation49"><![CDATA[$D^{\{i\}(j)}(x) - i \lambda$]]></tex-math></inline-formula>, are given by [cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-7">2.7</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-8">2.8</xref>)]
<disp-formula id="ptz160M2-10"><label>(2.10)</label><tex-math notation="LaTeX" id="Equation11"><![CDATA[$$\begin{align}
x_{\underset{2}{1}}
&=
\frac{
\det{(G^{\{i\}})} - \widetilde{D}_{ij} + \widetilde{D}_{ik}
\pm
\sqrt{ {\cal K}\big( \det{(G^{\{i\}})},\widetilde{D}_{ij},\widetilde{D}_{ik} \big) + i \, \lambda \, S_G}
}{2 \, \det{(G^{\{i\}})}} ,
\label{eqroot12}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation50"><![CDATA[${\cal K}$]]></tex-math></inline-formula> is the K&#x00E4;ll&#x00E9;n function,
<disp-formula id="ptz160M2-11"><label>(2.11)</label><tex-math notation="LaTeX" id="Equation12"><![CDATA[$$\begin{equation}
{\cal K}(x,y,z) = x^2 + y^2 + z^2 - 2 \, x \, y - 2 \, x \, z - 2 \, y \, z ,
\label{eqkallenfunc}
\end{equation}$$]]></tex-math></disp-formula>
and <inline-formula><tex-math notation="LaTeX" id="ImEquation51"><![CDATA[$S_G = \mbox{sign}(\det{(G^{\{i\}})})$]]></tex-math></inline-formula>. Then, we introduce <inline-formula><tex-math notation="LaTeX" id="ImEquation52"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation53"><![CDATA[$K(x_1,x_2)$]]></tex-math></inline-formula> defined by
<disp-formula id="ptz160M2-12"><label>(2.12)</label><tex-math notation="LaTeX" id="Equation13"><![CDATA[$$\begin{align}
K(x_1,x_2)
&= \int^1_0 dx \, \frac{1}{(x - x_1) \, (x - x_2)} ,
\label{eqcompk1} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M2-13"><label>(2.13)</label><tex-math notation="LaTeX" id="Equation14"><![CDATA[$$\begin{align}
J(x_1,x_2)
&= \int^1_0 dx \,
\frac{\ln\left( (x-x_1) \, (x-x_2) \right)}{(x - x_1) \, (x - x_2)} .
\label{eqcompj1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As will become clear in the forthcoming paragraph on the origin of infrared singularities, only the case with <inline-formula><tex-math notation="LaTeX" id="ImEquation54"><![CDATA[$m_j^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation55"><![CDATA[$m_k^2$]]></tex-math></inline-formula> both real matters in practice, which makes the explicit calculation of <inline-formula><tex-math notation="LaTeX" id="ImEquation56"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula> somewhat simpler.<sup><xref ref-type="fn" rid="FN5">5</xref></sup> The latter is provided in Appendix <xref ref-type="sec" rid="SEC7">B</xref>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation57"><![CDATA[$x$]]></tex-math></inline-formula> integration in the function <inline-formula><tex-math notation="LaTeX" id="ImEquation58"><![CDATA[$K(x_1,x_2)$]]></tex-math></inline-formula> straightforwardly gives
<disp-formula id="ptz160M2-14"><label>(2.14)</label><tex-math notation="LaTeX" id="Equation15"><![CDATA[$$\begin{align}
K(x_1,x_2)
&= \frac{1}{x_1-x_2} \,
\left[
\ln \left( \frac{x_1-1}{x_1} \right)
-
\ln \left( \frac{x_2-1}{x_2} \right)
\right]\!.
\label{eqcompk11}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Thus, <inline-formula><tex-math notation="LaTeX" id="ImEquation59"><![CDATA[$W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, \widetilde{D}_{ik})$]]></tex-math></inline-formula> reads
<disp-formula id="ptz160M2-15"><label>(2.15)</label><tex-math notation="LaTeX" id="Equation16"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, \widetilde{D}_{ik}) \notag \\
&=
\frac{1}{\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{\det{(G^{\{i\}})}}
\left\{
\left[
1 - \varepsilon \, \ln \left(\frac{\det{(G^{\{i\}})}}{2} - i \, \lambda \right)
\right] \, K(x_1,x_2)
- \varepsilon \, J(x_1,x_2)
\right\}\!.
\label{eqcompintlog1}
\end{align}$$]]></tex-math></disp-formula></p>
</sec>
<sec id="SEC2.1.2"><title>(b) One and only one of <inline-formula><tex-math notation="LaTeX" id="ImEquation60"><![CDATA[$m_j^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation61"><![CDATA[$m_k^2$]]></tex-math></inline-formula> vanishes</title>
<p>Let us assume that the vanishing internal mass is <inline-formula><tex-math notation="LaTeX" id="ImEquation62"><![CDATA[$m_j^2$]]></tex-math></inline-formula>. <inline-formula><tex-math notation="LaTeX" id="ImEquation63"><![CDATA[$D^{\{i\}(j)}(x)$]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M2-16"><label>(2.16)</label><tex-math notation="LaTeX" id="Equation17"><![CDATA[$$\begin{equation}
D^{\{i\}(j)}(x)
= x \, \big( G^{\{i\}(j)} \, x - 2 \, V^{\{i\}(j)} \big) .
\label{eqnewquad1}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>From Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-6">2.6</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-8">2.8</xref>), <inline-formula><tex-math notation="LaTeX" id="ImEquation64"><![CDATA[$W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)$]]></tex-math></inline-formula> is thus of the form
<disp-formula id="ptz160M2-17"><label>(2.17)</label><tex-math notation="LaTeX" id="Equation18"><![CDATA[$$\begin{align}
W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \, \int^1_0 dx \, x^{-1-\varepsilon} \, (a \, x + z)^{-1-\varepsilon} ,
\label{eqdefyint1}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation65"><![CDATA[$a = \det{(G^{\{i\}})}$]]></tex-math></inline-formula> is real and <inline-formula><tex-math notation="LaTeX" id="ImEquation66"><![CDATA[$z = - \det{(G^{\{i\}})} + \widetilde{D}_{ij} - i \, \lambda$]]></tex-math></inline-formula> is complex. As <inline-formula><tex-math notation="LaTeX" id="ImEquation67"><![CDATA[$z$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation68"><![CDATA[$a \, x + z$]]></tex-math></inline-formula> have imaginary parts of the same sign, <inline-formula><tex-math notation="LaTeX" id="ImEquation69"><![CDATA[$(a \, x + z)^{-1-\varepsilon}$]]></tex-math></inline-formula> can be split as follows:
<disp-formula id="ptz160UM2"><tex-math notation="LaTeX" id="Equation19"><![CDATA[$$(a \, x + z)^{-1-\varepsilon}
=
z^{-1-\varepsilon} \, \left( 1 + \frac{a}{z} \, x \right)^{-1-\varepsilon} .$$]]></tex-math></disp-formula></p>
<p>The right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M2-17">2.17</xref>) involves the Gauss hypergeometric function <inline-formula><tex-math notation="LaTeX" id="ImEquation70"><![CDATA[$_{2}F_{1}$]]></tex-math></inline-formula>:
<disp-formula id="ptz160M2-18"><label>(2.18)</label><tex-math notation="LaTeX" id="Equation20"><![CDATA[$$\begin{equation}
W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0) =
- \, \frac{2^{\varepsilon}}{\varepsilon^2} \, \Gamma(1+\varepsilon) \, z^{-1-\varepsilon} \,
_{2}F_{1}
\left( 1+\varepsilon,-\varepsilon;1-\varepsilon;- \, \frac{a}{z} \right)\!.
\label{eqdefyint2}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>We use the identity [<xref ref-type="bibr" rid="B13">13</xref>]
<disp-formula id="ptz160UM3"><tex-math notation="LaTeX" id="Equation21"><![CDATA[$$\begin{align*}
_{2}F_{1}(a,b;c;w) &= \frac{\Gamma(c) \, \Gamma(c-a-b)}{\Gamma(c-a) \, \Gamma(c-b)} \; _{2}F_{1}(a,b;a+b-c+1;1-w) \\
&\quad {} + (1-w)^{c-a-b} \, \frac{\Gamma(c) \, \Gamma(a+b-c)}{\Gamma(a) \, \Gamma(b)} \; _{2}F_{1}(c-a,c-b;c-a-b+1;1-w)
\end{align*}$$]]></tex-math></disp-formula>
and the Pfaff identity
<disp-formula id="ptz160UM4"><tex-math notation="LaTeX" id="Equation22"><![CDATA[$$_{2}F_{1}(a,b;c;w)
=
(1-w)^{-b} \, _{2}F_{1} \bigg( c-a,b;c;\frac{w}{w-1} \bigg)$$]]></tex-math></disp-formula>
to rewrite
<disp-formula id="ptz160M2-19"><label>(2.19)</label><tex-math notation="LaTeX" id="Equation23"><![CDATA[$$\begin{align}
&_{2}F_{1}
\left( 1+\varepsilon,-\varepsilon;1-\varepsilon;- \, \frac{a}{z} \right)
\notag\\
&=
2 \, \frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\left( - \frac{a}{z} \right)^{\varepsilon}
- \left( \frac{a+z}{z} \right)^{-\varepsilon} \,
\left( - \frac{a}{z} \right)^{2 \, \varepsilon} \,
_{2}F_{1}
\left( - 2 \, \varepsilon, - \varepsilon; 1-\varepsilon; \frac{a+z}{a} \right)\!.
\label{eqsplit2F1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Performing a Taylor expansion in <inline-formula><tex-math notation="LaTeX" id="ImEquation71"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> we get
<disp-formula id="ptz160UM5"><tex-math notation="LaTeX" id="Equation24"><![CDATA[$$_{2}F_{1}
\left( - 2 \, \varepsilon, - \, \varepsilon ;1 - \varepsilon; \tau \right)
=
1 + 2 \, \varepsilon^2 \, \mbox{Li}_2(\tau) ,$$]]></tex-math></disp-formula>
and splitting <inline-formula><tex-math notation="LaTeX" id="ImEquation72"><![CDATA[$\ln( (a+z)/z) = \ln(a+z) - \ln(z)$]]></tex-math></inline-formula>, we rewrite <inline-formula><tex-math notation="LaTeX" id="ImEquation73"><![CDATA[$W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)$]]></tex-math></inline-formula> as
<disp-formula id="ptz160M2-20"><label>(2.20)</label><tex-math notation="LaTeX" id="Equation25"><![CDATA[$$\begin{align}
W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)
&= \frac{2^{\varepsilon}}{\varepsilon^2} \, \Gamma(1+\varepsilon) \, \frac{1}{z} \,
\left\{
\left( a+z \right)^{-\varepsilon} \,
\left( - \frac{a}{z} \right)^{2 \, \varepsilon} \,
\left[ 1 + 2 \, \varepsilon^2 \, \mbox{Li}_2 \left( \frac{a+z}{a} \right) \right]
\right.
\notag \\
&\qquad \qquad \qquad \qquad {}
-
\left.
2 \; \frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\left( z \right)^{-\varepsilon} \,
\left( - \frac{a}{z} \right)^{\varepsilon}
\right\}\!.
\label{eqdefyint3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Making explicit that <inline-formula><tex-math notation="LaTeX" id="ImEquation74"><![CDATA[$z = - \det{(G^{\{i\}})} + \widetilde{D}_{ij} - i \, \lambda$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation75"><![CDATA[$a+z = \widetilde{D}_{ij} - i \, \lambda$]]></tex-math></inline-formula>, we get
<disp-formula id="ptz160M2-21"><label>(2.21)</label><tex-math notation="LaTeX" id="Equation26"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0 ) \notag \\
&=
- \frac{2^{\varepsilon}}{\varepsilon^2} \, \Gamma(1+\varepsilon) \; \frac{1}{\det{(G^{\{i\}})} - \widetilde{D}_{ij}} \notag \\
&\quad {} \times
\left\{
\left( \frac{\det{(G^{\{i\}})}}{\det{(G^{\{i\}})} - \widetilde{D}_{ij} + i \, \lambda } \right)^{2 \, \varepsilon} \, \, \left(\widetilde{D}_{ij} - i \, \lambda \right)^{-\varepsilon} \, \left[ 1 + 2 \, \varepsilon^2 \, \mbox{Li}_2 \left( \frac{\widetilde{D}_{ij} - i \, \lambda}{\det{(G^{\{i\}})}} \right) \right]
\right. \notag \\
&\qquad \qquad {} - \left.
2 \, \frac{\Gamma^2(1-\varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \, \left( \frac{\det{(G^{\{i\}})}}{\det{(G^{\{i\}})} - \widetilde{D}_{ij} + i \, \lambda } \right)^{\varepsilon} \, \left( \widetilde{D}_{ij} - \det{(G^{\{i\}})} - i \, \lambda \right)^{- \varepsilon}
\vphantom{\mbox{Li}_2 \left( \frac{- {\cal S}_{kk} - i \, \lambda}{G^{\{i\}(j)}_{kk}} \right)} \right\}\!.
\label{eqcompwcasb4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>This formula is manifestly well behaved as <inline-formula><tex-math notation="LaTeX" id="ImEquation76"><![CDATA[$\widetilde{D}_{ij} \rightarrow 0$]]></tex-math></inline-formula> (<inline-formula><tex-math notation="LaTeX" id="ImEquation77"><![CDATA[$m_k^2 \rightarrow 0$]]></tex-math></inline-formula>), yet it is not handy to expand around <inline-formula><tex-math notation="LaTeX" id="ImEquation78"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula>. A more practical alternative may be obtained as follows. First, we use the identities relating <inline-formula><tex-math notation="LaTeX" id="ImEquation79"><![CDATA[$\mbox{Li}_2(1-w)$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation80"><![CDATA[$\mbox{Li}_2(w)$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation81"><![CDATA[$\mbox{Li}_2(1/w)$]]></tex-math></inline-formula> to change the argument of the <inline-formula><tex-math notation="LaTeX" id="ImEquation82"><![CDATA[$\mbox{Li}_2$]]></tex-math></inline-formula> function, and the following relations:
<disp-formula id="ptz160M2-22"><label>(2.22)</label><tex-math notation="LaTeX" id="Equation27"><![CDATA[$$\begin{align}
\ln \left( \frac{a}{z} \right) &= \ln \left( - \frac{a}{z} \right) - i \, \pi \, S(a \, z) , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M2-23"><label>(2.23)</label><tex-math notation="LaTeX" id="Equation28"><![CDATA[$$\begin{align}
\ln \left( \frac{a+z}{a} \right) &= \ln \left( -\frac{a+z}{a} \right) + i \, \pi \, S(a \, z) , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M2-24"><label>(2.24)</label><tex-math notation="LaTeX" id="Equation29"><![CDATA[$$\begin{align}
\ln \left( -\frac{a+z}{a} \right) &= \ln \left( \frac{a+z}{z} \right) - \ln \left( - \frac{a}{z} \right)\!,
\label{eqrelalamormoilenoeud}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M2-25"><label>(2.25)</label><tex-math notation="LaTeX" id="Equation30"><![CDATA[$$\begin{equation}
S(z) = \mbox{sign}\left( \Im(z) \right)\!,
\label{eqdeffuncS0}
\end{equation}$$]]></tex-math></disp-formula>
so that the <inline-formula><tex-math notation="LaTeX" id="ImEquation83"><![CDATA[$\mbox{Li}_2$]]></tex-math></inline-formula> function can be rewritten as
<disp-formula id="ptz160M2-26"><label>(2.26)</label><tex-math notation="LaTeX" id="Equation31"><![CDATA[$$\begin{align}
\mbox{Li}_2 \left( \frac{a+z}{a} \right) &= \mbox{Li}_2 \left( - \frac{a}{z} \right) - \frac{\pi^2}{6} - \frac{1}{2} \, \ln^2 \left( - \frac{a}{z} \right) + \ln \left( \frac{a+z}{z} \right) \, \ln \left( - \frac{a}{z} \right)\!.
\label{eqchangdilog1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Secondly, we Taylor expand around <inline-formula><tex-math notation="LaTeX" id="ImEquation84"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> the <inline-formula><tex-math notation="LaTeX" id="ImEquation85"><![CDATA[$(-a/z)$]]></tex-math></inline-formula> terms in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-20">2.20</xref>). We thus get
<disp-formula id="ptz160M2-27"><label>(2.27)</label><tex-math notation="LaTeX" id="Equation32"><![CDATA[$$\begin{align}
W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)
&= -\,\frac{2^{\varepsilon}}{\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{z} \,
\left[
\frac{2}{\varepsilon} \, (z)^{-\varepsilon}
-
\frac{1}{\varepsilon} \, (a+z)^{-\varepsilon}
-
2 \, \varepsilon \, \mbox{Li}_2 \left( - \frac{a}{z} \right)
\right]\!,
\label{eqdefyint6}
\end{align}$$]]></tex-math></disp-formula>
i.e., making explicit <inline-formula><tex-math notation="LaTeX" id="ImEquation86"><![CDATA[$z$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation87"><![CDATA[$a$]]></tex-math></inline-formula> in terms of <inline-formula><tex-math notation="LaTeX" id="ImEquation88"><![CDATA[$\det{(G^{\{i\}})}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation89"><![CDATA[$\widetilde{D}_{ij}$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M2-28"><label>(2.28)</label><tex-math notation="LaTeX" id="Equation33"><![CDATA[$$\begin{align}
W(\det{(G^{\{i\}})},\widetilde{D}_{ij}, 0)
&= \frac{1}{\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{\det{(G^{\{i\}})} - \widetilde{D}_{ij}} \notag \\
&\quad {} \times
\left\{
\frac{2}{\varepsilon} \,
\left[ \frac{1}{2} \, \big( \widetilde{D}_{ij} - \det{(G^{\{i\}})} \big) - i \, \lambda \right]^{-\varepsilon} -
\frac{1}{\varepsilon} \, \left( \frac{\widetilde{D}_{ij}}{2} - i \, \lambda \right)^{-\varepsilon}
\right.
\notag \\
&\qquad \quad {} -
\left.
2 \, \varepsilon \,
\mbox{Li}_2
\left(
\frac{\det{(G^{\{i\}})}}{\det{(G^{\{i\}})} - \widetilde{D}_{ij} + i \, \lambda }
\right)
\right\}\!,
\label{eqdirei3n3}
\end{align}$$]]></tex-math></disp-formula>
which is both well behaved when <inline-formula><tex-math notation="LaTeX" id="ImEquation90"><![CDATA[$\widetilde{D}_{ij} \rightarrow 0$]]></tex-math></inline-formula> (<inline-formula><tex-math notation="LaTeX" id="ImEquation91"><![CDATA[$m_k^2 \rightarrow 0$]]></tex-math></inline-formula>) and more compact.</p>
</sec>
<sec id="SEC2.1.3"><title>(c) Both <inline-formula><tex-math notation="LaTeX" id="ImEquation92"><![CDATA[$m_j^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation93"><![CDATA[$m_k^2$]]></tex-math></inline-formula> vanish</title>
<p>The function <inline-formula><tex-math notation="LaTeX" id="ImEquation94"><![CDATA[$D^{\{i\}(j)}(x)$]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M2-29"><label>(2.29)</label><tex-math notation="LaTeX" id="Equation34"><![CDATA[$$\begin{equation}
D^{\{i\}(j)}(x) = - \, G^{\{i\}(j)} \, x \, (1-x)
\label{eqnewquad2}
\end{equation}$$]]></tex-math></disp-formula>
and we immediately get
<disp-formula id="ptz160M2-30"><label>(2.30)</label><tex-math notation="LaTeX" id="Equation35"><![CDATA[$$\begin{align}
W(\det{(G^{\{i\}})},0,0)
&= -\frac{1}{\varepsilon^2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}
{\Gamma(1 - 2 \, \varepsilon)} \,
\left( - \, \frac{\det{(G^{\{i\}})}}{2} - i \, \lambda \right)^{-1-\varepsilon} .
\label{eqdirei3n4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation95"><![CDATA[$\widetilde{D}_{ij} \rightarrow 0$]]></tex-math></inline-formula> (<inline-formula><tex-math notation="LaTeX" id="ImEquation96"><![CDATA[$m_k^2 \rightarrow 0$]]></tex-math></inline-formula>), Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-28">2.28</xref>) or (<xref ref-type="disp-formula" rid="ptz160M2-21">2.21</xref>) smoothly become Eq. (<xref ref-type="disp-formula" rid="ptz160M2-30">2.30</xref>), as expected.</p>
</sec>
</sec>
<sec id="SEC2.2"><title>2.2. Practical implementation of the preceding cases and explicit examples</title>
<p>The various cases reviewed above may or may not be involved in a specific computation because some coefficients weighing the <inline-formula><tex-math notation="LaTeX" id="ImEquation97"><![CDATA[$W(\det{(G^{\{i\}})},\widetilde{D}_{ij},\widetilde{D}_{ik})$]]></tex-math></inline-formula> may vanish. In particular, as seen in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-5">2.5</xref>), when <inline-formula><tex-math notation="LaTeX" id="ImEquation98"><![CDATA[$\Delta_2=0$]]></tex-math></inline-formula>, the three-point function in dimension <inline-formula><tex-math notation="LaTeX" id="ImEquation99"><![CDATA[$4 - 2 \, \varepsilon$]]></tex-math></inline-formula> is the sum of three two-point functions in dimension <inline-formula><tex-math notation="LaTeX" id="ImEquation100"><![CDATA[$2 - 2 \, \varepsilon$]]></tex-math></inline-formula>. These two-point functions correspond to the three distinct pinchings of the internal propagators of the three-point function. At first sight, one should worry that some of these two-point functions in low dimensions may badly diverge due to a threshold singularity which is, however, not present in the three-point function! For example, one of the pinchings of a three-point function having IR/collinear singularities would lead to a two-point function with the external legs on the mass shell of one of the propagators whereas the other propagator is massless. This would lead to a polynomial <inline-formula><tex-math notation="LaTeX" id="ImEquation101"><![CDATA[$D^{\{i\}(j)}(x) \propto x^2$]]></tex-math></inline-formula> or <inline-formula><tex-math notation="LaTeX" id="ImEquation102"><![CDATA[$(1-x)^2$]]></tex-math></inline-formula>. Fortunately the corresponding <inline-formula><tex-math notation="LaTeX" id="ImEquation103"><![CDATA[$\overline{b}$]]></tex-math></inline-formula> coefficients weighting such pathological terms identically vanish, and the discussion which follows, illustrated with examples, elucidates why this happens. Let us define <inline-formula><tex-math notation="LaTeX" id="ImEquation104"><![CDATA[$s_i = p_i^2$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation105"><![CDATA[$i=1,2,3$]]></tex-math></inline-formula>.</p>
<sec id="SEC2.2.1"><title>Case 1</title>
<p>A soft divergence occurs when the kinematic matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation106"><![CDATA[${\cal S}$]]></tex-math></inline-formula> has a vanishing line (and corresponding column). This happens whenever a massless propagator connects two vertices in which external momenta enter on the mass shells of the two other propagators. As the external momenta are real, this case can occur only when the non-vanishing internal masses are real. Let us assume, cf. <xref ref-type="fig" rid="F1">Fig. 1</xref>, that the internal mass squared <inline-formula><tex-math notation="LaTeX" id="ImEquation107"><![CDATA[$m_1^2$]]></tex-math></inline-formula> vanishes whereas the external four-momenta <inline-formula><tex-math notation="LaTeX" id="ImEquation108"><![CDATA[$p_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation109"><![CDATA[$p_2$]]></tex-math></inline-formula> satisfy the mass shell conditions <inline-formula><tex-math notation="LaTeX" id="ImEquation110"><![CDATA[$s_1=m_3^2$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation111"><![CDATA[$s_2=m_2^2$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation112"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix has the following texture:
<disp-formula id="ptz160M2-31"><label>(2.31)</label><tex-math notation="LaTeX" id="Equation36"><![CDATA[$$\begin{equation}
{\cal S}^{\,{\rm soft}} =
\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & -2 \, m_2^2 & s_3 - m_2^2 - m_3^2 \\
0 & s_3 - m_2^2 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!.
\label{eqcalssoft}
\end{equation}$$]]></tex-math></disp-formula></p>
<fig id="F1" orientation="portrait" position="float"><label>Fig. 1.</label><caption><p>The triangle picturing the one-loop three-point function.</p></caption>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" orientation="portrait" position="float" mimetype="image" xlink:href="ptz160f1.tif"/></fig>
<p>If one singles out row and column <inline-formula><tex-math notation="LaTeX" id="ImEquation113"><![CDATA[$1$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation114"><![CDATA[${\cal S}^{\,{\rm soft}}$]]></tex-math></inline-formula>, the two-component vector <inline-formula><tex-math notation="LaTeX" id="ImEquation115"><![CDATA[$V^{(1)}$]]></tex-math></inline-formula> is readily seen to vanish and so do the coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation116"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation117"><![CDATA[$\overline{b}_3$]]></tex-math></inline-formula> which are proportional to the two components of <inline-formula><tex-math notation="LaTeX" id="ImEquation118"><![CDATA[$(G^{(1)})^{-1} \cdot V^{(1)}$]]></tex-math></inline-formula>; thus, only <inline-formula><tex-math notation="LaTeX" id="ImEquation119"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> differs from zero, cf. Eq. (2.15) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>].</p>
<p>To illustrate this point, let us consider the case where <inline-formula><tex-math notation="LaTeX" id="ImEquation120"><![CDATA[$m_1=0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation121"><![CDATA[$m_2=m_3=m$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation122"><![CDATA[$s_1=s_2=m^2$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation123"><![CDATA[$s_3$]]></tex-math></inline-formula> is arbitrary. In this case, since <inline-formula><tex-math notation="LaTeX" id="ImEquation124"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> is the only non-vanishing coefficient, the polynomial <inline-formula><tex-math notation="LaTeX" id="ImEquation125"><![CDATA[$D^{\{1\}(2)}(x)$]]></tex-math></inline-formula> involved in <inline-formula><tex-math notation="LaTeX" id="ImEquation126"><![CDATA[$I_3^n$]]></tex-math></inline-formula> corresponds to the one appearing in the two-point function obtained by pinching the internal line with four-momentum <inline-formula><tex-math notation="LaTeX" id="ImEquation127"><![CDATA[$q_1$]]></tex-math></inline-formula> (cf. <xref ref-type="fig" rid="F1">Fig. 1</xref>). This polynomial involves two masses (equal here), and this example corresponds to case (a) of the previous section. In this simple case, the two roots of the polynomial <inline-formula><tex-math notation="LaTeX" id="ImEquation128"><![CDATA[$D^{\{1\}(2)}(x)$]]></tex-math></inline-formula> are given by
<disp-formula id="ptz160M2-32"><label>(2.32)</label><tex-math notation="LaTeX" id="Equation37"><![CDATA[$$\begin{equation}
x_{1,2} = \frac{1}{2} \pm \frac{1}{2} \, \sqrt{ 1 - \frac{4 \, ( m^2 - i \,\lambda)}{s_3}}
\label{eqroot12ex}
\end{equation}$$]]></tex-math></disp-formula>
with the property
<disp-formula id="ptz160UM6"><tex-math notation="LaTeX" id="Equation38"><![CDATA[$$ 1 - x_1 = x_2 .$$]]></tex-math></disp-formula></p>
<p>Injecting this property into Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-14">2.14</xref>) and (<xref ref-type="disp-formula" rid="ptz160M7-8">B.8</xref>), we get for the functions <inline-formula><tex-math notation="LaTeX" id="ImEquation129"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation130"><![CDATA[$K(x_1,x_2)$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-15">2.15</xref>):
<disp-formula id="ptz160M2-33"><label>(2.33)</label><tex-math notation="LaTeX" id="Equation39"><![CDATA[$$\begin{align}
J(x_1,x_2) &= \frac{2}{x_1-x_2} \, \left\{ \mbox{Li}_2 \left( \frac{x_2}{x_1} \right) - \mbox{Li}_2 \left( \frac{x_1}{x_2} \right) \right. \notag \\
&\quad + \left. \ln \left( -\frac{x_2}{x_1} \right) \, \ln \left( - (x_1-x_2)^2 \right) \right\}\!, \label{eqcompj6} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M2-34"><label>(2.34)</label><tex-math notation="LaTeX" id="Equation40"><![CDATA[$$\begin{align}
K(x_1,x_2) &= \frac{2}{x_1-x_2} \, \ln \left( -\frac{x_2}{x_1} \right)\!. \label{eqcompk2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We check numerically that we recover the result of Ref. [<xref ref-type="bibr" rid="B4">4</xref>].</p>
</sec>
<sec id="SEC2.2.2"><title>Case 2</title>
<p>A collinear divergence occurs when two internal masses vanish whereas the external four-momentum which enters into the vertex connecting the two adjacent massless propagators is lightlike (massless collinear splitting at this vertex). Note that the non-vanishing internal mass can be real or complex. Let us assume that the labels of the two massless propagators are <inline-formula><tex-math notation="LaTeX" id="ImEquation131"><![CDATA[$1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation132"><![CDATA[$2$]]></tex-math></inline-formula>, with <inline-formula><tex-math notation="LaTeX" id="ImEquation133"><![CDATA[$s_2=0$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation134"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix has the following texture:
<disp-formula id="ptz160M2-35"><label>(2.35)</label><tex-math notation="LaTeX" id="Equation41"><![CDATA[$$\begin{equation}
{\cal S}^{\,{\rm coll}} =
\left(
\begin{array}{ccc}
0 & 0 & s_1 - m_3^2 \\
0 & 0 & s_3 - m_3^2 \\
s_1 - m_3^2 & s_3 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!.
\label{eqcalscoll}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>If one singles out row and column <inline-formula><tex-math notation="LaTeX" id="ImEquation135"><![CDATA[$3$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation136"><![CDATA[${\cal S}^{\,{\rm coll}}$]]></tex-math></inline-formula>, the Gram matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation137"><![CDATA[$G^{(3)}$]]></tex-math></inline-formula> and the vector <inline-formula><tex-math notation="LaTeX" id="ImEquation138"><![CDATA[$V^{(3)}$]]></tex-math></inline-formula> read
<disp-formula id="ptz160M2-36"><label>(2.36)</label><tex-math notation="LaTeX" id="Equation42"><![CDATA[$$\begin{align}
G^{(3)} &= \left(
\begin{array}{cc}
2 \, s_1 & s_1+s_3\\
s_1+s_3 & 2 \, s_3
\end{array}
\right)\!,
\label{eqdefg3coll} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M2-37"><label>(2.37)</label><tex-math notation="LaTeX" id="Equation43"><![CDATA[$$\begin{align}
V^{(3)} &= \left(
\begin{array}{c}
s_1 + m_3^2 \\
s_3 + m_3^2
\end{array}
\right)\!.
\label{eqdefv3coll}
\end{align}$$]]></tex-math></disp-formula></p>
<p>A simple calculation yields
<disp-formula id="ptz160M2-38"><label>(2.38)</label><tex-math notation="LaTeX" id="Equation44"><![CDATA[$$\begin{align}
\sum_{i \in S_3 \setminus \{3\}} \,
\left[ \big( G^{(3)} \big) ^{-1} \cdot V^{(3)} \right]_i
&= 1 ,
\label{eqsumgm1v}
\end{align}$$]]></tex-math></disp-formula>
so that <inline-formula><tex-math notation="LaTeX" id="ImEquation139"><![CDATA[$\overline{b}_3$]]></tex-math></inline-formula> given by Eq. (2.15) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] vanishes, whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation140"><![CDATA[$\overline{b}_{1}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation141"><![CDATA[$\overline{b}_{2}$]]></tex-math></inline-formula> generically differ from zero.</p>
<p>Let us compute <inline-formula><tex-math notation="LaTeX" id="ImEquation142"><![CDATA[$I_3^n$]]></tex-math></inline-formula> for this specific case. As <inline-formula><tex-math notation="LaTeX" id="ImEquation143"><![CDATA[$\overline{b}_3 = 0$]]></tex-math></inline-formula>, the relevant polynomials <inline-formula><tex-math notation="LaTeX" id="ImEquation144"><![CDATA[$D^{\{i\}(j)}(x)$]]></tex-math></inline-formula> are those of the two-point functions obtained in the two pinching configurations (cf. <xref ref-type="fig" rid="F1">Fig. 1</xref>) where either the internal line with four-momentum <inline-formula><tex-math notation="LaTeX" id="ImEquation145"><![CDATA[$q_1$]]></tex-math></inline-formula> or the one with four-momentum <inline-formula><tex-math notation="LaTeX" id="ImEquation146"><![CDATA[$q_2$]]></tex-math></inline-formula> is pinched. As these two lines are associated with vanishing masses and the third propagator is associated with a non-vanishing mass, the two polynomials both have one vanishing mass, which corresponds to case (b) of the previous section. Starting with Eq. (<xref ref-type="disp-formula" rid="ptz160M2-5">2.5</xref>), <inline-formula><tex-math notation="LaTeX" id="ImEquation147"><![CDATA[$I_3^n$]]></tex-math></inline-formula> reads
<disp-formula id="ptz160M2-39"><label>(2.39)</label><tex-math notation="LaTeX" id="Equation45"><![CDATA[$$\begin{align}
I_3^n
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left[
\frac{\overline{b}_1}{\det(G)} \,
\int^1_0 dx \,
\big( D^{\{1\}(2)}(x) - i \, \lambda \big)^{-1-\varepsilon}
\right.
\notag \\
&
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
\left.
\, + \frac{\overline{b}_2}{\det(G)} \,
\int^1_0 dx \,
\big( D^{\{2\}(3)}(x) - i \, \lambda \big)^{-1-\varepsilon} \right]\!.
\label{eqdirei3n5}
\end{align}$$]]></tex-math></disp-formula></p>
<p>It is better to change <inline-formula><tex-math notation="LaTeX" id="ImEquation148"><![CDATA[$x \leftrightarrow 1-x$]]></tex-math></inline-formula> in the second integral in order to have a polynomial of the type in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-16">2.16</xref>) and write <inline-formula><tex-math notation="LaTeX" id="ImEquation149"><![CDATA[$I_3^n$]]></tex-math></inline-formula> as
<disp-formula id="ptz160M2-40"><label>(2.40)</label><tex-math notation="LaTeX" id="Equation46"><![CDATA[$$\begin{align}
I_3^n
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left[
\,
\frac{\overline{b}_1}{\det(G)} \,
\int_0^1 dx \,
\big( D^{\{1\}(2)}(x) - i \, \lambda \big)^{-1-\varepsilon}
\right.
\notag \\
&
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
\left.
+ \,
\frac{\overline{b}_2}{\det(G)} \,
\int_0^1 dx \,
\big( D^{\{2\}(1)}(x) - i \, \lambda \big)^{-1-\varepsilon}
\right] \notag \\
&= \frac{\overline{b}_1}{\det(G)} \, W( \det{(G^{\{1\}})},\widetilde{D}_{12},0 ) + \frac{\overline{b}_2}{\det(G)} \, W( \det{(G^{\{2\}})},\widetilde{D}_{21},0 ) .
\label{eqdirei3n6}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Noting that <inline-formula><tex-math notation="LaTeX" id="ImEquation150"><![CDATA[$\overline{b}_1 = (s_3-m_3^2) \, (s_1-s_3)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation151"><![CDATA[$\overline{b}_2 = (s_1-m_3^2) \, (s_3-s_1)$]]></tex-math></inline-formula>, determining <inline-formula><tex-math notation="LaTeX" id="ImEquation152"><![CDATA[$\det{(G^{\{1\}})}$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation153"><![CDATA[$\det{(G^{\{2\}})}$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation154"><![CDATA[$\widetilde{D}_{12}$]]></tex-math></inline-formula> from the <inline-formula><tex-math notation="LaTeX" id="ImEquation155"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix elements, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-8">2.8</xref>), and directly applying the result of Eq. (<xref ref-type="disp-formula" rid="ptz160M2-28">2.28</xref>), we get
<disp-formula id="ptz160M2-41"><label>(2.41)</label><tex-math notation="LaTeX" id="Equation47"><![CDATA[$$\begin{align}
I_3^n
&= \frac{\Gamma(1+\varepsilon)}{s_1-s_3} \,
\left\{ - \, \frac{1}{\varepsilon^2}
\left[
\left( {} - s_3+m_3^2 - i \, \lambda \right)^{-\varepsilon}
\;\; - \;\;
\left( {} -s_1+m_3^2 -i \, \lambda \right)^{-\varepsilon}
\right]
\right.
\notag \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
+
\left.
\mbox{Li}_2 \left( \frac{s_3}{s_3-m_3^2+i \, \lambda} \right)
\; - \;\;
\mbox{Li}_2 \left( \frac{s_1}{s_1-m_3^2+i \, \lambda} \right) \;
\right\}\!.
\label{eqdirei3n8}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using the Landen identity of Eq. (<xref ref-type="disp-formula" rid="ptz160M7-3">B.3</xref>) we recover the formula in Eq. (4.8) of Ref. [<xref ref-type="bibr" rid="B4">4</xref>] after some algebra.</p>
</sec>
<sec id="SEC2.2.3"><title>Case 3</title>
<p>Both a soft and a collinear divergence may occur at the same time, thereby proceeding from both cases 1 and 2 above.</p>
<p>Let us take the example of case 2 and specify <inline-formula><tex-math notation="LaTeX" id="ImEquation156"><![CDATA[$s_1 = m_3^2$]]></tex-math></inline-formula>. The texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation157"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix becomes
<disp-formula id="ptz160M2-42"><label>(2.42)</label><tex-math notation="LaTeX" id="Equation48"><![CDATA[$$\begin{equation}
{\cal S}^{\,{\rm cs}} =
\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & s_3 - m_3^2 \\
0 & s_3 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!.
\label{eqcalscs}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>Only <inline-formula><tex-math notation="LaTeX" id="ImEquation158"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> does not vanish, so <inline-formula><tex-math notation="LaTeX" id="ImEquation159"><![CDATA[$I_3^n$]]></tex-math></inline-formula> reads simply
<disp-formula id="ptz160M2-43"><label>(2.43)</label><tex-math notation="LaTeX" id="Equation49"><![CDATA[$$\begin{align}
I_3^n
&= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\frac{\overline{b}_1}{\det(G)} \,
\int^1_0 dx \,
\big( D^{\{1\}(2)}(x) - i \, \lambda \big)^{-1-\varepsilon} \notag \\
&= \frac{\overline{b}_1}{\det(G)} \, W( \det{(G^{\{1\}})},\widetilde{D}_{12},0 ) .
\label{eqdirei3n9}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using Eq. (<xref ref-type="disp-formula" rid="ptz160M2-28">2.28</xref>) and expressing <inline-formula><tex-math notation="LaTeX" id="ImEquation160"><![CDATA[$\det{(G^{\{1\}})}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation161"><![CDATA[$\widetilde{D}_{12}$]]></tex-math></inline-formula> in terms of the <inline-formula><tex-math notation="LaTeX" id="ImEquation162"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix elements, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-8">2.8</xref>), we get
<disp-formula id="ptz160M2-44"><label>(2.44)</label><tex-math notation="LaTeX" id="Equation50"><![CDATA[$$\begin{align}
I_3^n &=
\frac{\Gamma(1+\varepsilon)}{m_3^2-s_3} \,
\left\{
- \, \frac{1}{\varepsilon^2}
\left( {} - s_3+m_3^2 - i \, \lambda \right)^{-\varepsilon}
+ \frac{1}{2 \, \varepsilon^2} \,
\left( m_3^2 - i \, \lambda \right)^{-\varepsilon}
+ \mbox{Li}_2 \left( \frac{s_3}{s_3-m_3^2+i \, \lambda} \right)
\right\}\!.
\label{eqdirei3n10}
\end{align}$$]]></tex-math></disp-formula></p>
<p>After some algebra, we recover<sup><xref ref-type="fn" rid="FN6">6</xref></sup> the formula in Eq. (4.11) of Ref. [<xref ref-type="bibr" rid="B4">4</xref>].</p>
</sec>
</sec>
<sec id="SEC2.3"><title>2.3. Indirect way</title>
<p>This subsection provides the calculation according to the &#x201C;indirect way.&#x201D; The results presented are valid for both real and complex masses; unless explicitly specified the <inline-formula><tex-math notation="LaTeX" id="ImEquation163"><![CDATA[$i \, \lambda$]]></tex-math></inline-formula> prescription is kept, bearing in mind that it is ineffective in the case of complex masses. Let us start from Eq. (<xref ref-type="disp-formula" rid="ptz160M2-3">2.3</xref>). The <inline-formula><tex-math notation="LaTeX" id="ImEquation164"><![CDATA[$x$]]></tex-math></inline-formula> integration is traded for a <inline-formula><tex-math notation="LaTeX" id="ImEquation165"><![CDATA[$\rho$]]></tex-math></inline-formula> integration in a way very similar to the four-dimensional case (see Sect. 2.2.2 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]) and we get
<disp-formula id="ptz160M2-45"><label>(2.45)</label><tex-math notation="LaTeX" id="Equation51"><![CDATA[$$\begin{align}
I_3^n
&= \sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G)} \,
\sum_{j \in S_3 \setminus \{i\}} \, \frac{\overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}} \,
L_3^n \big( 0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} \big)
\label{eqdef3n3}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M2-46"><label>(2.46)</label><tex-math notation="LaTeX" id="Equation52"><![CDATA[$$\begin{align}
L_3^n \big( 0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} \big)
&= \kappa_{_{IR}} \,
\int^{+\infty}_0 \frac{d \xi}{\xi^{\nu} - i \, \lambda}
\notag\\
&\quad {} \quad {}
\times
\int^{+\infty}_0
\frac{d \rho}{
(\xi^{\nu} + \rho^2 - \Delta_1^{\{i\}} - i \, \lambda)
(\xi^{\nu} + \rho^2+ \widetilde{D}_{ij} - i \, \lambda)^{1/2}
}
\label{eqlijsoft1}
\end{align}$$]]></tex-math></disp-formula>
and
<disp-formula id="ptz160UM7"><tex-math notation="LaTeX" id="Equation53"><![CDATA[$$ \kappa_{_{IR}} =2^{\varepsilon} \, \frac{\Gamma(1+\varepsilon)}{(1-\varepsilon)} \,
\frac{1}{B(1+\varepsilon,1-\varepsilon)}, \quad \nu = \frac{1}{1-\varepsilon} .$$]]></tex-math></disp-formula></p>
<p>To handle the cases with soft and/or collinear divergences, the two relevant configurations are (1) <inline-formula><tex-math notation="LaTeX" id="ImEquation166"><![CDATA[$\widetilde{D}_{ij} \neq 0$]]></tex-math></inline-formula> and (2) <inline-formula><tex-math notation="LaTeX" id="ImEquation167"><![CDATA[$\widetilde{D}_{ij} = 0$]]></tex-math></inline-formula>. Note that when both <inline-formula><tex-math notation="LaTeX" id="ImEquation168"><![CDATA[$\Delta_2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation169"><![CDATA[$\Delta_1^{\{i\}}$]]></tex-math></inline-formula> vanish, <inline-formula><tex-math notation="LaTeX" id="ImEquation170"><![CDATA[$L_3^n ( 0, 0, \widetilde{D}_{ij} )$]]></tex-math></inline-formula> is weighting a vanishing <inline-formula><tex-math notation="LaTeX" id="ImEquation171"><![CDATA[$\overline{b}$]]></tex-math></inline-formula>, and therefore it will not be considered (cf. Sect. <xref ref-type="sec" rid="SEC2.2">2.2</xref>).</p>
<p>The <inline-formula><tex-math notation="LaTeX" id="ImEquation172"><![CDATA[$\rho$]]></tex-math></inline-formula> integration can be done using Appendix A as in the four-dimensional case; as the outcome of this integration, we shall distinguish two cases depending on the sign of <inline-formula><tex-math notation="LaTeX" id="ImEquation173"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}})$]]></tex-math></inline-formula>.</p>
<sec id="SEC2.3.1"><title>(1) <inline-formula><tex-math notation="LaTeX" id="ImEquation174"><![CDATA[$\widetilde{D}_{ij} \neq 0$]]></tex-math></inline-formula></title>
<p><italic>(1a)</italic>&#x02002;<inline-formula><tex-math notation="LaTeX" id="ImEquation175"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}}) > 0$]]></tex-math></inline-formula> This case covers real masses in particular. After the <inline-formula><tex-math notation="LaTeX" id="ImEquation176"><![CDATA[$\rho$]]></tex-math></inline-formula> integration, <inline-formula><tex-math notation="LaTeX" id="ImEquation177"><![CDATA[$L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )$]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M2-47"><label>(2.47)</label><tex-math notation="LaTeX" id="Equation54"><![CDATA[$$\begin{align}
&L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
\notag\\
&\quad = \kappa_{_{IR}} \,\int^{+\infty}_0 d \xi \,
\int^{1}_0 d z \,
\frac{1}
{
(\xi^{\nu} - i \, \lambda)
(\xi^{\nu} - (1-z^2) \, \Delta_1^{\{i\}} + z^2 \, \widetilde{D}_{ij} - i \, \lambda)
} .
\label{eqlijsoft3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The <inline-formula><tex-math notation="LaTeX" id="ImEquation178"><![CDATA[$\xi$]]></tex-math></inline-formula> integration is performed first, using Eq. (<xref ref-type="disp-formula" rid="ptz160M6-4">A.4</xref>) of Appendix <xref ref-type="sec" rid="SEC6">A</xref>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation179"><![CDATA[$L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )$]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M2-48"><label>(2.48)</label><tex-math notation="LaTeX" id="Equation55"><![CDATA[$$\begin{align}
L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
&= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\int^1_0 dz
\big(
z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}} - i \, \lambda
\big)^{-1-\varepsilon} .
\label{eqlijsoft40}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Expanding<sup><xref ref-type="fn" rid="FN7">7</xref></sup> the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M2-48">2.48</xref>) around <inline-formula><tex-math notation="LaTeX" id="ImEquation180"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> then gives
<disp-formula id="ptz160M2-49"><label>(2.49)</label><tex-math notation="LaTeX" id="Equation56"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},\widetilde{D}_{ij} )
&= 2^{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left[
- \, \frac{1}{\varepsilon} \,
\int^1_0
\frac{dz}
{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}} - i \, \lambda}
\right.
\notag\\
& \quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
\left.
+ \int^1_0 dz \,
\frac{\ln(z^2 \,(\widetilde{D}_{ij}+\Delta_1^{\{i\}})-\Delta_1^{\{i\}} - i \,\lambda)}
{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}} - i \, \lambda}
\right]\!.
\label{eqlijsoft41}
\end{align}$$]]></tex-math></disp-formula></p>
<p><italic>(1b)</italic>&#x02002;<inline-formula><tex-math notation="LaTeX" id="ImEquation181"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}}) < 0$]]></tex-math></inline-formula> This case occurs only with complex masses, in a way such that the <inline-formula><tex-math notation="LaTeX" id="ImEquation182"><![CDATA[$i \, \lambda$]]></tex-math></inline-formula> prescriptions are overshadowed and ineffective; the latter are therefore dropped in this part. After <inline-formula><tex-math notation="LaTeX" id="ImEquation183"><![CDATA[$\rho$]]></tex-math></inline-formula> integration performed as in the four-dimensional case, we get, see Eq. (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) of Appendix <xref ref-type="sec" rid="SEC6">A</xref>,
<disp-formula id="ptz160M2-50"><label>(2.50)</label><tex-math notation="LaTeX" id="Equation57"><![CDATA[$$\begin{align}
&L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
\notag \\
&\quad{}=
- \kappa_{_{IR}} \,\int^{+\infty}_0
\frac{d \xi}{\xi^{\nu} - i \, \lambda} \,
\left[
i \int^{+\infty}_0
\frac{dz}{- \xi^{\nu} + \widetilde{D}_{ij} \, z^2 + (1+z^2) \, \Delta_1^{\{i\}}}
\right.
\notag \\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {} \quad {}\quad {}
\quad {}\quad {}\quad {}
\left.
+ \int^{+\infty}_1 \frac{dz}
{\xi^{\nu} + \widetilde{D}_{ij} \, z^2 - (1-z^2) \, \Delta_1^{\{i\}}}
\right]\!.
\label{eqlijsoft5}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using the identity in Eq. (<xref ref-type="disp-formula" rid="ptz160M6-4">A.4</xref>), <inline-formula><tex-math notation="LaTeX" id="ImEquation184"><![CDATA[$L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )$]]></tex-math></inline-formula> reads
<disp-formula id="ptz160M2-51"><label>(2.51)</label><tex-math notation="LaTeX" id="Equation58"><![CDATA[$$\begin{align}
L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
&\quad{} = \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left[
- \, i \int^{+\infty}_0 dz
\big(
-z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}
\big)^{-1-\varepsilon}
\right.
\notag \\
&\quad {}\quad {}\quad {}\quad {} \quad {} \quad {} \quad {}
+
\left.
\int^{+\infty}_1 dz \;
\big(\,
z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}
\big)^{-1-\varepsilon}
\;
\right]\!.
\label{eqlijsoft60}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Expanding the terms in the square bracket in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-51">2.51</xref>) around <inline-formula><tex-math notation="LaTeX" id="ImEquation185"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> then gives
<disp-formula id="ptz160M2-52"><label>(2.52)</label><tex-math notation="LaTeX" id="Equation59"><![CDATA[$$\begin{align}
&L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},\widetilde{D}_{ij} )
\notag \\
&= 2^{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left\{
\;
\frac{1}{\varepsilon} \,
\left[
\; i \, \int^{+\infty}_0
\frac{dz}{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) + \Delta_1^{\{i\}}}
+
\int^{+\infty}_1
\frac{dz}{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}}
\;
\right]
\right.
\notag \\
&\quad \quad \quad \quad \quad \quad \quad {}
-i \int^{+\infty}_0 dz \,
\frac{\ln(-z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}})}
{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) + \Delta_1^{\{i\}}}
\notag\\
&\quad \quad \quad \quad \quad \quad \quad {}
\left.
- \;\; \int^{+\infty}_1 dz \,
\frac{\ln(z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}})}
{z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}}
\right\}\!.
\label{eqlijsoft61}
\end{align}$$]]></tex-math></disp-formula></p>
<p>For Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-52">2.52</xref>), the <inline-formula><tex-math notation="LaTeX" id="ImEquation186"><![CDATA[$z$]]></tex-math></inline-formula> integration can then be performed using Appendix <xref ref-type="sec" rid="SEC10">E</xref>.</p>
<p>Since the cuts of <inline-formula><tex-math notation="LaTeX" id="ImEquation187"><![CDATA[$( z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) -\Delta_1^{\{i\}})^{-1-\varepsilon}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation188"><![CDATA[$\ln ( z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}} )$]]></tex-math></inline-formula> are the same, the discussion carried to extend the &#x201C;indirect way&#x201D; to the general complex mass case holds likewise here (cf. Sect. 2 of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]). Equation (<xref ref-type="disp-formula" rid="ptz160M2-51">2.51</xref>) can be rewritten as
<disp-formula id="ptz160M2-53"><label>(2.53)</label><tex-math notation="LaTeX" id="Equation60"><![CDATA[$$\begin{align}
&L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
\notag\\
&= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\left\{ \int_{0}^{i \, \infty} + \int_{+\infty}^{1} \right\} dz
\big(
\, z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}
\big)^{-1-\varepsilon} .
\label{eqlijsoft62}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In Eq. (<xref ref-type="disp-formula" rid="ptz160M2-53">2.53</xref>) let us add a vanishing contribution along the &#x201C;contour at <inline-formula><tex-math notation="LaTeX" id="ImEquation189"><![CDATA[$\infty$]]></tex-math></inline-formula>&#x201D; in the north-east quadrant <inline-formula><tex-math notation="LaTeX" id="ImEquation190"><![CDATA[$\{\operatorname{Re}(z) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation191"><![CDATA[$\operatorname{Im}(z) > 0\}$]]></tex-math></inline-formula> so as to concatenate the two contributions. The connected contour thus obtained can in turn be deformed into a finite contour <inline-formula><tex-math notation="LaTeX" id="ImEquation192"><![CDATA[$\widehat{(0,1)}_{i,j}$]]></tex-math></inline-formula> stretched from 0 to 1 as pictured in <xref ref-type="fig" rid="F2">Fig. 2</xref>, thereby unifying Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-48">2.48</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-53">2.53</xref>):
<disp-formula id="ptz160M2-54"><label>(2.54)</label><tex-math notation="LaTeX" id="Equation61"><![CDATA[$$\begin{align}
L_3^n (0, \Delta_{1}^{\{i\}}, \widetilde{D}_{ij} )
&= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\int_{\widehat{(0,1)}_{i,j}} dz
\big(
\, z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}
\big)^{-1-\varepsilon} .
\label{eqlijsoft63}
\end{align}$$]]></tex-math></disp-formula></p>
<fig id="F2" orientation="portrait" position="float"><label>Fig. 2.</label><caption><p>Location of the relevant discontinuity cut <inline-formula><tex-math notation="LaTeX" id="ImEquation193"><![CDATA[${\cal C}_{i,j}$]]></tex-math></inline-formula> with respect to the two half straight lines <inline-formula><tex-math notation="LaTeX" id="ImEquation194"><![CDATA[$[0, + i \infty[$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation195"><![CDATA[$[1,+\infty[$]]></tex-math></inline-formula>, and deformation of the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation196"><![CDATA[$\widehat{(0,1)}$]]></tex-math></inline-formula> partly wrapping the extremity of the cut.</p></caption>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" orientation="portrait" position="float" mimetype="image" xlink:href="ptz160f2.tif"/></fig>
</sec>
<sec id="SEC2.3.2"><title>(2) <inline-formula><tex-math notation="LaTeX" id="ImEquation197"><![CDATA[$\widetilde{D}_{ij} = 0$]]></tex-math></inline-formula></title>
<p>Here again we shall distinguish two cases depending on the sign of <inline-formula><tex-math notation="LaTeX" id="ImEquation198"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}})$]]></tex-math></inline-formula>.</p>
<p><italic>(2a)</italic>&#x02002;<inline-formula><tex-math notation="LaTeX" id="ImEquation199"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}}) > 0$]]></tex-math></inline-formula> This case covers real masses in particular. The calculation initially amounts to setting <inline-formula><tex-math notation="LaTeX" id="ImEquation200"><![CDATA[$\widetilde{D}_{ij} = 0$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-48">2.48</xref>), which becomes
<disp-formula id="ptz160M2-55"><label>(2.55)</label><tex-math notation="LaTeX" id="Equation62"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},0 )
&= - \, \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
( - \Delta_1^{\{i\}} - i \, \lambda )^{-1-\varepsilon} \,
\int^1_0 dz \left(1-z^2 \right)^{-1-\varepsilon} .
\label{eqlijsoft7}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation201"><![CDATA[$z$]]></tex-math></inline-formula> is performed using Eq. (<xref ref-type="disp-formula" rid="ptz160M8-6">C.6</xref>), and <inline-formula><tex-math notation="LaTeX" id="ImEquation202"><![CDATA[$L_{3}^{n}(0,\Delta_{1}^{\{i\}},0) $]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M2-56"><label>(2.56)</label><tex-math notation="LaTeX" id="Equation63"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},0 )
&= \frac{1}{\varepsilon^2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma(1-\varepsilon)^2}{\Gamma(1-2 \, \varepsilon)} \,
( - 2 \, \Delta_1^{\{i\}} - i \, \lambda )^{-1-\varepsilon} .
\label{eqlijsoft8}
\end{align}$$]]></tex-math></disp-formula></p>
<p><italic>(2b)</italic>&#x02002;<inline-formula><tex-math notation="LaTeX" id="ImEquation203"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}}) < 0$]]></tex-math></inline-formula> Here again, this case occurs only with complex masses, with the <inline-formula><tex-math notation="LaTeX" id="ImEquation204"><![CDATA[$i \, \lambda$]]></tex-math></inline-formula> prescriptions overshadowed and therefore dropped. The calculation initially amounts to setting <inline-formula><tex-math notation="LaTeX" id="ImEquation205"><![CDATA[$\widetilde{D}_{ij} = 0$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-51">2.51</xref>), which becomes
<disp-formula id="ptz160M2-57"><label>(2.57)</label><tex-math notation="LaTeX" id="Equation64"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},0 )
&= \frac{2^{\varepsilon} \, \Gamma(1+\varepsilon)}{\varepsilon} \,
\left[
- i \, \big( - \Delta_1^{\{i\}} \big)^{-1-\varepsilon} \,
\int^{+\infty}_0 dz \left(1+z^2 \right)^{-1-\varepsilon}
\right.
\notag\\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\left.
- \big( \Delta_1^{\{i\}} \big)^{-1-\varepsilon} \,
\int^{+\infty}_1 dz \left(z^2-1 \right)^{-1-\varepsilon}
\right]\!.
\label{eqlijsoft9}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The <inline-formula><tex-math notation="LaTeX" id="ImEquation206"><![CDATA[$z$]]></tex-math></inline-formula> integrals are computed in Appendix <xref ref-type="sec" rid="SEC8">C</xref> [Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-4">C.4</xref>) and (<xref ref-type="disp-formula" rid="ptz160M8-5">C.5</xref>)]. Hence, for <inline-formula><tex-math notation="LaTeX" id="ImEquation207"><![CDATA[$L_{3}^{n}(0,\Delta_{1}^{\{i\}},0)$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M2-58"><label>(2.58)</label><tex-math notation="LaTeX" id="Equation65"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},0 )
&=
- \, \frac{2^{-\varepsilon}}{2 \, \varepsilon^2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^{2}(1- \varepsilon)}{\Gamma(1 - 2 \,\varepsilon)} \,
\frac{1}{\cos(\pi \, \varepsilon)}
\notag \\
&
\;\;\;\;\;\;\;\; {} \times
\left[
i \, \sin(\pi \, \varepsilon) \,
\big( - \Delta_1^{\{i\}} \big)^{-1-\varepsilon} \,
+
\big( \Delta_1^{\{i\}} \big)^{-1-\varepsilon} \,
\right]\!.
\label{eqlijsoft10}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Since <inline-formula><tex-math notation="LaTeX" id="ImEquation208"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i\}}) < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation209"><![CDATA[$( \Delta_1^{\{i\}} )^{-1-\varepsilon}$]]></tex-math></inline-formula> may be rewritten as <inline-formula><tex-math notation="LaTeX" id="ImEquation210"><![CDATA[$- \, {e}^{\, i \, \pi \, \varepsilon}\, ( - \, \Delta_1^{\{i\}} )^{-1-\varepsilon}$]]></tex-math></inline-formula>, so that <inline-formula><tex-math notation="LaTeX" id="ImEquation211"><![CDATA[$L_{3}^{n}(0,\Delta_{1}^{\{i\}},\widetilde{D}_{ij})$]]></tex-math></inline-formula> simplifies into
<disp-formula id="ptz160M2-59"><label>(2.59)</label><tex-math notation="LaTeX" id="Equation66"><![CDATA[$$\begin{align}
L_{3}^{n} ( 0,\Delta_{1}^{\{i\}},0 )
&=
\frac{1}{\varepsilon^2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^{2}(1- \varepsilon)}{\Gamma(1 - 2 \,\varepsilon)} \,
\big( - 2 \, \Delta_1^{\{i\}} \big)^{-1-\varepsilon} ,
\label{eqlijsoft10bis}
\end{align}$$]]></tex-math></disp-formula>
which coincides with Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>).</p>
</sec>
</sec>
<sec id="SEC2.4"><title>2.4. &#x201C;Direct&#x201D;&#x2013;&#x201C;indirect&#x201D; equivalence</title>
<p>We now show the equivalence between the &#x201C;indirect way&#x201D; and the &#x201C;direct way&#x201D; starting from the integral representation of <inline-formula><tex-math notation="LaTeX" id="ImEquation212"><![CDATA[$L_3^n(0,\Delta_1^{\{i\}},\widetilde{D}_{ij})$]]></tex-math></inline-formula> and disregarding the fact that some <inline-formula><tex-math notation="LaTeX" id="ImEquation213"><![CDATA[$\widetilde{D}_{ij}$]]></tex-math></inline-formula> may or may not vanish. Thus, <inline-formula><tex-math notation="LaTeX" id="ImEquation214"><![CDATA[$I_3^n$]]></tex-math></inline-formula> now reads
<disp-formula id="ptz160M2-60"><label>(2.60)</label><tex-math notation="LaTeX" id="Equation67"><![CDATA[$$\begin{align}
I_3^n
&= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G)} \,
\sum_{j \in S_3 \setminus \{i\}} \, \frac{\overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}}
\notag \\
&\quad \quad \quad \quad
\times
\int_{\widehat{(0,1)}_{i,j}} dz \,
\big(
\, z^2 \, (\widetilde{D}_{ij}+\Delta_1^{\{i\}}) - \Delta_1^{\{i\}}
\big)^{-1-\varepsilon} .
\label{eqlijsoft64}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Sticking to the general complex mass case, we perform the change of variable <inline-formula><tex-math notation="LaTeX" id="ImEquation215"><![CDATA[$s = \overline{b}_{j}^{\{i\}} \, z$]]></tex-math></inline-formula> in such a way that the two integrands corresponding to the sum over <inline-formula><tex-math notation="LaTeX" id="ImEquation216"><![CDATA[$j$]]></tex-math></inline-formula> (at fixed <inline-formula><tex-math notation="LaTeX" id="ImEquation217"><![CDATA[$i$]]></tex-math></inline-formula>) in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-60">2.60</xref>) are the same (use Eqs. (2.36) and (2.38) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]). Specifying the two elements of <inline-formula><tex-math notation="LaTeX" id="ImEquation218"><![CDATA[$S_3 \setminus \{i\}$]]></tex-math></inline-formula> to be <inline-formula><tex-math notation="LaTeX" id="ImEquation219"><![CDATA[$k \equiv 1 + ((i+1)$]]></tex-math></inline-formula> modulo <inline-formula><tex-math notation="LaTeX" id="ImEquation220"><![CDATA[$3)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation221"><![CDATA[$l \equiv 1 + (i$]]></tex-math></inline-formula> modulo <inline-formula><tex-math notation="LaTeX" id="ImEquation222"><![CDATA[$3)$]]></tex-math></inline-formula>, the two integrals are concatenated into a single one integrated along the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation223"><![CDATA[${\cal I}^{(i)}_{k,l} \equiv -\overline{b}_{k}^{\{i\}}\widehat{(0,1)}_{i,k} \cup \overline{b}_{l}^{\{i\}}\widehat{(0,1)}_{i,l}$]]></tex-math></inline-formula> in the complex <inline-formula><tex-math notation="LaTeX" id="ImEquation224"><![CDATA[$s$]]></tex-math></inline-formula>-plane:
<disp-formula id="ptz160M2-61"><label>(2.61)</label><tex-math notation="LaTeX" id="Equation68"><![CDATA[$$\begin{align}
I_3^n
&= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G) \det{(G^{\{i\}})}} \,
\notag \\
&\quad \quad \quad
\times
\int_{{\cal I}^{(i)}_{k,l}} ds
\left(
\frac{s^2 + \det{({\cal S}^{\{i\}})}}{\det{(G^{\{i\}})}}
\right)^{-1-\varepsilon} .
\label{eqlijsoft65}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As in the case <inline-formula><tex-math notation="LaTeX" id="ImEquation225"><![CDATA[$\Delta_2 \ne 0$]]></tex-math></inline-formula> treated in Ref. [<xref ref-type="bibr" rid="B2">2</xref>], the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation226"><![CDATA[${\cal I}^{(i)}_{k,l}$]]></tex-math></inline-formula> can be deformed into the straight line <inline-formula><tex-math notation="LaTeX" id="ImEquation227"><![CDATA[$\big[ -\overline{b}_{k}^{\{i\}},\overline{b}_{l}^{\{i\}} \big]$]]></tex-math></inline-formula> as depicted in <xref ref-type="fig" rid="F3">Fig. 3</xref>:
<disp-formula id="ptz160M2-62"><label>(2.62)</label><tex-math notation="LaTeX" id="Equation69"><![CDATA[$$\begin{align}
I_3^n &= - \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G) \det{(G^{\{i\}})}} \,
\notag \\
&\quad
\times
\int_{- \overline{b}_{k}^{\{i\}}}^{\overline{b}_{l}^{\{i\}}} ds
\left(
\frac{s^2 + \det{({\cal S}^{\{i\}})}}{\det{(G^{\{i\}})}}
\right)^{-1-\varepsilon} .
\label{eqlijsoft66}
\end{align}$$]]></tex-math></disp-formula></p>
<fig id="F3" orientation="portrait" position="float"><label>Fig. 3.</label><caption><p>Example of a contour deformation involving a triangle with one distorted side, for which no cut crosses the straight base <inline-formula><tex-math notation="LaTeX" id="ImEquation228"><![CDATA[$[ -\overline{b}_{k}^{\{i\}},\overline{b}_{l}^{\{i\}}]$]]></tex-math></inline-formula>.</p></caption>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" orientation="portrait" position="float" mimetype="image" xlink:href="ptz160f3.tif"/></fig>
<p>Performing the change of variable <inline-formula><tex-math notation="LaTeX" id="ImEquation229"><![CDATA[$s = -\overline{b}_{k}^{\{i\}} - \det{(G^{\{i\}})} \,u$]]></tex-math></inline-formula> and following the procedure given by Eqs. (2.45)&#x2013; (2.47) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] leads to
<disp-formula id="ptz160UM8"><tex-math notation="LaTeX" id="Equation70"><![CDATA[$$\begin{align}
I_3^n &= \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\sum_{i \in S_3} \, \frac{\overline{b}_i}{\det(G)} \,
\int_{0}^{1} du \big( D^{\{i\} \, (l)}(u) \big)^{-1-\varepsilon},
\nonumber
\end{align}$$]]></tex-math></disp-formula>
namely, Eq. (<xref ref-type="disp-formula" rid="ptz160M2-5">2.5</xref>).</p>
<p>It is instructive to recover the results of the &#x201C;direct way&#x201D; from those of the &#x201C;indirect way&#x201D; using for the latter the closed form formulae. These formulae can be obtained for the case <inline-formula><tex-math notation="LaTeX" id="ImEquation230"><![CDATA[$\widetilde{D}_{ij} \ne 0$]]></tex-math></inline-formula> by using results of Appendix <xref ref-type="sec" rid="SEC10">E</xref> and for <inline-formula><tex-math notation="LaTeX" id="ImEquation231"><![CDATA[$\widetilde{D}_{ij} = 0$]]></tex-math></inline-formula> using Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>). This exercise is performed in great detail in Appendix <xref ref-type="sec" rid="SEC9">D</xref>.</p>
</sec>
</sec>
<sec id="SEC3"><title>3. Four-point function with infrared divergences</title>
<p>The one-loop four-point function is depicted in <xref ref-type="fig" rid="F4">Fig. 4</xref>. In the case where infrared (IR) divergences appear, these can be regularized by dimensional regularization shifting the spacetime dimension <inline-formula><tex-math notation="LaTeX" id="ImEquation232"><![CDATA[$n = 4 - 2 \varepsilon$]]></tex-math></inline-formula> slightly above 4 (<inline-formula><tex-math notation="LaTeX" id="ImEquation233"><![CDATA[$\varepsilon < 0$]]></tex-math></inline-formula>). The Feynman parametrization of <inline-formula><tex-math notation="LaTeX" id="ImEquation234"><![CDATA[$I_4^n$]]></tex-math></inline-formula> reads
<disp-formula id="ptz160M3-1"><label>(3.1)</label><tex-math notation="LaTeX" id="Equation71"><![CDATA[$$\begin{eqnarray}
I_4^n
& = &
\Gamma\left(2 + \varepsilon \right)
\int_0^1 \, \prod_{i=1}^4 dz_i\, \delta\bigg(1- \sum_{i=1}^4 z_i\bigg)
\left( -\frac{1}{2} \, Z^{\;{\rm T}} \cdot
{\cal S} \cdot Z - i \, \lambda \right)^{-2 - \varepsilon} ,
\label{eqstartingpointir}
\end{eqnarray}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation235"><![CDATA[$Z$]]></tex-math></inline-formula> is a column four-vector whose components are the <inline-formula><tex-math notation="LaTeX" id="ImEquation236"><![CDATA[$z_{i}$]]></tex-math></inline-formula>. The power <inline-formula><tex-math notation="LaTeX" id="ImEquation237"><![CDATA[$- 2 - \varepsilon$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-1">3.1</xref>) is not an integer; notwithstanding this, the tricks and techniques elaborated in Sect. 3 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] can be used with a slight adaptation. Let us sketch the different steps for this special case.</p>
<fig id="F4" orientation="portrait" position="float"><label>Fig. 4.</label><caption><p>The box picturing the one-loop four-point function.</p></caption>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" orientation="portrait" position="float" mimetype="image" xlink:href="ptz160f4.tif"/></fig>
<sec id="SEC3.1"><title>3.1. Computation of <inline-formula><tex-math notation="LaTeX" id="ImEquation238"><![CDATA[$I_4^{n}$]]></tex-math></inline-formula></title>
<p>We make use of the identity in Eq. (2.26) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] to shift the power of the denominator in the integrand, choosing <inline-formula><tex-math notation="LaTeX" id="ImEquation239"><![CDATA[$\mu = 5/2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation240"><![CDATA[$\nu = 2/(1-2 \varepsilon)$]]></tex-math></inline-formula> so that <inline-formula><tex-math notation="LaTeX" id="ImEquation241"><![CDATA[$I_4^{n}$]]></tex-math></inline-formula> is recast as
<disp-formula id="ptz160M3-2"><label>(3.2)</label><tex-math notation="LaTeX" id="Equation72"><![CDATA[$$\begin{eqnarray}
I_4^{n}
& = &
\frac{2^{3+\varepsilon}}{B(2 + \varepsilon,1/2-\varepsilon)} \,
\frac{\Gamma(2 + \varepsilon)}{(1-2 \, \varepsilon)} \,
\nonumber \\
&& \mbox{} \times
\int_0^{+\infty} d \xi \,
\int_{\Sigma_{bcd}}
\frac{dx_b \, d x_c \, d x_d}
{(D^{(a)}(x_b,x_c,x_d) + \xi^{\nu} - i \, \lambda)^{5/2}} .
\label{eqI4b1}
\end{eqnarray}$$]]></tex-math></disp-formula></p>
<p>Step 1 is very similar to the case <inline-formula><tex-math notation="LaTeX" id="ImEquation242"><![CDATA[$n=4$]]></tex-math></inline-formula>, and we get
<disp-formula id="ptz160M3-3"><label>(3.3)</label><tex-math notation="LaTeX" id="Equation73"><![CDATA[$$\begin{eqnarray}
I_4^{n}
& = &
\frac{2^{3+\varepsilon}}{3 \, B(2 +\varepsilon,1/2 - \varepsilon)} \,
\frac{\Gamma(2 +\varepsilon)}{(1-2 \, \varepsilon)} \,
\nonumber \\
&& \mbox{} \times
\sum_{i=1}^{4} \, \frac{\overline{b}_i}{\det{(G)}} \,
\int_0^{+\infty}
d \xi \, \frac{1}{\Delta_3-\xi^{\nu} + i \, \lambda}
\nonumber\\
& &
\quad{} \quad{} \quad{} \quad{}
\times \int_{\Sigma_{kl}}
\frac{dx_k \, d x_l}
{(D^{\{i\} \, (i^{\prime})}(x_k,x_l)
+ \xi^{\nu} - i \, \lambda)^{3/2}} ,
\label{eqI4b2}
\end{eqnarray}$$]]></tex-math></disp-formula>
with the same notational conventions as in Eqs. (3.6) and (3.17) of Sect. <xref ref-type="sec" rid="SEC3">3</xref> in Ref. [<xref ref-type="bibr" rid="B1">1</xref>]. Likewise, steps 2 and 3 are identical to Sect. 3 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] except that the power of the variable <inline-formula><tex-math notation="LaTeX" id="ImEquation243"><![CDATA[$\xi$]]></tex-math></inline-formula> is <inline-formula><tex-math notation="LaTeX" id="ImEquation244"><![CDATA[$\nu$]]></tex-math></inline-formula> instead of 2, and will not be repeated here. Regarding step 4, the integration is performed over the variables <inline-formula><tex-math notation="LaTeX" id="ImEquation245"><![CDATA[$\xi$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation246"><![CDATA[$\rho$]]></tex-math></inline-formula>, then <inline-formula><tex-math notation="LaTeX" id="ImEquation247"><![CDATA[$\sigma$]]></tex-math></inline-formula> in the corresponding <inline-formula><tex-math notation="LaTeX" id="ImEquation248"><![CDATA[$L_4^n(\Delta_3,\Delta_2^{\{i\}},\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula>, now given by
<disp-formula id="ptz160M3-4"><label>(3.4)</label><tex-math notation="LaTeX" id="Equation74"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,\Delta_2^{\{i\}},\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})
&=
\kappa \,\int^{+\infty}_0 d \xi \, \int^{+\infty}_0 d \rho \,
\int^{+\infty}_0 d \sigma
\frac{1}{\xi^{\nu} - \Delta_3 - i \, \lambda} \,
\nonumber \\
&\quad
\mbox{} \times \frac{1}{\xi^{\nu} + \rho^2 - \Delta_2^{\{i\}} - i \, \lambda} \,
\frac{1}{\xi^{\nu} + \rho^2 + \sigma^2 - \Delta_1^{\{i,j\}} - i \, \lambda} \notag \\
&\quad {} \times \frac{1}{(\widetilde{D}_{ijk} + \xi^{\nu} + \rho^2 + \sigma^2 - i \, \lambda)^{1/2}} ,
\label{eqdefnl1}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160UM9"><tex-math notation="LaTeX" id="Equation75"><![CDATA[$$\kappa =
\frac{2^{4+\varepsilon}}
{3 \, B(2+\varepsilon,1/2-\varepsilon) \, B(3/2,1/2) \, B(1,1/2)}
\frac{\Gamma(2+\varepsilon)}{(1-2 \, \varepsilon)} ,$$]]></tex-math></disp-formula>
reminiscent of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-35">3.35</xref>) of Sect. <xref ref-type="sec" rid="SEC3">3</xref> in Ref. [<xref ref-type="bibr" rid="B1">1</xref>]. Infrared divergences in the four-point function are not dominant Landau-type singularities but subleading ones. Such an infrared divergence corresponds to the vanishing determinant <inline-formula><tex-math notation="LaTeX" id="ImEquation249"><![CDATA[$\det{({\cal S}^{\{i\}})}$]]></tex-math></inline-formula> of some reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation250"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix (or some <inline-formula><tex-math notation="LaTeX" id="ImEquation251"><![CDATA[$\Delta_2^{\{i\}}$]]></tex-math></inline-formula>) associated with some three-point functions which are obtained from the four-point function considered by one pinching [<xref ref-type="bibr" rid="B7">7</xref>]. The various cases of vanishing kinematic matrices associated with three-point functions plagued with infrared soft or collinear singularities have been evoked in Sect. <xref ref-type="sec" rid="SEC2">2</xref>. Let us note that the method developed in Sect. 3 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] is still valid because we never divide by <inline-formula><tex-math notation="LaTeX" id="ImEquation252"><![CDATA[$\Delta_2^{\{i\}}$]]></tex-math></inline-formula> per se but by <inline-formula><tex-math notation="LaTeX" id="ImEquation253"><![CDATA[$\Delta_2^{\{i\}}- \xi^{\nu}- \rho^2$]]></tex-math></inline-formula>. The divergences will show up when performing the integrations over the parameters of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-4">3.4</xref>). Note also that if we face a case when IR divergences arise when there are some non-vanishing internal masses, only some of the contributions (let us call them sectors) &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation254"><![CDATA[$i,j,k$]]></tex-math></inline-formula>,&#x201D; not all, are plagued with IR divergences; the other ones, which correspond to <inline-formula><tex-math notation="LaTeX" id="ImEquation255"><![CDATA[$\Delta_2^{\{i\}} \ne 0$]]></tex-math></inline-formula>, will be treated as in the <inline-formula><tex-math notation="LaTeX" id="ImEquation256"><![CDATA[$n=4$]]></tex-math></inline-formula> case. Also, in any IR-divergent sector &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation257"><![CDATA[$i,j,k$]]></tex-math></inline-formula>&#x201D; for which <inline-formula><tex-math notation="LaTeX" id="ImEquation258"><![CDATA[$\Delta_2^{\{i\}}=0$]]></tex-math></inline-formula>, we do not have to sum over all three sub-sectors <inline-formula><tex-math notation="LaTeX" id="ImEquation259"><![CDATA[$j \in S_{4} \setminus \{i\}$]]></tex-math></inline-formula> because, as in the IR-divergent three-point function case, some of the <inline-formula><tex-math notation="LaTeX" id="ImEquation260"><![CDATA[$\overline{b}_{i}^{\{j\}}$]]></tex-math></inline-formula> vanish.</p>
<p>Let us consider a sector &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation261"><![CDATA[$i,j,k$]]></tex-math></inline-formula>&#x201D; which diverges in the IR region. We have to distinguish two cases: (1) when <inline-formula><tex-math notation="LaTeX" id="ImEquation262"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation263"><![CDATA[$\widetilde{D}_{ijk} \ne 0$]]></tex-math></inline-formula>, in which case there are only soft divergences; (2) when both <inline-formula><tex-math notation="LaTeX" id="ImEquation264"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation265"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula>, in which case there are collinear or both soft and collinear divergences. In this section we will treat both real and complex mass cases. Some of the cases correspond to real or complex masses, others to complex masses only. For those corresponding to real or complex masses, we keep the imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation266"><![CDATA[$- i \, \lambda$]]></tex-math></inline-formula>, explicitly bearing in mind that with complex masses this <inline-formula><tex-math notation="LaTeX" id="ImEquation267"><![CDATA[$- i \, \lambda$]]></tex-math></inline-formula> is ineffective. Let us stress in passing that we also keep an infinitesimal prescription <inline-formula><tex-math notation="LaTeX" id="ImEquation268"><![CDATA[$- i \, \lambda$]]></tex-math></inline-formula> in the pole term where we put <inline-formula><tex-math notation="LaTeX" id="ImEquation269"><![CDATA[$\Delta_{2}^{\{i\}} = 0$]]></tex-math></inline-formula>.</p>
</sec>
<sec id="SEC3.2"><title>3.2. <inline-formula><tex-math notation="LaTeX" id="ImEquation270"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation271"><![CDATA[$\widetilde{D}_{ijk} \ne 0$]]></tex-math></inline-formula></title>
<p><disp-formula id="ptz160M3-5"><label>(3.5)</label><tex-math notation="LaTeX" id="Equation76"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) &=
\kappa \,\int^{+\infty}_0 d \xi \, \int^{+\infty}_0 d \rho \,
\int^{+\infty}_0 d \sigma
\frac{1}
{(\xi^{\nu} - \Delta_3 - i \, \lambda) \, (\xi^{\nu} + \rho^2 - i \, \lambda)}
\nonumber \\
&\quad {} \quad {}
\times
\frac{1}
{(\xi^{\nu} + \rho^2 + \sigma^2 - \Delta_1^{\{i,j\}} - i \, \lambda) \,
(\widetilde{D}_{ijk} + \xi^{\nu} + \rho^2 + \sigma^2 - i \, \lambda)^{1/2}} .
\label{eqdefnlir2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the complex mass case, <inline-formula><tex-math notation="LaTeX" id="ImEquation272"><![CDATA[$\operatorname{Im}(\Delta_3)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation273"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}})$]]></tex-math></inline-formula> have arbitrary signs and <inline-formula><tex-math notation="LaTeX" id="ImEquation274"><![CDATA[$\operatorname{Im}(\widetilde{D}_{ijk})$]]></tex-math></inline-formula> is negative, so we have to distinguish between different cases.</p>
<p>Let us define the function <inline-formula><tex-math notation="LaTeX" id="ImEquation275"><![CDATA[$M_2(\xi^{\nu})$]]></tex-math></inline-formula> as
<disp-formula id="ptz160M3-6"><label>(3.6)</label><tex-math notation="LaTeX" id="Equation77"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) &= \kappa \,\int^{+\infty}_0 d \xi \, \frac{1}{\xi^{\nu} - \Delta_3 - i \, \lambda} \, M_2(\xi^{\nu}) .
\label{eqdefM2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The integrations over <inline-formula><tex-math notation="LaTeX" id="ImEquation276"><![CDATA[$\sigma$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation277"><![CDATA[$\rho$]]></tex-math></inline-formula> are identical to those appearing in the massive cases, we thus borrow the results derived in Sect. 3.4 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] and in Sect. 3.1 of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] with <inline-formula><tex-math notation="LaTeX" id="ImEquation278"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula>, explicitly: <sup><xref ref-type="fn" rid="FN8">8</xref></sup>
<list list-type="simple">
<list-item><p>for <inline-formula><tex-math notation="LaTeX" id="ImEquation279"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) \geq 0$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M3-7"><label>(3.7)</label><tex-math notation="LaTeX" id="Equation78"><![CDATA[$$\begin{align}
M_2(\xi^{\nu})
&=
\frac{1}{2} \, B(1/2,1/2) \,
\int^1_0 \frac{d u}{u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}}) -\Delta_1^{\{i,j\}}}
\notag \\
&\quad {} \times
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}}
-
\frac{1}
{\big(\xi^{\nu}+ u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}})
-
\Delta_1^{\{i,j\}} - i \, \lambda \big)^{1/2}}
\right]\!;
\label{eqisigir3}
\end{align}$$]]></tex-math></disp-formula></p></list-item>
<list-item><p>for <inline-formula><tex-math notation="LaTeX" id="ImEquation280"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) < 0$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M3-8"><label>(3.8)</label><tex-math notation="LaTeX" id="Equation79"><![CDATA[$$\begin{align}
M_2(\xi^{\nu})
&=
- \frac{1}{2} \, B(1/2,1/2) \,
\left\{
i \, \int^{+\infty}_0
\frac{d u}{u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}}) +\Delta_1^{\{i,j\}}}
\right.
\notag \\
&\quad {} \times
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}} -
\frac{1}{\big(\xi^{\nu}- u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}})
-\Delta_1^{\{i,j\}} \big)^{1/2}}
\right]
\notag \\
&\quad {}
+ \int^{+\infty}_1
\frac{d u}{u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}}) -\Delta_1^{\{i,j\}}}
\notag \\
&\quad {} \times
\left.
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}}
-
\frac{1}
{\big(\xi^{\nu}+ u^2 \, (\widetilde{D}_{ijk}+\Delta_1^{\{i,j\}})-\Delta_1^{\{i,j\}}
\big)^{1/2}}
\right]
\right\}\!.
\label{eqisigir4}
\end{align}$$]]></tex-math></disp-formula></p></list-item>
</list></p>
<p>The integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation281"><![CDATA[$\xi$]]></tex-math></inline-formula> to obtain <inline-formula><tex-math notation="LaTeX" id="ImEquation282"><![CDATA[$L_4^n$]]></tex-math></inline-formula> is of the &#x201C;second kind&#x201D;&#x2014; cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) and (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation283"><![CDATA[$\nu = 2/(1 - 2 \, \varepsilon)$]]></tex-math></inline-formula>. Let us go through the different cases with respect to the sign of the imaginary part of <inline-formula><tex-math notation="LaTeX" id="ImEquation284"><![CDATA[$\Delta_3$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation285"><![CDATA[$\Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>. Sticking with the notation of Sect. 3 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>], we introduce
<disp-formula id="ptz160UM10"><tex-math notation="LaTeX" id="Equation80"><![CDATA[$$\begin{align*}
P_{ijk} &= \widetilde{D}_{ijk} + \Delta_1^{\{i,j\}} , \\
R_{ij} &= - \, \Delta_1^{\{i,j\}} , \\
T &= - \, \Delta_3 .
\end{align*}$$]]></tex-math></disp-formula></p>
<p>The strategy for computing the different integrals is the same for all cases, and very similar to Sect. 3 of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]. We display the successive steps for the first case, and we give the final results for the three others.</p>
<sec id="SEC3.2.1"><title>3.2.1. <inline-formula><tex-math notation="LaTeX" id="ImEquation286"><![CDATA[$\operatorname{Im}(\Delta_3) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation287"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) > 0$]]></tex-math></inline-formula></title>
<p>We start with Eq. (<xref ref-type="disp-formula" rid="ptz160M3-7">3.7</xref>) for <inline-formula><tex-math notation="LaTeX" id="ImEquation288"><![CDATA[$M_2(\xi^{\nu})$]]></tex-math></inline-formula> and use Eq. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) for the <inline-formula><tex-math notation="LaTeX" id="ImEquation289"><![CDATA[$\xi$]]></tex-math></inline-formula> integration to get
<disp-formula id="ptz160M3-9"><label>(3.9)</label><tex-math notation="LaTeX" id="Equation81"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})
&
= F(\varepsilon) \, \int^1_0 \frac{d u}{u^2 \, P_{ijk} + R_{ij}} \notag \\
&\qquad \qquad \quad {} \times \left[
\int^1_0 \frac{d z}
{\left( (1-z^2) \, T - i \, \lambda \right)^{1+\varepsilon}}
\right.
\label{eqlijkir10} \\
&\qquad \qquad \qquad \quad {}
-
\left.
\int^1_0 \frac{d z}
{\left(
z^2 \, (u^2 \, P_{ijk} + R_{ij})+ (1-z^2) \, T - i \, \lambda
\right)^{1+\varepsilon}
}
\right]\!,
\notag
\end{align}$$]]></tex-math></disp-formula>
where
<disp-formula id="ptz160M3-10"><label>(3.10)</label><tex-math notation="LaTeX" id="Equation82"><![CDATA[$$\begin{align}
F(\varepsilon)
&= 2^{1+\varepsilon} \, \Gamma(1+\varepsilon) .
\label{eqdeffepsilon}
\end{align}$$]]></tex-math></disp-formula></p>
<p>To help the reader, we will define some steps in a similar way to Sect. 3.4 of Ref. [<xref ref-type="bibr" rid="B1">1</xref>].</p>
<p><italic>Step 1</italic> We change <inline-formula><tex-math notation="LaTeX" id="ImEquation290"><![CDATA[$u=\sqrt{y/x}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation291"><![CDATA[$z = \sqrt{x}$]]></tex-math></inline-formula> and exchange the <inline-formula><tex-math notation="LaTeX" id="ImEquation292"><![CDATA[$y$]]></tex-math></inline-formula> and the <inline-formula><tex-math notation="LaTeX" id="ImEquation293"><![CDATA[$x$]]></tex-math></inline-formula> integration so that
<disp-formula id="ptz160M3-11"><label>(3.11)</label><tex-math notation="LaTeX" id="Equation83"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})
&=
- \frac{F(\varepsilon)}{4} \,
\int^1_0 \frac{d y}{\sqrt{y}} \, \int^1_y d x \,
\frac{1}{y \, P_{ijk} + x \, R_{ij}} \notag \\
&\qquad \qquad {} \times \left[
\frac{1}{\left( y \, P_{ijk} + x \, (R_{ij} - T) + T
- i \, \lambda \right)^{1+\varepsilon}}
\right.
\notag \\
&\qquad \qquad \qquad {}
-
\left.
\frac{1}{\left( (1-x) \, T - i \, \lambda \right)^{1+\varepsilon}}
\right]\!.
\label{eqlijkir12}
\end{align}$$]]></tex-math></disp-formula></p>
<p><italic>Step 2</italic> We set <inline-formula><tex-math notation="LaTeX" id="ImEquation294"><![CDATA[$y = u^2$]]></tex-math></inline-formula> and perform a partial fraction decomposition on the variable <inline-formula><tex-math notation="LaTeX" id="ImEquation295"><![CDATA[$x$]]></tex-math></inline-formula> to write <inline-formula><tex-math notation="LaTeX" id="ImEquation296"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> in the following form:
<disp-formula id="ptz160M3-12"><label>(3.12)</label><tex-math notation="LaTeX" id="Equation84"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\
&=
- \frac{F(\varepsilon)}{2} \,
\int^1_0 \frac{du}{u^2 \, P_{ijk} T + R_{ij} (T - i \, \lambda)}
\int^1_{u^2} d x \,
\notag \\
&\qquad \qquad {}
\times
\Biggl\{
(T - R_{ij}) \,
\left[
u^2 \, P_{ijk} + x \, (R_{ij} - T) + T - i \, \lambda
\right]^{-1-\varepsilon} \notag \\
&\qquad \qquad \qquad {}
-
T \,
\left[ (1-x) \, T - i \, \lambda \right]^{-1-\varepsilon}
+ \frac{R_{ij}}{u^2 \, P_{ijk} + x \, R_{ij}}
\notag \\
&\qquad {} \qquad \qquad {}
\times
\Biggl[
\left[
u^2 \, P_{ijk} + x \, (R_{ij} - T) + T - i \, \lambda
\right]^{-\varepsilon}
-
\left[ (1-x) \, T - i \, \lambda \right]^{-\varepsilon} \,
\Biggr]
\Biggr\} .
\label{eqlijkir131}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The last line of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-12">3.12</xref>) provides a contribution of order <inline-formula><tex-math notation="LaTeX" id="ImEquation297"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> only; for the computation of the one-loop four-point function it can thus be dropped.<sup><xref ref-type="fn" rid="FN9">9</xref></sup></p>
<p><italic>Step 3</italic> Thus, ignoring these terms, the integration in <inline-formula><tex-math notation="LaTeX" id="ImEquation298"><![CDATA[$x$]]></tex-math></inline-formula> readily provides
<disp-formula id="ptz160M3-13"><label>(3.13)</label><tex-math notation="LaTeX" id="Equation85"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})
&=
\frac{2^{\varepsilon}}{T} \, \frac{\Gamma(1+\varepsilon)}{\varepsilon}
\int^1_0
\frac{d u}{u^2 \, P_{ijk} + R_{ij} - i \, \lambda \, \sigma_0} \,
\label{eqlijkir14} \\
&
\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\times
\Bigl\{
\left[
u^2 \, (P_{ijk} + R_{ij} - T) + T - i \, \lambda
\right]^{-\varepsilon}
\notag\\
&
\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
-
\left[
u^2 \, P_{ijk} + R_{ij} - i \, \lambda
\right]^{-\varepsilon}
- (T - i \, \lambda)^{-\varepsilon} \, (1-u^2)^{-\varepsilon}
\Bigr\} ,
\notag
\end{align}$$]]></tex-math></disp-formula>
where we have introduced <inline-formula><tex-math notation="LaTeX" id="ImEquation299"><![CDATA[$\sigma_0 = \mbox{sign}(R_{ij}/T)$]]></tex-math></inline-formula> in the real mass case. In the complex mass case all &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation300"><![CDATA[$-i \lambda$]]></tex-math></inline-formula>&#x201D; and &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation301"><![CDATA[$- i \sigma_0 \, \lambda$]]></tex-math></inline-formula>&#x201D; contour prescriptions are ineffective and irrelevant, and all three terms in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-13">3.13</xref>) can be straightforwardly expanded in powers of <inline-formula><tex-math notation="LaTeX" id="ImEquation302"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>. In contrast the real mass case requires a more cautious treatment, which we elaborate below. We note that when <inline-formula><tex-math notation="LaTeX" id="ImEquation303"><![CDATA[$u^2 \to - R_{ij}/P_{ijk}$]]></tex-math></inline-formula>,
<disp-formula id="ptz160UM11"><tex-math notation="LaTeX" id="Equation86"><![CDATA[$$\begin{align}
u^2 \, (P_{ijk} + R_{ij} - T) + T & \to (T - R_{ij}) \frac{\widetilde{D}_{ijk}}{P_{ijk}}
\neq 0 ,
\notag\\
1-u^2 & \to \frac{\widetilde{D}_{ijk}}{P_{ijk}} \neq 0 ,
\notag
\end{align}$$]]></tex-math></disp-formula>
i.e. the pole at <inline-formula><tex-math notation="LaTeX" id="ImEquation304"><![CDATA[$u^2 = - R_{ij}/P_{ijk}$]]></tex-math></inline-formula> is distinct from the branch points of the first and third functions in the numerator. Therefore we can readily perform an expansion around <inline-formula><tex-math notation="LaTeX" id="ImEquation305"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> for the first and third terms as
<disp-formula id="ptz160UM12"><tex-math notation="LaTeX" id="Equation87"><![CDATA[$$\begin{align}
\frac{1}{\varepsilon} \,
\left[
u^2 \, (P_{ijk} + R_{ij} - T) + T - i \, \lambda
\right]^{-\varepsilon}
& =
\frac{1}{\varepsilon} -
\ln
\left[
u^2 \, (P_{ijk} + R_{ij} - T) + T - i \, \lambda
\right] + {\cal O}(\varepsilon) ,
\notag\\
\frac{1}{\varepsilon} \,
(T - i \, \lambda)^{-\varepsilon} \, (1-u^2)^{-\varepsilon}
& =
- \frac{1}{\varepsilon} -
\ln(T - i \, \lambda) - \ln(1-u^2) + {\cal O}(\varepsilon) .
\notag
\end{align}$$]]></tex-math></disp-formula></p>
<p>The <inline-formula><tex-math notation="LaTeX" id="ImEquation306"><![CDATA[$1/\varepsilon$]]></tex-math></inline-formula> poles cancel between these two contributions, leaving only logarithms. On the other hand, the contribution coming from the second term requires some care since pole and branch point coincide. If this singularity lies outside the integration region the contour prescription for the pole is irrelevant and can be dropped. If the singularity lies inside <inline-formula><tex-math notation="LaTeX" id="ImEquation307"><![CDATA[$[0,1]$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation308"><![CDATA[$\sigma_0=+$]]></tex-math></inline-formula>, the contour prescription for the pole and cut are the same, there is no pinching. The integral over <inline-formula><tex-math notation="LaTeX" id="ImEquation309"><![CDATA[$u$]]></tex-math></inline-formula> can be performed after an expansion in <inline-formula><tex-math notation="LaTeX" id="ImEquation310"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> using Appendix <xref ref-type="sec" rid="SEC10">E</xref>, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M10-8">E.8</xref>). If the singularity lies inside <inline-formula><tex-math notation="LaTeX" id="ImEquation311"><![CDATA[$[0,1]$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation312"><![CDATA[$\sigma_0 = -$]]></tex-math></inline-formula>, however, a pinching occurs at the singular point in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation313"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula>. A too early expansion in powers of <inline-formula><tex-math notation="LaTeX" id="ImEquation314"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> before performing the integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation315"><![CDATA[$u$]]></tex-math></inline-formula> would lead to a divergence order by order in <inline-formula><tex-math notation="LaTeX" id="ImEquation316"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>. On the other hand, we note that for <inline-formula><tex-math notation="LaTeX" id="ImEquation317"><![CDATA[$u^2 = - R_{ij}/P_{ijk}$]]></tex-math></inline-formula> the numerator of the second term, i.e. the pole residue, is <inline-formula><tex-math notation="LaTeX" id="ImEquation318"><![CDATA[$(- \, i \, \lambda)^{-\varepsilon} \to 0$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation319"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula> for any fixed <inline-formula><tex-math notation="LaTeX" id="ImEquation320"><![CDATA[$\varepsilon <0$]]></tex-math></inline-formula>. Therefore, up to terms vanishing <inline-formula><tex-math notation="LaTeX" id="ImEquation321"><![CDATA[$\propto \lambda^{- \varepsilon}$]]></tex-math></inline-formula>, we can make the replacement
<disp-formula id="ptz160UM13"><tex-math notation="LaTeX" id="Equation88"><![CDATA[$$\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, \lambda \right]^{-\varepsilon}}
{u^2 \, P_{ijk} + R_{ij} + i \, \lambda}
\to
\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, \lambda \right]^{-\varepsilon}}
{u^2 \, P_{ijk} + R_{ij} - i \, \lambda} .$$]]></tex-math></disp-formula></p>
<p>Details are provided in Appendix <xref ref-type="sec" rid="SEC11">F</xref>. Finally, we perform a partial expansion around <inline-formula><tex-math notation="LaTeX" id="ImEquation322"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> to get
<disp-formula id="ptz160M3-14"><label>(3.14)</label><tex-math notation="LaTeX" id="Equation89"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) = \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon) \notag \\
& {} \times
\left\{
\int^1_0 \frac{d u }{u^2 P_{ijk} + R_{ij} - i \lambda \sigma_0}
\left[
\ln
\left(
(T - i \lambda) (1-u^2)
\right) -
\ln
\left(
u^2 (P_{ijk} + R_{ij} - T) + T - i \lambda
\right)
\right]
\right.
\notag\\
&\qquad{}
\left.
-
\frac{1}{\varepsilon} \,
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} - i \, \lambda}
+
\int^1_0 d u \,
\frac{\ln(u^2 \, P_{ijk} + R_{ij} - i \, \lambda)}
{u^2 \, P_{ijk} + R_{ij} - i \, \lambda}
\right\}\!.
\label{eqlijkir15}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Note that the imaginary part of the argument of each logarithm in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-14">3.14</xref>) keeps a constant sign when <inline-formula><tex-math notation="LaTeX" id="ImEquation323"><![CDATA[$u$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation324"><![CDATA[$[0,1]$]]></tex-math></inline-formula>. This is obviously true for the real mass case, and in the case of complex masses this is easily verified keeping in mind the assumptions that <inline-formula><tex-math notation="LaTeX" id="ImEquation325"><![CDATA[$\operatorname{Im}(T) < 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation326"><![CDATA[$\operatorname{Im}(R_{ij}) < 0$]]></tex-math></inline-formula>. For the sake of coherence with respect to Ref. [<xref ref-type="bibr" rid="B2">2</xref>], the pole residue contributions will be added and subtracted for the two logarithms making up the first term inside the curly brackets of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-14">3.14</xref>), which is recast in the form
<disp-formula id="ptz160M3-15"><label>(3.15)</label><tex-math notation="LaTeX" id="Equation90"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) = \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon) \notag \\
& {} \times
\left\{
-
\frac{1}{\varepsilon} \,
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} - i \, \lambda}
+
\int^1_0 d u \,
\frac{\ln(u^2 \, P_{ijk} + R_{ij} - i \, \lambda)}
{u^2 \, P_{ijk} + R_{ij} - i \, \lambda}
\right.
\notag \\
&\qquad {}
+ \int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} - i \, \lambda \, \sigma_0} \notag\\
&\qquad \quad{}
\times
\left[ \ln \left( (T - i \lambda) \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} - i \, \lambda\right) \right.
\notag\\
&\qquad \quad{}
-
\left. \ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T - i \, \lambda \right) + \ln \left( \frac{(P_{ijk}+R_{ij}) \, (T - R_{ij})}{P_{ijk}} - i \, \lambda \right) \right]
\notag\\
&\qquad {}
\left.
-
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} - i \, \lambda \, \sigma_0} \,
\ln \left( \frac{T - R_{ij} - i \, \lambda \, \sigma_1}{T - i \, \lambda \, \sigma_1} \right)
\right\}\!,
\label{eqlijkir15r}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation327"><![CDATA[$\sigma_1 = \mbox{sign}( (P_{ijk} + R_{ij})/P_{ijk} )$]]></tex-math></inline-formula> for the real mass case and
<disp-formula id="ptz160M3-16"><label>(3.16)</label><tex-math notation="LaTeX" id="Equation91"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) = \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon) \notag \\
& {} \times
\left\{
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} } \,
\left[ \ln \left( T \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} \right) \right]
\right.
\notag \\
&\qquad {} \;
-
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} } \,
\left[ \ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T \right) - \ln \left( \frac{(P_{ijk}+R_{ij}) \, (T - R_{ij})}{P_{ijk}} \right) \right]
\notag\\
&\qquad {} \;
-
\frac{1}{\varepsilon} \,
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} }
+
\int^1_0 d u \,
\frac{\ln(u^2 \, P_{ijk} + R_{ij} )}
{u^2 \, P_{ijk} + R_{ij} }
\notag\\
&\qquad {} \;
\left.
-
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} } \,
\left[ \ln \left( \frac{T - R_{ij}}{T} \right) + \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) \right]
\right\}
\label{eqlijkir15c}
\end{align}$$]]></tex-math></disp-formula>
for the complex mass case. The relevant integrals are given in Appendices <xref ref-type="sec" rid="SEC10">E</xref> of this paper, E of Ref. [<xref ref-type="bibr" rid="B1">1</xref>], and B of Ref. [<xref ref-type="bibr" rid="B2">2</xref>].</p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M3-16">3.16</xref>) matches Eq. (3.19) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] obtained in the corresponding general complex mass case 1(a), considering the latter in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation328"><![CDATA[$\operatorname{Re}(\Delta_{2}^{\{i\}}) = \operatorname{Re}(Q_{i}+T) \to 0$]]></tex-math></inline-formula> while keeping an infinitesimal positive imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation329"><![CDATA[$\operatorname{Im}(\Delta_{2}^{\{i\}}) = \lambda$]]></tex-math></inline-formula>. One then formally gets
<disp-formula id="ptz160UM14"><tex-math notation="LaTeX" id="Equation92"><![CDATA[$$\begin{multline}
\text{RHS}
\text{ of 2, [Eq. (3.19)]}
\to \frac{1}{T} \,
\left\{
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} } \,
\left[ \ln \left( T \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} \right) \right.
\right.
\notag \\
\quad \qquad -
\left. \ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T \right) + \ln \left( \frac{(P_{ijk}+R_{ij}) \, (T - R_{ij})}{P_{ijk}} \right) \right]
\notag\\
\quad -
\ln\left( Q_i + T \right) \,
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} }
+
\int^1_0 d u \,
\frac{\ln(u^2 \, P_{ijk} + R_{ij} )}
{u^2 \, P_{ijk} + R_{ij} }
\notag\\
\left.
-
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij} } \,
\left[ \ln \left( \frac{T - R_{ij}}{T} \right) + \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) \right]
\right\}\!,
\notag
\end{multline}$$]]></tex-math></disp-formula>
where the divergent term <inline-formula><tex-math notation="LaTeX" id="ImEquation330"><![CDATA[$\ln ( Q_{i}+T )$]]></tex-math></inline-formula> corresponds to the &#x201C;dressed pole&#x201D; <inline-formula><tex-math notation="LaTeX" id="ImEquation331"><![CDATA[$(2 \, e^{- \gamma_{E}})^{\varepsilon}/\varepsilon$]]></tex-math></inline-formula>.</p>
</sec>
<sec id="SEC3.2.2"><title>3.2.2. <inline-formula><tex-math notation="LaTeX" id="ImEquation332"><![CDATA[$\operatorname{Im}(\Delta_3) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation333"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) < 0$]]></tex-math></inline-formula></title>
<p>Compared to the previous case, one uses Eq. (<xref ref-type="disp-formula" rid="ptz160M3-8">3.8</xref>) for <inline-formula><tex-math notation="LaTeX" id="ImEquation334"><![CDATA[$M_2(\xi^{\nu})$]]></tex-math></inline-formula> and the <inline-formula><tex-math notation="LaTeX" id="ImEquation335"><![CDATA[$\xi$]]></tex-math></inline-formula> integration is carried out with the help of Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) and (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) depending on the different terms. Then, performing the first step described in Sect. <xref ref-type="sec" rid="SEC3.2.1">3.2.1</xref>, we obtain
<disp-formula id="ptz160M3-17"><label>(3.17)</label><tex-math notation="LaTeX" id="Equation93"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})
\notag \\[-1pt]
&= - \frac{F(\varepsilon)}{4} \left\{ i \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^1 \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, \left( T (1-x) \right)^{-1-\varepsilon} \right. \notag \\[-1pt]
&\qquad \qquad \quad {} + e^{i \, \pi \, \varepsilon} \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^{\infty} \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, \left( - y \, P_{ijk} + x \, R_{ij} - T \, (1+x) \right)^{-1-\varepsilon} \notag \\[-1pt]
&\qquad \qquad \quad {} + i \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_1^{\infty} \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, \left( - y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \notag \\[-1pt]
&\qquad \qquad \quad {} + \int_1^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^1 \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( T \, (1-x) \right)^{-1-\varepsilon} \right. \notag \\[-1pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {} - \left. \left( y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \right] \notag \\[-1pt]
&\qquad \qquad \quad {} + \int_0^{1} \frac{dy}{\sqrt{y}} \, \int_0^y \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( T \, (1-x) \right)^{-1-\varepsilon} \right. \notag \\[-1pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {} - \left. \left. \left( y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \right] \vphantom{\frac{dx}{y \, P_{ijk} - x \, R_{ij}}} \right\}\!.
\label{eqlijkir160}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We set <inline-formula><tex-math notation="LaTeX" id="ImEquation336"><![CDATA[$y = u^2$]]></tex-math></inline-formula>, perform a partial fraction decomposition on the variable <inline-formula><tex-math notation="LaTeX" id="ImEquation337"><![CDATA[$x$]]></tex-math></inline-formula>, and expand around <inline-formula><tex-math notation="LaTeX" id="ImEquation338"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation339"><![CDATA[$x$]]></tex-math></inline-formula> integration is readily done, and <inline-formula><tex-math notation="LaTeX" id="ImEquation340"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> is written as
<disp-formula id="ptz160M3-18"><label>(3.18)</label><tex-math notation="LaTeX" id="Equation94"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\[-1pt]
&= - \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon)
\left\{
- \frac{T^{-\varepsilon}}{\varepsilon} \,
\left[
i \, \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} - R_{ij}}
+ \int^{+\infty}_1 \,
\frac{d u }{u^2 \, P_{ijk} + R_{ij}}
\right]
\right.
\notag \\[-1pt]
&\qquad \qquad \qquad \qquad {}
+ i \, \int^{+\infty}_0 \,
\frac{d u}{u^2 \, P_{ijk} - R_{ij}} \,
\left[
\ln \left(\frac{u^2 \, P_{ijk} - R_{ij}}{u^2 \, P_{ijk}} \right)
-
\ln \left( \frac{R_{ij}-T}{R_{ij}} \right)
\right]
\notag \\[-1pt]
&\qquad \qquad \qquad \qquad {}
- \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\left[
\ln \left(\frac{- u^2 \, P_{ijk}}{ R_{ij}} \right)
-
\ln \left( \frac{P_{ijk} \, u^2 + T}{T-R_{ij}} \right)
\right]
\notag \\[-1pt]
&\qquad \qquad \qquad \qquad {}
+ \int^{+\infty}_1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\ln \left(\frac{u^2 \, P_{ijk} + R_{ij}}{u^2 \, P_{ijk} + T} \right)
\notag \\[-1pt]
&\qquad \qquad \qquad \qquad {}
+
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\left[
\ln \left(\frac{u^2 \, (P_{ijk} + R_{ij} - T) + T}{u^2 \, P_{ijk} + T} \right) \right. \notag \\[-1pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {} - \left. \left. \ln(1-u^2)
\vphantom{\ln \left(\frac{u^2 \, (P_{ijk} + R_{ij} - T) + T}{u^2 \, P_{ijk} + T} \right)}
\right]
\right\}\!.\label{eqlijkir16}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In this case we have <inline-formula><tex-math notation="LaTeX" id="ImEquation341"><![CDATA[$\operatorname{Im}(R_{ij}) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation342"><![CDATA[$\operatorname{Im}(P_{ijk}) < 0$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation343"><![CDATA[$\operatorname{Im}(R_{ij}-T) > 0$]]></tex-math></inline-formula>. Furthermore,
<list list-type="bullet">
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation344"><![CDATA[$u^2 \, P_{ijk} - R_{ij} = (1+u^2) \, \Delta_1^{\{i,j\}} + u^2 \, \widetilde{D}_{ijk}$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation345"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} - R_{ij}) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation346"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation347"><![CDATA[$u^2 \, P_{ijk} + R_{ij} = u^2 \, \widetilde{D}_{ijk} + (u^2 - 1) \, \Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation348"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} + R_{ij}) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation349"><![CDATA[$u \in [1,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation350"><![CDATA[$u^2 \, P_{ijk} + T = u^2 \, (\widetilde{D}_{ijk} + \Delta_1^{\{i,j\}}) - \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation351"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} + T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation352"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation353"><![CDATA[$u^2 \, (P_{ijk}+R_{ij}-T) + T = u^2 \, \widetilde{D}_{ijk} - (1-u^2) \, \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation354"><![CDATA[$\operatorname{Im}(u^2 \, (P_{ijk}+R_{ij}-T) + T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation355"><![CDATA[$u \in [0,1]$]]></tex-math></inline-formula>.</p></list-item>
</list></p>
<p>Rearrangements and simplifications similar to those done in the massive case 2(a) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] can be performed, which lead to the following alternative expression:
<disp-formula id="ptz160M3-19"><label>(3.19)</label>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160m1.gif"/>
</disp-formula></p>
<p>The contour <inline-formula><inline-graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160inline1.gif"/></inline-formula> can be deformed into a contour <inline-formula><tex-math notation="LaTeX" id="ImEquation356"><![CDATA[$\widehat{(0,1)}^{+}$]]></tex-math></inline-formula> stretched from 0 to 1 and which eventually wraps from above the cut of <inline-formula><tex-math notation="LaTeX" id="ImEquation357"><![CDATA[$\ln(u^2 \, P_{ijk} + R_{ij})$]]></tex-math></inline-formula> emerging from the branch point <inline-formula><tex-math notation="LaTeX" id="ImEquation358"><![CDATA[$u_{0} = \sqrt{- R_{ij}/P_{ijk}}$]]></tex-math></inline-formula>, whenever the latter lies in the &#x201C;north-east&#x201D; quadrant <inline-formula><tex-math notation="LaTeX" id="ImEquation359"><![CDATA[$\{\operatorname{Re}(u) > 0, \operatorname{Im}(u) > 0\}$]]></tex-math></inline-formula>.</p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M3-19">3.19</xref>) can be compared with Eq. (3.37) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] obtained in the corresponding general complex mass case 2(a) considered in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation360"><![CDATA[$\operatorname{Re}(\Delta_{2}^{\{i\}}) = \operatorname{Re}(Q_{i}+T) \to 0$]]></tex-math></inline-formula> while keeping an infinitesimal positive imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation361"><![CDATA[$\operatorname{Im}(\Delta_{2}^{\{i\}}) = \lambda$]]></tex-math></inline-formula>. Whereas it appeared convenient to formulate Eq. (3.37) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] in terms of manifestly vanishing pole residues, this is no longer the case for Eq. (<xref ref-type="disp-formula" rid="ptz160M3-19">3.19</xref>) since the pole has also become the branch point of <inline-formula><tex-math notation="LaTeX" id="ImEquation362"><![CDATA[$\ln(u^2 \, P_{ijk} + (R_{ij} +Q_{i} +T))$]]></tex-math></inline-formula> in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation363"><![CDATA[$(Q_{i}+T) \to 0$]]></tex-math></inline-formula>. For the purpose of comparison, the <inline-formula><tex-math notation="LaTeX" id="ImEquation364"><![CDATA[$\eta$]]></tex-math></inline-formula> functions introduced in Eq. (3.37) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] containing <inline-formula><tex-math notation="LaTeX" id="ImEquation365"><![CDATA[$Q_{i} + T$]]></tex-math></inline-formula> will thus be made explicit in terms of constant logarithms, part of which then cancel against the constant logarithms which were subtracted so as to build the explicitly vanishing pole residues. We get
<disp-formula id="ptz160UM15"><tex-math notation="LaTeX" id="Equation95"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}\text{RHS of 2, [Eq. (3.37)]} \to \frac{1}{T} \notag \\
& {} \times \Bigg\{
-
\ln \left( Q_i + T \right) \,
\int_{\widehat{(0,1)}^{+}} \, \frac{d u}{u^2 \, P_{ijk} + R_{ij}}
\quad + \quad
\int_{\widehat{(0,1)}^{+}} d u \,
\frac{\ln(u^2 \, P_{ijk} + R_{ij})}{u^2 \, P_{ijk} + R_{ij}}
\notag\\
&\qquad {} +
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\Bigg[
\ln \left( T \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} \right)
\notag\\
&\qquad {}
-
\ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T \right) + \ln \left( \frac{(P_{ijk} + R_{ij}) \, (T - R_{ij})}{P_{ijk}} \right)
\notag\\
&\qquad {}
- \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right)\Biggr]
-
\int_{\widehat{(0,1)}^{+}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\ln \left( \frac{T - R_{ij}}{T} \right)
\Biggl\} ,
\notag
\end{align}$$]]></tex-math></disp-formula>
where the divergent term <inline-formula><tex-math notation="LaTeX" id="ImEquation366"><![CDATA[$\ln ( Q_{i}+T )$]]></tex-math></inline-formula> corresponds to the &#x201C;dressed pole&#x201D; <inline-formula><tex-math notation="LaTeX" id="ImEquation367"><![CDATA[$(2 \, e^{- \gamma_{E}})^{\varepsilon}/\varepsilon$]]></tex-math></inline-formula>.</p>
</sec>
<sec id="SEC3.2.3"><title>3.2.3. <inline-formula><tex-math notation="LaTeX" id="ImEquation368"><![CDATA[$\operatorname{Im}(\Delta_3) < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation369"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) > 0$]]></tex-math></inline-formula></title>
<p>In this case we start with Eq. (<xref ref-type="disp-formula" rid="ptz160M3-7">3.7</xref>) for <inline-formula><tex-math notation="LaTeX" id="ImEquation370"><![CDATA[$M_2(\xi^{\nu})$]]></tex-math></inline-formula> and use Eq. (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) for all the <inline-formula><tex-math notation="LaTeX" id="ImEquation371"><![CDATA[$\xi$]]></tex-math></inline-formula> integrations. We give the result for <inline-formula><tex-math notation="LaTeX" id="ImEquation372"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> after the intermediate step 1 of Sect. <xref ref-type="sec" rid="SEC3.2.1">3.2.1</xref>:
<disp-formula id="ptz160M3-20"><label>(3.20)</label><tex-math notation="LaTeX" id="Equation96"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\[2pt]
&= \frac{F(\varepsilon)}{4} \left\{ i \, e^{- i \, \pi \, \varepsilon} \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_y^{\infty} \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( y \, P_{ijk} + x \, R_{ij} - T \, (1+x) \right)^{-1-\varepsilon} \right. \right. \notag \\[2pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {} - \left. \left( - T \, (1+x) \right)^{-1-\varepsilon} \right] \notag \\[2pt]
&\qquad \qquad \quad {} + \int_1^{\infty} \frac{dy}{\sqrt{y}} \, \int_y^{\infty} \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \right. \notag \\[2pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {} - \left. \left( T \, (1-x) \right)^{-1-\varepsilon} \right] \notag \\[2pt]
&\qquad \qquad \quad {} + \int_0^{1} \frac{dy}{\sqrt{y}} \, \int_1^{\infty} \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \right. \notag \\[2pt]
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {} - \left. \left. \left( T \, (1-x) \right)^{-1-\varepsilon} \right] \vphantom{\frac{dx}{y \, P_{ijk} + x \, R_{ij}}} \right\}\!.
\label{eqeqlijkir170}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Then, performing steps 2 and 3 above yields
<disp-formula id="ptz160M3-21"><label>(3.21)</label><tex-math notation="LaTeX" id="Equation97"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\[2pt]
&= \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon)
\left\{
- \frac{(-T)^{-\varepsilon}}{\varepsilon} \,
\int^{1}_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}}
\right.
\notag \\[2pt]
&\qquad \qquad \qquad \quad {}
+ i \, \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} - R_{ij}} \,
\left[
\ln \left( \frac{u^2 \, (P_{ijk} + R_{ij}-T) - T}{R_{ij}-T} \right)
-
\ln \left( u^2+1 \right) \right]
\notag \\[2pt]
&\qquad \qquad \qquad \quad {}
+ \int^{+\infty}_1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\left[
\ln \left(\frac{u^2 \, (P_{ijk} + R_{ij}-T) + T}{R_{ij}-T} \right)
-
\ln \left( u^2 -1 \right)
\right]
\notag \\[2pt]
&\qquad \qquad \qquad \quad {}
+
\left.
\int^1_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\ln \left( \frac{u^2 \, P_{ijk} + R_{ij}}{R_{ij} - T} \right)
\right\}\!.
\label{eqlijkir17}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In this case we have <inline-formula><tex-math notation="LaTeX" id="ImEquation373"><![CDATA[$\operatorname{Im}(R_{ij}-T) < 0$]]></tex-math></inline-formula>. Furthermore,
<list list-type="bullet">
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation374"><![CDATA[$u^2 \, P_{ijk} + R_{ij} = u^2 \, \widetilde{D}_{ijk} - (1 - u^2) \, \Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation375"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} + R_{ij}) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation376"><![CDATA[$u \in [0,1]$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation377"><![CDATA[$u^2 \, (P_{ijk}+R_{ij}-T) - T = u^2 \, \widetilde{D}_{ijk} + (1+u^2) \, \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation378"><![CDATA[$\operatorname{Im}(u^2 \, (P_{ijk}+R_{ij}-T) - T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation379"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation380"><![CDATA[$u^2 \, (P_{ijk}+R_{ij}-T) + T = u^2 \, \widetilde{D}_{ijk} + (u^2-1) \, \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation381"><![CDATA[$\operatorname{Im}(u^2 \, (P_{ijk}+R_{ij}-T) + T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation382"><![CDATA[$u \in [1,\infty[$]]></tex-math></inline-formula>.</p></list-item>
</list></p>
<p>Rearrangements and simplifications similar to those done in the massive case 1(c) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] can be performed, leading to the following alternative expression:
<disp-formula id="ptz160M3-22"><label>(3.22)</label>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160m2.gif"/>
</disp-formula></p>
<p>The deformation of the contour <inline-formula><inline-graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160inline2.gif"/></inline-formula> into a contour <inline-formula><tex-math notation="LaTeX" id="ImEquation383"><![CDATA[$\widehat{(0,1)}^{+}$]]></tex-math></inline-formula> stretched from 0 to 1 may eventually wrap from above the cut of <inline-formula><tex-math notation="LaTeX" id="ImEquation384"><![CDATA[$\ln(u^2 \, (P_{ijk} + R_{ij} -T) +T)$]]></tex-math></inline-formula> emerging from the branch point <inline-formula><tex-math notation="LaTeX" id="ImEquation385"><![CDATA[$u_{0} = \sqrt{- T/(P_{ijk} + R_{ij} -T)}$]]></tex-math></inline-formula>, whenever the latter lies in the &#x201C;north-east&#x201D; quadrant.</p>
<p>Similar to the previous case, Eq. (<xref ref-type="disp-formula" rid="ptz160M3-22">3.22</xref>) matches Eq. (3.33) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] obtained in the corresponding general complex mass case 1(c) considered in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation386"><![CDATA[$\operatorname{Re}(\Delta_{2}^{\{i\}}) = \operatorname{Re}(Q_{i}+T) \to 0$]]></tex-math></inline-formula> while keeping an infinitesimal positive imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation387"><![CDATA[$\operatorname{Im}(\Delta_{2}^{\{i\}}) = \lambda$]]></tex-math></inline-formula>.</p>
</sec>
<sec id="SEC3.2.4"><title>3.2.4. <inline-formula><tex-math notation="LaTeX" id="ImEquation388"><![CDATA[$\operatorname{Im}(\Delta_3) < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation389"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) < 0$]]></tex-math></inline-formula></title>
<p>One uses Eq. (<xref ref-type="disp-formula" rid="ptz160M3-8">3.8</xref>) for <inline-formula><tex-math notation="LaTeX" id="ImEquation390"><![CDATA[$M_2(\xi^{\nu})$]]></tex-math></inline-formula> and the <inline-formula><tex-math notation="LaTeX" id="ImEquation391"><![CDATA[$\xi$]]></tex-math></inline-formula> integration is carried out with the help of Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) and (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>), depending on the terms. Then, step 1 of Sect. <xref ref-type="sec" rid="SEC3.2.1">3.2.1</xref> leads to
<disp-formula id="ptz160M3-23"><label>(3.23)</label><tex-math notation="LaTeX" id="Equation98"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\
&= \frac{F(\varepsilon)}{4} \, \left\{ i \, \left(-T\right)^{-1-\varepsilon} \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_1^{\infty} \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, (x-1)^{-1-\varepsilon} \right. \notag \\
&\qquad \qquad \quad {} - e^{-i \, \pi \, \varepsilon} \, \left( - T \right)^{-1-\varepsilon} \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^{\infty} \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, (1+x)^{-1-\varepsilon} \notag \\
&\qquad \qquad \quad {} + i \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^{1} \, \frac{dx}{y \, P_{ijk} - x \, R_{ij}} \, \left( - y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \notag \\
&\qquad \qquad \quad {} - i \, e^{- i \, \pi \, \varepsilon} \, \int_0^{\infty} \frac{dy}{\sqrt{y}} \, \int_0^{y} \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( y \, P_{ijk} + x \, R_{ij} - T \, (1+x) \right)^{-1-\varepsilon} \right. \notag \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {} - \left. \left( -T \, (1+x) \right)^{-1-\varepsilon} \right] \notag \\
&\qquad \qquad \quad {} - \int_1^{\infty} \frac{dy}{\sqrt{y}} \, \int_1^{y} \, \frac{dx}{y \, P_{ijk} + x \, R_{ij}} \, \left[ \left( y \, P_{ijk} + x \, R_{ij} + T \, (1-x) \right)^{-1-\varepsilon} \right. \notag \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {} - \left. \left. \left( T \, (1-x) \right)^{-1-\varepsilon} \right] \vphantom{\frac{dx}{y \, P_{ijk} - x \, R_{ij}}} \right\}\!.
\label{eqlijkir180}
\end{align}$$]]></tex-math></disp-formula></p>
<p>At the end of step 5 we get
<disp-formula id="ptz160M3-24"><label>(3.24)</label><tex-math notation="LaTeX" id="Equation99"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\
&=
- \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon)
\left\{
- \frac{(-T)^{-\varepsilon}}{\varepsilon} \,
\left[
i \, \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} - R_{ij}}
+ \int^{+\infty}_1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}}
\right]
\right.
\notag \\
&\qquad \qquad \qquad \quad {}
+ i \, \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} - R_{ij}} \,
\left[
\ln \left(\frac{R_{ij} - u^2 \, P_{ijk}}{R_{ij}} \right)
-
\ln \left( \frac{P_{ijk} \, u^2 - T}{u^2 \, P_{ijk}} \right)
\right]
\notag \\
&\qquad \qquad \qquad \quad {}
- i \, \int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} - R_{ij}} \,
\left[
\ln \left( \frac{u^2 \, (P_{ijk} + R_{ij} - T) - T}{u^2 \, P_{ijk} - T} \right)
-
\ln \left( u^2+1 \right)
\right]
\notag \\
&\qquad \qquad \qquad \quad {}
-
\int^{+\infty}_1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\left[
\ln
\left(
\frac{u^2 \, (P_{ijk} + R_{ij} - T) + T}{u^2 \, P_{ijk} + R_{ij}}
\right)
- \ln(u^2-1)
\right]
\notag \\
&\qquad \qquad \qquad \quad {}
-
\left.
\int^{+\infty}_0 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\ln \left( \frac{- u^2 \, P_{ijk}}{ R_{ij}} \right)
\right\}\!.
\label{eqlijkir18}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In this case we have <inline-formula><tex-math notation="LaTeX" id="ImEquation392"><![CDATA[$\operatorname{Im}(R_{ij}) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation393"><![CDATA[$\operatorname{Im}(P_{ijk}) < 0$]]></tex-math></inline-formula>. Furthermore,
<list list-type="bullet">
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation394"><![CDATA[$u^2 \, P_{ijk} - R_{ij} = u^2 \, \widetilde{D}_{ijk} + (u^2+1) \, \Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation395"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} - R_{ij}) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation396"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation397"><![CDATA[$u^2 \, P_{ijk} + R_{ij} = u^2 \, \widetilde{D}_{ijk} + (u^2- 1) \, \Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation398"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} + R_{ij}) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation399"><![CDATA[$u \in [1,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation400"><![CDATA[$u^2 \, P_{ijk} - T = u^2 \, (\widetilde{D}_{ijk} + \Delta_1^{\{i,j\}}) + \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation401"><![CDATA[$\operatorname{Im}(u^2 \, P_{ijk} - T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation402"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation403"><![CDATA[$u^2 \, (P_{ijk}+R_{ij}-T) - T = u^2 \, \widetilde{D}_{ijk} + (u^2+1) \, \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation404"><![CDATA[$\operatorname{Im}(u^2 \, (P_{ijk}+R_{ij}-T) - T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation405"><![CDATA[$u \in [0,\infty[$]]></tex-math></inline-formula>;</p></list-item>
<list-item><p><inline-formula><tex-math notation="LaTeX" id="ImEquation406"><![CDATA[$u^2 \, (P_{ijk}+R_{ij}-T) + T = u^2 \, \widetilde{D}_{ijk} + (u^2-1) \, \Delta_3$]]></tex-math></inline-formula>, and thus <inline-formula><tex-math notation="LaTeX" id="ImEquation407"><![CDATA[$\operatorname{Im}(u^2 \, (P_{ijk}+R_{ij}-T) + T) < 0$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation408"><![CDATA[$u \in [1,\infty[$]]></tex-math></inline-formula>.</p></list-item>
</list></p>
<p>Rearrangements and simplifications similar to those done in the massive case 2(c) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] can be performed, leading to the following alternative expression:
<disp-formula id="ptz160M3-25"><label>(3.25)</label>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160m3.gif"/>
</disp-formula></p>
<p>As in the previous cases, the contours <inline-formula><inline-graphic xmlns:xlink="http://www.w3.org/1999/xlink" mimetype="image" xlink:href="ptz160inline3.gif"/></inline-formula> can be deformed into contours <inline-formula><tex-math notation="LaTeX" id="ImEquation409"><![CDATA[$\widehat{(0,1)}^{+}_{1,2}$]]></tex-math></inline-formula> stretched from 0 to 1. They eventually wrap from above the cuts of <inline-formula><tex-math notation="LaTeX" id="ImEquation410"><![CDATA[$\ln(u^2 \, P_{ijk} + R_{ij})$]]></tex-math></inline-formula> emerging from the branch point <inline-formula><tex-math notation="LaTeX" id="ImEquation411"><![CDATA[$\sqrt{- R_{ij}/P_{ijk}}$]]></tex-math></inline-formula> and of <inline-formula><tex-math notation="LaTeX" id="ImEquation412"><![CDATA[$\ln(u^2 \, (P_{ijk} + R_{ij} -T) +T)$]]></tex-math></inline-formula> emerging from the branch point <inline-formula><tex-math notation="LaTeX" id="ImEquation413"><![CDATA[$\sqrt{- T/(P_{ijk} + R_{ij} -T)}$]]></tex-math></inline-formula>, respectively, whenever either or both of these branch points lie in the &#x201C;north-east&#x201D; quadrant.</p>
<p>Similar to the previous case, Eq. (<xref ref-type="disp-formula" rid="ptz160M3-25">3.25</xref>) matches Eq. (3.41) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] obtained in the corresponding general complex mass case 2(c) considered in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation414"><![CDATA[$\operatorname{Re}(\Delta_{2}^{\{i\}}) = \operatorname{Re}(Q_{i}+T) \to 0$]]></tex-math></inline-formula> while keeping an infinitesimal positive imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation415"><![CDATA[$\operatorname{Im}(\Delta_{2}^{\{i\}}) = \lambda$]]></tex-math></inline-formula>.</p>
</sec>
</sec>
<sec id="SEC3.3"><title>3.3. <inline-formula><tex-math notation="LaTeX" id="ImEquation416"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation417"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula></title>
<p>In this case we have <inline-formula><tex-math notation="LaTeX" id="ImEquation418"><![CDATA[$P_{ijk} = - R_{ij}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation419"><![CDATA[$Q_i = -T$]]></tex-math></inline-formula>, so that Eqs. (<xref ref-type="disp-formula" rid="ptz160M3-7">3.7</xref>) and (<xref ref-type="disp-formula" rid="ptz160M3-8">3.8</xref>) become
<list list-type="bullet">
<list-item><p>for <inline-formula><tex-math notation="LaTeX" id="ImEquation420"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}} > 0)$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M3-26"><label>(3.26)</label><tex-math notation="LaTeX" id="Equation100"><![CDATA[$$\begin{align}
M_2(\xi^{\nu})
&=
\frac{1}{2} \, B(1/2,1/2) \,
\int^1_0 \frac{d z}{\Delta_1^{\{i,j\}} \, (z^2 - 1)}
\notag \\
&\quad {} \times
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}}
-
\frac{1}{\big(\xi^{\nu}+ \Delta_1^{\{i,j\}} \, (z^2-1)
- i \, \lambda \big)^{1/2}}
\right]\!;
\label{eqisigir3p}
\end{align}$$]]></tex-math></disp-formula></p></list-item>
<list-item><p>for <inline-formula><tex-math notation="LaTeX" id="ImEquation421"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}} < 0)$]]></tex-math></inline-formula>,
<disp-formula id="ptz160M3-27"><label>(3.27)</label><tex-math notation="LaTeX" id="Equation101"><![CDATA[$$\begin{align}
M_2(\xi^{\nu})
&=
- \frac{1}{2} \, B(1/2,1/2) \,
\left\{
i \, \int^{+\infty}_0 \frac{d z}{\Delta_1^{\{i,j\}} \, (z^2+1)}
\right.
\notag \\
&\quad {}
\times
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}}
-
\frac{1}{\big(\xi^{\nu}- \Delta_1^{\{i,j\}} \, (1+z^2) \big)^{1/2}}
\right]
\notag \\
&\quad {}
+ \int^{+\infty}_1 \frac{d z}{\Delta_1^{\{i,j\}} \, (z^2-1)}
\notag \\
&\quad {}
\times
\left.
\left[
\frac{1}{\left(\xi^{\nu}-i \, \lambda \right)^{1/2}}
-
\frac{1}{\big(\xi^{\nu}+ \Delta_1^{\{i,j\}} \, (z^2-1) \big)^{1/2}}
\right]
\right\}\!.
\label{eqisigir4p}
\end{align}$$]]></tex-math></disp-formula></p></list-item>
</list></p>
<p>Here again, the <inline-formula><tex-math notation="LaTeX" id="ImEquation422"><![CDATA[$\xi$]]></tex-math></inline-formula> integration will be of the type in Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) or (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) (cf. Appendix <xref ref-type="sec" rid="SEC6">A</xref>). Let us go through the different cases according to the signs of <inline-formula><tex-math notation="LaTeX" id="ImEquation423"><![CDATA[$\operatorname{Im}(\Delta_3)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation424"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}})$]]></tex-math></inline-formula>.</p>
<sec id="SEC3.3.1"><title>3.3.1. <inline-formula><tex-math notation="LaTeX" id="ImEquation425"><![CDATA[$\operatorname{Im}(\Delta_3) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation426"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) > 0$]]></tex-math></inline-formula></title>
<p>We proceed along the two first steps described in Sect. <xref ref-type="sec" rid="SEC3.2.1">3.2.1</xref>. We borrow the result obtained in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-11">3.11</xref>), set <inline-formula><tex-math notation="LaTeX" id="ImEquation427"><![CDATA[$P_{ijk} = -R_{ij}$]]></tex-math></inline-formula>, and make the change of variables <inline-formula><tex-math notation="LaTeX" id="ImEquation428"><![CDATA[$x = y + (1-y) \, v$]]></tex-math></inline-formula>. The quantity <inline-formula><tex-math notation="LaTeX" id="ImEquation429"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula> now reads
<disp-formula id="ptz160M3-28"><label>(3.28)</label><tex-math notation="LaTeX" id="Equation102"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) &= - \frac{F(\varepsilon)}{4 \, R_{ij}} \,
\int^1_0 \frac{d v}{v} \,
\left[
\frac{1}{[ v \, R_{ij} +(1-v) \, T \, - i \, \lambda]^{1+\varepsilon}}
-
\frac{1}{[(1-v) \, T \, - i \, \lambda]^{1+\varepsilon}}
\right]
\notag\\
& \quad {}\quad {}\quad {} \quad {}
\times
\int^1_0 d y \, y^{-1/2} \, (1-y)^{-1-\varepsilon} .
\label{eqisigir8p}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The integrals over <inline-formula><tex-math notation="LaTeX" id="ImEquation430"><![CDATA[$y$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation431"><![CDATA[$v$]]></tex-math></inline-formula> are unnested and are computed easily using Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-3">C.3</xref>), (<xref ref-type="disp-formula" rid="ptz160M8-6">C.6</xref>), and (<xref ref-type="disp-formula" rid="ptz160M8-10">C.10</xref>). Notice that <inline-formula><tex-math notation="LaTeX" id="ImEquation432"><![CDATA[$\operatorname{Im}(v \, R_{ij} +(1-v) \, T \, - i \, \lambda)$]]></tex-math></inline-formula> never changes its sign when <inline-formula><tex-math notation="LaTeX" id="ImEquation433"><![CDATA[$v$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation434"><![CDATA[$[0,1]$]]></tex-math></inline-formula>; this is obviously true in the real mass case, and due to the fact that <inline-formula><tex-math notation="LaTeX" id="ImEquation435"><![CDATA[$\operatorname{Im}(T)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation436"><![CDATA[$\operatorname{Im}(R_{ij})$]]></tex-math></inline-formula> have imaginary parts of the same sign in the complex mass case. Inserting the explicit results for the <inline-formula><tex-math notation="LaTeX" id="ImEquation437"><![CDATA[$y$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation438"><![CDATA[$v$]]></tex-math></inline-formula> integrals, we get
<disp-formula id="ptz160M3-29"><label>(3.29)</label><tex-math notation="LaTeX" id="Equation103"><![CDATA[$$\begin{eqnarray}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)
& = &
\frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\nonumber \\
& & {} \times
\left[
\frac{1}{\varepsilon^2} \, (2 \, R_{ij} - i \, \lambda)^{-\varepsilon}
+
\mbox{Li}_2 \left( \frac{T-R_{ij}}{T - i \, \lambda} \right)
- \frac{\pi^2}{6}
\right]\!.
\label{eqdefnl6}
\end{eqnarray}$$]]></tex-math></disp-formula></p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M3-29">3.29</xref>) explicitly displays the singularity of <inline-formula><tex-math notation="LaTeX" id="ImEquation439"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation440"><![CDATA[$\varepsilon \to 0$]]></tex-math></inline-formula>. In Eq. (<xref ref-type="disp-formula" rid="ptz160M3-29">3.29</xref>) we use the property <inline-formula><tex-math notation="LaTeX" id="ImEquation441"><![CDATA[$\mbox{Li}_2(z) + \mbox{Li}_2(1-z) = \pi^2/6 - \ln(z) \, \ln(1-z)$]]></tex-math></inline-formula> to obtain the following alternative form suitable for further comparisons:
<disp-formula id="ptz160M3-30"><label>(3.30)</label><tex-math notation="LaTeX" id="Equation104"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)
&=
\frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\nonumber \\
&\quad {} \times
\left\{
\frac{1}{\varepsilon^2} \, (2 \, R_{ij} - i \lambda)^{-\varepsilon}
- \mbox{Li}_2 \left( \frac{R_{ij}- i \lambda}{T- i \lambda} \right)
\right.
\notag\\
& \quad {} \quad {}
\left.
-
\left[
\ln \left(R_{ij}- i \lambda \right) - \ln \left( T- i \lambda \right)
\right] \, \ln \left( \frac{T - R_{ij}}{T- i \lambda} \right)
\right\}\!.
\label{eqdefnl6bis}
\end{align}$$]]></tex-math></disp-formula></p>
</sec>
<sec id="SEC3.3.2"><title>3.3.2. <inline-formula><tex-math notation="LaTeX" id="ImEquation442"><![CDATA[$\operatorname{Im}(\Delta_3) > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation443"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) < 0$]]></tex-math></inline-formula></title>
<p>The starting point is Eq. (<xref ref-type="disp-formula" rid="ptz160M3-17">3.17</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation444"><![CDATA[$P_{ijk} = - R_{ij}$]]></tex-math></inline-formula>. In the first line of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-17">3.17</xref>) we rescale <inline-formula><tex-math notation="LaTeX" id="ImEquation445"><![CDATA[$y = x \, u^2$]]></tex-math></inline-formula> so that the double integral factorizes into a product of two unnested integrals over <inline-formula><tex-math notation="LaTeX" id="ImEquation446"><![CDATA[$u$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation447"><![CDATA[$x$]]></tex-math></inline-formula>. The integrals of the second and third lines of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-17">3.17</xref>) yield no divergences when <inline-formula><tex-math notation="LaTeX" id="ImEquation448"><![CDATA[$\varepsilon \rightarrow 0$]]></tex-math></inline-formula>, so we take <inline-formula><tex-math notation="LaTeX" id="ImEquation449"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> in them. We then make the change of variables <inline-formula><tex-math notation="LaTeX" id="ImEquation450"><![CDATA[$x = y - (y-1) \, v$]]></tex-math></inline-formula> in the penultimate line, and <inline-formula><tex-math notation="LaTeX" id="ImEquation451"><![CDATA[$x = y -(1-y) \, v$]]></tex-math></inline-formula> in the last line. We obtain
<disp-formula id="ptz160M3-31"><label>(3.31)</label><tex-math notation="LaTeX" id="Equation105"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&= \frac{F(\varepsilon)}{4 \, R_{ij}} \,
\left\{
i \, T^{-1-\varepsilon} \, \int_0^{+\infty} \frac{2 \, du}{1+u^2} \,
\int_{0}^1 \, \frac{dx}{\sqrt{x}} \, (1-x)^{-1-\varepsilon}
\right.
\notag \\
&\quad {}
+
\quad {}
\int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \int_0^{+\infty}
\frac{dx}{(y+x) \; ( (x+y) \, R_{ij} - T \, (1+x) )}
\notag \\
&\quad {}
+ i \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \,
\int_1^{+\infty} \frac{dx}{(y+x) \; ( (x+y) \, R_{ij} + T \, (1-x) )}
\label{eqdefnl92} \\
&\quad {}
+ \int_1^{+\infty} \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon} \,
\int_1^{\frac{y}{y-1}} \, \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (v-1)^{-1-\varepsilon}
-
\left( - v \, R_{ij} - T \, (1-v) \right)^{-1-\varepsilon}
\right]
\notag \\
&\quad {}
\left.
+ \int_0^{1} \frac{dy}{\sqrt{y}} \, (1-y)^{-1-\varepsilon} \,
\int_0^{\, \frac{y}{1-y}} \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (1+v)^{-1-\varepsilon}
-
\left( - v \, R_{ij} + T \, (1+v) \right)^{-1-\varepsilon}
\right]
\right\}\!. \notag
\end{align}$$]]></tex-math></disp-formula></p>
<p>Let us compute the different terms. Using Eq. (<xref ref-type="disp-formula" rid="ptz160M8-3">C.3</xref>), the first integral is readily given by
<disp-formula id="ptz160M3-32"><label>(3.32)</label><tex-math notation="LaTeX" id="Equation106"><![CDATA[$$\begin{align}
\int_0^{+\infty} \frac{2 \, du}{1+u^2} \,
\int_{0}^1 \, \frac{dx}{\sqrt{x}} \, (1-x)^{-1-\varepsilon}
&=
\pi \, B \left( \frac{1}{2}, - \, \varepsilon \right)\!.
\label{eqfirstintyv}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the second and third lines of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-31">3.31</xref>), the <inline-formula><tex-math notation="LaTeX" id="ImEquation452"><![CDATA[$x$]]></tex-math></inline-formula> integration is easily performed after a partial fraction decomposition on the <inline-formula><tex-math notation="LaTeX" id="ImEquation453"><![CDATA[$x$]]></tex-math></inline-formula> variable. We get
<disp-formula id="ptz160M3-33"><label>(3.33)</label><tex-math notation="LaTeX" id="Equation107"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}
\int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \int_0^{+\infty}
\frac{dx}{(y+x) \; ( (x+y) \, R_{ij} - T \, (1+x) )}
\notag \\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
= \frac{1}{T} \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{y-1}
\left[ \ln \left( \frac{y \, R_{ij} -T}{R_{ij} - T} \right) - \ln(y) \right]\!,
\label{eqsecondintyv} \\
\hspace{2em}&\hspace{-2em}
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-34"><label>(3.34)</label><tex-math notation="LaTeX" id="Equation108"><![CDATA[$$\begin{align}
\int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \int_1^{+\infty}
\frac{dx}{(y+x) \; ( (x+y) \, R_{ij} + T \, (1-x) )}
\notag \\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
= \frac{1}{T} \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{y+1} \,
\ln \left( \frac{R_{ij}}{R_{ij} - T} \right)\!.
\label{eqthirdintyv}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We write the last two integrals of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-31">3.31</xref>) as
<disp-formula id="ptz160M3-35"><label>(3.35)</label><tex-math notation="LaTeX" id="Equation109"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}
\int_1^{+\infty} \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon} \,
\int_1^{\frac{y}{y-1}} \, \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (v-1)^{-1-\varepsilon}
-
\left( - v \, R_{ij} - T \, (1-v) \right)^{-1-\varepsilon}
\right]
\notag \\
&= \int_1^{+\infty} \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon} \,
\left[ E_1(y)-E_1(1^{+})\right] \,
\; + \; E_1(1^{+}) \,
B \left( \frac{1}{2} + \varepsilon, - \, \varepsilon \right)
\label{eqfourthintyv} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-36"><label>(3.36)</label><tex-math notation="LaTeX" id="Equation110"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}
\int_0^{1} \frac{dy}{\sqrt{y}} \, (1-y)^{-1-\varepsilon} \,
\int_0^{\, \frac{y}{1-y}} \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (1+v)^{-1-\varepsilon}
-
\left( - v \, R_{ij} + T \, (1+v) \right)^{-1-\varepsilon}
\right]
\notag \\
&= \int_0^{1} \frac{dy}{\sqrt{y}} \, (1-y)^{-1-\varepsilon} \,
\left[ E_2(y)-E_2(1^{-}) \right]
\; + \; E_2(1^{-}) \, B\left( \frac{1}{2}, - \, \varepsilon \right)\!,
\label{eqfifthintyv}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M3-37"><label>(3.37)</label><tex-math notation="LaTeX" id="Equation111"><![CDATA[$$\begin{align}
E_1(y)
&= \int_1^{\frac{y}{y-1}} \, \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (v-1)^{-1-\varepsilon}
-
\left( - v \, R_{ij} - T \, (1-v) \right)^{-1-\varepsilon}
\right]\!,
\label{eqdeffunce1} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-38"><label>(3.38)</label><tex-math notation="LaTeX" id="Equation112"><![CDATA[$$\begin{align}
E_2(y)
&= \int_0^{\, \frac{y}{1-y}} \frac{dv}{v} \,
\left[
T^{-1-\varepsilon} \, (1+v)^{-1-\varepsilon}
-
\left( - v \, R_{ij} + T \, (1+v) \right)^{-1-\varepsilon}
\right]\!.
\label{eqdeffunce2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>When <inline-formula><tex-math notation="LaTeX" id="ImEquation454"><![CDATA[$y \to 1^{+}$]]></tex-math></inline-formula>, the upper bound of the integral defining the function <inline-formula><tex-math notation="LaTeX" id="ImEquation455"><![CDATA[$E_1(y)$]]></tex-math></inline-formula> goes to <inline-formula><tex-math notation="LaTeX" id="ImEquation456"><![CDATA[$+ \infty$]]></tex-math></inline-formula> and likewise for <inline-formula><tex-math notation="LaTeX" id="ImEquation457"><![CDATA[$E_2(y)$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation458"><![CDATA[$y \to 1^{-}$]]></tex-math></inline-formula>. The quantities <inline-formula><tex-math notation="LaTeX" id="ImEquation459"><![CDATA[$E_1(y)-E_1(1^{+})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation460"><![CDATA[$E_2(y)-E_2(1^{-})$]]></tex-math></inline-formula> are given by integrals between <inline-formula><tex-math notation="LaTeX" id="ImEquation461"><![CDATA[$y/(y-1)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation462"><![CDATA[$+\infty$]]></tex-math></inline-formula> and between <inline-formula><tex-math notation="LaTeX" id="ImEquation463"><![CDATA[$y/(1-y)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation464"><![CDATA[$+\infty$]]></tex-math></inline-formula>, respectively. As, in <inline-formula><tex-math notation="LaTeX" id="ImEquation465"><![CDATA[$E_1(y)-E_1(1^{+})$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation466"><![CDATA[$y$]]></tex-math></inline-formula> is greater than <inline-formula><tex-math notation="LaTeX" id="ImEquation467"><![CDATA[$1$]]></tex-math></inline-formula> and so is the lower bound <inline-formula><tex-math notation="LaTeX" id="ImEquation468"><![CDATA[$y/(y-1)$]]></tex-math></inline-formula>, the singular support <inline-formula><tex-math notation="LaTeX" id="ImEquation469"><![CDATA[$v=1$]]></tex-math></inline-formula> of the distribution <inline-formula><tex-math notation="LaTeX" id="ImEquation470"><![CDATA[$(v-1)^{-1-\varepsilon}$]]></tex-math></inline-formula> lies outside the range of integration, and thus the first term of the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-37">3.37</xref>) can be taken at <inline-formula><tex-math notation="LaTeX" id="ImEquation471"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula>. In <inline-formula><tex-math notation="LaTeX" id="ImEquation472"><![CDATA[$E_2(y)-E_2(1^{-})$]]></tex-math></inline-formula>, the integrand is non-singular and the first term of the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-38">3.38</xref>) can also be taken at <inline-formula><tex-math notation="LaTeX" id="ImEquation473"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula>. These two terms give
<disp-formula id="ptz160M3-39"><label>(3.39)</label><tex-math notation="LaTeX" id="Equation113"><![CDATA[$$\begin{align}
\int_1^{+\infty} \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon} \,
\left[ E_1(y)-E_1(1^{+}) \right]
&= - \, \frac{1}{T} \, \int_1^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{y-1} \,
\ln \left( \frac{y \, R_{ij} - T}{R_{ij} - T} \right)\!,
\label{eqfourthintyv1t} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-40"><label>(3.40)</label><tex-math notation="LaTeX" id="Equation114"><![CDATA[$$\begin{align}
\int_0^{1} \frac{dy}{\sqrt{y}} \, (1-y)^{-1-\varepsilon} \,
\left[ E_2(y)-E_2(1^{-}) \right]
&= - \, \frac{1}{T} \, \int_0^1 \frac{dy}{\sqrt{y}} \, \frac{1}{y-1} \,
\ln \left( \frac{y \, R_{ij} - T}{R_{ij} - T} \right)\!.
\label{eqfifthintyv1t}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The combined right-hand sides of Eqs. (<xref ref-type="disp-formula" rid="ptz160M3-39">3.39</xref>) and (<xref ref-type="disp-formula" rid="ptz160M3-40">3.40</xref>) cancel against the first term in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-33">3.33</xref>). The expressions of <inline-formula><tex-math notation="LaTeX" id="ImEquation474"><![CDATA[$E_1(1^{+})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation475"><![CDATA[$E_2(1^{-})$]]></tex-math></inline-formula> are computed using Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-13">C.13</xref>) and (<xref ref-type="disp-formula" rid="ptz160M8-15">C.15</xref>):
<disp-formula id="ptz160M3-41"><label>(3.41)</label><tex-math notation="LaTeX" id="Equation115"><![CDATA[$$\begin{align}
E_1(1^{+})
&= - \, K_4(T-R_{ij},-T)
\notag \\
&= \frac{1}{T} \,
\left\{
- \, \frac{1}{\varepsilon} \, \left( -R_{ij} \right)^{-\varepsilon}
+ \ln (T) - \ln (T-R_{ij})
\right.
\notag \\
&\quad {} \quad {} \quad {}
\left.
+ \, \varepsilon \,
\left[
\mbox{Li}_2 \left( \frac{R_{ij}}{T} \right)
+ \ln \left( -R_{ij} \right)\, \ln \left( \frac{T-R_{ij}}{T} \right)
\right]
\right\}\!,
\label{eqdefe3de1}\\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-42"><label>(3.42)</label><tex-math notation="LaTeX" id="Equation116"><![CDATA[$$\begin{align}
E_2(1^{-})
&= - \, K_3(T-R_{ij},T)
\notag \\
&= \frac{1}{T} \, \ln \left( \frac{T-R_{ij}}{T} \right) \,
\left[ 1 - \varepsilon \, \ln(T) \right] .
\label{eqdefe2de1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Putting everything together, we get
<disp-formula id="ptz160M3-43"><label>(3.43)</label><tex-math notation="LaTeX" id="Equation117"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&= \frac{F(\varepsilon)}{4 \, R_{ij} \, T} \,
\left\{
i \, \pi \, B \left( \frac{1}{2}, - \, \varepsilon \right) \,
\left( 1 - \varepsilon \ln(T) \right)
\right.
\notag\\
& \quad {} \quad {} \quad {} \quad {} \quad {}
- \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{y-1} \, \ln(y)
+ i \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{y+1} \,
\ln \left( \frac{R_{ij}}{R_{ij}-T} \right)
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {}
+ B \left( \frac{1}{2}, - \, \varepsilon \right) \,
\left[
\ln \left( \frac{T - R_{ij}}{T} \right) \,
\left( 1 - \varepsilon \, \ln(T) \right)
\right]
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {}
+ B \left( \frac{1}{2} + \varepsilon, - \, \varepsilon \right)
\left[
- \, \frac{1}{\varepsilon} \left( -R_{ij} \right)^{-\varepsilon}
+ \ln (T) - \ln (T-R_{ij})
\right.
\notag \\
&\quad \quad \quad \quad \quad \quad \quad
\quad \quad \quad \quad \qquad {}
+ \left.
\left.
\varepsilon
\left[
\mbox{Li}_2\left( \frac{R_{ij}}{T} \right)
+ \ln \left( -R_{ij} \right)\, \ln \left( \frac{T-R_{ij}}{T} \right)
\right] \right]
\right\}\!.
\label{eqdefnl93}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Extracting the Euler Beta functions from Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-2">C.2</xref>), (<xref ref-type="disp-formula" rid="ptz160M8-5">C.5</xref>), (<xref ref-type="disp-formula" rid="ptz160M8-3">C.3</xref>), and (<xref ref-type="disp-formula" rid="ptz160M8-6">C.6</xref>), and using the fact that <inline-formula><tex-math notation="LaTeX" id="ImEquation476"><![CDATA[$\ln(- \, R_{ij}) = \ln(R_{ij}) - i \, \pi$]]></tex-math></inline-formula>, Eq. (<xref ref-type="disp-formula" rid="ptz160M3-43">3.43</xref>) can be cast in the form
<disp-formula id="ptz160M3-44"><label>(3.44)</label><tex-math notation="LaTeX" id="Equation118"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&= \frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1-2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\notag\\
& \quad {} \times
\left\{
\frac{1}{\varepsilon^2} \, \left( 2 \, R_{ij} \right)^{-\varepsilon}
- \mbox{Li}_2\left( \frac{R_{ij}}{T} \right)
- \left[ \ln \left(R_{ij}\right) - \ln \left( T \right) \right] \,
\ln \left( \frac{T - R_{ij}}{T} \right)
\right\}\!.
\label{eqdefnl95}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Equations (<xref ref-type="disp-formula" rid="ptz160M3-44">3.44</xref>) and (<xref ref-type="disp-formula" rid="ptz160M3-30">3.30</xref>) have the same analytic expression.</p>
</sec>
<sec id="SEC3.3.3"><title>3.3.3. <inline-formula><tex-math notation="LaTeX" id="ImEquation477"><![CDATA[$\operatorname{Im}(\Delta_3) < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation478"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) > 0$]]></tex-math></inline-formula></title>
<p>We start with Eq. (<xref ref-type="disp-formula" rid="ptz160M3-20">3.20</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation479"><![CDATA[$P_{ijk} = - R_{ij}$]]></tex-math></inline-formula>. We then set <inline-formula><tex-math notation="LaTeX" id="ImEquation480"><![CDATA[$x = y + (1+y) \, v$]]></tex-math></inline-formula> in the first integral of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-20">3.20</xref>), <inline-formula><tex-math notation="LaTeX" id="ImEquation481"><![CDATA[$x = y + (y-1) \, v$]]></tex-math></inline-formula> in the second integral, and <inline-formula><tex-math notation="LaTeX" id="ImEquation482"><![CDATA[$x = y + (1-y) \, v$]]></tex-math></inline-formula> in the third one, and we get
<disp-formula id="ptz160M3-45"><label>(3.45)</label><tex-math notation="LaTeX" id="Equation119"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)
&= \frac{F(\varepsilon)}{4 R_{ij}}
\left\{
i \text{e}^{-i \pi \varepsilon} \int^{+\infty}_0 \frac{d y}{\sqrt{y}}
(1+y)^{-1-\varepsilon}
\right.
\notag\\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\times
\int^{+\infty}_0 \frac{d v}{v}
\left[
\frac{1}{[ v R_{ij} - (1+v) T]^{1+\varepsilon}}
-
\frac{1}{[- (1+v) \, T]^{1+\varepsilon}}
\right]
\notag \\
&\quad {} \quad {}\quad {}\quad {}\quad {}\quad {}
+ \quad {}
\int^{+\infty}_1 \frac{d y}{\sqrt{y}} (y-1)^{-1-\varepsilon}
\notag\\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\times
\int^{+\infty}_0 \frac{d v}{v}
\left[
\frac{1}{[ v R_{ij} - (1+v) T]^{1+\varepsilon}}
-
\frac{1}{[- (1+v) T]^{1+\varepsilon}}
\right]
\notag \\
& \quad {} \quad {}\quad {}\quad {}\quad {}\quad {}
+ \quad {}
\int^{1}_0 \frac{d y}{\sqrt{y}} (1-y)^{-1-\varepsilon}
\notag\\
& \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\left.
\times
\int^{+\infty}_1 \frac{d v}{v}
\left[
\frac{1}{[ v R_{ij} + (1-v) T ]^{1+\varepsilon}}
-
\frac{1}{[(1-v) \, T]^{1+\varepsilon}}
\right]
\right\}\!.
\label{eqisigir8s}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Here again, the integrals over <inline-formula><tex-math notation="LaTeX" id="ImEquation483"><![CDATA[$y$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation484"><![CDATA[$v$]]></tex-math></inline-formula> are unnested. Since the signs of <inline-formula><tex-math notation="LaTeX" id="ImEquation485"><![CDATA[$\operatorname{Im}(R_{ij})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation486"><![CDATA[$\operatorname{Im}(T)$]]></tex-math></inline-formula> are mutually opposite, let us note that the imaginary parts of each of the terms raised to the power <inline-formula><tex-math notation="LaTeX" id="ImEquation487"><![CDATA[$1+\varepsilon$]]></tex-math></inline-formula> in denominators in the <inline-formula><tex-math notation="LaTeX" id="ImEquation488"><![CDATA[$v$]]></tex-math></inline-formula> integrals remain constant over the corresponding ranges of integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation489"><![CDATA[$v$]]></tex-math></inline-formula>. The first two integrals on <inline-formula><tex-math notation="LaTeX" id="ImEquation490"><![CDATA[$v$]]></tex-math></inline-formula> of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-45">3.45</xref>) are given by Eq. (<xref ref-type="disp-formula" rid="ptz160M8-13">C.13</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation491"><![CDATA[$A = R_{ij} - T$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation492"><![CDATA[$B = - T$]]></tex-math></inline-formula>, while the last one is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M8-15">C.15</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation493"><![CDATA[$A^{\prime} = R_{ij} - T$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation494"><![CDATA[$B^{\prime} = T$]]></tex-math></inline-formula>. As for the <inline-formula><tex-math notation="LaTeX" id="ImEquation495"><![CDATA[$y$]]></tex-math></inline-formula> integration, they can be read from Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-1">C.1</xref>)&#x2013; (<xref ref-type="disp-formula" rid="ptz160M8-6">C.6</xref>). All the ingredients combine into:
<disp-formula id="ptz160M3-46"><label>(3.46)</label><tex-math notation="LaTeX" id="Equation120"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&=
\frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\nonumber \\
&\quad {} \times
\left\{
\frac{1}{\varepsilon^2} \, (2 \, R_{ij})^{-\varepsilon}
- \mbox{Li}_2 \left( \frac{R_{ij}}{T} \right) -
\left[
\ln \left( R_{ij} \right) - \ln \left( - \, T \right) \, - i \, \pi
\right] \,
\ln \left( \frac{T-R_{ij}}{T} \right)
\right\}\!.
\label{eqdefnl8}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the present case <inline-formula><tex-math notation="LaTeX" id="ImEquation496"><![CDATA[$\operatorname{Im}(- \, R_{ij})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation497"><![CDATA[$\operatorname{Im}(T)> 0$]]></tex-math></inline-formula>, so that <inline-formula><tex-math notation="LaTeX" id="ImEquation498"><![CDATA[$\ln \left( - \,T \right) = \ln \left( T \right) - i \, \pi$]]></tex-math></inline-formula> and thus Eq. (<xref ref-type="disp-formula" rid="ptz160M3-46">3.46</xref>) can be rewritten as
<disp-formula id="ptz160M3-47"><label>(3.47)</label><tex-math notation="LaTeX" id="Equation121"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)
&=
\frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\label{eqdefnl8bis} \\
&\quad {} \times
\left\{
\frac{1}{\varepsilon^2} \, (2 \, R_{ij})^{-\varepsilon}
- \mbox{Li}_2 \left( \frac{R_{ij}}{T} \right) -
\left[ \ln \left( R_{ij} \right) - \ln \left( T \right) \right] \,
\ln \left( \frac{T - R_{ij}}{T} \right)
\right\}\!,
\nonumber
\end{align}$$]]></tex-math></disp-formula>
i.e. again, the same analytic form as the previous two cases, cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M3-30">3.30</xref>) and (<xref ref-type="disp-formula" rid="ptz160M3-44">3.44</xref>).</p>
</sec>
<sec id="SEC3.3.4"><title>3.3.4. <inline-formula><tex-math notation="LaTeX" id="ImEquation499"><![CDATA[$\operatorname{Im}(\Delta_3) < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation500"><![CDATA[$\operatorname{Im}(\Delta_1^{\{i,j\}}) < 0$]]></tex-math></inline-formula></title>
<p>The starting point is now Eq. (<xref ref-type="disp-formula" rid="ptz160M3-23">3.23</xref>) with <inline-formula><tex-math notation="LaTeX" id="ImEquation501"><![CDATA[$P_{ijk} = - R_{ij}$]]></tex-math></inline-formula>. In the first two integrals we rescale <inline-formula><tex-math notation="LaTeX" id="ImEquation502"><![CDATA[$y = x \, u^2$]]></tex-math></inline-formula> to factorize the double integral into a product of two unnested integrals over <inline-formula><tex-math notation="LaTeX" id="ImEquation503"><![CDATA[$u$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation504"><![CDATA[$x$]]></tex-math></inline-formula>. Using the results given in Eqs. (<xref ref-type="disp-formula" rid="ptz160M8-1">C.1</xref>) and (<xref ref-type="disp-formula" rid="ptz160M8-2">C.2</xref>), these unnested integrals are readily performed to yield
<disp-formula id="ptz160M3-48"><label>(3.48)</label><tex-math notation="LaTeX" id="Equation122"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&= \frac{F(\varepsilon)}{4 \, R_{ij}} \,
\left\{
i \, \pi \, \left( -T \right)^{-1-\varepsilon}
\left[
- i \, \text{e}^{- i \, \pi \, \varepsilon} \,
B \left( \frac{1}{2}, \frac{1}{2}+ \varepsilon \right)
- \,
B \left( \frac{1}{2}+ \varepsilon, - \, \varepsilon \right)
\right]
\right.
\notag \\
&\quad {} \quad {}
- i \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \int_0^1 \frac{dx}{y+x}
\left( R_{ij} \, (x+y) + T \, (1-x) \right)^{-1-\varepsilon}
\notag \\
&\quad {}\quad {}
+ i \, \text{e}^{- i \, \pi \, \varepsilon} \,
\int_0^{+\infty} \frac{dy}{\sqrt{y}} \int_0^y \frac{dx}{y-x}
\notag \\
&\qquad \qquad \qquad {} \times
\left[
\left( - R_{ij} \, (y-x) - T \, (1+x) \right)^{-1-\varepsilon}
- \left( -T \right)^{-1-\varepsilon} \, (1+x)^{-1-\varepsilon} \right]
\notag \\
&\quad {}\quad {}
+ \int_1^{+\infty} \frac{dy}{\sqrt{y}} \int_1^y \frac{dx}{y-x}
\notag \\
&\qquad \qquad \qquad {} \times
\left.
\left[
\left( - R_{ij} \, (y-x) + T \, (1-x) \right)^{-1-\varepsilon}
-
\left( -T \right)^{-1-\varepsilon} \, (x-1)^{-1-\varepsilon}
\right]
\vphantom{\frac{1}{2}} \right\}\!.
\label{eqdefnl101}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In the three remaining integrals in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-48">3.48</xref>), the first two remain finite when <inline-formula><tex-math notation="LaTeX" id="ImEquation505"><![CDATA[$\varepsilon \to 0$]]></tex-math></inline-formula> and are thus computed in this limit, which yields
<disp-formula id="ptz160M3-49"><label>(3.49)</label><tex-math notation="LaTeX" id="Equation123"><![CDATA[$$\begin{align}
&\int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \int_0^1 \frac{dx}{y+x}
\left( R_{ij} \, (x+y) + T \, (1-x) \right)^{-1}
\notag\\
& \quad {} \quad {} \quad {} \quad {}
=
\frac{1}{T} \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{1+y} \,
\left[ \ln\left( \frac{y \, R_{ij} + T}{R_{ij}} \right) - \ln(y) \right] ,
&
\label{i1}\\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-50"><label>(3.50)</label><tex-math notation="LaTeX" id="Equation124"><![CDATA[$$\begin{align}
& \int_0^{+\infty} \frac{dy}{\sqrt{y}} \int_0^y \frac{dx}{y-x}
\left[
\left(
- R_{ij} \, (y-x) - T \, (1+x) \right)^{-1-\varepsilon}
-
\left( -T \right)^{-1} \, (1+x)^{-1}
\right]
\notag\\
& \quad {} \quad {} \quad {} \quad {}
=
\frac{1}{T} \, \int_0^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{1}{1+y} \,
\ln \left( \frac{y \, R_{ij} + T}{T} \right)\!.
\label{i2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Making the change of variable <inline-formula><tex-math notation="LaTeX" id="ImEquation506"><![CDATA[$u=1/y$]]></tex-math></inline-formula> we readily see that
<disp-formula id="ptz160UM16"><tex-math notation="LaTeX" id="Equation125"><![CDATA[$$\int_{0}^{+\infty} \frac{dy}{\sqrt{y}} \, \frac{\ln(y)}{1+y}
\, = \,
- \int_{0}^{+\infty} \frac{du}{\sqrt{u}} \, \frac{\ln(u)}{1+u}
\, = \, 0 ;$$]]></tex-math></disp-formula>
the combination <inline-formula><tex-math notation="LaTeX" id="ImEquation507"><![CDATA[$\{- \, i \times \text{Eq. } \text{(3.49)} + \, i \times \text{Eq. } \text{(3.50)}\}$]]></tex-math></inline-formula> thus gives <inline-formula><tex-math notation="LaTeX" id="ImEquation508"><![CDATA[$i \, \pi \, \ln(R_{ij}/T)$]]></tex-math></inline-formula>. In the third integral, we make the change of variable <inline-formula><tex-math notation="LaTeX" id="ImEquation509"><![CDATA[$x = y - (y-1) \, v$]]></tex-math></inline-formula> to get
<disp-formula id="ptz160UM17"><tex-math notation="LaTeX" id="Equation126"><![CDATA[$$ \int_1^{+\infty} \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon} \,
\int_0^{1} \frac{dv}{v}
\left[
\left( -v \, R_{ij} + T \, (v-1) \right)^{-1-\varepsilon}
-
\left( -T \right)^{-1-\varepsilon} \, (1-v)^{-1-\varepsilon}
\right]\!.$$]]></tex-math></disp-formula></p>
<p>The <inline-formula><tex-math notation="LaTeX" id="ImEquation510"><![CDATA[$v$]]></tex-math></inline-formula> integration is performed using the result of <inline-formula><tex-math notation="LaTeX" id="ImEquation511"><![CDATA[$K_2(- R_{ij}, - T)$]]></tex-math></inline-formula>, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M8-10">C.10</xref>). After some algebra, Eq. (<xref ref-type="disp-formula" rid="ptz160M3-48">3.48</xref>) thus reads
<disp-formula id="ptz160M3-51"><label>(3.51)</label><tex-math notation="LaTeX" id="Equation127"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&= \frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1 - \varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\left[
\frac{1}{\varepsilon^2} \, \left( 2 \, R_{ij} \right)^{-\varepsilon}
+ \mbox{Li}_2\left( \frac{T - R_{ij}}{T} \right) - \frac{\pi^2}{6}
\right]\!.
\label{eqdefnl103}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M3-51">3.51</xref>) is identical to Eq. (<xref ref-type="disp-formula" rid="ptz160M3-29">3.29</xref>); since <inline-formula><tex-math notation="LaTeX" id="ImEquation512"><![CDATA[$\operatorname{Im}(R_{ij})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation513"><![CDATA[$\operatorname{Im}(T)$]]></tex-math></inline-formula> have the same sign as in Sect. <xref ref-type="sec" rid="SEC3.3.1">3.3.1</xref>, Eq. (<xref ref-type="disp-formula" rid="ptz160M3-51">3.51</xref>) can be recast into a form identical to eq. (<xref ref-type="disp-formula" rid="ptz160M3-30">3.30</xref>):
<disp-formula id="ptz160M3-52"><label>(3.52)</label><tex-math notation="LaTeX" id="Equation128"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0) \notag \\
&=
\frac{1}{2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1 - \varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \,
\frac{1}{R_{ij} \, T} \,
\notag\\
& \quad {} \times
\left\{
\frac{1}{\varepsilon^2} \, \left( 2 \, R_{ij} \right)^{-\varepsilon}
- \mbox{Li}_2 \left( \frac{R_{ij}}{T} \right)
- \left[
\ln \left( R_{ij} \right) - \ln \left( T \right)
\right] \,
\ln \left( \frac{T-R_{ij}}{T} \right)
\right\}\!.
\label{eqdefnl103bis}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In summary, in all four cases <inline-formula><tex-math notation="LaTeX" id="ImEquation514"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula> takes the same analytical form. Compared with the multiplicity of forms met in the general complex mass case, and still with the diverse cases met in Sect. <xref ref-type="sec" rid="SEC3.2">3.2</xref>, this simplification comes from the coalescence of the pole and branch points all at the value 1, which is the end-point singularity causing the appearance of the soft and collinear singularity in all four cases.</p>
</sec>
</sec>
<sec id="SEC3.4"><title>3.4. Consistency checks and explicit examples</title>
<p>In Refs. [<xref ref-type="bibr" rid="B7">7</xref>,<xref ref-type="bibr" rid="B8">8</xref>] the infrared structure of any IR-divergent <inline-formula><tex-math notation="LaTeX" id="ImEquation515"><![CDATA[$N$]]></tex-math></inline-formula>-point one-loop integral was shown to be carried by IR-divergent three-point one-loop functions resulting from appropriate iterated pinchings. In the following this feature is explicitly verified for the formulae obtained in this article for the four-point functions, when compared with those for the three-point functions, where the latter are formulated most conveniently according to the so-called &#x201C;indirect way&#x201D; for this purpose.</p>
<p>In the case of purely soft divergence, i.e. whenever some <inline-formula><tex-math notation="LaTeX" id="ImEquation516"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation517"><![CDATA[$\widetilde{D}_{ijk} \ne 0$]]></tex-math></inline-formula>, the various cases of Eqs. (<xref ref-type="disp-formula" rid="ptz160M3-14">3.14</xref>), (<xref ref-type="disp-formula" rid="ptz160M3-19">3.19</xref>), (<xref ref-type="disp-formula" rid="ptz160M3-22">3.22</xref>), and (<xref ref-type="disp-formula" rid="ptz160M3-25">3.25</xref>) can be encompassed in one single formula. For this purpose let us introduce the following notation, where for any complex <inline-formula><tex-math notation="LaTeX" id="ImEquation518"><![CDATA[$Q$]]></tex-math></inline-formula> we denote <inline-formula><tex-math notation="LaTeX" id="ImEquation519"><![CDATA[$Q_{\rm R} \equiv \operatorname{Re}(Q)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation520"><![CDATA[$Q_{\rm I} \equiv \operatorname{Im}(Q)$]]></tex-math></inline-formula>.</p>
<p><disp-formula id="ptz160M3-53"><label>(3.53)</label><tex-math notation="LaTeX" id="Equation129"><![CDATA[$$\begin{align}
\int_{_{\widetilde{(0,1)}}} du \, F(u)
&=
\left\{
\begin{array}{lcl}
\int_0^{^{- i \, S_{\!_{A}} \, \infty}} du \, F(u) +
\int_{_{+\infty}}^{^1} du \, F(u)
& \mbox{if} & 0 < - \, B_{\rm I}/A_{\rm I} <1 \\
& \mbox{and} & A_{\rm I} \, [ A_{\rm R} \, B_{\rm I} - A_{\rm I} \, B_{\rm R} ] > 0 , \\
\int_{_{0}}^{^{1}} du \, F(u) & & \text{otherwise},
\end{array}
\right.
\label{eqdefwtilde01}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation521"><![CDATA[$S_{A} = \mbox{sign}(A_{\rm I})$]]></tex-math></inline-formula>, whether
<disp-formula id="ptz160UM18"><tex-math notation="LaTeX" id="Equation130"><![CDATA[$$F(u) \; = \; \frac{\ln(A \, u^2 + B) - \ln(A \, u_0^2 + B)}{u^2 - u_0^2}
\quad {} \mbox{with} \quad {} u_0^2 \neq - \, \frac{B}{A}$$]]></tex-math></disp-formula>
or
<disp-formula id="ptz160UM19"><tex-math notation="LaTeX" id="Equation131"><![CDATA[$$ F(u) \; = \; (A \, u^2 + B)^{-1-\varepsilon} .$$]]></tex-math></disp-formula></p>
<p>The condition &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation522"><![CDATA[$ 0 < - \, B_{\rm I}/A_{\rm I} <1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation523"><![CDATA[$A_{\rm I} \, [ A_{\rm R} \, B_{\rm I} - A_{\rm I} \, B_{\rm R} ] > 0$]]></tex-math></inline-formula>&#x201D; is the condition for the discontinuity cuts of <inline-formula><tex-math notation="LaTeX" id="ImEquation524"><![CDATA[$\ln (A \, u^2 + B)$]]></tex-math></inline-formula> to cross the real axis between <inline-formula><tex-math notation="LaTeX" id="ImEquation525"><![CDATA[$0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation526"><![CDATA[$1$]]></tex-math></inline-formula> (cf. Appendix D of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]). This enables us to rewrite <inline-formula><tex-math notation="LaTeX" id="ImEquation527"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> in a generic way in the various cases as
<disp-formula id="ptz160M3-54"><label>(3.54)</label><tex-math notation="LaTeX" id="Equation132"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) = \frac{2^{\varepsilon}}{T} \, \Gamma(1+\varepsilon) \notag \\
& {} \times
\left\{
- \frac{1}{\varepsilon} \,
\int_{\widetilde{(0,1)}} d u \,
\left( u^2 \, P_{ijk} + R_{ij} \right)^{-1-\varepsilon} - U \big( \Delta_3,\Delta_1^{\{ij\}},\widetilde{D}_{ijk} \big)
\right.
\notag \\
&\qquad {}
\;\; + \int_{\widetilde{(0,1)}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\left[
\vphantom{\frac{d u }{u^2 \, P_{ijk} + R_{ij}}}
\ln \left( T \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} \right) \right.
\notag \\
&\qquad \qquad{}
- \left.
\ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T \right) + \ln \left( \frac{(P_{ijk} + R_{ij}) \, (T - R_{ij})}{P_{ijk}} \right)
\right]
\notag \\
&\qquad {}
\left.
- \int_0^1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \, \left[ \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) + \ln \left( \frac{T - R_{ij}}{T} \right) \right]
\vphantom{\frac{d u }{u^2 \, P_{ijk} + R_{ij}}}
\right\}\!,
\label{eqnewverLijk0}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M3-55"><label>(3.55)</label><tex-math notation="LaTeX" id="Equation133"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}U(\Delta_3,\Delta_1^{\{ij\}},\widetilde{D}_{ijk}) \notag \\
&= \left\{
\begin{array}{lcl}
0 & \mbox{if} & \operatorname{Im}(\Delta_3) > 0 \; \operatorname{Im}(\Delta_1^{\{ij\}}) > 0 , \\
\int_{_{\Gamma^{+}}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\ln \left( \frac{T - R_{ij}}{T} \right) &
\mbox{if} & \operatorname{Im}(\Delta_3) > 0 \; \operatorname{Im}(\Delta_1^{\{ij\}}) < 0 , \\
\int_{_{\Gamma^{+}}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) &
\mbox{if} & \operatorname{Im}(\Delta_3) < 0 \; \operatorname{Im}(\Delta_1^{\{ij\}}) > 0 , \\
\int_{_{\Gamma^{+}}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \, \left[
\ln \left( \frac{T - R_{ij}}{T} \right) \right. & & \\
\qquad {} + \left. \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) \right] & \mbox{if} & \operatorname{Im}(\Delta_3) < 0 \; \operatorname{Im}(\Delta_1^{\{ij\}}) < 0 ,
\end{array}
\right.
\label{eqdeffuncu}
\end{align}$$]]></tex-math></disp-formula>
where the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation528"><![CDATA[$\Gamma^{+}$]]></tex-math></inline-formula> is the closed contour encircling the &#x201C;north-east&#x201D; quadrant <italic>clockwise</italic> (cf. Sect. 3.1 of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]). Depending on the location of the cuts of <inline-formula><tex-math notation="LaTeX" id="ImEquation529"><![CDATA[$ \left( u^2 \, P_{ijk} + R_{ij} \right)^{-1-\varepsilon}$]]></tex-math></inline-formula>, the first term of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-54">3.54</xref>) is the same as those which appear in Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-48">2.48</xref>) or (<xref ref-type="disp-formula" rid="ptz160M2-51">2.51</xref>); <inline-formula><tex-math notation="LaTeX" id="ImEquation530"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> can thus be written as
<disp-formula id="ptz160M3-56"><label>(3.56)</label><tex-math notation="LaTeX" id="Equation134"><![CDATA[$$\begin{align}
L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) &= \frac{1}{T} \, L_3^n(0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) + \tilde{L}_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) ,
\label{decomp-ir}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M3-57"><label>(3.57)</label><tex-math notation="LaTeX" id="Equation135"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}\tilde{L}_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) \notag \\
&=
\frac{1}{T} \,
\left\{
\vphantom{\frac{d u }{u^2 \, P_{ijk} + R_{ij}}}
\int_{\widetilde{(0,1)}} \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \,
\Bigg[
\ln \left( T \, (1-u^2) \right) - \ln \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}} \right) \right.
\notag \\
&\qquad \qquad{}
-
\ln \left( u^2 \, (P_{ijk} + R_{ij} - T) + T \right) + \ln \left( \frac{(P_{ijk} + R_{ij}) \, (T - R_{ij})}{P_{ijk}} \right)
\Biggr]
\notag \\
&\qquad \qquad {}
- \int_0^1 \, \frac{d u }{u^2 \, P_{ijk} + R_{ij}} \, \left[ \eta \left( \frac{T \, (P_{ijk} + R_{ij})}{P_{ijk}}, \frac{T - R_{ij}}{T} \right) + \ln \left( \frac{T - R_{ij}}{T} \right) \right]
\notag \\
&\qquad \qquad {}
- \left. U \big( \Delta_3,\Delta_1^{\{ij\}}, \widetilde{D}_{ijk} \big)
\vphantom{\frac{d u }{u^2 \, P_{ijk} + R_{ij}}}
\right\}\!,
\label{eqnewvertLijk1}
\end{align}$$]]></tex-math></disp-formula>
and <inline-formula><tex-math notation="LaTeX" id="ImEquation531"><![CDATA[$L_3^n(0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-48">2.48</xref>) or Eq. (<xref ref-type="disp-formula" rid="ptz160M2-53">2.53</xref>) depending on the sign of the imaginary part of <inline-formula><tex-math notation="LaTeX" id="ImEquation532"><![CDATA[$\Delta_1^{\{i,j\}}$]]></tex-math></inline-formula>.</p>
<p>For the cases where <inline-formula><tex-math notation="LaTeX" id="ImEquation533"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation534"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula>, a unique formula for <inline-formula><tex-math notation="LaTeX" id="ImEquation535"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula> was found above, whatever the signs of <inline-formula><tex-math notation="LaTeX" id="ImEquation536"><![CDATA[$\operatorname{Im}(\Delta_3)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation537"><![CDATA[$\operatorname{Im}(\Delta_1^{\{ij\}})$]]></tex-math></inline-formula>. The decomposition of the form in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-56">3.56</xref>) stills holds: the IR-divergent part of <inline-formula><tex-math notation="LaTeX" id="ImEquation538"><![CDATA[$L_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula> is the same as in the three-point case, cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-59">2.59</xref>), whereas now,
<disp-formula id="ptz160M3-58"><label>(3.58)</label><tex-math notation="LaTeX" id="Equation136"><![CDATA[$$\begin{align}
\tilde{L}_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)
&= - \frac{1}{2 \, R_{ij} \, T} \,
\left[
\mbox{Li}_2\left( \frac{R_{ij}}{T} \right) + \left[ \ln(R_{ij}) - \ln(T) \right] \,
\ln \left( \frac{T - R_{ij}}{T} \right)
\right]\!.
\label{eqnewvertLijk2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Coming back to <inline-formula><tex-math notation="LaTeX" id="ImEquation539"><![CDATA[$I_4^n$]]></tex-math></inline-formula> and using the results of Sect. <xref ref-type="sec" rid="SEC2.3">2.3</xref>, we have
<disp-formula id="ptz160M3-59"><label>(3.59)</label><tex-math notation="LaTeX" id="Equation137"><![CDATA[$$\begin{align}
I_4^n
&= \sum_{i \in S_4} \frac{\overline{b}_i}{\det{(G)}} \, \sum_{j \in S_4 \setminus \{i\}}
\frac{\overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}} \, \frac{W\big(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, \widetilde{D}_{ijl}\big)}{T}
\notag \\
&\quad {}
+ \sum_{i \in S_4} \frac{\overline{b}_i}{\det{(G)}} \, \sum_{j \in S_4 \setminus \{i\}}
\frac{\overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}} \, \sum_{k \in S_4 \setminus \{i,j\}}
\frac{\overline{b}_{k}^{\{i,j\}}}{\det{(G^{\{i,j\}})}} \, \tilde{L}_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},\widetilde{D}_{ijk}) ,
\label{eqI4nb3}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation540"><![CDATA[$l \in S_4 \setminus \{i,j,k\}$]]></tex-math></inline-formula>. The first term on the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M3-59">3.59</xref>) is nothing but the combination of the three-point functions as in Refs. [<xref ref-type="bibr" rid="B7">7</xref>,<xref ref-type="bibr" rid="B8">8</xref>] decomposed according to the so-called &#x201C;direct way&#x201D; made explicit in Sect.. <xref ref-type="sec" rid="SEC2.1">2.1</xref>. Note that when <inline-formula><tex-math notation="LaTeX" id="ImEquation541"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula>, the quantity <inline-formula><tex-math notation="LaTeX" id="ImEquation542"><![CDATA[$\tilde{L}_4^n(\Delta_3,0,\Delta_1^{\{i,j\}},0)$]]></tex-math></inline-formula>, which does not actually depend on <inline-formula><tex-math notation="LaTeX" id="ImEquation543"><![CDATA[$k$]]></tex-math></inline-formula>, factors out from the sum over <inline-formula><tex-math notation="LaTeX" id="ImEquation544"><![CDATA[$k$]]></tex-math></inline-formula>, which then yields trivially <inline-formula><tex-math notation="LaTeX" id="ImEquation545"><![CDATA[$-1$]]></tex-math></inline-formula>. The function <inline-formula><tex-math notation="LaTeX" id="ImEquation546"><![CDATA[$W(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, \widetilde{D}_{ijl})$]]></tex-math></inline-formula> has been defined by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-6">2.6</xref>). Its arguments have one extra subscript, tracing back the extra pinching which was involved compared with the three-point case. We recap here the different results concerning this function:
<disp-formula id="ptz160M3-60"><label>(3.60)</label><tex-math notation="LaTeX" id="Equation138"><![CDATA[$$\begin{equation}
W\big(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, \widetilde{D}_{ijl}\big) = \frac{2^{\varepsilon}}{\varepsilon} \, \Gamma(1+\varepsilon) \, \int^1_0 dx \,
\big( D^{\{i,j\}(k)}(x) - i \, \lambda \big)^{-1-\varepsilon} ,
\label{eqdefwi01}
\end{equation}$$]]></tex-math></disp-formula>
where
<disp-formula id="ptz160M3-61"><label>(3.61)</label><tex-math notation="LaTeX" id="Equation139"><![CDATA[$$\begin{equation}
D^{\{i,j\}(k)}(x)
=
G^{\{i,j\}(k)} \, x^2 - 2 \, V^{\{i,j\}(k)} \, x - C^{\{i,j\}(k)} ,
\label{eqremd11}
\end{equation}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M3-62"><label>(3.62)</label><tex-math notation="LaTeX" id="Equation140"><![CDATA[$$\begin{align}
G^{\{i,j\}(k)}
&= - {\cal S}_{ll} + 2 \, {\cal S}_{kl} - {\cal S}_{kk} = \det{(G^{\{i,j\}})} ,
\notag \\
V^{\{i,j\}(k)}
&= {\cal S}_{kl} - {\cal S}_{kk} = \frac{1}{2} \, \big[ \det{(G^{\{i,j\}})} - \widetilde{D}_{ijk} + \widetilde{D}_{ijl} \big] ,
\label{eqremd21}\\
C^{\{i,j\}(k)}
&= {\cal S}_{kk} = - \widetilde{D}_{ijl} .
\notag
\end{align}$$]]></tex-math></disp-formula></p>
<p>Note that <inline-formula><tex-math notation="LaTeX" id="ImEquation547"><![CDATA[$W(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, \widetilde{D}_{ijl})$]]></tex-math></inline-formula> is symmetric under the exchange of <inline-formula><tex-math notation="LaTeX" id="ImEquation548"><![CDATA[$i$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation549"><![CDATA[$j$]]></tex-math></inline-formula>.</p>
<p><list list-type="bullet">
<list-item><p>If <inline-formula><tex-math notation="LaTeX" id="ImEquation550"><![CDATA[$\widetilde{D}_{ijk}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation551"><![CDATA[$\widetilde{D}_{ijl}$]]></tex-math></inline-formula> both differ from zero, <inline-formula><tex-math notation="LaTeX" id="ImEquation552"><![CDATA[$W\big(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, \widetilde{D}_{ijl}\big)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-15">2.15</xref>).</p></list-item>
<list-item><p>If only <inline-formula><tex-math notation="LaTeX" id="ImEquation553"><![CDATA[$\widetilde{D}_{ijl}$]]></tex-math></inline-formula> vanishes, <inline-formula><tex-math notation="LaTeX" id="ImEquation554"><![CDATA[$W\big(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk}, 0\big)$]]></tex-math></inline-formula> is read from Eq. (<xref ref-type="disp-formula" rid="ptz160M2-28">2.28</xref>).</p></list-item>
<list-item><p>If <inline-formula><tex-math notation="LaTeX" id="ImEquation555"><![CDATA[$\widetilde{D}_{ijk}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation556"><![CDATA[$\widetilde{D}_{ijl}$]]></tex-math></inline-formula> both vanish, <inline-formula><tex-math notation="LaTeX" id="ImEquation557"><![CDATA[$W\big(\det{(G^{\{i,j\}})},0,0\big)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-30">2.30</xref>).</p></list-item>
</list></p>
<p>For practical purposes let us stress that we do not have to compute a dedicated formula for each IR four-point case as in Ref. [<xref ref-type="bibr" rid="B4">4</xref>], where 16 cases were distinguished. Indeed, for each contribution labelled by the index <inline-formula><tex-math notation="LaTeX" id="ImEquation558"><![CDATA[$i$]]></tex-math></inline-formula>, we merely distinguish two cases: either <inline-formula><tex-math notation="LaTeX" id="ImEquation559"><![CDATA[$\Delta_2^{\{i\}} \ne 0$]]></tex-math></inline-formula>, for which we use the generic formula suited to the massive case, or <inline-formula><tex-math notation="LaTeX" id="ImEquation560"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula>, for which we use the appropriate formula suited to the IR case at hand. The massive case is split depending on the vanishing of <inline-formula><tex-math notation="LaTeX" id="ImEquation561"><![CDATA[$\operatorname{Im}(\Delta_3)$]]></tex-math></inline-formula> (namely, if one internal mass squared has an imaginary part different from zero). The IR case is also divided in two cases: <inline-formula><tex-math notation="LaTeX" id="ImEquation562"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation563"><![CDATA[$\widetilde{D}_{ijk} \ne 0$]]></tex-math></inline-formula>. The latter is further separated according to the fact that one or several internal masses squared have a non-vanishing imaginary part. All these cases are depicted on the decision tree presented in <xref ref-type="fig" rid="F5">Fig. 5</xref>; for each case the appropriate formula is given. Notice that when <inline-formula><tex-math notation="LaTeX" id="ImEquation564"><![CDATA[$\operatorname{Im}(\Delta_3) \ne 0$]]></tex-math></inline-formula>, it may appear that <inline-formula><tex-math notation="LaTeX" id="ImEquation565"><![CDATA[$\Delta_2^{\{i\}}$]]></tex-math></inline-formula> and/or <inline-formula><tex-math notation="LaTeX" id="ImEquation566"><![CDATA[$\Delta_1^{\{i,j\}}$]]></tex-math></inline-formula> are real; in this case they must be understood as having a &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation567"><![CDATA[$+ i \, \lambda$]]></tex-math></inline-formula>&#x201D; prescription. The expense paid by the present method is a possible proliferation of dilogarithms, a counteraction against which would require some extra work. This point will be commented on in more detail in the examples studied below.</p>
<fig id="F5" orientation="portrait" position="float"><label>Fig. 5.</label><caption><p>Decision tree to compute <inline-formula><tex-math notation="LaTeX" id="ImEquation568"><![CDATA[$L_4^n(\Delta_3,\Delta_2^{\{i\}},\Delta_1^{\{i,j\}},\widetilde{D}_{ijk})$]]></tex-math></inline-formula> for a given sector labelled by <inline-formula><tex-math notation="LaTeX" id="ImEquation569"><![CDATA[$i$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation570"><![CDATA[$j$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation571"><![CDATA[$k$]]></tex-math></inline-formula>.</p></caption>
<graphic xmlns:xlink="http://www.w3.org/1999/xlink" orientation="portrait" position="float" mimetype="image" xlink:href="ptz160f5.tif"/></fig>
<sec id="SEC3.4.1"><title>3.4.1. Two opposite external masses</title>
<p>In this case all internal masses are zero and two opposite external legs (say 1 and 3) have non-lightlike four-momenta, the two others being lightlike, i.e. <inline-formula><tex-math notation="LaTeX" id="ImEquation572"><![CDATA[$p_1^2 \neq 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation573"><![CDATA[$p_2^2 = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation574"><![CDATA[$p_3^2 \neq 0$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation575"><![CDATA[$p_4^2 = 0$]]></tex-math></inline-formula> (our convention is depicted in <xref ref-type="fig" rid="F4">Fig. 4</xref>). The texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation576"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is<sup><xref ref-type="fn" rid="FN10">10</xref></sup>
<disp-formula id="ptz160M3-63"><label>(3.63)</label><tex-math notation="LaTeX" id="Equation141"><![CDATA[$$\begin{align}
{\cal S} &= \left(
\begin{array}{cccc}
0 & 0 & s_{23} & s_1 \\
0 & 0 & s_3 & s_{12} \\
s_{23} & s_3 & 0 & 0 \\
s_1 & s_{12} & 0 & 0
\end{array}
\right)\!,
\label{eqsmatexempl1}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation577"><![CDATA[$s_i = p_i^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation578"><![CDATA[$s_{ij} = (p_i + p_j)^2$]]></tex-math></inline-formula> (all the momenta are taken ingoing). All the contributions <inline-formula><tex-math notation="LaTeX" id="ImEquation579"><![CDATA[$i$]]></tex-math></inline-formula> are such that <inline-formula><tex-math notation="LaTeX" id="ImEquation580"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation581"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula>. In this case, <inline-formula><tex-math notation="LaTeX" id="ImEquation582"><![CDATA[$R_{ij} = - \Delta_1^{\{i,j\}}$]]></tex-math></inline-formula> is symmetric under the exchange of <inline-formula><tex-math notation="LaTeX" id="ImEquation583"><![CDATA[$i$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation584"><![CDATA[$j$]]></tex-math></inline-formula>. The four-point function is given, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M3-29">3.29</xref>), by
<disp-formula id="ptz160M3-64"><label>(3.64)</label><tex-math notation="LaTeX" id="Equation142"><![CDATA[$$\begin{align}
I_4^n
&= \frac{1}{2 \, \det{({\cal S})}} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \,
\sum_{i \in S_4} \, \sum_{j > i} \frac{1}{\Delta_1^{\{i,j\}}} \,
\bigg(
\frac{\overline{b}_i \, \overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}}
+
\frac{\overline{b}_j \, \overline{b}_{i}^{\{j\}}}{\det{(G^{\{j\}})}}
\bigg)
\notag \\
&\quad {}
\times \,
\left[
\frac{1}{\varepsilon^2} \,
\big( - 2 \, \Delta_1^{\{i,j\}} \big)^{- \varepsilon}
+ \mbox{Li}_2 \left( \frac{T - R_{ij}}{T - i \, \lambda} \right)
- \frac{\pi^2}{6}
\right]\!.
\label{eqI4nexemple1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Due to the hollow texture of the reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation585"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrices, a bunch of <inline-formula><tex-math notation="LaTeX" id="ImEquation586"><![CDATA[$\overline{b}_{j}^{\{i\}}$]]></tex-math></inline-formula> coefficients vanish. Equation (<xref ref-type="disp-formula" rid="ptz160M3-64">3.64</xref>) thus simplifies into
<disp-formula id="ptz160M3-65"><label>(3.65)</label><tex-math notation="LaTeX" id="Equation143"><![CDATA[$$\begin{align}
I_4^n
&= \frac{1}{2 \, \det{({\cal S})}} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \,
\sum_{i=1}^{2} \, \sum_{j=3}^{4} \frac{1}{\Delta_1^{\{i,j\}}} \,
\bigg(
\frac{\overline{b}_i \, \overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}}
+
\frac{\overline{b}_j \, \overline{b}_{i}^{\{j\}}}{\det{(G^{\{j\}})}}
\bigg)
\notag \\
&\quad {}
\times \,
\left[ \frac{1}{\varepsilon^2} \,
\big( - 2 \, \Delta_1^{\{i,j\}} \big)^{- \varepsilon}
+ \mbox{Li}_2\left( 1 - \frac{R_{ij} - i \, \lambda}{T - i \, \lambda} \right)
- \frac{\pi^2}{6}
\right]\!.
\label{eqI4nexemple2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Expressing the <inline-formula><tex-math notation="LaTeX" id="ImEquation587"><![CDATA[$\overline{b}_{i}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation588"><![CDATA[$\overline{b}_{j}^{\{i\}}$]]></tex-math></inline-formula> coefficients as well as the various determinants as functions of the <inline-formula><tex-math notation="LaTeX" id="ImEquation589"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix elements, we get
<disp-formula id="ptz160M3-66"><label>(3.66)</label><tex-math notation="LaTeX" id="Equation144"><![CDATA[$$\begin{align}
I_4^n
&= \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \, \frac{2}{d} \,
\notag\\
& \quad {}
\left\{
\vphantom{\frac{s_1}{s_1}}
(- s_1 - i \, \lambda)^{-\varepsilon} +
(- s_3 - i \, \lambda)^{-\varepsilon} -
(- s_{12} - i \, \lambda)^{-\varepsilon} -
(- s_{23} - i \, \lambda)^{-\varepsilon}
\right.
\notag \\
&\quad {} \quad {}
- \mbox{Li}_2
\left( 1 - \frac{s_{12} + i \, \lambda}{d/\Sigma + i \, \lambda} \right)
- \mbox{Li}_2
\left( 1 - \frac{s_{23} + i \, \lambda}{d/\Sigma + i \, \lambda} \right)
\notag \\
&\quad {}\quad {}
\left.
+ \mbox{Li}_2
\left( 1 - \frac{s_{1} + i \, \lambda}{d/\Sigma + i \, \lambda} \right)
+ \mbox{Li}_2
\left( 1 - \frac{s_{3} + i \, \lambda}{d/\Sigma + i \, \lambda} \right)
\right\}\!,
\label{eqI4nexemple3}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M3-67"><label>(3.67)</label><tex-math notation="LaTeX" id="Equation145"><![CDATA[$$\begin{align}
d &= s_1 \, s_3 - s_{12} \, s_{23} , \label{eqdefdI4n} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M3-68"><label>(3.68)</label><tex-math notation="LaTeX" id="Equation146"><![CDATA[$$\begin{align}
\Sigma &= s_1 + s_3 - s_{12} - s_{23} .
\label{eqdefSigmaI4n}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M3-68">3.68</xref>) involves four dilogarithms, as in Refs. [<xref ref-type="bibr" rid="B14">14</xref>,<xref ref-type="bibr" rid="B15">15</xref>]. The arguments of the dilogarithms in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-66">3.66</xref>) vs. in Ref. [<xref ref-type="bibr" rid="B14">14</xref>] are seemingly different, namely Ref. [<xref ref-type="bibr" rid="B14">14</xref>] involves <inline-formula><tex-math notation="LaTeX" id="ImEquation590"><![CDATA[$\mbox{Li}_2 ( 1 - (w+i \, \lambda)/(d/\Sigma) )$]]></tex-math></inline-formula> whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation591"><![CDATA[$\mbox{Li}_2 ( 1 - (w+i \, \lambda)/(d/\Sigma + i \, \lambda) )$]]></tex-math></inline-formula> appears in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-66">3.66</xref>), where <inline-formula><tex-math notation="LaTeX" id="ImEquation592"><![CDATA[$w$]]></tex-math></inline-formula> stands for <inline-formula><tex-math notation="LaTeX" id="ImEquation593"><![CDATA[$s_1$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation594"><![CDATA[$s_3$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation595"><![CDATA[$s_{12}$]]></tex-math></inline-formula>, or <inline-formula><tex-math notation="LaTeX" id="ImEquation596"><![CDATA[$s_{23}$]]></tex-math></inline-formula>. However, the <inline-formula><tex-math notation="LaTeX" id="ImEquation597"><![CDATA[$i \, \lambda$]]></tex-math></inline-formula> prescriptions matter only when the real parts of the arguments of the dilogarithms are greater than 1, i.e. whenever <inline-formula><tex-math notation="LaTeX" id="ImEquation598"><![CDATA[$w \, \Sigma/d < 0$]]></tex-math></inline-formula>, in which case the signs of <inline-formula><tex-math notation="LaTeX" id="ImEquation599"><![CDATA[$(d/\Sigma) - w$]]></tex-math></inline-formula> and of <inline-formula><tex-math notation="LaTeX" id="ImEquation600"><![CDATA[$\Sigma/d$]]></tex-math></inline-formula>, which respectively control the signs of the <inline-formula><tex-math notation="LaTeX" id="ImEquation601"><![CDATA[$i \lambda$]]></tex-math></inline-formula> prescriptions in either case, are the same: the two results in Eq. (<xref ref-type="disp-formula" rid="ptz160M3-66">3.66</xref>) and in Ref. [<xref ref-type="bibr" rid="B14">14</xref>] are actually identical.</p>
<p>This is to be compared with the formula given in Ref. [<xref ref-type="bibr" rid="B4">4</xref>]. This was taken from Ref. [<xref ref-type="bibr" rid="B15">15</xref>] and involves five dilogarithms instead of four. The authors of Ref. [<xref ref-type="bibr" rid="B15">15</xref>] used the so-called Mantel identity, which entails nine dilogarithms, to prove that the four-dilogarithm and five-dilogarithm results are actually equivalent. The Mantel identity happens to be a corollary of the Hill identity, the former being derived by applying the latter three times to some suitable combinations of variables.<sup><xref ref-type="fn" rid="FN11">11</xref></sup> The continuation of the Hill identity to any two arbitrary complex variables, however, requires additional combinations of <inline-formula><tex-math notation="LaTeX" id="ImEquation602"><![CDATA[$\eta$]]></tex-math></inline-formula> functions handling the mismatch between the various discontinuities of the dilogarithms involved, and these <inline-formula><tex-math notation="LaTeX" id="ImEquation603"><![CDATA[$\eta$]]></tex-math></inline-formula> functions are often skipped in the literature.<sup><xref ref-type="fn" rid="FN12">12</xref></sup> An even busier modification is then required for the Mantel identity. This drove of <inline-formula><tex-math notation="LaTeX" id="ImEquation604"><![CDATA[$\eta$]]></tex-math></inline-formula> functions makes the analytical check of the equivalence between the four-dilogarithm and five-dilogarithm expressions extremely awkward in general, and to our understanding this drove of <inline-formula><tex-math notation="LaTeX" id="ImEquation605"><![CDATA[$\eta$]]></tex-math></inline-formula> functions was not accounted for in Ref. [<xref ref-type="bibr" rid="B15">15</xref>]. We did perform numerical tests accounting for these <inline-formula><tex-math notation="LaTeX" id="ImEquation606"><![CDATA[$\eta$]]></tex-math></inline-formula> functions, which verified the equivalence for the configurations probed.</p>
</sec>
<sec id="SEC3.4.2"><title>3.4.2. A simple case with one internal mass</title>
<p>In this example, with the convention depicted in <xref ref-type="fig" rid="F4">Fig. 4</xref>, propagator 3 (with four-momentum <inline-formula><tex-math notation="LaTeX" id="ImEquation607"><![CDATA[$q_3$]]></tex-math></inline-formula>) is taken massive, the others being massless, the two external legs 1 and 2 have lightlike four-momenta, that of the external leg 3 is on the mass shell of propagator 3, and the external leg 4 has a non-lightlike four-momentum; i.e. <inline-formula><tex-math notation="LaTeX" id="ImEquation608"><![CDATA[$p_1^2=p_2^2=0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation609"><![CDATA[$p_3^2=m_3^2$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation610"><![CDATA[$p_4^2\neq0$]]></tex-math></inline-formula>. With the same notation as the preceding example, the texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation611"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is
<disp-formula id="ptz160M3-69"><label>(3.69)</label><tex-math notation="LaTeX" id="Equation147"><![CDATA[$$\begin{align}
{\cal S} &= \left(
\begin{array}{cccc}
0 & 0 & s_{23}-m_3^2 & 0 \\
0 & 0 & 0 & s_{12} \\
s_{23} - m_3^2 & 0 & - 2 \, m_3^2 & s_4 - m_3^2 \\
0 & s_{12} & s_4 - m_3^2 & 0
\end{array}
\right)\!.
\label{eqsmatexempl2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In this case, the sector <inline-formula><tex-math notation="LaTeX" id="ImEquation612"><![CDATA[$i=1$]]></tex-math></inline-formula> has no soft or collinear divergence and is computed using the massive formula. The other sectors correspond to the cases <inline-formula><tex-math notation="LaTeX" id="ImEquation613"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation614"><![CDATA[$\widetilde{D}_{ijk} \ne 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation615"><![CDATA[$\Delta_2^{\{i\}} = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation616"><![CDATA[$\widetilde{D}_{ijk} = 0$]]></tex-math></inline-formula>. The four-point amplitude can be cast in a divergent part and a finite one. The divergent part is given by
<disp-formula id="ptz160M3-70"><label>(3.70)</label><tex-math notation="LaTeX" id="Equation148"><![CDATA[$$\begin{align}
\left( I^n_4 \right)_{\rm div} &= \sum_{i \in S_4 \setminus \{1\}} \, \frac{\overline{b}_i}{\det{(G)}} \, \sum_{j \in S_4 \setminus \{i\}} \, \frac{\overline{b}_{j}^{\{i\}}}{\det{(G^{\{i\}})}} \, \frac{W\big(\det{(G^{\{i,j\}})},\widetilde{D}_{ijk},\widetilde{D}_{ijl}\big)}{T} .
\label{eqdefdivpartI44}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As several <inline-formula><tex-math notation="LaTeX" id="ImEquation617"><![CDATA[$\overline{b}_{j}^{\{i\}}$]]></tex-math></inline-formula> vanish due to the hollow texture of reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation618"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrices, we actually have to compute<sup><xref ref-type="fn" rid="FN13">13</xref></sup>
<disp-formula id="ptz160M3-71"><label>(3.71)</label><tex-math notation="LaTeX" id="Equation149"><![CDATA[$$\begin{align}
\left( I^n_4 \right)_{\rm div} &= \frac{\overline{b}_2}{\det{({\cal S})}} \, \left[ \frac{\overline{b}_{1}^{\{2\}}}{\det{(G^{\{2\}})}} \, W\big( \det{(G^{\{1,2\}})},\widetilde{D}_{124},0 \big) + \frac{\overline{b}_{4}^{\{2\}}}{\det{(G^{\{2\}})}} \, W\big( \det{(G^{\{2,4\}})},\widetilde{D}_{124},0 \big) \right] \notag \\
&\quad {} + \frac{\overline{b}_3}{\det{({\cal S})}} \, \frac{\overline{b}_{1}^{\{3\}}}{\det{(G^{\{3\}})}} \, W\big( \det{(G^{\{1,3\}})},0,0 \big) \notag \\
&\quad {} + \frac{\overline{b}_4}{\det{({\cal S})}} \, \frac{\overline{b}_{2}^{\{4\}}}{\det{(G^{\{4\}})}} \, W\big( \det{(G^{\{2,4\}})},\widetilde{D}_{124},0 \big) ,
\label{eqdefdivpartI441}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation619"><![CDATA[$W\big( \det{(G^{\{1,2\}})},\widetilde{D}_{124},0 \big)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation620"><![CDATA[$W\big( \det{(G^{\{2,4\}})},\widetilde{D}_{124},0 \big)$]]></tex-math></inline-formula> are given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-28">2.28</xref>), and <inline-formula><tex-math notation="LaTeX" id="ImEquation621"><![CDATA[$W\big( \det{(G^{\{1,3\}})},0,0 \big)$]]></tex-math></inline-formula> by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-30">2.30</xref>). We get:
<disp-formula id="ptz160M3-72"><label>(3.72)</label><tex-math notation="LaTeX" id="Equation150"><![CDATA[$$\begin{align}
\left( I^n_4 \right)_{\rm div}
&=
\frac{1}{\varepsilon} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1 - 2 \, \varepsilon)} \,
\frac{1}{s_{12} \, (s_{23} - m_3^2)} \,
\notag\\
& \quad {}
\left[
\;
\frac{2}{\varepsilon} \,
\left( - s_{23} + m_3^2 - i \, \lambda \right)^{-\varepsilon}
+
\frac{1}{\varepsilon} \,
\left( - s_{12} - i \, \lambda \right)^{-\varepsilon}
\right.
\notag \\
&\quad {}
- \frac{1}{\varepsilon} \, \left( - s_4 + m_3^2 - i \, \lambda \right)^{-\varepsilon} - \frac{1}{2 \, \varepsilon} \left( m_3^2 - i \, \lambda \right)^{-\varepsilon} \notag \\
&\quad {}
\left.
+ \varepsilon \,
\mbox{Li}_2\left( \frac{s_4}{s_4 - m_3^2 + i \, \lambda} \right)
- 2 \, \varepsilon \,
\mbox{Li}_2 \left( \frac{s_{23}}{s_{23} - m_3^2 + i \, \lambda} \right)
+ \varepsilon \, \frac{\pi^2}{12} \;
\right]\!.
\label{eqdefdivpartI442}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Expanding Eq. (<xref ref-type="disp-formula" rid="ptz160M3-72">3.72</xref>) in <inline-formula><tex-math notation="LaTeX" id="ImEquation622"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>, we recover the results of Eq. (4.27) in Ref. [<xref ref-type="bibr" rid="B4">4</xref>] taken from Ref. [<xref ref-type="bibr" rid="B18">18</xref>] for the terms proportional to <inline-formula><tex-math notation="LaTeX" id="ImEquation623"><![CDATA[$1/\varepsilon^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation624"><![CDATA[$1/\varepsilon$]]></tex-math></inline-formula>. Concerning the finite part, we obtain a host of terms for which it is cumbersome to verify analytically that they do reduce to the finite part of Eq. (4.27) of Ref. [<xref ref-type="bibr" rid="B4">4</xref>]. We verified numerically that they do indeed.</p>
</sec>
</sec>
</sec>
<sec id="SEC4"><title>4. Summary and outlook</title>
<p>In this article we have presented an extension of a novel approach developed in the companion articles of Refs. [<xref ref-type="bibr" rid="B1">1</xref>,<xref ref-type="bibr" rid="B2">2</xref>] to the case of vanishing internal masses involving soft and/or collinear divergences. For this latter case, the method remains very similar to the massive cases: the three- and four-point functions are split into &#x201C;sectors&#x201D; whose coefficients are expressed in terms of algebraic kinematical invariants involved in reduction algorithms. Each &#x201C;sector&#x201D; may diverge or not when the IR regulator is sent to zero, yielding to a simple decision tree to compute the relevant integrals. This avoids the computation of the numerous different integrals over Feynman parameters, as is usually done in the literature. This extension also applies to general kinematics beyond those relevant for one-loop collider processes, offering a potential application to the calculation of two-loop processes using one-loop (generalized) <inline-formula><tex-math notation="LaTeX" id="ImEquation625"><![CDATA[$N$]]></tex-math></inline-formula>-point functions as building blocks, as discussed in the introduction of Ref. [<xref ref-type="bibr" rid="B1">1</xref>].</p>
<p>One drawback of the present method is the proliferation of dilogarithms in the expression of the four-point function computed in closed form. This requires some extra work to be better apprehended, in order to counteract it. But as the method used here is the same as in the real mass case, up to slight modifications, any solution found for the latter case can be applied in the infrared-divergent case. This issue will be addressed in a future article.</p>
<p>The last goal is to provide the generalized one-loop building blocks entering as integrands in the computation of two-loop three- and four-point functions by means of an extra numerical double integration. In this respect, let us mention that the expansion around <inline-formula><tex-math notation="LaTeX" id="ImEquation626"><![CDATA[$\varepsilon = 0$]]></tex-math></inline-formula> of the results given in this article has been truncated in order to keep only the divergent and the constant terms. This is sufficient for any one-loop computation, but may not be enough for two-loop applications of the method. This article already contains a lot of results, so the expansion around <inline-formula><tex-math notation="LaTeX" id="ImEquation627"><![CDATA[$\varepsilon = 0$]]></tex-math></inline-formula> at the necessary orders is postponed to a future work.</p>
</sec>
<sec id="SEC5"><title>In memoriam</title>
<p>Various ideas and techniques used in this work were initiated by Prof. Shimizu after a visit to LAPTh. He explained to us his ideas about the numerical computation of scalar two-loop three- and four-point functions, he shared his notes partly in English, partly in Japanese with us, and he encouraged us to push this project forward. J.Ph. G. would like to thank Shimizu-sensei for giving him a taste of the Japanese culture and for his kindness.</p>
</sec>
</body>
<back>
<ack id="ack1">
<title>Acknowledgements</title>
<p>We would like to thank P. Aurenche for his support throughout this project and for a careful reading of the manuscript.</p>
</ack>
<sec>
<title>Funding</title>
<p>Open Access funding: SCOAP<inline-formula><tex-math notation="LaTeX" id="ImEquation628"><![CDATA[$^3$]]></tex-math></inline-formula>.</p>
</sec>
<app-group>
<app><title/>
<sec id="SEC6"><title>Appendix A. Two basic integrals</title>
<p>In what follows, <inline-formula><tex-math notation="LaTeX" id="ImEquation629"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation630"><![CDATA[$B$]]></tex-math></inline-formula> are assumed dimensionless and complex valued, the signs of their real parts are unknown, and the signs of their imaginary parts may or may not be the same.</p>
<sec id="SEC6.1"><title>A.1. First kind</title>
<p>The computation of the three- and four-point functions in a spacetime of arbitrary dimensions involves the following extension of the case treated in Appendix D of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]:
<disp-formula id="ptz160M6-1"><label>(A.1)</label><tex-math notation="LaTeX" id="Equation151"><![CDATA[$$\begin{equation}
K(\nu) = \int^{\infty}_0 \frac{d \xi}{(\xi^{\nu}+A) \, (\xi^{\nu}+B)} .
\label{eqdefk1ext}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>After partial fraction decomposition, the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M6-1">A.1</xref>) becomes
<disp-formula id="ptz160M6-2"><label>(A.2)</label><tex-math notation="LaTeX" id="Equation152"><![CDATA[$$\begin{equation}
K(\nu) = \frac{1}{B-A} \,
\int^{\infty}_0 d \xi \,
\left[ \frac{1}{\xi^{\nu}+A} - \frac{1}{\xi^{\nu}+B} \right]\!.
\label{eqdefk2ext}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>Let us assume that <inline-formula><tex-math notation="LaTeX" id="ImEquation631"><![CDATA[$\nu > 1$]]></tex-math></inline-formula>, such that the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M6-1">A.1</xref>) can be split into the difference of two convergent integrals at infinity, which can be separately computed using Appendix B of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] with <inline-formula><tex-math notation="LaTeX" id="ImEquation632"><![CDATA[$\mu =1$]]></tex-math></inline-formula>. <inline-formula><tex-math notation="LaTeX" id="ImEquation633"><![CDATA[$K(\nu)$]]></tex-math></inline-formula> thus reads
<disp-formula id="ptz160M6-3"><label>(A.3)</label><tex-math notation="LaTeX" id="Equation153"><![CDATA[$$\begin{equation}
K(\nu)
=
\frac{1}{B-A} \,
\frac{1}{\nu} \, B \left(1 - \frac{1}{\nu}, \frac{1}{\nu} \right) \,
\left[ A^{\frac{1}{\nu} - 1} - B^{\frac{1}{\nu} - 1} \right]
\label{eqdefk3ext}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>regardless of the signs of <inline-formula><tex-math notation="LaTeX" id="ImEquation634"><![CDATA[$\operatorname{Im}(A)$]]></tex-math></inline-formula> vs. <inline-formula><tex-math notation="LaTeX" id="ImEquation635"><![CDATA[$\operatorname{Im}(B)$]]></tex-math></inline-formula>. In the case of the three-point function <inline-formula><tex-math notation="LaTeX" id="ImEquation636"><![CDATA[$\nu = 1/(1-\varepsilon)$]]></tex-math></inline-formula>, which is slightly less than <inline-formula><tex-math notation="LaTeX" id="ImEquation637"><![CDATA[$1$]]></tex-math></inline-formula> for <inline-formula><tex-math notation="LaTeX" id="ImEquation638"><![CDATA[$\varepsilon < 0$]]></tex-math></inline-formula>, and the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M6-3">A.3</xref>) can be analytically continued in <inline-formula><tex-math notation="LaTeX" id="ImEquation639"><![CDATA[$\nu$]]></tex-math></inline-formula> as long as <inline-formula><tex-math notation="LaTeX" id="ImEquation640"><![CDATA[$\nu \neq -1/n$]]></tex-math></inline-formula> or <inline-formula><tex-math notation="LaTeX" id="ImEquation641"><![CDATA[$\nu \neq 1/(n+1)$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation642"><![CDATA[$n$]]></tex-math></inline-formula> an arbitrary positive integer. So, the result of Eq. (<xref ref-type="disp-formula" rid="ptz160M6-3">A.3</xref>) can be used in the case where the dimension of the spacetime is shifted by a small positive amount from <inline-formula><tex-math notation="LaTeX" id="ImEquation643"><![CDATA[$n=4$]]></tex-math></inline-formula> to <inline-formula><tex-math notation="LaTeX" id="ImEquation644"><![CDATA[$n=4 - 2 \, \varepsilon$]]></tex-math></inline-formula>. We also note that the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation645"><![CDATA[$\nu \to 1$]]></tex-math></inline-formula> of <inline-formula><tex-math notation="LaTeX" id="ImEquation646"><![CDATA[$K(\nu)$]]></tex-math></inline-formula> leads to Eq. (D4) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>].</p>
<p>A practical case met in Sect. <xref ref-type="sec" rid="SEC2">2</xref> is <inline-formula><tex-math notation="LaTeX" id="ImEquation647"><![CDATA[$A = - i \, \lambda$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation648"><![CDATA[$B$]]></tex-math></inline-formula> remaining an arbitrary complex number with an imaginary part different from <inline-formula><tex-math notation="LaTeX" id="ImEquation649"><![CDATA[$0$]]></tex-math></inline-formula>. In the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation650"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula>, Eq. (<xref ref-type="disp-formula" rid="ptz160M6-3">A.3</xref>) becomes
<disp-formula id="ptz160M6-4"><label>(A.4)</label><tex-math notation="LaTeX" id="Equation154"><![CDATA[$$\begin{align}
\int^{+\infty}_0
\frac{d \xi}{(\xi^{\nu} - i \, \lambda)\, (\xi^{\nu} + B)}
&= - \frac{1}{\varepsilon} \, (1-\varepsilon) \, \Gamma(1+\varepsilon) \, \Gamma(1-\varepsilon)
\, B^{-1-\varepsilon} .
\label{eqmodifk}
\end{align}$$]]></tex-math></disp-formula></p>
<p>This is the well-known fact that the two limits <inline-formula><tex-math notation="LaTeX" id="ImEquation651"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation652"><![CDATA[$\varepsilon \to 0$]]></tex-math></inline-formula> do not commute.</p>
</sec>
<sec id="SEC6.2"><title>A.2. Second kind</title>
<p>The most general case for the integral
<disp-formula id="ptz160M6-5"><label>(A.5)</label><tex-math notation="LaTeX" id="Equation155"><![CDATA[$$\begin{equation}
J(\nu)
=
\int^{+\infty}_0
\frac{d \xi}{\left(\xi^{\nu}+A \right) \, \sqrt{\xi^{\nu}+B}}
\label{eqdefj1}
\end{equation}$$]]></tex-math></disp-formula>
has been treated in Appendix A of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]; let us just recap the results. Two cases can be distinguished according to the signs of the imaginary parts of <inline-formula><tex-math notation="LaTeX" id="ImEquation653"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation654"><![CDATA[$B$]]></tex-math></inline-formula>.</p>
<p><list list-type="simple">
<list-item><p>(1) <inline-formula><tex-math notation="LaTeX" id="ImEquation655"><![CDATA[$\operatorname{Im}(A)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation656"><![CDATA[$\operatorname{Im}(B)$]]></tex-math></inline-formula> of the same sign:
<disp-formula id="ptz160M6-6"><label>(A.6)</label><tex-math notation="LaTeX" id="Equation156"><![CDATA[$$\begin{equation}
J(\nu)
=
\frac{1}{\nu} \,
B \left( \frac{3}{2}- \frac{1}{\nu},\frac{1}{\nu} \right) \,
\int^1_0 dz \, \left( (1-z^2) \, A + z^2 \, B \right)^{-3/2 + 1/\nu} .
\label{eqdeffuncj2}
\end{equation}$$]]></tex-math></disp-formula></p></list-item>
<list-item><p>(2) <inline-formula><tex-math notation="LaTeX" id="ImEquation657"><![CDATA[$\operatorname{Im}(A)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation658"><![CDATA[$\operatorname{Im}(B)$]]></tex-math></inline-formula> of opposite signs:
<disp-formula id="ptz160M6-7"><label>(A.7)</label><tex-math notation="LaTeX" id="Equation157"><![CDATA[$$\begin{align}
J(\nu)
&= - \, \frac{1}{\nu} \,
B \left( \frac{3}{2}-\frac{1}{\nu},\frac{1}{\nu} \right) \notag \\
&\quad {} \times
\left[
e^{- i \, S_B \, \pi/\nu} \,
\int^{+\infty}_0 dz \,
\left( B \, z^2 - (1+z^2) \, A \right)^{-3/2+1/\nu}
\right.
\notag \\
&
\;\;\;\;\;\;\;\;\;\;\;
+
\left.
\int^{+\infty}_{1} dz \,
\left( B \, z^2 + (1-z^2) \, A \right)^{-3/2+1/\nu}
\right]\!,
\label{eqdeffuncj7}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation659"><![CDATA[$S_B = \mbox{sign}\left( \operatorname{Im}\left( B \right) \right)$]]></tex-math></inline-formula>.</p></list-item>
</list></p>
<p>Remember that the two cases (1) and (2) can be reunified by seeing Eq. (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) as an analytic continuation in <inline-formula><tex-math notation="LaTeX" id="ImEquation660"><![CDATA[$A$]]></tex-math></inline-formula> of Eq. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>), which possibly requires a deformation of the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation661"><![CDATA[$[0,1]$]]></tex-math></inline-formula> originally drawn along the real axis in Eq. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>); see Appendix A of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] for more details.</p>
</sec>
</sec>
<sec id="SEC7"><title>Appendix B. The function <inline-formula><tex-math notation="LaTeX" id="ImEquation662"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula></title>
<p>This appendix computes the function
<disp-formula id="ptz160UM20"><tex-math notation="LaTeX" id="Equation158"><![CDATA[$$J(x_1,x_2)
= \int^1_0 dx \,
\frac{\ln\left( (x-x_1) \, (x-x_2) \right)}{(x - x_1) \, (x - x_2)} .$$]]></tex-math></disp-formula></p>
<p>Using partial fraction decomposition, <inline-formula><tex-math notation="LaTeX" id="ImEquation663"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula> can be written as
<disp-formula id="ptz160M7-1"><label>(B.1)</label><tex-math notation="LaTeX" id="Equation159"><![CDATA[$$\begin{align}
&J(x_1,x_2)
\notag \\
&= \frac{1}{x_1-x_2}
\left[
\int^1_0 dx \, \frac{\ln(x-x_1)}{x-x_1}
- \int^1_0 dx \, \frac{\ln(x-x_2)}{x-x_2}
\right.
\notag \\
&\quad {} \quad {} \quad {}
+ \int^1_0 dx \, \frac{\ln(x-x_2) - \ln(x_1-x_2)}{x-x_1}
\; - \;
\int^1_0 dx \, \frac{\ln(x-x_1) - \ln(x_2-x_1)}{x-x_2}
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {}
+
\left.
\ln(x_1-x_2) \, \int^1_0 dx \, \frac{d x}{x-x_1} \;\; - \;\;
\ln(x_2-x_1) \, \int^1_0 dx \, \frac{d x}{x-x_2}
\right]\!.
\label{eqcompj2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>With the help of Appendix E of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] (see also Appendix B of Ref. [<xref ref-type="bibr" rid="B19">19</xref>]), we immediately get
<disp-formula id="ptz160M7-2"><label>(B.2)</label><tex-math notation="LaTeX" id="Equation160"><![CDATA[$$\begin{align}
\int^1_0 dx \, \frac{\ln(x-x_1) - \ln(x_2-x_1)}{x-x_2}
&= R^{\prime}(x_1,x_2) \notag \\
& = \mbox{Li}_2 \left( \frac{x_2}{x_2-x_1} \right)
- \mbox{Li}_2 \left( \frac{x_2-1}{x_2-x_1} \right)
\notag \\
&\quad {}
+ \eta \left(-x_1, \frac{1}{x_2-x_1} \right) \,
\ln \left( \frac{x_2}{x_2-x_1} \right)
\notag \\
&\quad {}
- \eta \left( 1-x_1, \frac{1}{x_2-x_1} \right) \,
\ln \left( \frac{x_2-1}{x_2-x_1} \right)\!,
\label{eqappbth1}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation664"><![CDATA[$R^{\prime}(x_1,x_2)$]]></tex-math></inline-formula> is given by Eq. (E15) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]<sup><xref ref-type="fn" rid="FN14">14</xref></sup> Since <inline-formula><tex-math notation="LaTeX" id="ImEquation665"><![CDATA[$x_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation666"><![CDATA[$x_2$]]></tex-math></inline-formula> have imaginary parts of opposite signs, all the <inline-formula><tex-math notation="LaTeX" id="ImEquation667"><![CDATA[$\eta$]]></tex-math></inline-formula> functions vanish in Eq. (<xref ref-type="disp-formula" rid="ptz160M7-2">B.2</xref>). Then, using the Landen identity,
<disp-formula id="ptz160M7-3"><label>(B.3)</label><tex-math notation="LaTeX" id="Equation161"><![CDATA[$$\begin{equation}
\mbox{Li}_2(z) + \mbox{Li}_2 \left( \frac{z}{z-1} \right)
= - \, \frac{1}{2} \, \ln^2(1-z) ,
\label{eqlanden}
\end{equation}$$]]></tex-math></disp-formula>
we can rewrite Eq. (<xref ref-type="disp-formula" rid="ptz160M7-2">B.2</xref>) as
<disp-formula id="ptz160M7-4"><label>(B.4)</label><tex-math notation="LaTeX" id="Equation162"><![CDATA[$$\begin{align}
&\int^1_0 dx \,
\frac{\ln(x-x_1) - \ln(x_2-x_1)}{x-x_2}
\notag\\
& \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
=
\mbox{Li}_2 \left( \frac{x_2-1}{x_1-1} \right)
- \mbox{Li}_2 \left( \frac{x_2}{x_1} \right)
\notag\\
& \quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+ \frac{1}{2}
\left[
\ln^2 \left( x_1-1 \right) \vphantom{\frac{x_1-1}{x_1}}
- \ln^2 \left( x_1 \right) - 2 \, \ln(x_1-x_2) \,
\ln \left( \frac{x_1-1}{x_1} \right)
\right]\!.
\label{eqappbth2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Substituting into Eq. (<xref ref-type="disp-formula" rid="ptz160M7-1">B.1</xref>) and easily computing the remaining <inline-formula><tex-math notation="LaTeX" id="ImEquation668"><![CDATA[$x$]]></tex-math></inline-formula> integrals, we get
<disp-formula id="ptz160M7-5"><label>(B.5)</label><tex-math notation="LaTeX" id="Equation163"><![CDATA[$$\begin{align}
&J(x_1,x_2)
\notag\\
&= \frac{1}{x_1-x_2}
\left\{
\;\;
\frac{1}{2} \, \left[ \ln^2(1-x_1) - \ln^2(-x_1) \right]
-
\frac{1}{2} \, \left[ \ln^2(1-x_2) - \ln^2(-x_2) \right]
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+ \mbox{Li}_2 \left( \frac{x_1-1}{x_2-1} \right)
- \mbox{Li}_2 \left( \frac{x_1}{x_2} \right)
- \mbox{Li}_2 \left( \frac{x_2-1}{x_1-1} \right)
+ \mbox{Li}_2 \left( \frac{x_2}{x_1} \right)
\notag\\
&
\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+
\frac{1}{2}
\left[
\ln^2 \left( x_2-1 \right) \vphantom{\frac{x_2-1}{x_2}}
-
\ln^2 \left( x_2 \right)
-
2\, \ln(x_2-x_1) \, \ln \left( \frac{x_2-1}{x_2} \right)
\right]
\notag \\
& \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
-
\frac{1}{2}
\left[
\ln^2 \left( x_1-1 \right) \vphantom{\frac{x_1-1}{x_1}}
-
\ln^2 \left( x_1 \right)
-
2\, \ln(x_1-x_2) \, \ln \left( \frac{x_1-1}{x_1} \right)
\right]
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+
\left.
\ln(x_1-x_2) \, \ln \left( \frac{x_1-1}{x_1} \right)
-
\ln(x_2-x_1) \, \ln \left( \frac{x_2-1}{x_2} \right)
\;\;
\right\}\!.
\label{eqcompj3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The identity
<disp-formula id="ptz160UM21"><tex-math notation="LaTeX" id="Equation164"><![CDATA[$$\ln^2(1-x_1) - \ln^2(-x_1) - \ln^2(x_1-1)+\ln^2(x_1)
= - 2 \, i \, \pi \, S(x_1) \, \ln \left( \frac{x_1-1}{x_1} \right)\!,$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation669"><![CDATA[$S(x_1)$]]></tex-math></inline-formula> is the sign of the imaginary part of <inline-formula><tex-math notation="LaTeX" id="ImEquation670"><![CDATA[$x_1$]]></tex-math></inline-formula>, allows us to simplify Eq. (<xref ref-type="disp-formula" rid="ptz160M7-5">B.5</xref>):
<disp-formula id="ptz160M7-6"><label>(B.6)</label><tex-math notation="LaTeX" id="Equation165"><![CDATA[$$\begin{align}
&J(x_1,x_2)
= \frac{1}{x_1-x_2}
\left\{
\mbox{Li}_2 \left( \frac{x_1-1}{x_2-1} \right)
-
\mbox{Li}_2 \left( \frac{x_1}{x_2} \right)
-
\mbox{Li}_2 \left( \frac{x_2-1}{x_1-1} \right)
+
\mbox{Li}_2 \left( \frac{x_2}{x_1} \right)
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
\quad {} \quad {} \quad {} \quad {}
+ \left[ 2\, \ln(x_1-x_2) - i \, \pi \, S(x_1) \right] \,
\ln \left( \frac{x_1-1}{x_1} \right)
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
\quad {} \quad {} \quad {} \quad {}\left.
- \left[ 2\, \ln(x_2-x_1) - i \, \pi \, S(x_2) \right] \,
\ln \left( \frac{x_2-1}{x_2} \right)
\right\}\!.
\label{eqcompj4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Then, we use the identity relating <inline-formula><tex-math notation="LaTeX" id="ImEquation671"><![CDATA[$\mbox{Li}_2 (z)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation672"><![CDATA[$\mbox{Li}_2(1/z)$]]></tex-math></inline-formula> [<xref ref-type="bibr" rid="B13">13</xref>], and also the relation
<disp-formula id="ptz160M7-7"><label>(B.7)</label><tex-math notation="LaTeX" id="Equation166"><![CDATA[$$\begin{equation}
2 \, \ln \left( x_1 - x_2 \right) - i \, \pi \, S(x_1) = 2 \, \ln \left( x_2 - x_1 \right) - i \, \pi \, S(x_2) = \ln \left( - (x_1 - x_2)^2 \right)\!,
\label{eqrelsimp1}
\end{equation}$$]]></tex-math></disp-formula>
whose validity relies on the fact that <inline-formula><tex-math notation="LaTeX" id="ImEquation673"><![CDATA[$x_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation674"><![CDATA[$x_2$]]></tex-math></inline-formula> have imaginary parts of opposite signs. This yields
<disp-formula id="ptz160M7-8"><label>(B.8)</label><tex-math notation="LaTeX" id="Equation167"><![CDATA[$$\begin{align}
&J(x_1,x_2)
\notag\\
&= \frac{1}{x_1-x_2}
\left\{
\;\;
2 \, \mbox{Li}_2 \left( \frac{x_1-1}{x_2-1} \right) -
2 \, \mbox{Li}_2 \left( \frac{x_1}{x_2} \right)
+ \frac{1}{2} \, \ln^2 \left( - \frac{x_1-1}{x_2-1} \right)
- \frac{1}{2} \, \ln^2 \left( - \frac{x_1}{x_2} \right)
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+
\left.
\ln \left( - (x_1-x_2)^2 \right) \,
\left[
\ln \left( \frac{x_1-1}{x_1} \right) - \ln \left( \frac{x_2-1}{x_2} \right)
\right]
\;\;
\right\}\!.
\label{eqcompj5}
\end{align}$$]]></tex-math></disp-formula></p>
</sec>
<sec id="SEC8"><title>Appendix C. Herbarium of utilitarian integrals</title>
<p>This appendix collects a bunch of integrals appearing in the three- and four-point cases to help the reader.</p>
<p>The first series appears under the following forms and can be computed in terms of the Euler Beta function:
<disp-formula id="ptz160M8-1"><label>(C.1)</label><tex-math notation="LaTeX" id="Equation168"><![CDATA[$$\begin{align}
\int^{+\infty}_0 dz \left(1+z^2 \right)^{-1-\varepsilon} & = \frac{1}{2} \, \int^{+\infty}_0 \frac{dy}{\sqrt{y}} \, (1+y)^{-1-\varepsilon}
= \frac{1}{2} \, B \left( \frac{1}{2},\frac{1}{2} + \varepsilon \right)\!,
\label{firstinty0} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M8-2"><label>(C.2)</label><tex-math notation="LaTeX" id="Equation169"><![CDATA[$$\begin{align}
\int^{+\infty}_1 dz \left(z^2-1 \right)^{-1-\varepsilon} & = \frac{1}{2} \, \int^{+\infty}_1 \frac{dy}{\sqrt{y}} \, (y-1)^{-1-\varepsilon}
=
\frac{1}{2} \,B \left( \frac{1}{2} + \varepsilon, - \, \varepsilon \right)\!,
\label{secondinty0} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M8-3"><label>(C.3)</label><tex-math notation="LaTeX" id="Equation170"><![CDATA[$$\begin{align}
\int^1_0 dz \left(1-z^2 \right)^{-1-\varepsilon} & = \frac{1}{2} \,\int^{1}_0 \frac{dy}{\sqrt{y}} \, (1-y)^{-1-\varepsilon}
\quad {}
= \frac{1}{2} \, B \left( \frac{1}{2}, - \, \varepsilon \right)\!.
\label{thirdinty0}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using the duplication formula for the Gamma functions [<xref ref-type="bibr" rid="B13">13</xref>], the <inline-formula><tex-math notation="LaTeX" id="ImEquation675"><![CDATA[$z$]]></tex-math></inline-formula> integrals computed in closed form read
<disp-formula id="ptz160M8-4"><label>(C.4)</label><tex-math notation="LaTeX" id="Equation171"><![CDATA[$$\begin{align}
\int^{+\infty}_0 dz \left(1+z^2 \right)^{-1-\varepsilon}
&= \frac{\tan(\pi \, \varepsilon)}{2\varepsilon} \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1- 2\, \varepsilon)} \,
2^{-2 \, \varepsilon} ,
\label{firstinty} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M8-5"><label>(C.5)</label><tex-math notation="LaTeX" id="Equation172"><![CDATA[$$\begin{align}
\int^{+\infty}_1 dz \left(z^2-1 \right)^{-1-\varepsilon}
&= - \frac{1}{2 \, \varepsilon} \, \frac{1}{\cos(\pi \, \varepsilon)}
\frac{\Gamma^{2}(1-\varepsilon)}{\Gamma(1-2 \, \varepsilon)} \,
2^{- 2 \, \varepsilon} ,
\label{secondinty} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M8-6"><label>(C.6)</label><tex-math notation="LaTeX" id="Equation173"><![CDATA[$$\begin{align}
\int^1_0 dz \left(1-z^2 \right)^{-1-\varepsilon}
&=
- \frac{1}{2 \, \varepsilon} \, 2^{-2 \, \varepsilon} \,
\frac{\Gamma(1-\varepsilon)^2}{\Gamma(1-2 \, \varepsilon)} .
\label{thirdinty}
\end{align}$$]]></tex-math></disp-formula></p>
<p>For the four-point functions in the real mass case, or in the complex mass case when <inline-formula><tex-math notation="LaTeX" id="ImEquation676"><![CDATA[$\mbox{sign}(\operatorname{Im}(R_{ij})) = \mbox{sign}(\operatorname{Im}(T))$]]></tex-math></inline-formula>, the following integral needs to be evaluated:
<disp-formula id="ptz160M8-7"><label>(C.7)</label><tex-math notation="LaTeX" id="Equation174"><![CDATA[$$\begin{align}
K_2(R_{ij},T)
&= \int^1_0 \frac{d v}{v} \,
\left[
\frac{1}{[ v \, R_{ij} + (1-v) \, T \, - i \, \lambda ]^{1+\varepsilon}}
-
\frac{1}{[(1-v) \, T \, - i \, \lambda]^{1+\varepsilon}}
\right]\!.
\label{eqdefK2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Note that in both cases, <inline-formula><tex-math notation="LaTeX" id="ImEquation677"><![CDATA[$\operatorname{Im}(v \, R_{ij} + (1-v) \, T \, - i \, \lambda)$]]></tex-math></inline-formula> has a constant sign when <inline-formula><tex-math notation="LaTeX" id="ImEquation678"><![CDATA[$v$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation679"><![CDATA[$[0,1]$]]></tex-math></inline-formula>. After a partial fraction decomposition with respect to the variable <inline-formula><tex-math notation="LaTeX" id="ImEquation680"><![CDATA[$v$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation681"><![CDATA[$K_2(R_{ij},T)$]]></tex-math></inline-formula> can be written as
<disp-formula id="ptz160M8-8"><label>(C.8)</label><tex-math notation="LaTeX" id="Equation175"><![CDATA[$$\begin{align}
K_2(R_{ij},T)
&=
\frac{1}{T} \, \int^{1}_{0} dv \,
\biggl\{
- (R_{ij}-T) \,
\left[ R_{ij} \, v + T \, (1-v) - i \, \lambda \right]^{-1-\varepsilon}
-
(T - i \, \lambda)^{-\varepsilon} \, (1-v)^{-1-\varepsilon}
\notag \\
&\quad {}\quad {}\quad {} \quad {} \quad {}\quad {}
+
\frac{1}{v} \,
\left[
\left( R_{ij} \,v + T \, (1-v) - i \, \lambda \right)^{- \varepsilon}
- (T - i \, \lambda)^{- \varepsilon}
\right]
\notag \\
&\quad {}\quad {}\quad {} \quad {}\quad {}\quad {}
+ \frac{(T - i \, \lambda)^{-\varepsilon}}{v} \,
\left[ 1 - (1-v)^{- \varepsilon} \right]
\biggr\} .
\label{eqsecondontv}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The first two terms of Eq. (<xref ref-type="disp-formula" rid="ptz160M8-8">C.8</xref>), which yield a <inline-formula><tex-math notation="LaTeX" id="ImEquation682"><![CDATA[$1/\varepsilon$]]></tex-math></inline-formula> pole, are integrated in closed form, whereas the last three terms of Eq. (<xref ref-type="disp-formula" rid="ptz160M8-8">C.8</xref>), which are not divergent, can be expanded around <inline-formula><tex-math notation="LaTeX" id="ImEquation683"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> up to order <inline-formula><tex-math notation="LaTeX" id="ImEquation684"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>:
<disp-formula id="ptz160M8-9"><label>(C.9)</label><tex-math notation="LaTeX" id="Equation176"><![CDATA[$$\begin{align}
K_2(R_{ij},T)
&= \frac{1}{T} \,
\biggl[
\;\;\;\; \frac{1}{\varepsilon} \, (R_{ij} - i \, \lambda)^{-\varepsilon}
\notag \\
&\quad {} \quad {}\quad {}
- \; \varepsilon \, \int^{1}_{0} \frac{dv}{v} \,
\left[
\ln \left( R_{ij} \,v + T \, (1-v) - i \, \lambda \right)
- \ln \left(T - i \, \lambda \right)
\right]
\notag \\
&\quad {} \quad {}\quad {}
+ \; \varepsilon \, (T - i \, \lambda)^{-\varepsilon} \,
\int^1_0 \frac{dv}{v} \, \ln(1-v)
\;
\biggr] .
\label{eqsecondontv1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Since <inline-formula><tex-math notation="LaTeX" id="ImEquation685"><![CDATA[$\mbox{sign}(\operatorname{Im}(R_{ij} \, v + T \, (1-v) - i \, \lambda)) = \mbox{sign}(\operatorname{Im}(T - i \, \lambda))$]]></tex-math></inline-formula> when <inline-formula><tex-math notation="LaTeX" id="ImEquation686"><![CDATA[$v \in [0,1]$]]></tex-math></inline-formula>, the logarithms in the first integral can be combined together. The last integration is performed explicitly to give
<disp-formula id="ptz160M8-10"><label>(C.10)</label><tex-math notation="LaTeX" id="Equation177"><![CDATA[$$\begin{align}
K_2(R_{ij},T)
&= \frac{1}{T} \,
\biggl[
\frac{1}{\varepsilon} \, (R_{ij} - i \, \lambda)^{-\varepsilon}
+ \varepsilon \, \mbox{Li}_2 \left( \frac{T - R_{ij}}{T - i \, \lambda} \right)
- \varepsilon \, \frac{\pi^2}{6} \biggr] .
\label{eqsecondontv3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>With respect to the preceding case, two new integrals show up when <inline-formula><tex-math notation="LaTeX" id="ImEquation687"><![CDATA[$\mbox{sign}(\operatorname{Im}(T)) \ne \mbox{sign}(\operatorname{Im}(R_{ij}))$]]></tex-math></inline-formula>:
<disp-formula id="ptz160M8-11"><label>(C.11)</label><tex-math notation="LaTeX" id="Equation178"><![CDATA[$$\begin{align}
K_3(A,B)
&= \int^{+\infty}_0 \frac{dv}{v} \,
\left[ (A \, v + B)^{-1-\varepsilon} - (B \, (1+v))^{-1-\varepsilon} \right]\!,
\label{eqdefk3text} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M8-12"><label>(C.12)</label><tex-math notation="LaTeX" id="Equation179"><![CDATA[$$\begin{align}
K_4(A^{\prime},B^{\prime})
&= \int^{+\infty}_1 \frac{dv}{v} \,
\left[ (A^{\prime} \, v + B^{\prime})^{-1-\varepsilon} - (B^{\prime} \, (1-v))^{-1-\varepsilon} \right]\!,
\label{eqdefk4text}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation688"><![CDATA[$\mbox{sign}(\operatorname{Im}(A \, v + B))$]]></tex-math></inline-formula> [resp. <inline-formula><tex-math notation="LaTeX" id="ImEquation689"><![CDATA[$\mbox{sign}(\operatorname{Im}(A^{\prime} \, v + B^{\prime}))$]]></tex-math></inline-formula>] keeps a constant sign when <inline-formula><tex-math notation="LaTeX" id="ImEquation690"><![CDATA[$v$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation691"><![CDATA[$[0, + \infty[$]]></tex-math></inline-formula> (resp. <inline-formula><tex-math notation="LaTeX" id="ImEquation692"><![CDATA[$[1, + \infty[$]]></tex-math></inline-formula>). To compute <inline-formula><tex-math notation="LaTeX" id="ImEquation693"><![CDATA[$K_3(A,B)$]]></tex-math></inline-formula>, we expand the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M8-11">C.11</xref>) around <inline-formula><tex-math notation="LaTeX" id="ImEquation694"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula> to get
<disp-formula id="ptz160M8-13"><label>(C.13)</label><tex-math notation="LaTeX" id="Equation180"><![CDATA[$$\begin{align}
K_3(A,B)
&= \frac{1}{B} \,
\ln \left( \frac{B}{A} \right)
\left[ 1 - \varepsilon \, \ln(B) \right]\!.
\label{eqresulK3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>To compute <inline-formula><tex-math notation="LaTeX" id="ImEquation695"><![CDATA[$K_4(A^{\prime},B^{\prime})$]]></tex-math></inline-formula>, we expand <inline-formula><tex-math notation="LaTeX" id="ImEquation696"><![CDATA[$(A^{\prime} \, v + B^{\prime})^{-1-\varepsilon}$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M8-12">C.12</xref>) around <inline-formula><tex-math notation="LaTeX" id="ImEquation697"><![CDATA[$\varepsilon=0$]]></tex-math></inline-formula>, keeping in mind that <inline-formula><tex-math notation="LaTeX" id="ImEquation698"><![CDATA[$\mbox{sign}(\operatorname{Im}(A^{\prime}+B^{\prime})) = \mbox{sign}(\operatorname{Im}(A^{\prime}))$]]></tex-math></inline-formula>, and we get
<disp-formula id="ptz160M8-14"><label>(C.14)</label><tex-math notation="LaTeX" id="Equation181"><![CDATA[$$\begin{align}
K_4(A^{\prime},B^{\prime})
&=
\frac{1}{B^{\prime}}
\left\{
- \, \frac{1}{\varepsilon} (-B^{\prime})^{\varepsilon}
+ \left( \ln(A^{\prime}+B^{\prime}) - \ln(A^{\prime}) \right)
\right.
\notag\\
& \quad {}\quad {}\quad {}
+ \left. \varepsilon \,
\left[
\mbox{Li}_2 \left( - \, \frac{B^{\prime}}{A^{\prime}} \right) - \frac{\pi^2}{6}
-
\frac{1}{2}
\left( \ln^2(A^{\prime}+B^{\prime}) - \ln^2(A^{\prime}) \right)
\right]
\right\}\!.
\label{eqresulK4proto}
\end{align}$$]]></tex-math></disp-formula></p>
<p>With the additional assumption that <inline-formula><tex-math notation="LaTeX" id="ImEquation699"><![CDATA[$\mbox{sign}(\operatorname{Im}(A^{\prime})) = - \mbox{sign}(\operatorname{Im}(B^{\prime}))$]]></tex-math></inline-formula>, and after some algebra, <inline-formula><tex-math notation="LaTeX" id="ImEquation700"><![CDATA[$K_4(A^{\prime},B^{\prime})$]]></tex-math></inline-formula> can be recast in an alternative, more useful, form:
<disp-formula id="ptz160M8-15"><label>(C.15)</label><tex-math notation="LaTeX" id="Equation182"><![CDATA[$$\begin{align}
K_4(A^{\prime},B^{\prime})
&=
\frac{1}{B^{\prime}}
\left\{
- \, \frac{1}{\varepsilon} (A^{\prime}+B^{\prime})^{\varepsilon}
+ \left( \ln(-B^{\prime}) - \ln(A^{\prime}) \right)
\right.
\notag\\
& \quad {}\quad {}\quad {}
+ \left. \varepsilon \,
\left[
\mbox{Li}_2 \left( \frac{A^{\prime}+B^{\prime}}{B^{\prime}} \right) +
\ln \left( A^{\prime}+B^{\prime} \right) \,
\ln \left( - \, \frac{A^{\prime}}{B^{\prime}} \right)
\right]
\right\}\!.
\label{eqresulK4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Notice that this additional assumption is always fulfilled in the cases met.</p>
</sec>
<sec id="SEC9"><title>Appendix D. Detailed comparisons with the &#x201C;direct way&#x201D; (Sect. <xref ref-type="sec" rid="SEC2">2</xref>)</title>
<p>Our present goal is to check that the &#x201C;indirect way&#x201D; leads to the same results for infrared-divergent three-point functions as the &#x201C;direct way.&#x201D; By explicitly performing the sum over the <inline-formula><tex-math notation="LaTeX" id="ImEquation701"><![CDATA[$j$]]></tex-math></inline-formula> index in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-45">2.45</xref>), we recover the results of Sect. <xref ref-type="sec" rid="SEC2">2</xref>. This part is not necessary for the understanding of the method proposed in this article and can be skipped in a first reading.</p>
<sec id="SEC9.1"><title>D.1. Real mass case</title>
<p>Let us start with the real mass case. We successively revisit the examples examined in Sect. (<xref ref-type="sec" rid="SEC2.2">2.2</xref>).</p>
<p><italic>Occurrence of a soft divergence</italic> Let us recap the texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation702"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix [cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-31">2.31</xref>)]:
<disp-formula id="ptz160M9-1"><label>(D.1)</label><tex-math notation="LaTeX" id="Equation183"><![CDATA[$$\begin{equation}
{\cal S} =
\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & -2 \, m_2^2 & s_3 - m_2^2 - m_3^2 \\
0 & s_3 - m_2^2 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!;
\label{eqcalssoft_ver}
\end{equation}$$]]></tex-math></disp-formula>
thus <inline-formula><tex-math notation="LaTeX" id="ImEquation703"><![CDATA[$\det({\cal S}) = 0$]]></tex-math></inline-formula>. Let us single out row and column 1. We readily see that the vector <inline-formula><tex-math notation="LaTeX" id="ImEquation704"><![CDATA[$V^{(1)}$]]></tex-math></inline-formula> vanishes, and so do the coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation705"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation706"><![CDATA[$\overline{b}_3$]]></tex-math></inline-formula>; thus <inline-formula><tex-math notation="LaTeX" id="ImEquation707"><![CDATA[$\overline{b}_1 = \det{(G)}$]]></tex-math></inline-formula> since the coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation708"><![CDATA[$\overline{b}_i$]]></tex-math></inline-formula> fulfill <inline-formula><tex-math notation="LaTeX" id="ImEquation709"><![CDATA[$\sum_{i \in S_3} \overline{b}_i = \det{(G)}$]]></tex-math></inline-formula>. The three-point function is thus given by [cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-45">2.45</xref>)]
<disp-formula id="ptz160M9-2"><label>(D.2)</label><tex-math notation="LaTeX" id="Equation184"><![CDATA[$$\begin{align}
I_3^n
&=
\frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{12} \big)
+ \frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{13} \big) .
\label{eq_verif_ir1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The only relevant reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation710"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is
<disp-formula id="ptz160M9-3"><label>(D.3)</label><tex-math notation="LaTeX" id="Equation185"><![CDATA[$$\begin{equation}
{\cal S}^{\{1\}} =
\left(
\begin{array}{cc}
-2 \, m_2^2 & s_3 - m_2^2 - m_3^2 \\
s_3 - m_2^2 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!,
\label{eqcalssoft_ver1}
\end{equation}$$]]></tex-math></disp-formula>
whose determinant <inline-formula><tex-math notation="LaTeX" id="ImEquation711"><![CDATA[$\det ({\cal S}^{\{1\}}) = - {\cal K}(s_3,m_2^2,m_3^2)$]]></tex-math></inline-formula> involves the K&#x00E4;ll&#x00E9;n function given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-11">2.11</xref>). The associated Gram &#x201C;matrix&#x201D; degenerates into a single scalar,
<disp-formula id="ptz160M9-4"><label>(D.4)</label><tex-math notation="LaTeX" id="Equation186"><![CDATA[$$\begin{equation}
G^{\{1\}(2)} = \det{(G^{\{1\}})} = \left( 2 \, s_3 \right)\!.
\label{eqgrammat_ver1}
\end{equation}$$]]></tex-math></disp-formula>
One easily reads the <inline-formula><tex-math notation="LaTeX" id="ImEquation712"><![CDATA[$\overline{b}_{j}^{\{1\}}$]]></tex-math></inline-formula> coefficients from the reduced Gram matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation713"><![CDATA[$G^{\{1\}(2)}$]]></tex-math></inline-formula> and the vector <inline-formula><tex-math notation="LaTeX" id="ImEquation714"><![CDATA[$V^{\{1\}(2)}$]]></tex-math></inline-formula> (cf. Eq. (2.29) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]):
<disp-formula id="ptz160M9-5"><label>(D.5)</label><tex-math notation="LaTeX" id="Equation187"><![CDATA[$$\begin{align}
\overline{b}_{2}^{\{1\}} &= m_2^2 - m_3^2 - s_3, \quad \overline{b}_{3}^{\{1\}} = m_3^2 - m_2^2 - s_3 . \label{bbar21}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Since <inline-formula><tex-math notation="LaTeX" id="ImEquation715"><![CDATA[$\widetilde{D}_{12} = 2 \, m_3^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation716"><![CDATA[$\widetilde{D}_{13} = 2 \, m_2^2$]]></tex-math></inline-formula>, the quantities <inline-formula><tex-math notation="LaTeX" id="ImEquation717"><![CDATA[$L_3^n(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation718"><![CDATA[$L_3^n(0,\Delta_1^{\{1\}},\widetilde{D}_{13})$]]></tex-math></inline-formula> are given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>). The <inline-formula><tex-math notation="LaTeX" id="ImEquation719"><![CDATA[$\widetilde{D}_{12} + \Delta_1^{\{1\}}$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation720"><![CDATA[$\widetilde{D}_{13} + \Delta_1^{\{1\}}$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation721"><![CDATA[$\Delta_1^{\{1\}}$]]></tex-math></inline-formula> terms are given by Eqs. (2.36) and (2.38) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]:
<disp-formula id="ptz160M9-6"><label>(D.6)</label><tex-math notation="LaTeX" id="Equation188"><![CDATA[$$\begin{align}
\widetilde{D}_{12} + \Delta_1^{\{1\}}
&= \frac{(s_3 + m_3^2 - m_2^2)^2}{2 \, s_3} ,
\label{eqdelta1pd12_ver} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M9-7"><label>(D.7)</label><tex-math notation="LaTeX" id="Equation189"><![CDATA[$$\begin{align}
\widetilde{D}_{13} + \Delta_1^{\{1\}}
&= \frac{(s_3 + m_2^2 - m_3^2)^2}{2 \, s_3} ,
\label{eqdelta1pd13_ver} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M9-8"><label>(D.8)</label><tex-math notation="LaTeX" id="Equation190"><![CDATA[$$\begin{align}
\Delta_1^{\{1\}}
&= \frac{{\cal K}(s_3,m_2^2,m_3^2)}{2 \, s_3} .
\label{eqdelta1_ver}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The roots of the denominator of Eq. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>) are such that
<disp-formula id="ptz160M9-9"><label>(D.9)</label><tex-math notation="LaTeX" id="Equation191"><![CDATA[$$\begin{align}
(\bar{z}^{12})^2
&= \frac{\Delta_1^{\{1\}} + i \, \lambda}{\widetilde{D}_{12} + \Delta_1^{\{1\}}}
=
\frac{{\cal K}(s_3,m_2^2,m_3^2) + i \, \lambda \, \sigma_s}
{(s_3 + m_3^2 - m_2^2)^2} ,
\label{eqroot_ver12} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M9-10"><label>(D.10)</label><tex-math notation="LaTeX" id="Equation192"><![CDATA[$$\begin{align}
(\bar{z}^{13})^2
&=
\frac{\Delta_1^{\{1\}} + i \, \lambda}{\widetilde{D}_{13} + \Delta_1^{\{1\}}}
= \frac{{\cal K}(s_3,m_2^2,m_3^2) + i \, \lambda \, \sigma_s}
{(s_3 + m_2^2 - m_3^2)^2} ,
\label{eqroot_ver13}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation722"><![CDATA[$\sigma_s = \mbox{sign}(s_3)$]]></tex-math></inline-formula>. We specify
<disp-formula id="ptz160M9-11"><label>(D.11)</label><tex-math notation="LaTeX" id="Equation193"><![CDATA[$$\begin{align}
\bar{z}^{12}
= \frac{\sqrt{{\cal K}(s_3,m_2^2,m_3^2) + i \, \lambda \, \sigma_s}}
{s_3 + m_3^2 - m_2^2} ,
& \quad {}
\bar{z}^{13}
= \frac{\sqrt{{\cal K}(s_3,m_2^2,m_3^2) + i \, \lambda \, \sigma_s}}
{s_3 + m_2^2 - m_3^2} .
\label{eqroot_ver13r}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We introduce two new quantities, <inline-formula><tex-math notation="LaTeX" id="ImEquation723"><![CDATA[$\tilde{x}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation724"><![CDATA[$\tilde{x}_2$]]></tex-math></inline-formula>, which are the two roots of the equation <inline-formula><tex-math notation="LaTeX" id="ImEquation725"><![CDATA[$D^{\{1\}\, (2)}(x) = 0$]]></tex-math></inline-formula> appearing in the &#x201C;direct way&#x201D;:
<disp-formula id="ptz160M9-12"><label>(D.12)</label><tex-math notation="LaTeX" id="Equation194"><![CDATA[$$\begin{align}
\tilde{x}_1
&=
\frac{s_3 + m_2^2 - m_3^2 + \sqrt{{\cal K}(s_3,m_2^2,m_3^2)
+ i \, \lambda \, \sigma_s}}{2 \, s_3} ,
\label{eqxtilde1def} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M9-13"><label>(D.13)</label><tex-math notation="LaTeX" id="Equation195"><![CDATA[$$\begin{align}
\tilde{x}_2
&= \frac{s_3 + m_2^2 - m_3^2 - \sqrt{{\cal K}(s_3,m_2^2,m_3^2)
+ i \, \lambda \, \sigma_s}}{2 \, s_3} .
\label{eqxtilde2def}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Notice that the quantities <inline-formula><tex-math notation="LaTeX" id="ImEquation726"><![CDATA[$1-\tilde{x}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation727"><![CDATA[$1-\tilde{x}_2$]]></tex-math></inline-formula> are the roots of the equation <inline-formula><tex-math notation="LaTeX" id="ImEquation728"><![CDATA[$D^{\{1\} \, (3)}(x) = 0$]]></tex-math></inline-formula>. The quantities <inline-formula><tex-math notation="LaTeX" id="ImEquation729"><![CDATA[$\bar{z}^{12}$]]></tex-math></inline-formula>, one of the roots of the equation <inline-formula><tex-math notation="LaTeX" id="ImEquation730"><![CDATA[$(\widetilde{D}_{12} + \Delta_1^{\{1\}}) \, z^2 - \Delta_1^{\{1\}} -i \, \lambda = 0$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation731"><![CDATA[$\bar{z}^{13}$]]></tex-math></inline-formula>, a root of the equation <inline-formula><tex-math notation="LaTeX" id="ImEquation732"><![CDATA[$(\widetilde{D}_{13} + \Delta_1^{\{1\}}) \, z^2 - \Delta_1^{\{1\}} -i \, \lambda = 0$]]></tex-math></inline-formula>, can be related to the roots <inline-formula><tex-math notation="LaTeX" id="ImEquation733"><![CDATA[$\tilde{x}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation734"><![CDATA[$\tilde{x}_2$]]></tex-math></inline-formula> by the following relations:
<disp-formula id="ptz160M9-14"><label>(D.14)</label><tex-math notation="LaTeX" id="Equation196"><![CDATA[$$\begin{align}
\bar{z}^{12}
= \frac{\tilde{x}_1 - \tilde{x}_2}{2 - \tilde{x}_1 - \tilde{x}_2} ,
&\quad {}
\bar{z}^{13}
= \frac{\tilde{x}_1 - \tilde{x}_2}{\tilde{x}_1 + \tilde{x}_2} . \label{eqroot_ver13r1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Putting things together, <inline-formula><tex-math notation="LaTeX" id="ImEquation735"><![CDATA[$I_3^n$]]></tex-math></inline-formula> can be written as
<disp-formula id="ptz160M9-15"><label>(D.15)</label><tex-math notation="LaTeX" id="Equation197"><![CDATA[$$\begin{align}
I_3^n
&=
- \frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}
{2 \, \sqrt{{\cal K}(s_3,m_2^2,m_3^2) + i \, \lambda \, \sigma_s}} \,
\left\{
- \frac{1}{\varepsilon} \,
\left[
\ln \left( \frac{\bar{z}^{12}-1}{\bar{z}^{12}+1} \right) +
\ln \left( \frac{\bar{z}^{13}-1}{\bar{z}^{13}+1} \right)
\right]
\right.
\notag \\
&\quad
+
\left.
\bar{H}_{0,1}
\big( \widetilde{D}_{12}+\Delta_1^{\{1\}},-\Delta_1^{\{1\}}- i \, \lambda \big)
+
\bar{H}_{0,1}
\big( \widetilde{D}_{13}+\Delta_1^{\{1\}},-\Delta_1^{\{1\}}- i \, \lambda \big)
\vphantom{\frac{\bar{z}^{12}-1}{\bar{z}^{12}+1}} \right\}\!,
\label{eq_verif_ir2}
\end{align}$$]]></tex-math></disp-formula>
with
<disp-formula id="ptz160M9-16"><label>(D.16)</label><tex-math notation="LaTeX" id="Equation198"><![CDATA[$$\begin{equation}
\bar{H}_{0,1}
\left( A,B \right)
=
2 \, A \, \sqrt{ - \frac{B}{A} } \,
H_{0,1}
\left( A,B \right)\!,
\label{eqdefhbarij}
\end{equation}$$]]></tex-math></disp-formula>
which is effectively the content between the curly brackets of Eq. (<xref ref-type="disp-formula" rid="ptz160M10-8">E.8</xref>) in Appendix <xref ref-type="sec" rid="SEC10">E</xref>. Expressing all the arguments of the logarithms and dilogarithms in Eq. (<xref ref-type="disp-formula" rid="ptz160M10-8">E.8</xref>) in terms of <inline-formula><tex-math notation="LaTeX" id="ImEquation736"><![CDATA[$\tilde{x}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation737"><![CDATA[$\tilde{x}_2$]]></tex-math></inline-formula>, we get
<disp-formula id="ptz160M9-17"><label>(D.17)</label><tex-math notation="LaTeX" id="Equation199"><![CDATA[$$\begin{align}
I_3^n
&=
- \, \frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}
{(\tilde{x}_1 - \tilde{x}_2) \, \det{(G^{\{1\}})}} \,
\notag\\
& \quad {} \times
\left\{
- \, \frac{1}{\varepsilon} \,
\left[
\ln \left( - \, \frac{1 - \tilde{x}_1}{1 - \tilde{x}_2} \right)
+
\ln \left( - \, \frac{\tilde{x}_2}{\tilde{x}_1} \right)
\right]
\right.
\notag \\
&\qquad \quad {}
+ \ln \left( - \, \frac{1 - \tilde{x}_1}{1- \tilde{x}_2} \right) \,
\left[
\; \ln
\left(
\frac{(2 - \tilde{x}_1 - \tilde{x}_2)^2}{4} \, \det{(G^{\{1\}})} + i \, \lambda \, \sigma_s
\right)
\right.
\notag \\
&\qquad \qquad \quad \quad \quad \quad \quad \quad \quad {}
\left.
+
\frac{1}{2}
\left(
\ln \left( \frac{4 \, (1-\tilde{x}_1) \, (1-\tilde{x}_2)}{(2-\tilde{x}_1-\tilde{x}_2)^2} \right)
+
\ln \left( - \, \frac{4 \, (\tilde{x}_1 - \tilde{x}_2)^2}{(2 - \tilde{x}_1 - \tilde{x}_2)^2} \right)
\right)
\right]
\notag \\
&\qquad \quad {}
+
\ln \left( - \, \frac{\tilde{x}_2}{\tilde{x}_1} \right) \,
\left[
\ln
\left(
\frac{(\tilde{x}_1 + \tilde{x}_2)^2}{4} \, \det{(G^{\{1\}})} + i \, \lambda \, \sigma_s
\right)
\right.
\notag \\
&\qquad \qquad \quad \quad \quad \quad \quad \quad {}
\left.
+ \frac{1}{2} \,
\left( \ln \left( \frac{4 \, \tilde{x}_1 \, \tilde{x}_2}{(\tilde{x}_1+\tilde{x}_2)^2} \right)
+ \ln \left( - \frac{4 \, (\tilde{x}_1 - \tilde{x}_2)^2}{(\tilde{x}_1 + \tilde{x}_2)^2} \right)
\right)
\right]
\notag \\
&\qquad \quad {} +
\left.
\mbox{Li}_2 \left( \frac{1-\tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
- \mbox{Li}_2 \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1 - \tilde{x}_2} \right)
+ \mbox{Li}_2 \left( \frac{\tilde{x}_1}{\tilde{x}_1 - \tilde{x}_2} \right)
- \mbox{Li}_2 \left( - \, \frac{\tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
\right\}\!.
\label{eq_verif_ir3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As <inline-formula><tex-math notation="LaTeX" id="ImEquation738"><![CDATA[$\operatorname{Im}(\tilde{x}_1)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation739"><![CDATA[$\operatorname{Im}(\tilde{x}_2)$]]></tex-math></inline-formula> have opposite signs, Eq. (<xref ref-type="disp-formula" rid="ptz160M9-17">D.17</xref>) can be rearranged as
<disp-formula id="ptz160M9-18"><label>(D.18)</label><tex-math notation="LaTeX" id="Equation200"><![CDATA[$$\begin{align}
I_3^n
&=
- \,
\frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}{(\tilde{x}_1 - \tilde{x}_2) \, \det{(G^{\{1\}})}}
\notag\\
& \quad {} \times
\left\{
- \, \frac{1}{\varepsilon} \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]
\right.
\notag \\
&\qquad \quad {}
+
\left[
\ln \big( \det{(G^{\{1\}})} + i \, \lambda \, \sigma_s \big)
+
\frac{1}{2} \, \ln \left( - \, (\tilde{x}_1 - \tilde{x}_2)^2 \right)
\right] \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]
\notag \\
&\qquad \quad {}
+ \frac{1}{2} \, \ln \left( - \, \frac{1 - \tilde{x}_1}{1 - \tilde{x}_2} \right) \,
\ln \left( (1-\tilde{x}_1) \, (1-\tilde{x}_2) \right)
+ \frac{1}{2} \, \ln \left( - \, \frac{\tilde{x}_2}{\tilde{x}_1} \right) \,
\ln ( \tilde{x}_1 \, \tilde{x}_2 )
\notag \\
&\qquad \quad {}
- \frac{1}{2} \, \ln^2 \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1 - \tilde{x}_2} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{1 - \tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
- \frac{1}{2} \, \ln^2 \left( \frac{- \, \tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{\tilde{x}_1}{\tilde{x}_1 - \tilde{x}_2} \right)
\notag \\
&\qquad \quad {} +
\left.
2 \, \mbox{Li}_2 \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_2 - 1} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{1 - \tilde{x}_2}{\tilde{x}_1 - 1} \right)
- 2 \, \mbox{Li}_2 \left( \frac{\tilde{x}_1}{\tilde{x}_2} \right)
- \frac{1}{2} \, \ln^2 \left( - \,\frac{\tilde{x}_2}{\tilde{x}_1} \right) \right\}\!.
\label{eq_verif_ir4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using the relation between <inline-formula><tex-math notation="LaTeX" id="ImEquation740"><![CDATA[$\ln(z)$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation741"><![CDATA[$\ln(-z)$]]></tex-math></inline-formula>, after some algebra the quantity
<disp-formula id="ptz160UM22"><tex-math notation="LaTeX" id="Equation201"><![CDATA[$$\begin{align*}
E &= \;\;\;
\frac{1}{2} \, \ln \left( - \, \frac{1 - \tilde{x}_1}{1 - \tilde{x}_2} \right) \,
\ln \left( (1-\tilde{x}_1) \, (1-\tilde{x}_2) \right)
+ \frac{1}{2} \, \ln \left( - \, \frac{\tilde{x}_2}{\tilde{x}_1} \right) \,
\ln ( \tilde{x}_1 \, \tilde{x}_2 )
\notag \\
&\quad
- \frac{1}{2} \, \ln^2 \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1 - \tilde{x}_2} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{1 - \tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
- \frac{1}{2} \, \ln^2 \left( \frac{- \, \tilde{x}_2}{\tilde{x}_1 - \tilde{x}_2} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{\tilde{x}_1}{\tilde{x}_1 - \tilde{x}_2} \right)
\notag
\end{align*}$$]]></tex-math></disp-formula>
can be rewritten as
<disp-formula id="ptz160M9-19"><label>(D.19)</label><tex-math notation="LaTeX" id="Equation202"><![CDATA[$$\begin{align}
E
&=
\frac{1}{2} \,
\left[ 2 \, \ln (\tilde{x}_1 - \tilde{x}_2) - i \, \pi \, S(\tilde{x}_1) \right] \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]
\notag \\
&=
\frac{1}{2} \, \ln \left( - \, (\tilde{x}_1 - \tilde{x}_2)^2 \right) \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]\!,
\label{eqaux_vere1}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation742"><![CDATA[$S(\tilde{x}_1) = \mbox{sign}(\operatorname{Im}(\tilde{x}_1))$]]></tex-math></inline-formula>. Substituting Eq. (<xref ref-type="disp-formula" rid="ptz160M9-19">D.19</xref>) into Eq. (<xref ref-type="disp-formula" rid="ptz160M9-18">D.18</xref>), we end up with
<disp-formula id="ptz160M9-20"><label>(D.20)</label><tex-math notation="LaTeX" id="Equation203"><![CDATA[$$\begin{align}
I_3^n
&=
-
\frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}
{(\tilde{x}_1 - \tilde{x}_2) \, \det{(G^{\{1\}})}} \,
\notag \\
&\quad {} \times
\left\{
- \frac{1}{\varepsilon} \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]
\right.
\notag \\
&\qquad \quad {}
+
\left[
\ln \big( \det{(G^{\{1\}})} + i \, \lambda \, \sigma_s \big) +
\ln \left( - (\tilde{x}_1 - \tilde{x}_2)^2 \right)
\right] \,
\left[
\ln \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_1} \right)
-
\ln \left( \frac{\tilde{x}_2 - 1}{\tilde{x}_2} \right)
\right]
\notag \\
&\qquad \quad {}
+ 2 \, \mbox{Li}_2 \left( \frac{\tilde{x}_1 - 1}{\tilde{x}_2 - 1} \right)
+ \frac{1}{2} \, \ln^2 \left( \frac{1 - \tilde{x}_2}{\tilde{x}_1 - 1} \right)
-
\left.
2 \, \mbox{Li}_2 \left( \frac{\tilde{x}_1}{\tilde{x}_2} \right) -
\frac{1}{2} \, \ln^2 \left( -\frac{\tilde{x}_2}{\tilde{x}_1} \right)
\right\}
\notag \\
&=
- \frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}{\varepsilon \, \det{(G^{\{1\}})}}
\,
\left\{
\left[ 1 - \varepsilon \, \ln \big( \det{(G^{\{1\}})} + i \, \lambda \, \sigma_s \big) \right]
\, K(\tilde{x}_1,\tilde{x}_2) - \varepsilon \, J(\tilde{x}_1,\tilde{x}_2)
\right\}\!,
\label{eq_verif_ir5}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation743"><![CDATA[$K(\tilde{x}_1,\tilde{x}_2)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-14">2.14</xref>) and <inline-formula><tex-math notation="LaTeX" id="ImEquation744"><![CDATA[$J(\tilde{x}_1,\tilde{x}_2)$]]></tex-math></inline-formula> by Eq. (<xref ref-type="disp-formula" rid="ptz160M7-8">B.8</xref>). Last, we note that the prescription &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation745"><![CDATA[$+ i \, \lambda \, \sigma_s$]]></tex-math></inline-formula>&#x201D; in Eq. (<xref ref-type="disp-formula" rid="ptz160M9-20">D.20</xref>) can be replaced by &#x201C;<inline-formula><tex-math notation="LaTeX" id="ImEquation746"><![CDATA[$- i \, \lambda$]]></tex-math></inline-formula>&#x201D; as it matters only when <inline-formula><tex-math notation="LaTeX" id="ImEquation747"><![CDATA[$\det{(G^{\{1\}})} < 0$]]></tex-math></inline-formula>. Thus, Eq. (<xref ref-type="disp-formula" rid="ptz160M9-20">D.20</xref>) is nothing but Eq. (<xref ref-type="disp-formula" rid="ptz160M2-15">2.15</xref>): the indirect and direct ways do indeed lead to the same result.</p>
<p><italic>Occurrence of a collinear divergence</italic> We recap the texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation748"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix in this case [cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-35">2.35</xref>)]:
<disp-formula id="ptz160M9-21"><label>(D.21)</label><tex-math notation="LaTeX" id="Equation204"><![CDATA[$$\begin{equation}
{\cal S} =
\left(
\begin{array}{ccc}
0 & 0 & s_1 - m_3^2 \\
0 & 0 & s_3 - m_3^2 \\
s_1 - m_3^2 & s_3 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!.
\label{eqcalscoll_ver}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>Obviously, we have <inline-formula><tex-math notation="LaTeX" id="ImEquation749"><![CDATA[$\det({\cal S}) = 0$]]></tex-math></inline-formula>. As explained in Sect. <xref ref-type="sec" rid="SEC2">2</xref>, Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-36">2.36</xref>) and (<xref ref-type="disp-formula" rid="ptz160M2-37">2.37</xref>), the coefficient <inline-formula><tex-math notation="LaTeX" id="ImEquation750"><![CDATA[$\overline{b}_3$]]></tex-math></inline-formula> vanishes whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation751"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation752"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> are different from zero. So, the three-point function is given by
<disp-formula id="ptz160M9-22"><label>(D.22)</label><tex-math notation="LaTeX" id="Equation205"><![CDATA[$$\begin{align}
I_3^n
&= \;\;\;
\frac{\overline{b}_1}{\det{(G)}} \,
\left[
\frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{12} \big) +
\frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{13} \big)
\right]
\nonumber \\
&\quad
+ \frac{\overline{b}_2}{\det{(G)}} \,
\left[
\frac{\overline{b}_{1}^{\{2\}}}{\det{(G^{\{2\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{2\}},\widetilde{D}_{21} \big) +
\frac{\overline{b}_{3}^{\{2\}}}{\det{(G^{\{2\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{2\}},\widetilde{D}_{23} \big)
\right]\!.
\label{eq_verif_irc1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As <inline-formula><tex-math notation="LaTeX" id="ImEquation753"><![CDATA[$\widetilde{D}_{12} = \widetilde{D}_{21} = 2 \, m_3^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation754"><![CDATA[$\widetilde{D}_{13} = \widetilde{D}_{23} = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation755"><![CDATA[$L_{3}^{n} ( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{12} )$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation756"><![CDATA[$L_{3}^{n} ( 0,\Delta_{1}^{\{2\}},\widetilde{D}_{12} )$]]></tex-math></inline-formula> are given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>) whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation757"><![CDATA[$L_{3}^{n} ( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{13} )$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation758"><![CDATA[$L_{3}^{n} ( 0,\Delta_{1}^{\{2\}},\widetilde{D}_{23} )$]]></tex-math></inline-formula> are given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>).</p>
<p>Let us first focus on the first term of the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M9-22">D.22</xref>). The relevant reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation759"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is
<disp-formula id="ptz160M9-23"><label>(D.23)</label><tex-math notation="LaTeX" id="Equation206"><![CDATA[$$\begin{equation}
{\cal S}^{\{1\}} =
\left(
\begin{array}{cc}
0 & s_3 - m_3^2 \\
s_3 - m_3^2 & - 2 \, m_3^2
\end{array}
\right)\!,
\label{eqcalscol_ver1}
\end{equation}$$]]></tex-math></disp-formula>
whose determinant is <inline-formula><tex-math notation="LaTeX" id="ImEquation760"><![CDATA[$\det{({\cal S}^{\{1\}})} = - (s_3- m_3^2)^2$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation761"><![CDATA[$1 \times 1$]]></tex-math></inline-formula> associated Gram matrix is <inline-formula><tex-math notation="LaTeX" id="ImEquation762"><![CDATA[$G^{\{1\}(2)} = ( 2 \, s_3)$]]></tex-math></inline-formula>, and the reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation763"><![CDATA[$\bar{b}$]]></tex-math></inline-formula> coefficients are given by
<disp-formula id="ptz160M9-24"><label>(D.24)</label><tex-math notation="LaTeX" id="Equation207"><![CDATA[$$\begin{align}
\frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}}
= - \, \frac{s_3 + m_3^2}{2 \, s_3} ,
& \quad {}
\frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}}
= - \, \frac{s_3 - m_3^2}{2 \, s_3} , \label{eqbbar_vercol2}
\end{align}$$]]></tex-math></disp-formula>
whereas
<disp-formula id="ptz160M9-25"><label>(D.25)</label><tex-math notation="LaTeX" id="Equation208"><![CDATA[$$\begin{align}
\Delta_1^{\{1\}}
= \frac{(s_3 - m_3^2)^2}{2 \, s_3} ,
& \quad {}
\widetilde{D}_{12} + \Delta_1^{\{1\}} = \frac{(s_3 + m_3^2)^2}{2 \, s_3} .
\label{eqdtildepdelta1_vercol}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The square of the root of the polynomial <inline-formula><tex-math notation="LaTeX" id="ImEquation764"><![CDATA[$(\widetilde{D}_{12} + \Delta_1^{\{1\}}) \, z^2 - \Delta_1^{\{1\}} - i \, \lambda$]]></tex-math></inline-formula> is
<disp-formula id="ptz160M9-26"><label>(D.26)</label><tex-math notation="LaTeX" id="Equation209"><![CDATA[$$\begin{equation}
\bar{z}^2 = \frac{(s_3 - m_3^2)^2}{(s_3 + m_3^2)^2} + i \, \lambda \, \sigma_s ,
\label{eqsqroot_vercol}
\end{equation}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation765"><![CDATA[$\sigma_s = \mbox{sign}(s_3)$]]></tex-math></inline-formula>. Since
<disp-formula id="ptz160UM23"><tex-math notation="LaTeX" id="Equation210"><![CDATA[$$\begin{equation}
\sqrt{\bar{z}^2}
= \left| \frac{s_3 - m_3^2}{s_3 + m_3^2} \right| + i \, \lambda \, \sigma_s ,
\nonumber
\end{equation}$$]]></tex-math></disp-formula>
a handier choice for further manipulations is instead
<disp-formula id="ptz160M9-27"><label>(D.27)</label><tex-math notation="LaTeX" id="Equation211"><![CDATA[$$\begin{equation}
\bar{z} =
\frac{s_3 - m_3^2}{s_3 + m_3^2} + i \, \lambda \, \sigma_s \, \sigma_r ,
\label{eqroot_vercol2}
\end{equation}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation766"><![CDATA[$\sigma_r = \mbox{sign}( (s_3 - m_3^2)/(s_3 + m_3^2) )$]]></tex-math></inline-formula>. Let us define
<disp-formula id="ptz160M9-28"><label>(D.28)</label><tex-math notation="LaTeX" id="Equation212"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{12} \big)
+ \frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},\widetilde{D}_{13} \big) ,
\label{eq_verif_irc20}
\end{align}$$]]></tex-math></disp-formula>
so we get
<disp-formula id="ptz160M9-29"><label>(D.29)</label><tex-math notation="LaTeX" id="Equation213"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{2^{\varepsilon} \, \Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)}
\notag\\
&\quad {} \times
\left[
- \, \frac{1}{\varepsilon} \,
\ln \left( -\frac{m_3^2}{s_3} + i \, \lambda \, \sigma_s \, \sigma_r \right)
+ \bar{H}_{0,1}
\big( \widetilde{D}_{12} + \Delta_1^{\{1\}},-\Delta_1^{\{1\}} - i \, \lambda \big)
\right.
\notag \\
&\qquad \quad {} -
\left.
\frac{1}{\varepsilon^2} \,
\frac{\Gamma(1 - \varepsilon)^2}{\Gamma(1 - 2 \, \varepsilon)} \,
\left( - \frac{2 \, (s_3 - m_3^2)^2}{s_3} - i \, \lambda \right)^{-\varepsilon}
\right]\!.
\label{eq_verif_irc2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using the definition of <inline-formula><tex-math notation="LaTeX" id="ImEquation767"><![CDATA[$\bar{H}_{0,1}(X,Y)$]]></tex-math></inline-formula>, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M9-16">D.16</xref>), and Eq. (<xref ref-type="disp-formula" rid="ptz160M10-8">E.8</xref>), we have
<disp-formula id="ptz160M9-30"><label>(D.30)</label><tex-math notation="LaTeX" id="Equation214"><![CDATA[$$\begin{align}
& \bar{H}_{0,1}(\widetilde{D}_{12} + \Delta_1^{\{1\}},-\Delta_1^{\{1\}} - i \, \lambda)
\notag\\
&=
\mbox{Li}_2
\left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)
- \mbox{Li}_2
\left(
- \, \frac{m_3^2}{s_3 - m_3^2} + i \, \lambda \, \sigma_s \, \sigma_r
\right)
\notag\\
&\quad {}
+
\ln
\left( - \frac{m_3^2}{s_3} + i \, \lambda \, \sigma_s \, \sigma_r \right)
\left[
\ln
\left( \frac{(s_3 + m_3^2)^2}{2 \, s_3} + i \, \lambda \, \sigma_s \right)
+ \frac{1}{2} \,
\ln
\left(
\frac{4 \, s_3 \, m_3^2}{(s_3 + m_3^2)^2} - i \, \lambda \sigma_s
\right)
\right.
\notag \\
&\qquad \qquad \qquad \qquad \quad \quad \quad \quad {} +
\left.
\frac{1}{2} \,
\ln
\left(
-\, \frac{4 \, (s_3 - m_3^2)^2}{(s_3 + m_3^2)^2} - i \, \lambda \, \sigma_s
\right)
\right]\!,
\label{eqresubarh_vercol1}
\end{align}$$]]></tex-math></disp-formula>
which can be rewritten as
<disp-formula id="ptz160M9-31"><label>(D.31)</label><tex-math notation="LaTeX" id="Equation215"><![CDATA[$$\begin{align}
&\bar{H}_{0,1}
\big( \widetilde{D}_{12} + \Delta_1^{\{1\}},-\Delta_1^{\{1\}} - i \, \lambda \big)
\notag\\
&=
\ln \left( - \frac{m_3^2}{s_3} + i \, \lambda \, \sigma_s \, \sigma_r \right)
\,
\left[
-\ln \left( 2 \, s_3 - i \, \lambda \, \sigma_s \right)
+ \frac{1}{2} \,
\ln \left( 4 \, s_3 \, m_3^2 - i \, \lambda \sigma_s \right)
\right.
\notag \\
&\quad \quad \quad \quad \quad \quad \quad \quad
\quad \quad \quad \quad {} +
\left.
\frac{1}{2} \,
\ln \left( - \, 4 \, (s_3 - m_3^2)^2 - i \, \lambda \, \sigma_s \right)
\right]
\notag \\
&\quad {}
+ 2 \,
\mbox{Li}_2
\left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)
- \frac{\pi^2}{6}
\notag \\
&\quad {}
+
\ln
\left(
- \, \frac{m_3^2}{s_3 - m_3^2} + i \, \lambda \, \sigma_s \, \sigma_r
\right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)\!.
\label{eqresubarh_vercol2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Putting Eq. (<xref ref-type="disp-formula" rid="ptz160M9-31">D.31</xref>) into Eq. (<xref ref-type="disp-formula" rid="ptz160M9-29">D.29</xref>), the <inline-formula><tex-math notation="LaTeX" id="ImEquation768"><![CDATA[$\ln(2)$]]></tex-math></inline-formula> drop out and we end with
<disp-formula id="ptz160M9-32"><label>(D.32)</label><tex-math notation="LaTeX" id="Equation216"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\notag\\
&\quad {} \times
\left\{
- \, \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
\ln \left( - \frac{(s_3 - m_3^2)^2}{s_3} - i \, \lambda \right)
-
\ln\left( -\frac{m_3^2}{s_3} + i \, \lambda \, \sigma_s \, \sigma_r \right)
\right]
\right.
\notag \\
&\qquad \quad {}
- \frac{1}{2} \,
\ln^2 \left( - \frac{(s_3 - m_3^2)^2}{s_3} - i \, \lambda \right)
+ 2 \,
\mbox{Li}_2
\left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)
\notag \\
&\qquad \quad {}
+ \ln \left( - \frac{m_3^2}{s_3} + i \, \lambda \, \sigma_s \, \sigma_r \right)
\,
\left[
\vphantom{\frac{1}{2}}-\ln \left( s_3 - i \, \lambda \, \sigma_s \right)
\right.
\notag \\
&\qquad \quad {}
\left.
+ \frac{1}{2} \,
\ln \left( s_3 \, m_3^2 - i \, \lambda \sigma_s \right)
+ \frac{1}{2} \,
\ln \left( - (s_3 - m_3^2)^2 - i \, \lambda \, \sigma_s \right)
\right]
\notag \\
&\qquad \quad {} +
\left.
\ln
\left(
- \, \frac{m_3^2}{s_3 - m_3^2} + i \, \lambda \, \sigma_s \, \sigma_r
\right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)
\right\}\!.
\label{eq_verif_irc3}
\end{align}$$]]></tex-math></disp-formula></p>
<p><inline-formula><tex-math notation="LaTeX" id="ImEquation769"><![CDATA[$\Sigma_3^n(s_3)$]]></tex-math></inline-formula> can be shown to be equal to the following quantity:
<disp-formula id="ptz160M9-33"><label>(D.33)</label><tex-math notation="LaTeX" id="Equation217"><![CDATA[$$\begin{align}
\Upsilon_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \, \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
2 \, \ln \left( - s_3 + m_3^2 - i \, \lambda \right)
-
\ln \left( m_3^2 - i \, \lambda \right)
\right]
\right.
\notag \\
&\quad
\left.
- \ln^2 \left( - s_3 + m_3^2 - i \, \lambda \right)
+ \frac{1}{2} \, \ln^2 \left( m_3^2 - i \, \lambda \right)
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)
\right\}\!.
\label{eq_verif_irc4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>To show that, one has to distinguish the following cases: (1) <inline-formula><tex-math notation="LaTeX" id="ImEquation770"><![CDATA[$s_3 < - \, m_3^2$]]></tex-math></inline-formula>, (2) <inline-formula><tex-math notation="LaTeX" id="ImEquation771"><![CDATA[$- \, m_3^2 < s_3 < 0$]]></tex-math></inline-formula>, (3) <inline-formula><tex-math notation="LaTeX" id="ImEquation772"><![CDATA[$0 < s_3 < m_3^2$]]></tex-math></inline-formula>, and (4) <inline-formula><tex-math notation="LaTeX" id="ImEquation773"><![CDATA[$m_3^2 < s_3$]]></tex-math></inline-formula>. For each of them, some tedious algebra performed on the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M9-32">D.32</xref>) shows that the two results are equal. Let us discuss in detail only the case <inline-formula><tex-math notation="LaTeX" id="ImEquation774"><![CDATA[$s_3 > m_3^2$]]></tex-math></inline-formula>, for which <inline-formula><tex-math notation="LaTeX" id="ImEquation775"><![CDATA[$\sigma_s = \sigma_r = +$]]></tex-math></inline-formula>. The argument of the dilogarithm in Eq. (<xref ref-type="disp-formula" rid="ptz160M9-32">D.32</xref>) has a real part greater than <inline-formula><tex-math notation="LaTeX" id="ImEquation776"><![CDATA[$1$]]></tex-math></inline-formula>, hence
<disp-formula id="ptz160UM24"><tex-math notation="LaTeX" id="Equation218"><![CDATA[$$\begin{equation}
\mbox{Li}_2
\left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \sigma_s \, \sigma_r \right)
=
\mbox{Li}_2
\left( \frac{s_3}{s_3 - m_3^2} - i \, \lambda \right)
=
\mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)\!.
\nonumber
\end{equation}$$]]></tex-math></disp-formula></p>
<p>The logarithms in Eq. (<xref ref-type="disp-formula" rid="ptz160M9-32">D.32</xref>) can be modified in such a way that their arguments are ratios of positive quantities; <inline-formula><tex-math notation="LaTeX" id="ImEquation777"><![CDATA[$\Sigma_3^n(s_3)$]]></tex-math></inline-formula> thus reads
<disp-formula id="ptz160M9-34"><label>(D.34)</label><tex-math notation="LaTeX" id="Equation219"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\notag\\
& \quad {}
\left\{
- \, \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
\ln \left( \frac{(s_3 - m_3^2)^2}{s_3} \right)
-
\ln \left( \frac{m_3^2}{s_3} \right) - 2 \, i \, \pi
\right]
\right.
\notag \\
&\quad
- \frac{1}{2} \,
\left( \ln \left( \frac{(s_3 - m_3^2)^2}{s_3} \right) - i \, \pi \right)^2
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)
\notag\\
&\quad
+ \left( \ln \left( \frac{m_3^2}{s_3} \right) + i \, \pi \right) \,
\notag \\
&\quad \quad
\times
\left[
\vphantom{\frac{1}{2}} -\ln \left( s_3 \right)
+ \frac{1}{2} \,
\left(
\ln \left( s_3 \, m_3^2 \right) + \ln \left( (s_3 - m_3^2)^2 \right)
- i \, \pi
\right)
\right]
\notag \\
&\quad
\left.
+
\left( \ln \left( \frac{m_3^2}{s_3 - m_3^2} \right) + i \, \pi \right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\right\}\!.
\label{eq_verif_irc5}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Splitting logarithms of ratios, we get
<disp-formula id="ptz160M9-35"><label>(D.35)</label><tex-math notation="LaTeX" id="Equation220"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&=
\frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \, \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
2 \, \left( \ln \left( s_3 - m_3^2 \right) - i \, \pi \right)
- \ln\left( m_3^2 \right)
\right]
\right.
\notag \\
&\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
-
\left(
\ln^2 \left( s_3 - m_3^2 \right)
- 2 \, i \, \pi \, \ln \left( s_3 - m_3^2 \right) - \pi^2
\right)
\notag \\
&\quad{} \quad {}\quad {}\quad {}\quad {}\quad {}\quad {}
\left.
+ \frac{1}{2} \, \ln^2(m_3^2)
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)
\right\}\!.
\label{eq_verif_irc6}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In Eq. (<xref ref-type="disp-formula" rid="ptz160M9-35">D.35</xref>), for the case at hand, we recognize Eq. (<xref ref-type="disp-formula" rid="ptz160M9-33">D.33</xref>). Similar handling can be performed in the other three cases to reach the same conclusion. We thus conclude that
<disp-formula id="ptz160M9-36"><label>(D.36)</label><tex-math notation="LaTeX" id="Equation221"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
& = \Upsilon_3(s_3)
\notag \\
&=
\frac{\Gamma(1 + \varepsilon)}{(m_3^2 - s_3)} \,
\left\{
- \, \frac{1}{\varepsilon^2} \,
\left( -s_3 + m_3^2 - i \, \lambda \right)^{-\varepsilon}
+ \frac{1}{2 \, \varepsilon^2} \,
\left( m_3^2 - i \, \lambda \right)^{-\varepsilon}
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
\left.
+ \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)
\right\}\!.
\label{eq_verif_irc7}
\end{align}$$]]></tex-math></disp-formula></p>
<p>So long for the first term of Eq. (<xref ref-type="disp-formula" rid="ptz160M9-22">D.22</xref>). The second term can be obtained from the first one by replacing <inline-formula><tex-math notation="LaTeX" id="ImEquation778"><![CDATA[$s_3$]]></tex-math></inline-formula> by <inline-formula><tex-math notation="LaTeX" id="ImEquation779"><![CDATA[$s_1$]]></tex-math></inline-formula>. The coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation780"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation781"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula>, as well as <inline-formula><tex-math notation="LaTeX" id="ImEquation782"><![CDATA[$\det{(G)}$]]></tex-math></inline-formula>, are easily extracted from the <inline-formula><tex-math notation="LaTeX" id="ImEquation783"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix, cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M9-21">D.21</xref>):
<disp-formula id="ptz160M9-37"><label>(D.37)</label><tex-math notation="LaTeX" id="Equation222"><![CDATA[$$\begin{align}
\overline{b}_1 \; = \; (s_3 - m_3^2) \, (s_1 - s_3) &, \quad {}
\overline{b}_2 \; = \; (s_1 - m_3^2) \, (s_3 - s_1) ,
\notag\\
\det{(G)} &= - (s_1 - s_3)^2 .
\label{eqdefbb1bb2detg}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Thus, we finally get
<disp-formula id="ptz160UM25"><tex-math notation="LaTeX" id="Equation223"><![CDATA[$$\begin{align}
I_3^n
&= \frac{\overline{b}_1}{\det{(G)}} \, \Sigma_3^n(s_3) + \frac{\overline{b}_2}{\det{(G)}} \,
\Sigma_3^n(s_1)
\notag \\
&= \frac{\Gamma(1 + \varepsilon)}{ (s_1 - s_3)} \,
\left\{
- \frac{1}{\varepsilon^2} \,
\left[
\left( -s_3 + m_3^2 - i \, \lambda \right)^{-\varepsilon}
- \left( -s_1 + m_3^2 - i \, \lambda \right)^{-\varepsilon}
\right]
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
+ \left.
\mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2 + i \, \lambda} \right)
- \mbox{Li}_2 \left( \frac{s_1}{s_1 - m_3^2 + i \, \lambda} \right)
\right\}\!,
\label{eq_verif_irc8}
\notag
\end{align}$$]]></tex-math></disp-formula>
which coincides with Eq. (<xref ref-type="disp-formula" rid="ptz160M2-41">2.41</xref>): the direct and indirect ways lead to the same result.</p>
<p><italic>Concomitant occurrence of soft and collinear divergences</italic> Here again, we start with a recap of the texture of the <inline-formula><tex-math notation="LaTeX" id="ImEquation784"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix:<sup><xref ref-type="fn" rid="FN15">15</xref></sup>
<disp-formula id="ptz160M9-38"><label>(D.38)</label><tex-math notation="LaTeX" id="Equation224"><![CDATA[$$\begin{equation}
{\cal S} =
\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & s_3 \\
0 & s_3 & 0
\end{array}
\right)\!.
\label{eqcalscollsof_ver}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>Obviously, <inline-formula><tex-math notation="LaTeX" id="ImEquation785"><![CDATA[$\det({\cal S}) = 0$]]></tex-math></inline-formula>. As in the first example, the coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation786"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation787"><![CDATA[$\overline{b}_3$]]></tex-math></inline-formula> vanish whereas <inline-formula><tex-math notation="LaTeX" id="ImEquation788"><![CDATA[$\overline{b}_1 = \det{(G)}$]]></tex-math></inline-formula>, and, as <inline-formula><tex-math notation="LaTeX" id="ImEquation789"><![CDATA[$\widetilde{D}_{12} = \widetilde{D}_{13} = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation790"><![CDATA[$L(0,\Delta_1^{\{1\}},0)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>), thus
<disp-formula id="ptz160M9-39"><label>(D.39)</label><tex-math notation="LaTeX" id="Equation225"><![CDATA[$$\begin{align}
I_3^n &=
\frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},0 \big)
+
\frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}} \,
L_{3}^{n} \big( 0,\Delta_{1}^{\{1\}},0 \big) .
\label{eq_verif_ircs1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The relevant reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation791"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is
<disp-formula id="ptz160M9-40"><label>(D.40)</label><tex-math notation="LaTeX" id="Equation226"><![CDATA[$$\begin{equation}
{\cal S}^{\{1\}} =
\left(
\begin{array}{cc}
0 & s_3 \\
s_3 & 0
\end{array}
\right)\!,
\label{eqcalscols_ver1}
\end{equation}$$]]></tex-math></disp-formula>
whose determinant is <inline-formula><tex-math notation="LaTeX" id="ImEquation792"><![CDATA[$\det{({\cal S}^{\{1\}})} = - s_3^2$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation793"><![CDATA[$1 \times 1$]]></tex-math></inline-formula> associated Gram &#x201C;matrix&#x201D; is <inline-formula><tex-math notation="LaTeX" id="ImEquation794"><![CDATA[$G^{\{1\}(2)} = ( 2 \, s_3)$]]></tex-math></inline-formula>, and the reduced <inline-formula><tex-math notation="LaTeX" id="ImEquation795"><![CDATA[$\bar{b}$]]></tex-math></inline-formula> coefficients are given by
<disp-formula id="ptz160M9-41"><label>(D.41)</label><tex-math notation="LaTeX" id="Equation227"><![CDATA[$$\begin{align}
\frac{\overline{b}_{2}^{\{1\}}}{\det{(G^{\{1\}})}}
&=
\frac{\overline{b}_{3}^{\{1\}}}{\det{(G^{\{1\}})}}
\; = \;
- \, \frac{1}{2} ,
\label{eqbbar_vercols2}
\end{align}$$]]></tex-math></disp-formula>
whereas
<disp-formula id="ptz160M9-42"><label>(D.42)</label><tex-math notation="LaTeX" id="Equation228"><![CDATA[$$\begin{align}
\Delta_1^{\{1\}} &= \frac{s_3}{2} . \label{eqdelat1_vercols}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We thus obtain
<disp-formula id="ptz160UM26"><tex-math notation="LaTeX" id="Equation229"><![CDATA[$$\begin{align}
I_3^n
&= - L_3^n\big( 0,\Delta_1^{\{1\}},0 \big) \notag \\
&= -\frac{1}{\varepsilon^2} \, \Gamma(1+\varepsilon) \,
\frac{\Gamma^2(1-\varepsilon)}{\Gamma(1-2 \, \varepsilon)} \,
\left(
- \,s_3 - i \, \lambda
\right)^{-1-\varepsilon}
\notag \\
&= W\big( \det{(G^{\{1\}})}, 0, 0 \big) ,
\notag
\end{align}$$]]></tex-math></disp-formula>
cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M2-5">2.5</xref>), (<xref ref-type="disp-formula" rid="ptz160M2-6">2.6</xref>), and (<xref ref-type="disp-formula" rid="ptz160M2-30">2.30</xref>), so the direct and indirect ways lead to the same results.</p>
</sec>
<sec id="SEC9.2"><title>D.2. Complex mass case</title>
<p>We now treat the complex mass case. As discussed in Sect. <xref ref-type="sec" rid="SEC2">2</xref>, the only relevant case is the collinear case where the non-vanishing internal mass, say <inline-formula><tex-math notation="LaTeX" id="ImEquation796"><![CDATA[$m_3^2$]]></tex-math></inline-formula>, is complex:<sup><xref ref-type="fn" rid="FN16">16</xref></sup> <inline-formula><tex-math notation="LaTeX" id="ImEquation797"><![CDATA[$m_3^2 = m_{\rm R}^2 + i \, m_{\rm I}^2$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation798"><![CDATA[$m_{\rm R}^2$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation799"><![CDATA[$m_{\rm I}^2$]]></tex-math></inline-formula> real and <inline-formula><tex-math notation="LaTeX" id="ImEquation800"><![CDATA[$m_{\rm R}^2 > 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation801"><![CDATA[$m_{\rm I}^2 < 0$]]></tex-math></inline-formula>.<sup><xref ref-type="fn" rid="FN17">17</xref></sup> The <inline-formula><tex-math notation="LaTeX" id="ImEquation802"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-21">D.21</xref>), and <inline-formula><tex-math notation="LaTeX" id="ImEquation803"><![CDATA[$I_3^n$]]></tex-math></inline-formula> by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-22">D.22</xref>). As in the real mass case, let us focus on the first line of Eq. (<xref ref-type="disp-formula" rid="ptz160M9-22">D.22</xref>); the second line can be obtained by changing <inline-formula><tex-math notation="LaTeX" id="ImEquation804"><![CDATA[$s_3$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation805"><![CDATA[$s_1$]]></tex-math></inline-formula>. To compute <inline-formula><tex-math notation="LaTeX" id="ImEquation806"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation807"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{13})$]]></tex-math></inline-formula>, we have to determine which formulae to use, depending on the sign of <inline-formula><tex-math notation="LaTeX" id="ImEquation808"><![CDATA[$\operatorname{Im}(\Delta_1^{\{1\}})$]]></tex-math></inline-formula>. The quantity <inline-formula><tex-math notation="LaTeX" id="ImEquation809"><![CDATA[$\Delta_1^{\{1\}}$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-25">D.25</xref>), which reads
<disp-formula id="ptz160M9-43"><label>(D.43)</label><tex-math notation="LaTeX" id="Equation230"><![CDATA[$$\begin{align}
\Delta_1^{\{1\}}
&= \frac{1}{2 \, s_3} \,
\left( (s_3 - m_{\rm R}^2)^2 - m_{\rm I}^4 - 2 \, i \, m_{\rm I}^2 \, (s_3 - m_{\rm R}^2) \right)\!.
\label{eqdelta1_vercolc}
\end{align}$$]]></tex-math></disp-formula></p>
<p>When <inline-formula><tex-math notation="LaTeX" id="ImEquation810"><![CDATA[$(s_3 - m_{\rm R}^2)/s_3 > 0$]]></tex-math></inline-formula>, i.e. either <inline-formula><tex-math notation="LaTeX" id="ImEquation811"><![CDATA[$s_3 > m_{\rm R}^2$]]></tex-math></inline-formula> or <inline-formula><tex-math notation="LaTeX" id="ImEquation812"><![CDATA[$s_3 < 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation813"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>) as in the real mass case, and when <inline-formula><tex-math notation="LaTeX" id="ImEquation814"><![CDATA[$(s_3 - m_{\rm R}^2)/s_3 < 0$]]></tex-math></inline-formula>, i.e. <inline-formula><tex-math notation="LaTeX" id="ImEquation815"><![CDATA[$0 < s_3 < m_{\rm R}^2$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation816"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-52">2.52</xref>). As discussed previously, there is no such dichotomy for <inline-formula><tex-math notation="LaTeX" id="ImEquation817"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{13})$]]></tex-math></inline-formula>, which is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>).</p>
<p>Let us first consider <inline-formula><tex-math notation="LaTeX" id="ImEquation818"><![CDATA[$s_3 > m_{\rm R}^2$]]></tex-math></inline-formula> or <inline-formula><tex-math notation="LaTeX" id="ImEquation819"><![CDATA[$s_3 < 0$]]></tex-math></inline-formula>. We define <inline-formula><tex-math notation="LaTeX" id="ImEquation820"><![CDATA[$\bar{z} = (s_3 - m_3^2)/(s_3 + m_3^2)$]]></tex-math></inline-formula>, reminiscent of Eq. (<xref ref-type="disp-formula" rid="ptz160M9-27">D.27</xref>), and <inline-formula><tex-math notation="LaTeX" id="ImEquation821"><![CDATA[$\Sigma_3^n(s_3)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-32">D.32</xref>), in which the infinitesimal imaginary parts <inline-formula><tex-math notation="LaTeX" id="ImEquation822"><![CDATA[$\propto \lambda$]]></tex-math></inline-formula> are dropped out except for arguments of logarithms which do not depend on <inline-formula><tex-math notation="LaTeX" id="ImEquation823"><![CDATA[$m_3^2$]]></tex-math></inline-formula>, namely
<disp-formula id="ptz160M9-44"><label>(D.44)</label><tex-math notation="LaTeX" id="Equation231"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)}
\left\{
- \, \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
\ln \left( - \, \frac{(s_3 - m_3^2)^2}{s_3} \right)
-
\ln \left( - \, \frac{m_3^2}{s_3} \right)
\right]
\right.
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}\quad {}
- \frac{1}{2} \, \ln^2 \left( - \, \frac{(s_3 - m_3^2)^2}{s_3} \right)
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2} \right)
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}\quad {}
+ \ln \left( - \, \frac{m_3^2}{s_3} \right)
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {} \quad {}
\times \left[
\vphantom{\frac{1}{2}}- \ln \left( s_3 - i \, \lambda \, \sigma_s \right)
+ \frac{1}{2} \,
\left[
\ln \left( s_3 \, m_3^2 \right) + \ln \left( - \, (s_3 - m_3^2)^2 \right)
\right]
\right]
\notag \\
&\quad {} \quad {} \quad {} \quad {} \quad {} \quad {}\quad {}
\left.
+ \ln \left( - \, \frac{m_3^2}{s_3 - m_3^2} \right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\right\}\!.
\label{eq_verif_ircc3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Logarithms of products and ratios in Eq. (<xref ref-type="disp-formula" rid="ptz160M9-44">D.44</xref>) are further split. For this purpose we use Eq. (<xref ref-type="disp-formula" rid="ptz160M10-1">E.1</xref>) as well as
<disp-formula id="ptz160UM27"><tex-math notation="LaTeX" id="Equation232"><![CDATA[$$\ln \left( (s_3 - m_3^2)^2 \right)
=
2 \, \ln(s_3 - m_3^2)
- 2 \, i \, \pi \, \theta \left( - s_3 + m_{\rm R}^2 \right)$$]]></tex-math></disp-formula>
and, for any real <inline-formula><tex-math notation="LaTeX" id="ImEquation824"><![CDATA[$a$]]></tex-math></inline-formula> and complex <inline-formula><tex-math notation="LaTeX" id="ImEquation825"><![CDATA[$b$]]></tex-math></inline-formula>,
<disp-formula id="ptz160UM28"><tex-math notation="LaTeX" id="Equation233"><![CDATA[$$\begin{align}
\ln \left( \frac{b}{a} \right)
&= - \ln \left( a + i \, \lambda \, S(b) \right) + \ln(b) ,
\notag\\
\ln(a \, b)
&= \;\;\;\, \ln \left( a - i \, \lambda \, S(b) \right) + \ln(b) .
\notag
\end{align}$$]]></tex-math></disp-formula></p>
<p>We also use the notation <inline-formula><tex-math notation="LaTeX" id="ImEquation826"><![CDATA[$S(z) = \mbox{sign}\left( \operatorname{Im}(z) \right)$]]></tex-math></inline-formula>. Let us note that <inline-formula><tex-math notation="LaTeX" id="ImEquation827"><![CDATA[$S\left( (s_3 - m_3^2)^2 \right) = \mbox{sign}(s_3 - m_{\rm R}^2) \equiv \sigma_p$]]></tex-math></inline-formula>, and that <inline-formula><tex-math notation="LaTeX" id="ImEquation828"><![CDATA[$\ln( s_3 - i \, \lambda \, \sigma_s)$]]></tex-math></inline-formula> is equivalent to <inline-formula><tex-math notation="LaTeX" id="ImEquation829"><![CDATA[$\ln( s_3 + i \, \lambda)$]]></tex-math></inline-formula>. Then, to compactify Eq. (<xref ref-type="disp-formula" rid="ptz160M9-44">D.44</xref>), we take advantage of the following relations:
<disp-formula id="ptz160M9-45"><label>(D.45)</label><tex-math notation="LaTeX" id="Equation234"><![CDATA[$$\begin{align}
\ln\left( - s_3 \pm i \, \lambda \right) &= \ln\left( |s_3| \right) \pm i \, \pi \, \theta(s_3) , \notag \\
\ln\left( s_3 \pm i \, \lambda \right) &= \ln\left( |s_3| \right) \pm i \, \pi \, \theta(- s_3) , \notag \\
\theta(\pm s_3) &= \frac{(1 \pm \sigma_s)}{2} , \notag \\
\theta(- s_3 + m_{\rm R}^2) &= \frac{1- \sigma_p}{2} .
\label{eqnewrelovers3}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Equation (<xref ref-type="disp-formula" rid="ptz160M9-44">D.44</xref>) can be rewritten as
<disp-formula id="ptz160M9-46"><label>(D.46)</label><tex-math notation="LaTeX" id="Equation235"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
2 \, \left( \ln \left( s_3 - m_3^2 \right) - i \, \pi \right)
- \ln \left( m_3^2 \right)
\right]
\right.
\notag \\
&\qquad \qquad \qquad \quad {}
- \frac{1}{2} \,
\left[
2 \, \ln \left( s_3 - m_3^2 \right)
- i \, \pi \, \left( 1 - \frac{\sigma_p}{2} + \frac{\sigma_p \, \sigma_s}{2} \right)
\right]^2
\notag \\
&\qquad \qquad \qquad \quad {}
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2} \right)
+ \left( \ln \left( m_3^2 \right) + i \, \pi \, \frac{1+\sigma_s}{2}
\right)
\notag \\
&\qquad \qquad \qquad \quad {}
\times
\left[
\frac{1}{2} \, \ln \left( m_3^2 \right)
+ \ln \left( s_3 - m_3^2 \right) - i \, \pi \, (3 - \sigma_s)
\right]
\notag \\
&\qquad \qquad \qquad \quad {}
+
\left(
\ln \left( m_3^2 \right) - \ln \left( s_3 - m_3^2 \right) + i \, \pi
\right) \,
\left(
i \, \pi \, \frac{1-\sigma_s}{2} - \ln \left( s_3 - m_3^2 \right)
\right) \notag \\
&\qquad \qquad \qquad \quad {}
+ \left. i \, \frac{\pi}{2} \, \ln\left( |s_3| \right) \, (1-\sigma_s) \, (1+\sigma_p)
\vphantom{\frac{1}{\varepsilon^2}}
\right\}\!.
\label{eq_verif_ircc4}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Let us treat the case where <inline-formula><tex-math notation="LaTeX" id="ImEquation830"><![CDATA[$s_3 > m_{\rm R}^2$]]></tex-math></inline-formula>. We have <inline-formula><tex-math notation="LaTeX" id="ImEquation831"><![CDATA[$\sigma_p = +$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation832"><![CDATA[$\sigma_s=+$]]></tex-math></inline-formula>. By expanding Eq. (<xref ref-type="disp-formula" rid="ptz160M9-46">D.46</xref>), we get
<disp-formula id="ptz160M9-47"><label>(D.47)</label><tex-math notation="LaTeX" id="Equation236"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1 + \varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
2 \, \left( \ln \left( s_3 - m_3^2 \right) - i \, \pi \right)
- \ln\left( m_3^2\right)
\right]
\right.
\notag \\
&\quad {} \quad {}\quad {} \quad {} \quad {}\quad {} \quad {}
-
\left(
\ln^2 \left( s_3 - m_3^2 \right)
- 2 \, i \, \pi \, \ln \left( s_3 - m^2 \right) - \pi^2
\right)
\notag \\
&\quad {} \quad {}\quad {} \quad {} \quad {}\quad {} \quad {}
\left.
+ \frac{1}{2} \, \ln \left( m_3^2 \right)
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3 - m_3^2} \right)
\right\}\!.
\label{eq_verif_ircc5}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As <inline-formula><tex-math notation="LaTeX" id="ImEquation833"><![CDATA[$\ln(s_3 - m_3^2) - i \, \pi = \ln(-s_3 + m_3^2)$]]></tex-math></inline-formula>, we recover <inline-formula><tex-math notation="LaTeX" id="ImEquation834"><![CDATA[$\Upsilon_3^n(s_3)$]]></tex-math></inline-formula> given by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-33">D.33</xref>). The same exercise can easily be done for the case <inline-formula><tex-math notation="LaTeX" id="ImEquation835"><![CDATA[$s_3 < 0$]]></tex-math></inline-formula>, leading to the same conclusion.</p>
<p>In the case where <inline-formula><tex-math notation="LaTeX" id="ImEquation836"><![CDATA[$0 < s_3 < m_{\rm R}^2$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation837"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-52">2.52</xref>), i.e.
<disp-formula id="ptz160M9-48"><label>(D.48)</label><tex-math notation="LaTeX" id="Equation237"><![CDATA[$$\begin{align}
L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})
&=
\frac{2^{\varepsilon} \, \Gamma(1+\varepsilon)}
{2 \, (\widetilde{D}_{12} + \Delta_1^{\{1\}}) \, \bar{z}} \,
\left[
\frac{1}{\varepsilon} \,
\left(
i \, \pi \, S(i \, \bar{z} )
+ \ln \left( \frac{1+\bar{z}}{1-\bar{z}} \right)
\right)
\right.
\notag \\
&\qquad \qquad \qquad \qquad \quad {}
- \bar{H}_{0,\infty}(- \widetilde{D}_{12} - \Delta_1^{\{1\}}, - \Delta_1^{\{1\}})
\notag \\
&\qquad \qquad \qquad \qquad \quad {}
- \left.
\bar{H}_{1,\infty}(\widetilde{D}_{12} + \Delta_1^{\{1\}}, - \Delta_1^{\{1\}})
\vphantom{\frac{1}{\varepsilon}} \right]\!,
\label{eqdefnewL12}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation838"><![CDATA[$\bar{z} = \sqrt{\Delta_1^{\{1\}}/(\widetilde{D}_{12}+\Delta_1^{\{1\}})}$]]></tex-math></inline-formula>. We have chosen for the root of the equation <inline-formula><tex-math notation="LaTeX" id="ImEquation839"><![CDATA[$(\widetilde{D}_{12} + \Delta_1^{\{1\}})\, z^2 + \Delta_1^{\{1\}} = 0$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation840"><![CDATA[$\tilde{z} = i \, \bar{z}$]]></tex-math></inline-formula>. The function <inline-formula><tex-math notation="LaTeX" id="ImEquation841"><![CDATA[$\bar{H}_{1,\infty}(x,y)$]]></tex-math></inline-formula> is given by the expression in curly brackets of Eq. (<xref ref-type="disp-formula" rid="ptz160M10-15">E.15</xref>), and <inline-formula><tex-math notation="LaTeX" id="ImEquation842"><![CDATA[$\bar{H}_{0,\infty}(x,y)$]]></tex-math></inline-formula> by the expression in square brackets in Eq. (<xref ref-type="disp-formula" rid="ptz160M10-19">E.19</xref>) of Appendix <xref ref-type="sec" rid="SEC10">E</xref>, i.e.
<disp-formula id="ptz160M9-49"><label>(D.49)</label><tex-math notation="LaTeX" id="Equation238"><![CDATA[$$\begin{align}
\bar{H}_{0,\infty}(- \widetilde{D}_{12} - \Delta_1^{\{1\}}, - \Delta_1^{\{1\}})
&= i \, \pi \, S(i \, \bar{z}) \,
\left[
2 \, \ln \left( 2 \, i \, \bar{z} \right)
+ \ln \big( - \widetilde{D}_{12} - \Delta_1^{\{1\}} \big)
\right.
\notag \\
&\qquad \qquad \qquad {}
+ \left.
\eta \big( -\widetilde{D}_{12} - \Delta_1^{\{1\}}, \bar{z}^2 \big)
\right]
+ \pi^2 ,
\label{eqbarh0inf1} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M9-50"><label>(D.50)</label><tex-math notation="LaTeX" id="Equation239"><![CDATA[$$\begin{align}
\bar{H}_{1,\infty}(\widetilde{D}_{12} + \Delta_1^{\{1\}}, - \Delta_1^{\{1\}})
&=
\ln \left( \frac{1 + \bar{z}}{1 - \bar{z}} \right) \,
\left[ \vphantom{\frac{\widetilde{D}_{12}}{\widetilde{D}_{12} + \Delta_1^{\{1\}}}}
\ln \big( \widetilde{D}_{12} + \Delta_1^{\{1\}} \big)
+ \frac{1}{2} \, \ln \left( 1 - \bar{z}^2 \right)
+ \ln \left( 2 \, \bar{z} \right)
\right.
\notag \\
&\qquad \qquad \qquad \quad {}
+ \left.
\eta
\left(
\widetilde{D}_{12} + \Delta_1^{\{1\}}, \frac{\widetilde{D}_{12}}{\widetilde{D}_{12} + \Delta_1^{\{1\}}}
\right)
\right]
+ \frac{\pi^2}{2}
\notag \\
&\quad {}
- i \, \pi \, S(\bar{z}) \, \ln \left( \frac{\bar{z} + 1}{ 2 \, \bar{z}} \right)
- \mbox{Li}_2 \left( \frac{\bar{z} + 1}{ 2 \, \bar{z}} \right)
+ \mbox{Li}_2 \left( \frac{\bar{z} - 1}{ 2 \, \bar{z}} \right)\!.
\label{eqbarh1inf1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>The quantities <inline-formula><tex-math notation="LaTeX" id="ImEquation843"><![CDATA[$\widetilde{D}_{12}+\Delta_1^{\{1\}}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation844"><![CDATA[$\Delta_1^{\{1\}}$]]></tex-math></inline-formula> can be expressed in terms of <inline-formula><tex-math notation="LaTeX" id="ImEquation845"><![CDATA[$s_3$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation846"><![CDATA[$m_3^2$]]></tex-math></inline-formula> with Eq. (<xref ref-type="disp-formula" rid="ptz160M9-25">D.25</xref>). The expression obtained contains the ratio <inline-formula><tex-math notation="LaTeX" id="ImEquation847"><![CDATA[$(s_3 - m_3^2)/(s_3 + m_3^2)$]]></tex-math></inline-formula> given by
<disp-formula id="ptz160M9-51"><label>(D.51)</label><tex-math notation="LaTeX" id="Equation240"><![CDATA[$$\begin{equation}
\frac{s_3 - m_3^2}{s_3 + m_3^2}
=
\frac{s_3^2 - m_{\rm R}^4 - m_{\rm I}^4 - 2 \, i \, m_{\rm I}^2 \, s_3}{(s_3 + m_{\rm R}^2)^2 + m_{\rm I}^4} ,
\label{eqratios3pmm2}
\end{equation}$$]]></tex-math></disp-formula>
which has a negative real part and a positive imaginary part for <inline-formula><tex-math notation="LaTeX" id="ImEquation848"><![CDATA[$0 < s_3 < m_{\rm R}^2$]]></tex-math></inline-formula>. This implies that
<disp-formula id="ptz160M9-52"><label>(D.52)</label><tex-math notation="LaTeX" id="Equation241"><![CDATA[$$\begin{align}
S \left( i \, \bar{z} \right) &= -1, \quad
S \left( \bar{z} \right) = +1 .
\label{eqvariouss1}
\end{align}$$]]></tex-math></disp-formula></p>
<p>In addition, since <inline-formula><tex-math notation="LaTeX" id="ImEquation849"><![CDATA[$\operatorname{Im}( (s_3 - m_3^2)^2 ) = - 2 \, m_{\rm I}^2 \, (s_3 - m_{\rm R}^2) < 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation850"><![CDATA[$\operatorname{Im}( (s_3 + m_3^2)^2 ) = 2 \, m_{\rm I}^2 \, (s_3 + m_{\rm R}^2) < 0$]]></tex-math></inline-formula>, one can show that
<disp-formula id="ptz160UM29"><tex-math notation="LaTeX" id="Equation242"><![CDATA[$$\eta
\left(
\frac{(s_3+m_3^2)^2}{2 \, s_3}, \frac{4 \, m_3^2 \, s_3}{(s_3+m_3^2)^2}
\right)
= \eta
\left(
- \frac{(s_3+m_3^2)^2}{2 \, s_3}, \frac{(s_3-m_3^2)^2}{(s_3+m_3^2)^2}
\right)
= 0 .$$]]></tex-math></disp-formula></p>
<p>Putting all these things together, <inline-formula><tex-math notation="LaTeX" id="ImEquation851"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{12})$]]></tex-math></inline-formula> becomes
<disp-formula id="ptz160M9-53"><label>(D.53)</label><tex-math notation="LaTeX" id="Equation243"><![CDATA[$$\begin{align}
\hspace{2em}&\hspace{-2em}L(0,\Delta_1^{\{1\}},\widetilde{D}_{12}) \notag \\
&=
\frac{2^{\varepsilon} \, \Gamma(1+\varepsilon) \, s_3}
{(s_3 + m_3^2) \, (s_3 - m_3^2)} \,
\left\{
\frac{1}{\varepsilon} \,
\left[ \ln \left( \frac{s_3}{m_3^2} \right) - i \, \pi \right] - \frac{10 \, \pi^2}{6}
+ i \, \pi \, \ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\right.
\notag \\
&\qquad \qquad \qquad \qquad \qquad \quad
+ i \, \pi \,
\left[
2 \, \ln \left( 2 \, i \, \frac{s_3-m_3^2}{s_3+m_3^2} \right)
+ \ln \left( - \frac{(s_2-m_3^2)^2}{2 \, s_3} \right)
\right]
\notag \\
&\qquad \qquad \qquad \qquad \qquad \quad
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3-m_3^2} \right)
+ \ln \left( -\frac{m_3^2}{s_3-m_3^2} \right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\notag \\
&\qquad \qquad \qquad \qquad \qquad \quad
- \ln \left( \frac{s_3}{m_3^2} \right) \,
\left[
\ln \left( \frac{(s_3+m_3^2)^2}{2 \, s_3} \right)
+ \frac{1}{2} \, \ln \left( \frac{4 \, m_3^2 \, s_3}{(s_3+m_3^2)^2} \right)
\right.
\notag \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad
+ \left.
\left.
\ln \left( 2 \, \frac{s_3 - m_3^2}{s_3 + m_3^2} \right)
\right]
\right\}\!.
\label{eqdefnewL1210}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Substituting Eq. (<xref ref-type="disp-formula" rid="ptz160M9-53">D.53</xref>) into Eq. (<xref ref-type="disp-formula" rid="ptz160M9-28">D.28</xref>) with the explicit values for <inline-formula><tex-math notation="LaTeX" id="ImEquation852"><![CDATA[$\overline{b}_{2}^{\{1\}}$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation853"><![CDATA[$\overline{b}_{3}^{\{1\}}$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation854"><![CDATA[$\det{(G^{\{1\}})}$]]></tex-math></inline-formula>, and using Eq. (<xref ref-type="disp-formula" rid="ptz160M2-56">2.56</xref>) for <inline-formula><tex-math notation="LaTeX" id="ImEquation855"><![CDATA[$L(0,\Delta_1^{\{1\}},\widetilde{D}_{13})$]]></tex-math></inline-formula>, we get
<disp-formula id="ptz160M9-54"><label>(D.54)</label><tex-math notation="LaTeX" id="Equation244"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1+\varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
\ln \left( \frac{s_3}{m_3^2} \right)
+
\ln \left( - \frac{(s_3 - m_3^2)^2}{s_3} \right) - i \, \pi
\right]
\right.
\notag \\
&\qquad \qquad \qquad \quad
- \frac{1}{2} \, \ln^2 \left( - \frac{(s_3 - m_3^2)^2}{s_3} \right) - \frac{3 \, \pi^2}{2}
+ i \, \pi \, \ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\notag \\
&\qquad \qquad \qquad \quad
+ i \, \pi \,
\left[
2 \, \ln \left( i \, \frac{s_3-m_3^2}{s_3+m_3^2} \right)
+ \ln \left( - \frac{(s_2-m_3^2)^2}{s_3} \right)
\right]
- \ln \left( \frac{s_3}{m_3^2} \right)
\notag \\
&\qquad \qquad \qquad \quad
\times
\left[
\ln \left( \frac{(s_3+m_3^2)^2}{s_3} \right)
+ \frac{1}{2} \, \ln \left( \frac{m_3^2 \, s_3}{(s_3+m_3^2)^2} \right)
+ \ln \left( \frac{s_3 - m_3^2}{s_3 + m_3^2} \right)
\right]
\notag \\
&\qquad \qquad \qquad \quad
\left.
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3-m_3^2} \right)
+ \ln \left( -\frac{m_3^2}{s_3-m_3^2} \right) \,
\ln \left( \frac{s_3}{s_3 - m_3^2} \right)
\right\}\!.
\label{eq_verif_ircc2}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Keeping in mind that <inline-formula><tex-math notation="LaTeX" id="ImEquation856"><![CDATA[$0 < s_3 < m_{\rm R}^2$]]></tex-math></inline-formula>, we split the logarithms and expand the terms to end with
<disp-formula id="ptz160M9-55"><label>(D.55)</label><tex-math notation="LaTeX" id="Equation245"><![CDATA[$$\begin{align}
\Sigma_3^n(s_3)
&= \frac{\Gamma(1+\varepsilon)}{2 \, (m_3^2 - s_3)} \,
\left\{
- \frac{1}{\varepsilon^2} + \frac{1}{\varepsilon} \,
\left[
2 \, \ln \left( s_3 - m_3^2 \right) - 2 \, i \, \pi - \ln \left( m_3^2 \right)
\right]
\right.
\notag \\
&\qquad \qquad \qquad \quad
- \left( \ln^2 \left( s_3 - m_3^2 \right) - \pi^2
- 2 \, i \, \pi \, \ln \left( s_3-m_3^2 \right) \right)
\notag \\
&\qquad \qquad \qquad \quad
+ \left.
\frac{1}{2} \, \ln^2 \left( m_3^2 \right)
+ 2 \, \mbox{Li}_2 \left( \frac{s_3}{s_3-m_3^2} \right)
\right\}\!,
\label{eq_verif_ircc3final}
\end{align}$$]]></tex-math></disp-formula>
and, using <inline-formula><tex-math notation="LaTeX" id="ImEquation857"><![CDATA[$\ln(s_3 - m_3^2) = \ln(-s_3 + m_3^2) + i \, \pi$]]></tex-math></inline-formula>, we again recover Eq. (<xref ref-type="disp-formula" rid="ptz160M9-33">D.33</xref>). We note that the same formula holds for both <inline-formula><tex-math notation="LaTeX" id="ImEquation858"><![CDATA[$\operatorname{Im}(\Delta_{1}^{\{1\}})>0$]]></tex-math></inline-formula>, i.e. either <inline-formula><tex-math notation="LaTeX" id="ImEquation859"><![CDATA[$s_3<0$]]></tex-math></inline-formula> or <inline-formula><tex-math notation="LaTeX" id="ImEquation860"><![CDATA[$s_3> m_{\rm R}^2$]]></tex-math></inline-formula>, and for <inline-formula><tex-math notation="LaTeX" id="ImEquation861"><![CDATA[$\operatorname{Im}(\Delta_{1}^{\{1\}})<0$]]></tex-math></inline-formula>, i.e. <inline-formula><tex-math notation="LaTeX" id="ImEquation862"><![CDATA[$0< s_3< m_{\rm R}^2$]]></tex-math></inline-formula>. This is because in the last case, the integration contour <inline-formula><tex-math notation="LaTeX" id="ImEquation863"><![CDATA[$\int_{0}^{+ i \infty} + \int_{+\infty}^{1}$]]></tex-math></inline-formula> can actually be deformed into <inline-formula><tex-math notation="LaTeX" id="ImEquation864"><![CDATA[$\int_{0}^{1}$]]></tex-math></inline-formula>, i.e. Eq. (<xref ref-type="disp-formula" rid="ptz160M2-52">2.52</xref>) can be deformed into Eq. (<xref ref-type="disp-formula" rid="ptz160M2-49">2.49</xref>) by means of the Cauchy theorem. Indeed, when <inline-formula><tex-math notation="LaTeX" id="ImEquation865"><![CDATA[$0< s_3< m_{\rm R}^2$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation866"><![CDATA[$\operatorname{Im}(z^2 \, (\widetilde{D}_{12} + \Delta_{1}^{\{1\}}) - \Delta_{1}^{\{1\}})$]]></tex-math></inline-formula> never vanishes as <inline-formula><tex-math notation="LaTeX" id="ImEquation867"><![CDATA[$z$]]></tex-math></inline-formula> spans the real interval <inline-formula><tex-math notation="LaTeX" id="ImEquation868"><![CDATA[$[0,1]$]]></tex-math></inline-formula> and hence the cut of <inline-formula><tex-math notation="LaTeX" id="ImEquation869"><![CDATA[$\ln(z^2 \, (\widetilde{D}_{12} + \Delta_{1}^{\{1\}}) - \Delta_{1}^{\{1\}})$]]></tex-math></inline-formula> in the half-plane <inline-formula><tex-math notation="LaTeX" id="ImEquation870"><![CDATA[$\{\operatorname{Re}(z) >0\}$]]></tex-math></inline-formula> entirely lies inside the &#x201C;south-east&#x201D; quadrant <inline-formula><tex-math notation="LaTeX" id="ImEquation871"><![CDATA[$\{\operatorname{Re}(z) >0, \operatorname{Im}(z) <0\}$]]></tex-math></inline-formula>.</p>
<p><inline-formula><tex-math notation="LaTeX" id="ImEquation872"><![CDATA[$\Sigma_3^n(s_1)$]]></tex-math></inline-formula> is read from Eq. (<xref ref-type="disp-formula" rid="ptz160M9-55">D.55</xref>) by replacing <inline-formula><tex-math notation="LaTeX" id="ImEquation873"><![CDATA[$s_3$]]></tex-math></inline-formula> by <inline-formula><tex-math notation="LaTeX" id="ImEquation874"><![CDATA[$s_1$]]></tex-math></inline-formula>, and the coefficients <inline-formula><tex-math notation="LaTeX" id="ImEquation875"><![CDATA[$\overline{b}_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation876"><![CDATA[$\overline{b}_2$]]></tex-math></inline-formula> as well as <inline-formula><tex-math notation="LaTeX" id="ImEquation877"><![CDATA[$\det{(G)}$]]></tex-math></inline-formula> are still given by Eq. (<xref ref-type="disp-formula" rid="ptz160M9-37">D.37</xref>). So, in the complex mass case the same result is obtained as in the real mass case for <inline-formula><tex-math notation="LaTeX" id="ImEquation878"><![CDATA[$I_3^n$]]></tex-math></inline-formula>, and this leads to the conclusion that for the case of complex masses, the &#x201C;direct way&#x201D; and the &#x201C;indirect way&#x201D; also coincide.</p>
</sec>
</sec>
<sec id="SEC10"><title>Appendix E. Basic integrals in terms of dilogarithms and logarithms</title>
<p>In the presence of vanishing internal masses, specific integrals of the type
<disp-formula id="ptz160UM30"><tex-math notation="LaTeX" id="Equation246"><![CDATA[$$H
=
\int^a_b du \,
\frac{\ln ( A \, u^2 + B)}{A \, u^2 + B}$$]]></tex-math></disp-formula>
for the contour <inline-formula><tex-math notation="LaTeX" id="ImEquation879"><![CDATA[$[0,1]$]]></tex-math></inline-formula>, as well as the two other contours <inline-formula><tex-math notation="LaTeX" id="ImEquation880"><![CDATA[$[0,+\infty[$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation881"><![CDATA[$[1,+\infty[$]]></tex-math></inline-formula>, are involved.</p>
<p>Here, we compute all the above types of integrals successively. The presentation is ordered according to the integration contours <inline-formula><tex-math notation="LaTeX" id="ImEquation882"><![CDATA[$(a,b)$]]></tex-math></inline-formula> considered. We last provide an extra load of backup integrals. This appendix often makes use of the identity
<disp-formula id="ptz160M10-1"><label>(E.1)</label><tex-math notation="LaTeX" id="Equation247"><![CDATA[$$\begin{align}
\ln(z) &= \ln(-z) + i \, \pi \, S(z) ,
\label{eqdeflnzlnmz}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation883"><![CDATA[$S(z)$]]></tex-math></inline-formula> is given by Eq. (<xref ref-type="disp-formula" rid="ptz160M2-25">2.25</xref>).</p>
<sec id="SEC10.1"><title>E.1. <inline-formula><tex-math notation="LaTeX" id="ImEquation884"><![CDATA[$H$]]></tex-math></inline-formula>-type integrals for the IR case</title>
<sec id="SEC10.1.1"><title>E.1.1. First kind</title>
<p><disp-formula id="ptz160UM31"><tex-math notation="LaTeX" id="Equation248"><![CDATA[$$ H_{0,1}(A,B) = \int^1_0 du \, \frac{\ln(A \, u^2 + B)}{A \, u^2 + B} .$$]]></tex-math></disp-formula></p>
<p>The cases of real masses (<inline-formula><tex-math notation="LaTeX" id="ImEquation885"><![CDATA[$A$]]></tex-math></inline-formula> real) and of complex masses (<inline-formula><tex-math notation="LaTeX" id="ImEquation886"><![CDATA[$\operatorname{Im}(A) \ne 0$]]></tex-math></inline-formula>) are treated all at once, considering <inline-formula><tex-math notation="LaTeX" id="ImEquation887"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation888"><![CDATA[$B$]]></tex-math></inline-formula> both complex yet such that sign<inline-formula><tex-math notation="LaTeX" id="ImEquation889"><![CDATA[$(\operatorname{Im}(A \, u^2 + B)$]]></tex-math></inline-formula> is kept constant when <inline-formula><tex-math notation="LaTeX" id="ImEquation890"><![CDATA[$u$]]></tex-math></inline-formula> spans the range <inline-formula><tex-math notation="LaTeX" id="ImEquation891"><![CDATA[$[0,1]$]]></tex-math></inline-formula>, as is always the case for all our needs (cf. Sects. <xref ref-type="sec" rid="SEC2">2</xref> and <xref ref-type="sec" rid="SEC3">3</xref>). We write
<disp-formula id="ptz160M10-2"><label>(E.2)</label><tex-math notation="LaTeX" id="Equation249"><![CDATA[$$\begin{equation}
H_{0,1}(A,B)
=
\frac{1}{A} \, \int^1_0 du \, \frac{C_A + \ln(u^2 - \bar{u}^2)}{u^2-\bar{u}^2} ,
\label{eqdefh01}
\end{equation}$$]]></tex-math></disp-formula>
where
<disp-formula id="ptz160M10-3"><label>(E.3)</label><tex-math notation="LaTeX" id="Equation250"><![CDATA[$$\begin{align}
C_A &= \ln(A - i \, \lambda \, S(-\bar{u}^2))
\quad \text{if} \; \operatorname{Im}(A)=0 ,
\label{eqdefca1a}\\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-4"><label>(E.4)</label><tex-math notation="LaTeX" id="Equation251"><![CDATA[$$\begin{align}
C_A &= \ln(A) + \eta(A,-\bar{u}^2)
\quad \text{otherwise} ,
\label{eqdefca1b}
\end{align}$$]]></tex-math></disp-formula>
and <inline-formula><tex-math notation="LaTeX" id="ImEquation892"><![CDATA[$\bar{u}^2 = - B/A$]]></tex-math></inline-formula>. The <inline-formula><tex-math notation="LaTeX" id="ImEquation893"><![CDATA[$\eta$]]></tex-math></inline-formula> function is given by Eq. (E6) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>]. The term <inline-formula><tex-math notation="LaTeX" id="ImEquation894"><![CDATA[$\ln(u^2-\bar{u}^2)$]]></tex-math></inline-formula> can be split without the <inline-formula><tex-math notation="LaTeX" id="ImEquation895"><![CDATA[$\eta$]]></tex-math></inline-formula> function since <inline-formula><tex-math notation="LaTeX" id="ImEquation896"><![CDATA[$\bar{u}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation897"><![CDATA[$-\bar{u}$]]></tex-math></inline-formula> have imaginary parts of opposite signs. Performing a partial fraction decomposition, we get
<disp-formula id="ptz160M10-5"><label>(E.5)</label><tex-math notation="LaTeX" id="Equation252"><![CDATA[$$\begin{align}
H_{0,1}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left[
C_A \, \int^1_0 du \, \left( \frac{1}{u-\bar{u}} - \frac{1}{u+\bar{u}} \right)
\right.
\notag \\
&\quad {}\quad {} \quad {} \quad {}\quad {}
+
\int^1_0 du \, \frac{\ln(u-\bar{u})}{u-\bar{u}}
-
\int^1_0 du \, \frac{\ln(u-\bar{u})}{u+\bar{u}}
\notag \\
&\quad {}\quad {} \quad {} \quad {}\quad {}
\left.
+
\int^1_0 du \, \frac{\ln(u+\bar{u})}{u-\bar{u}} -
\int^1_0 du \, \frac{\ln(u+\bar{u})}{u+\bar{u}}
\right]\!.
\label{eqdefh02}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We can rearrange the terms of the right-hand side of Eq. (<xref ref-type="disp-formula" rid="ptz160M10-5">E.5</xref>) in the following way:
<disp-formula id="ptz160M10-6"><label>(E.6)</label><tex-math notation="LaTeX" id="Equation253"><![CDATA[$$\begin{align}
H_{0,1}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left\{
C_A \,
\left[
\ln \left( \frac{\bar{u}-1}{\bar{u}} \right) -
\ln \left( \frac{\bar{u}+1}{\bar{u}} \right)
\right]
\right.
\notag \\
&\quad {}\quad {}\quad {}\quad {} +
\frac{1}{2} \,
\left[
\ln^2(1-\bar{u}) - \ln^2(-\bar{u}) -\ln^2(1+\bar{u}) + \ln^2(\bar{u})
\right]
\notag \\
&\quad {} \quad {}\quad {}\quad {}+
\int^1_0 du \, \frac{\ln(u+\bar{u}) - \ln(2 \, \bar{u})}{u-\bar{u}}
\;\;\; + \;\;\;
\int^1_0 du \, \frac{\ln(2 \, \bar{u})}{u-\bar{u}}
\notag \\
&\quad {}\quad {}\quad {}\quad {} -
\left.
\int^1_0 du \, \frac{\ln(u-\bar{u}) - \ln(-2 \, \bar{u})}{u+\bar{u}} -
\int^1_0 du \, \frac{\ln(-2 \, \bar{u})}{u+\bar{u}}
\;\;
\right\}\!.
\label{eqdefh03}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using Eq. (<xref ref-type="disp-formula" rid="ptz160M10-1">E.1</xref>) we can write Eq. (<xref ref-type="disp-formula" rid="ptz160M10-6">E.6</xref>) as
<disp-formula id="ptz160M10-7"><label>(E.7)</label><tex-math notation="LaTeX" id="Equation254"><![CDATA[$$\begin{align}
H_{0,1}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left\{
\ln \left( \frac{\bar{u}-1}{\bar{u}+1} \right) \,
\left[
C_A +
\frac{1}{2} \, \left[ \ln(\bar{u}-1) + \ln(\bar{u}+1) \right] +
i \, \pi \, S(-\bar{u}) + \ln(2 \, \bar{u})
\right]
\right.
\notag \\
&\quad {} \quad {}\quad {}\quad {}+
\left.
R^{\prime}(-\bar{u},\bar{u})
\vphantom{\ln \left( \frac{\Lambda - \bar{u}}{1 - \bar{u}} \right)}
\right\}\!,
\label{eqdefh04}
\end{align}$$]]></tex-math></disp-formula>
where the function <inline-formula><tex-math notation="LaTeX" id="ImEquation898"><![CDATA[$R^{\prime}$]]></tex-math></inline-formula> is defined in Eq. (E11) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>].<sup><xref ref-type="fn" rid="FN18">18</xref></sup> Using Eq. (E15) of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] with <inline-formula><tex-math notation="LaTeX" id="ImEquation899"><![CDATA[$y=-\bar{u}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation900"><![CDATA[$z=\bar{u}$]]></tex-math></inline-formula> and rearranging the term in square brackets, we get
<disp-formula id="ptz160M10-8"><label>(E.8)</label><tex-math notation="LaTeX" id="Equation255"><![CDATA[$$\begin{align}
H_{0,1}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left\{
\ln \left( \frac{\bar{u}-1}{\bar{u}+1} \right) \,
\left[
C_A +
\frac{1}{2} \,
\left[
\ln \left( 1 - \bar{u}^2 \right) + \ln \left( - 4 \, \bar{u}^2 \right)
\right]
\right]
\right.
\notag \\
&\quad {} \quad {}\quad {}\quad {} +
\left.
\mbox{Li}_2 \left( \frac{\bar{u}+1}{2 \, \bar{u}} \right) -
\mbox{Li}_2 \left( \frac{\bar{u}-1}{2 \, \bar{u}} \right)
\right\}\!.
\label{eqdefh05}
\end{align}$$]]></tex-math></disp-formula></p>
</sec>
<sec id="SEC10.1.2"><title>E.1.2. Second kind</title>
<p>With complex masses we need to also compute
<disp-formula id="ptz160M10-9"><label>(E.9)</label><tex-math notation="LaTeX" id="Equation256"><![CDATA[$$\begin{equation}
H_{1,\infty}(A,B) = \int^{\infty}_1 du \,
\frac{\ln(A \, u^2 + B)}{A \, u^2 + B} ,
\label{eqdefh11}
\end{equation}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation901"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation902"><![CDATA[$B$]]></tex-math></inline-formula> are complex yet such that <inline-formula><tex-math notation="LaTeX" id="ImEquation903"><![CDATA[$\mbox{sign}(\operatorname{Im}(A \, u^2 + B))$]]></tex-math></inline-formula> is kept constant while <inline-formula><tex-math notation="LaTeX" id="ImEquation904"><![CDATA[$u$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation905"><![CDATA[$[1,+\infty[$]]></tex-math></inline-formula>. The quantity <inline-formula><tex-math notation="LaTeX" id="ImEquation906"><![CDATA[$H_{1,\infty}(A,B)$]]></tex-math></inline-formula> can be written as
<disp-formula id="ptz160M10-10"><label>(E.10)</label><tex-math notation="LaTeX" id="Equation257"><![CDATA[$$\begin{equation}
H_{1,\infty}(A,B)
=
\frac{1}{A} \, \int^{\infty}_1 du \,
\frac{C^{\prime}_A + \ln(u^2 - \bar{u}^2)}{u^2 - \bar{u}^2} ,
\label{eqdefh12}
\end{equation}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation907"><![CDATA[$\bar{u}^2 = - B/A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation908"><![CDATA[$C^{\prime}_A$]]></tex-math></inline-formula> is given by
<disp-formula id="ptz160M10-11"><label>(E.11)</label><tex-math notation="LaTeX" id="Equation258"><![CDATA[$$\begin{align}
C^{\prime}_A &= \ln(A) + \eta(A,1-\bar{u}^2) .
\label{eqdefca1c}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We perform a partial fraction decomposition and, writing <inline-formula><tex-math notation="LaTeX" id="ImEquation909"><![CDATA[$H_{1,\infty}(A,B)$]]></tex-math></inline-formula> as a sum of terms which are individually divergent when <inline-formula><tex-math notation="LaTeX" id="ImEquation910"><![CDATA[$u \rightarrow \infty$]]></tex-math></inline-formula>, we face a situation similar to the one met in Sect. B.2 of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]. We proceed likewise, introducing a large <inline-formula><tex-math notation="LaTeX" id="ImEquation911"><![CDATA[$u$]]></tex-math></inline-formula>-cut-off <inline-formula><tex-math notation="LaTeX" id="ImEquation912"><![CDATA[$\Lambda$]]></tex-math></inline-formula> and write <inline-formula><tex-math notation="LaTeX" id="ImEquation913"><![CDATA[$H_{1,\infty}(A,B)$]]></tex-math></inline-formula> as
<disp-formula id="ptz160UM32"><tex-math notation="LaTeX" id="Equation259"><![CDATA[$$\begin{align}
H_{1,\infty}(A,B)
&= \frac{1}{2 \, A \, \bar{u}} \lim_{\Lambda \rightarrow + \infty}
{\cal H}_{1,\infty}^{\Lambda}(A,B) ,
\notag
\end{align}$$]]></tex-math></disp-formula>
where
<disp-formula id="ptz160M10-12"><label>(E.12)</label><tex-math notation="LaTeX" id="Equation260"><![CDATA[$$\begin{align}
{\cal H}_{1,\infty}^{\Lambda}(A,B)
&=
\left\{
C^{\prime}_A \,
\int^{\Lambda}_1 du \,
\left[ \frac{1}{u-\bar{u}} - \frac{1}{u+\bar{u}} \right]
\right.
\notag \\
& \quad {} \quad {} +
\int^{\Lambda}_1 du \, \frac{\ln(u-\bar{u})}{u-\bar{u}}
-
\int^{\Lambda}_1 du \, \frac{\ln(u+\bar{u})}{u+\bar{u}}
\notag \\
& \quad {} \quad {}
+
\int^{\Lambda}_1 du \, \frac{\ln(u+\bar{u})- \ln(2 \, \bar{u})}{u-\bar{u}}
\;\;\;+\;\;\;
\ln ( 2 \, \bar{u}) \, \int^{\Lambda}_1 \frac{du}{u - \bar{u}}
\notag \\
&\quad {} \quad {}
\left.
-
\int^{\Lambda}_1 du \, \frac{\ln(u-\bar{u}) - \ln(- 2 \, \bar{u})}{u+\bar{u}}
-
\; \ln ( - 2 \, \bar{u}) \, \int^{\Lambda}_1 \frac{du}{u + \bar{u}}
\right\}\!.
\label{eqdefh24}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We express <inline-formula><tex-math notation="LaTeX" id="ImEquation914"><![CDATA[${\cal H}_{1,\infty}^{\Lambda}(A,B)$]]></tex-math></inline-formula> in terms of the function <inline-formula><tex-math notation="LaTeX" id="ImEquation915"><![CDATA[$R^{\Lambda}(y,z) $]]></tex-math></inline-formula> defined by Eq. (B.9) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]:
<disp-formula id="ptz160M10-13"><label>(E.13)</label><tex-math notation="LaTeX" id="Equation261"><![CDATA[$$\begin{align}
{\cal H}_{1,\infty}^{\Lambda}(A,B)
&=
\left\{
\left[ C^{\prime}_A + \ln ( 2 \, \bar{u}) \right] \,
\ln \left( \frac{\Lambda - \bar{u}}{1 - \bar{u}} \right) -
\left[ C^{\prime}_A + \ln ( - 2 \, \bar{u}) \right] \,\ln \left( \frac{\Lambda + \bar{u}}{1 + \bar{u}} \right)
\right.
\notag \\
&\quad {} +
\frac{1}{2} \, \left[ \ln^2 \left( \Lambda - \bar{u} \right) - \ln^2 \left( 1 - \bar{u} \right) \right] -
\frac{1}{2} \, \left[ \ln^2 \left( \Lambda + \bar{u} \right) - \ln^2 \left( 1 + \bar{u} \right) \right]
\notag \\
&\quad {} +
\left.
R^{\Lambda}(-\bar{u},\bar{u}) - R^{\Lambda}(\bar{u},-\bar{u})
\vphantom{\ln \left( \frac{\Lambda - \bar{u}}{1 - \bar{u}} \right)}
\right\}\!.
\label{eqdefh25}
\end{align}$$]]></tex-math></disp-formula></p>
<p>Using Eq. (B.14) of Ref. [<xref ref-type="bibr" rid="B2">2</xref>] for the <inline-formula><tex-math notation="LaTeX" id="ImEquation916"><![CDATA[$R^{\Lambda}$]]></tex-math></inline-formula> terms, we take the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation917"><![CDATA[$\Lambda \rightarrow \infty$]]></tex-math></inline-formula>. The terms proportional to <inline-formula><tex-math notation="LaTeX" id="ImEquation918"><![CDATA[$\ln^2(\Lambda)$]]></tex-math></inline-formula> and those proportional to <inline-formula><tex-math notation="LaTeX" id="ImEquation919"><![CDATA[$\ln (\Lambda)$]]></tex-math></inline-formula> drop out, and we get
<disp-formula id="ptz160M10-14"><label>(E.14)</label><tex-math notation="LaTeX" id="Equation262"><![CDATA[$$\begin{align}
H_{1,\infty}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left\{ \vphantom{\mbox{Li}_2 \left( \frac{\bar{u} + 1}{2 \, \bar{u}} \right)}
\left[ C^{\prime}_A + \ln ( - 2 \, \bar{u}) \right] \,\ln \left( 1 + \bar{u} \right) -
\left[ C^{\prime}_A + \ln ( 2 \, \bar{u}) \right] \, \ln \left( 1 - \bar{u} \right)
\right.
\notag \\
&\quad {} \quad {}\quad {}\quad {}+
\frac{1}{2} \,
\left[
\ln^2 \left( 1 + \bar{u} \right) - \ln^2 \left( 1 - \bar{u} \right) +
\ln^2 \left( 2 \, \bar{u} \right) - \ln^2 \left( - 2 \, \bar{u} \right)
\right]
\notag \\
&\quad {}\quad {}\quad {} \quad {}
\left.
- \mbox{Li}_2 \left( \frac{\bar{u} + 1}{2 \, \bar{u}} \right)
+ \mbox{Li}_2 \left( \frac{\bar{u} - 1}{2 \, \bar{u}} \right)
\label{eqdefh26}
\right\}\!.
\end{align}$$]]></tex-math></disp-formula></p>
<p>Noting that
<disp-formula id="ptz160UM33"><tex-math notation="LaTeX" id="Equation263"><![CDATA[$$ \ln \left( \frac{1 + \bar{u}}{1 - \bar{u}} \right)
= \ln (1 + \bar{u}) - \ln (1 - \bar{u}) ,$$]]></tex-math></disp-formula></p>
<p>Eq. (<xref ref-type="disp-formula" rid="ptz160M10-14">E.14</xref>) becomes, after some algebra,
<disp-formula id="ptz160M10-15"><label>(E.15)</label><tex-math notation="LaTeX" id="Equation264"><![CDATA[$$\begin{align}
H_{1,\infty}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\left\{
\ln \left( \frac{1 + \bar{u}}{1 - \bar{u}} \right) \,
\left[ C^{\prime}_A + \frac{1}{2} \, \ln ( 1 - \bar{u}^2) + \ln ( 2 \, \bar{u}) \right] +
\frac{\pi^2}{2}
\right.
\notag \\
&\quad {} \quad {}\quad {} \quad {} -
\left.
i \, \pi \, S(\bar{u}) \, \ln \left( \frac{\bar{u}+1}{2 \, \bar{u}} \right)
-
\mbox{Li}_2 \left( \frac{\bar{u} + 1}{2 \, \bar{u}} \right)
+
\mbox{Li}_2 \left( \frac{\bar{u} - 1}{2 \, \bar{u}} \right)
\right\}\!.
\label{eqdefh27}
\end{align}$$]]></tex-math></disp-formula></p>
<p>We remark that the same combination of dilogarithms, up to a sign, appears in Eq. (<xref ref-type="disp-formula" rid="ptz160M10-15">E.15</xref>) and in <inline-formula><tex-math notation="LaTeX" id="ImEquation920"><![CDATA[$H_{0,1}(A,B)$]]></tex-math></inline-formula>, so that we can rewrite <inline-formula><tex-math notation="LaTeX" id="ImEquation921"><![CDATA[$H_{1,\infty}$]]></tex-math></inline-formula> as
<disp-formula id="ptz160M10-16"><label>(E.16)</label><tex-math notation="LaTeX" id="Equation265"><![CDATA[$$\begin{align}
H_{1,\infty}(A,B)
&= - \, H_{0,1}(A,B) +
\frac{1}{2 \, A \, \bar{u}}
\left\{ \vphantom{\frac{1+\bar{u}}{1-\bar{u}}}
i \, \pi \, S(\bar{u}) \, \left[ 2 \, \ln ( 2 \, \bar{u}) + C_A \right] + \pi^2 \right.
\notag \\
&\qquad {} +
\left. \left[ \eta(A,-\bar{u}^2) - \eta(A,1-\bar{u}^2) \right] \, \ln \left( \frac{1+\bar{u}}{1-\bar{u}} \right)
\right\}\!.
\label{eqdefh28}
\end{align}$$]]></tex-math></disp-formula></p>
</sec>
<sec id="SEC10.1.3"><title>E.1.3. Third kind</title>
<p>With complex masses a third kind of integrals also has to be considered:
<disp-formula id="ptz160M10-17"><label>(E.17)</label><tex-math notation="LaTeX" id="Equation266"><![CDATA[$$\begin{equation}
H_{0,\infty}(A,B) = \int^{\infty}_0 du \, \frac{\ln(A \, u^2 + B)}{A \, u^2 + B} ,
\label{eqdefh21}
\end{equation}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation922"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation923"><![CDATA[$B$]]></tex-math></inline-formula> are complex yet such that <inline-formula><tex-math notation="LaTeX" id="ImEquation924"><![CDATA[$\mbox{sign}(\operatorname{Im}(A \, u^2 + B))$]]></tex-math></inline-formula> is kept constant while <inline-formula><tex-math notation="LaTeX" id="ImEquation925"><![CDATA[$u$]]></tex-math></inline-formula> spans <inline-formula><tex-math notation="LaTeX" id="ImEquation926"><![CDATA[$[0,+\infty[$]]></tex-math></inline-formula>. The quantity <inline-formula><tex-math notation="LaTeX" id="ImEquation927"><![CDATA[$H_{0,\infty}(A,B)$]]></tex-math></inline-formula> can be split as
<disp-formula id="ptz160M10-18"><label>(E.18)</label><tex-math notation="LaTeX" id="Equation267"><![CDATA[$$\begin{equation}
H_{0,\infty}(A,B) = H_{0,1}(A,B) + H_{1,\infty}(A,B) .
\label{eqdefh22}
\end{equation}$$]]></tex-math></disp-formula></p>
<p>From Eq. (<xref ref-type="disp-formula" rid="ptz160M10-16">E.16</xref>), and remembering that the assumption on the sign of <inline-formula><tex-math notation="LaTeX" id="ImEquation928"><![CDATA[$\operatorname{Im}(A \, u^2 + B)$]]></tex-math></inline-formula> implies that <inline-formula><tex-math notation="LaTeX" id="ImEquation929"><![CDATA[$\eta(A,-\bar{u}^2) = \eta(A,1-\bar{u}^2)$]]></tex-math></inline-formula>, we immediately get
<disp-formula id="ptz160M10-19"><label>(E.19)</label><tex-math notation="LaTeX" id="Equation268"><![CDATA[$$\begin{align}
H_{0,\infty}(A,B)
&= \frac{1}{2 \, A \, \bar{u}}
\Bigl[
i \, \pi \, S(\bar{u}) \, \left[ 2 \, \ln ( 2 \, \bar{u}) + C_A \right] + \pi^2
\Bigr] .
\label{eqdefh23}
\end{align}$$]]></tex-math></disp-formula></p>
<p>As happened for <inline-formula><tex-math notation="LaTeX" id="ImEquation930"><![CDATA[$K^{C}_{0,\infty}(A,B)$]]></tex-math></inline-formula> (cf. Appendix <xref ref-type="sec" rid="SEC7">B</xref> of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]), <inline-formula><tex-math notation="LaTeX" id="ImEquation931"><![CDATA[$H_{0,\infty}(A,B)$]]></tex-math></inline-formula> contains only logarithmic terms.</p>
</sec>
</sec>
<sec id="SEC10.2"><title>E.2. An extra load of backup integrals</title>
<p>We also need the following load of simpler integrals:
<disp-formula id="ptz160M10-20"><label>(E.20)</label><tex-math notation="LaTeX" id="Equation269"><![CDATA[$$\begin{align}
W_1(u_0^2) &= \int^1_0 du \, \frac{\ln(1-u^2)}{u^2 - u_0^2} , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-21"><label>(E.21)</label><tex-math notation="LaTeX" id="Equation270"><![CDATA[$$\begin{align}
W_2(u_0^2) &= \int^1_0 du \, \frac{\ln(u)}{u^2 - u_0^2} , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-22"><label>(E.22)</label><tex-math notation="LaTeX" id="Equation271"><![CDATA[$$\begin{align}
W_3(u_0^2) &= \int^{\infty}_1 du \, \frac{\ln(u^2-1)}{u^2 - u_0^2} , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-23"><label>(E.23)</label><tex-math notation="LaTeX" id="Equation272"><![CDATA[$$\begin{align}
W_4(u_0^2) &= \int^{\infty}_0 du \, \frac{\ln(u)}{u^2 - u_0^2} , \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-24"><label>(E.24)</label><tex-math notation="LaTeX" id="Equation273"><![CDATA[$$\begin{align}
W_5(u_0^2) &= \int^{\infty}_0 du \, \frac{\ln(u^2+1)}{u^2 + u_0^2} .
\end{align}$$]]></tex-math></disp-formula></p>
<p>For all these integrals, <inline-formula><tex-math notation="LaTeX" id="ImEquation932"><![CDATA[$u_0^2$]]></tex-math></inline-formula> is assumed to be a complex number; this is indeed the case because these integrals appear in the computation of the four-point function in the IR case where <inline-formula><tex-math notation="LaTeX" id="ImEquation933"><![CDATA[$u_0^2$]]></tex-math></inline-formula> is either a complex number with an imaginary part <inline-formula><tex-math notation="LaTeX" id="ImEquation934"><![CDATA[$\propto \lambda$]]></tex-math></inline-formula> or a genuine complex number.</p>
<p>One might compute these integrals using specific values for <inline-formula><tex-math notation="LaTeX" id="ImEquation935"><![CDATA[$A$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation936"><![CDATA[$B$]]></tex-math></inline-formula> in <inline-formula><tex-math notation="LaTeX" id="ImEquation937"><![CDATA[$K^R_{0,1}(A,B)$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation938"><![CDATA[$K^C_{0,1}(A,B)$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation939"><![CDATA[$K^C_{1,\infty}(A,B)$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation940"><![CDATA[$K^C_{0,\infty}(A,B)$]]></tex-math></inline-formula> given in Appendices E of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] and B of Ref. [<xref ref-type="bibr" rid="B2">2</xref>]; however, these integrals are simple enough to be computed directly (we verified that the results can be retrieved using the K-type integrals after some transformation on the dilogarithms). We give here the results of these integrals without any details:
<disp-formula id="ptz160M10-25"><label>(E.25)</label><tex-math notation="LaTeX" id="Equation274"><![CDATA[$$\begin{align}
W_1(u_0^2)
&= \frac{1}{2 \, u_0} \,
\left[
\mbox{Li}_2 \left( \frac{2}{1+u_0} \right) -
\mbox{Li}_2 \left( \frac{2}{1-u_0} \right) -
2 \, \ln(2) \, \ln \left( \frac{u_0+1}{u_0-1} \right)
\right]\!,
\label{eqresw1} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-26"><label>(E.26)</label><tex-math notation="LaTeX" id="Equation275"><![CDATA[$$\begin{align}
W_2(u_0^2)
&= \frac{1}{2 \, u_0} \,
\left[
\mbox{Li}_2 \left( \frac{1}{u_0} \right) -
\mbox{Li}_2 \left( - \, \frac{1}{u_0} \right)
\right]\!,
\label{eqresw3} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-27"><label>(E.27)</label><tex-math notation="LaTeX" id="Equation276"><![CDATA[$$\begin{align}
W_3(u_0^2)
&= - W_1(u_0^2) + \frac{1}{2 \, u_0} \, i \, S(u_0) \, \pi \ln \left( 1 - u_0^2 \right)\!,
\label{eqresw2} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-28"><label>(E.28)</label><tex-math notation="LaTeX" id="Equation277"><![CDATA[$$\begin{align}
W_4(u_0^2)
&= \frac{1}{4 \, u_0} \, i \, S(u_0) \, \pi \, \ln(-u^2_0) ,
\label{eqresw4} \\
\end{align}$$]]></tex-math></disp-formula>
<disp-formula id="ptz160M10-29"><label>(E.29)</label><tex-math notation="LaTeX" id="Equation278"><![CDATA[$$\begin{align}
W_5(u_0^2)
&= \frac{\pi}{u_0} \, \ln(1+u_0) ,
\label{eqresw5}
\end{align}$$]]></tex-math></disp-formula>
with <inline-formula><tex-math notation="LaTeX" id="ImEquation941"><![CDATA[$u_0 = \sqrt{u_0^2}$]]></tex-math></inline-formula>.</p>
</sec>
</sec>
<sec id="SEC11"><title>Appendix F. Change of contour prescription for the pole in the IR four-point integral</title>
<p>This appendix legitimates the replacement
<disp-formula id="ptz160M11-1"><label>(F.1)</label><tex-math notation="LaTeX" id="Equation279"><![CDATA[$$\begin{equation}\label{subst1}
\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, \lambda \right]^{-\varepsilon}}
{u^2 \, P_{ijk} + R_{ij} + i \, \lambda}
\to
\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, \lambda \right]^{-\varepsilon}}
{u^2 \, P_{ijk} + R_{ij} - i \, \lambda}
\end{equation}$$]]></tex-math></disp-formula>
when <inline-formula><tex-math notation="LaTeX" id="ImEquation942"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula> for fixed <inline-formula><tex-math notation="LaTeX" id="ImEquation943"><![CDATA[$0 < - \varepsilon \ll 1$]]></tex-math></inline-formula>, whenever <inline-formula><tex-math notation="LaTeX" id="ImEquation944"><![CDATA[$P_{ijk}$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation945"><![CDATA[$R_{ij}$]]></tex-math></inline-formula> are both real with <inline-formula><tex-math notation="LaTeX" id="ImEquation946"><![CDATA[$0 < - \, R_{ij}/P_{ijk} < 1$]]></tex-math></inline-formula>. Intuitively, this replacement is based on the fact that for any function <inline-formula><tex-math notation="LaTeX" id="ImEquation947"><![CDATA[$f(u)$]]></tex-math></inline-formula> analytic along <inline-formula><tex-math notation="LaTeX" id="ImEquation948"><![CDATA[$[0,1]$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation949"><![CDATA[$0 < u_{0} < 1$]]></tex-math></inline-formula>,
<disp-formula id="ptz160UM34"><tex-math notation="LaTeX" id="Equation280"><![CDATA[$$\int_{0}^{1} du \,
\frac{f(u)}{u - u_{0} - i \, \lambda}
-
\int_{0}^{1} du \,
\frac{f(u)}{u - u_{0} + i \, \lambda}
\to
2 i \, \pi \, f(u_{0})
\;\;
\mbox{when} \;\; \lambda \to 0^{+} ,$$]]></tex-math></disp-formula>
which vanishes if <inline-formula><tex-math notation="LaTeX" id="ImEquation950"><![CDATA[$u_{0}$]]></tex-math></inline-formula> is a zero of <inline-formula><tex-math notation="LaTeX" id="ImEquation951"><![CDATA[$f(u)$]]></tex-math></inline-formula>. However, the situation is made trickier when <inline-formula><tex-math notation="LaTeX" id="ImEquation952"><![CDATA[$f(u) = (u-u_{0})^{-\varepsilon}$]]></tex-math></inline-formula> and thus has a branch point at <inline-formula><tex-math notation="LaTeX" id="ImEquation953"><![CDATA[$u = u_{0}$]]></tex-math></inline-formula> and a cut running along part of the interval of integration. To make the above argument apply, one could think of shifting the branch point and cut by a contour prescription <inline-formula><tex-math notation="LaTeX" id="ImEquation954"><![CDATA[$- i \, a \lambda$]]></tex-math></inline-formula> with <inline-formula><tex-math notation="LaTeX" id="ImEquation955"><![CDATA[$a>1$]]></tex-math></inline-formula> so as to pass either above or below the pole while remaining on the same side of the cut. We would then get a residue value <inline-formula><tex-math notation="LaTeX" id="ImEquation956"><![CDATA[$\propto \lambda^{\; -\varepsilon}$]]></tex-math></inline-formula> vanishing in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation957"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula>, keeping <inline-formula><tex-math notation="LaTeX" id="ImEquation958"><![CDATA[$\varepsilon < 0$]]></tex-math></inline-formula> fixed. The &#x201C;hierarchised lambdalogy&#x201D; underpinning this disentanglement of pole from branch point may look awkward to the rigorous reader; let us therefore back up this hand-waving argument more rigorously as follows.</p>
<p>We first perform a partial fraction decomposition of the pole term:
<disp-formula id="ptz160M11-2"><label>(F.2)</label><tex-math notation="LaTeX" id="Equation281"><![CDATA[$$\begin{align}
&{\frac{1}{u^2 \, P_{ijk} + R_{ij} + i \, s \, \lambda}} \nonumber \\
& = 
\frac{1}{P_{ijk}} \, \frac{1}{2 \, \sqrt{- R_{ij}/P_{ijk}}} \,
\left[
\frac{1}{u - (u_{0} - i \, s \lambda^{\prime})}
-
\frac{1}{u + (u_{0} - i \, s \lambda^{\prime})}
\right]\!,
\label{subst2}
\end{align}$$]]></tex-math></disp-formula>
where <inline-formula><tex-math notation="LaTeX" id="ImEquation959"><![CDATA[$s = \pm$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation960"><![CDATA[$u_{0} = \sqrt{- R_{ij}/P_{ijk}}$]]></tex-math></inline-formula> is assumed in <inline-formula><tex-math notation="LaTeX" id="ImEquation961"><![CDATA[$]0,1[$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation962"><![CDATA[$\lambda^{\prime} = \lambda/(2 u_{0} P_{ijk})$]]></tex-math></inline-formula>.<sup><xref ref-type="fn" rid="FN19">19</xref></sup> We focus on the pole at <inline-formula><tex-math notation="LaTeX" id="ImEquation963"><![CDATA[$+ u_{0}$]]></tex-math></inline-formula> in the decomposition of Eq. (<xref ref-type="disp-formula" rid="ptz160M11-2">F.2</xref>): since the pole at <inline-formula><tex-math notation="LaTeX" id="ImEquation964"><![CDATA[$-u_{0}$]]></tex-math></inline-formula> involved in the second term lies outside the integration region, the contour prescription for this pole is irrelevant and so is the corresponding pole term in the discussion. We then study the legitimacy of the replacement
<disp-formula id="ptz160M11-3"><label>(F.3)</label><tex-math notation="LaTeX" id="Equation282"><![CDATA[$$\begin{equation}\label{subst3}
\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, a \lambda \right]^{-\varepsilon}}
{u - u_{0} + i \, \lambda^{\prime}
}
\to
\int_{0}^{1} du \,
\frac{\left[ u^2 \, P_{ijk} + R_{ij} - i \, a \lambda\right]^{-\varepsilon}}
{u - u_{0} - i \, \lambda^{\prime}}
\end{equation}$$]]></tex-math></disp-formula>
(<inline-formula><tex-math notation="LaTeX" id="ImEquation965"><![CDATA[$a$]]></tex-math></inline-formula> being positive yet kept arbitrary). We consider <inline-formula><tex-math notation="LaTeX" id="ImEquation966"><![CDATA[$\delta \equiv \text{``left-hand side minus right-hand side''}$]]></tex-math></inline-formula> of Eq. (<xref ref-type="disp-formula" rid="ptz160M11-3">F.3</xref>), which can be written as
<disp-formula id="ptz160UM35"><tex-math notation="LaTeX" id="Equation283"><![CDATA[$$\delta =
\int_{0}^{1} \frac{du \, ( 2 \, i \, \lambda^{\prime})}
{(u - u_{0})^{2} + \lambda^{\prime \, 2}}
\left[
P_{ijk} \,
\left( (u - u_{0})(u + u_{0}) - i \, (2 u_{0} a) \, \lambda^{\prime} \right)
\right]^{-\varepsilon} .$$]]></tex-math></disp-formula></p>
<p>We make the change of variable <inline-formula><tex-math notation="LaTeX" id="ImEquation967"><![CDATA[$(u - u_{0}) = |\lambda^{\prime}| \, v$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation968"><![CDATA[$\delta$]]></tex-math></inline-formula> reads
<disp-formula id="ptz160M11-4"><label>(F.4)</label><tex-math notation="LaTeX" id="Equation284"><![CDATA[$$\begin{equation}\label{subst4}
\delta
=
2 \, i \, \sigma \,
\left( \frac{\lambda}{2 u_{0}} \right) ^{- \varepsilon}
\int_{-u_{0}/|\lambda^{\prime}| }^{(1- u_{0})/|\lambda^{\prime}|}
\frac{dv}{v^2+1}
\left[
\sigma v \left( \sigma \lambda^{\prime} v \, + 2 u_{0} \right)
- i (2 u_{0} a)
\right]^{-\varepsilon}
\end{equation}$$]]></tex-math></disp-formula>
(where <inline-formula><tex-math notation="LaTeX" id="ImEquation969"><![CDATA[$\sigma = \mbox{sign}(\lambda^{\prime}) = \mbox{sign}(P_{ijk})$]]></tex-math></inline-formula>). For any <inline-formula><tex-math notation="LaTeX" id="ImEquation970"><![CDATA[$b > 1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation971"><![CDATA[$|\lambda^{\prime}|$]]></tex-math></inline-formula> small enough we have, for all real <inline-formula><tex-math notation="LaTeX" id="ImEquation972"><![CDATA[$v$]]></tex-math></inline-formula>,
<disp-formula id="ptz160UM36"><tex-math notation="LaTeX" id="Equation285"><![CDATA[$$\left|
\sigma v ( \sigma \lambda^{\prime} v \, + 2 u_{0}) - i (2 u_{0} a)
\right|
< \left[ v^4 + (2u_{0})^2 b \, v^2 + (2 u_{0} a)^2 \right] ^{1/2} ,$$]]></tex-math></disp-formula>
and the integral
<disp-formula id="ptz160UM37"><tex-math notation="LaTeX" id="Equation286"><![CDATA[$$\int_{- \infty}^{+\infty} dv \,
\frac{[ v^4 + (2u_{0})^2 b \, v^2 + (2 u_{0} a)^2]^{- \varepsilon/2}}{v^2+1}$$]]></tex-math></disp-formula>
is convergent when <inline-formula><tex-math notation="LaTeX" id="ImEquation973"><![CDATA[$-\varepsilon>0$]]></tex-math></inline-formula> is small enough. This provides an &#x201C;integrable hat&#x201D; for the application of Lebesgue&#x2019;s theorem of dominated convergence. When <inline-formula><tex-math notation="LaTeX" id="ImEquation974"><![CDATA[$\lambda \to 0^{+}$]]></tex-math></inline-formula> keeping <inline-formula><tex-math notation="LaTeX" id="ImEquation975"><![CDATA[$-\varepsilon>0$]]></tex-math></inline-formula> fixed and small enough, the integral in Eq. (<xref ref-type="disp-formula" rid="ptz160M11-4">F.4</xref>) has the limit
<disp-formula id="ptz160UM38"><tex-math notation="LaTeX" id="Equation287"><![CDATA[$${\cal L} = (2 u_{0})^{- \varepsilon}
\int_{-\infty}^{+\infty}
\frac{dv}{v^2+1} \left( \sigma v - i a \right)^{-\varepsilon} ,$$]]></tex-math></disp-formula>
which is finite regardless of <inline-formula><tex-math notation="LaTeX" id="ImEquation976"><![CDATA[$a > 0$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation977"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> small enough (its actual value, readily computable using the residue theorem, is irrelevant for the conclusion). We thus see that <inline-formula><tex-math notation="LaTeX" id="ImEquation978"><![CDATA[$\delta \sim {\cal O}(\lambda^{- \varepsilon})$]]></tex-math></inline-formula>, as anticipated.</p>
</sec>
</app>
</app-group>
<fn-group>
<title>Footnotes</title>
<fn id="FN1"><p><sup>1</sup> As in Refs. [<xref ref-type="bibr" rid="B1">1</xref>,<xref ref-type="bibr" rid="B2">2</xref>], we assume that the elements of the kinematic matrix <inline-formula><tex-math notation="LaTeX" id="ImEquation979"><![CDATA[${\cal S}$]]></tex-math></inline-formula> have been made dimensionless by an appropriate rescaling.</p></fn>
<fn id="FN2"><p><sup>2</sup> This decomposition has been discovered before and used for different purposes, see Refs. [<xref ref-type="bibr" rid="B9">9</xref>&#x2013;<xref ref-type="bibr" rid="B12">12</xref>].</p></fn>
<fn id="FN3"><p><sup>3</sup> Remember that <inline-formula><tex-math notation="LaTeX" id="ImEquation980"><![CDATA[${\cal S}_{jk} = {\cal S}^{\{i\}}_{jk}$]]></tex-math></inline-formula> for <inline-formula><tex-math notation="LaTeX" id="ImEquation981"><![CDATA[$j,k \ne i$]]></tex-math></inline-formula>.</p></fn>
<fn id="FN4"><p><sup>4</sup> Here and below, only the terms in the <inline-formula><tex-math notation="LaTeX" id="ImEquation982"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>-expansion providing the divergent and finite terms in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation983"><![CDATA[$\varepsilon \to 0$]]></tex-math></inline-formula> are kept.</p></fn>
<fn id="FN5"><p><sup>5</sup> With real masses, <inline-formula><tex-math notation="LaTeX" id="ImEquation984"><![CDATA[$x_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation985"><![CDATA[$x_2$]]></tex-math></inline-formula> in Eq. (<xref ref-type="disp-formula" rid="ptz160M2-10">2.10</xref>) have imaginary parts of opposite signs. This simplifies splittings and recombinations of logarithms of ratios in the explicit calculation of the function <inline-formula><tex-math notation="LaTeX" id="ImEquation986"><![CDATA[$J(x_1,x_2)$]]></tex-math></inline-formula> computed in Appendix <xref ref-type="sec" rid="SEC7">B</xref>.</p></fn>
<fn id="FN6"><p><sup>6</sup> In contrast to Eq. (<xref ref-type="disp-formula" rid="ptz160M2-44">2.44</xref>), Eq. (4.11) of Ref. [<xref ref-type="bibr" rid="B4">4</xref>] contains a factor <inline-formula><tex-math notation="LaTeX" id="ImEquation987"><![CDATA[$\Gamma^2(1-\varepsilon)/\Gamma(1-2\varepsilon)$]]></tex-math></inline-formula> and an extra term <inline-formula><tex-math notation="LaTeX" id="ImEquation988"><![CDATA[$+\pi^2/12$]]></tex-math></inline-formula> inside the brackets. These, however, cancel against each other when performing the <inline-formula><tex-math notation="LaTeX" id="ImEquation989"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>-expansion at the appropriate order.</p></fn>
<fn id="FN7"><p><sup>7</sup> Remember that only the terms in the <inline-formula><tex-math notation="LaTeX" id="ImEquation990"><![CDATA[$\varepsilon$]]></tex-math></inline-formula>-expansion which provide the divergent and finite terms in the limit <inline-formula><tex-math notation="LaTeX" id="ImEquation991"><![CDATA[$\varepsilon \to 0$]]></tex-math></inline-formula> are kept.</p></fn>
<fn id="FN8"><p><sup>8</sup> The integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation992"><![CDATA[$\sigma$]]></tex-math></inline-formula> is of the &#x201C;second kind,&#x201D; cf. Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>) and (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>), while the integration over <inline-formula><tex-math notation="LaTeX" id="ImEquation993"><![CDATA[$\rho$]]></tex-math></inline-formula> is of the &#x201C;first kind,&#x201D; cf. Eq. (<xref ref-type="disp-formula" rid="ptz160M6-1">A.1</xref>). For both integrations, the power <inline-formula><tex-math notation="LaTeX" id="ImEquation994"><![CDATA[$\nu$]]></tex-math></inline-formula> appearing in Eqs. (<xref ref-type="disp-formula" rid="ptz160M6-1">A.1</xref>), (<xref ref-type="disp-formula" rid="ptz160M6-6">A.6</xref>), and (<xref ref-type="disp-formula" rid="ptz160M6-7">A.7</xref>) is taken equal to <inline-formula><tex-math notation="LaTeX" id="ImEquation995"><![CDATA[$2$]]></tex-math></inline-formula>.</p></fn>
<fn id="FN9"><p><sup>9</sup> Similar truncations of <inline-formula><tex-math notation="LaTeX" id="ImEquation996"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> expansions will be performed everywhere throughout this section. From the perspective of computing generalized one-loop building blocks to be used in computations beyond one-loop, one might be led to keep further terms evanescent with <inline-formula><tex-math notation="LaTeX" id="ImEquation997"><![CDATA[$\varepsilon$]]></tex-math></inline-formula> to the appropriate order, whenever such terms would hit <inline-formula><tex-math notation="LaTeX" id="ImEquation998"><![CDATA[$1/\varepsilon$]]></tex-math></inline-formula> poles generated by the extra integrations, cf. the introduction of [<xref ref-type="bibr" rid="B1">1</xref>].</p></fn>
<fn id="FN10"><p><sup>10</sup> Remember that the elements of the <inline-formula><tex-math notation="LaTeX" id="ImEquation999"><![CDATA[${\cal S}$]]></tex-math></inline-formula> matrix are defined by <inline-formula><tex-math notation="LaTeX" id="ImEquation1000"><![CDATA[${\cal S}_{ij}=(q_i-q_j)^2-m_i^2-m_j^2$]]></tex-math></inline-formula>, where the internal momenta <inline-formula><tex-math notation="LaTeX" id="ImEquation1001"><![CDATA[$q_i$]]></tex-math></inline-formula> of the Feynman diagram depicted in <xref ref-type="fig" rid="F4">Fig. 4</xref> are such that <inline-formula><tex-math notation="LaTeX" id="ImEquation1002"><![CDATA[$q_2 - q_1 = p_2$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation1003"><![CDATA[$q_3 - q_2 = p_3$]]></tex-math></inline-formula>, <inline-formula><tex-math notation="LaTeX" id="ImEquation1004"><![CDATA[$q_4 - q_3 = p_4$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation1005"><![CDATA[$q_1 - q_4 = p_1$]]></tex-math></inline-formula>.</p></fn>
<fn id="FN11"><p><sup>11</sup> See Ref. [<xref ref-type="bibr" rid="B16">16</xref>], Chap. 1, pp. 2&#x2013;3 and Chap. 2, pp. 17&#x2013;18.</p></fn>
<fn id="FN12"><p><sup>12</sup> See, however, Ref. [<xref ref-type="bibr" rid="B17">17</xref>].</p></fn>
<fn id="FN13"><p><sup>13</sup> We will use the following properties: <inline-formula><tex-math notation="LaTeX" id="ImEquation1006"><![CDATA[$\det{(G^{\{i,j\}})}$]]></tex-math></inline-formula> is symmetric under the permutation <inline-formula><tex-math notation="LaTeX" id="ImEquation1007"><![CDATA[$i \leftrightarrow j$]]></tex-math></inline-formula>, and <inline-formula><tex-math notation="LaTeX" id="ImEquation1008"><![CDATA[$\widetilde{D}_{ijk}$]]></tex-math></inline-formula> is symmetric under any permutation of the set <inline-formula><tex-math notation="LaTeX" id="ImEquation1009"><![CDATA[$\{i,j,k\}$]]></tex-math></inline-formula>.</p></fn>
<fn id="FN14"><p><sup>14</sup> The subtlety discussed in Appendix E of Ref. [<xref ref-type="bibr" rid="B1">1</xref>] does not show up here because <inline-formula><tex-math notation="LaTeX" id="ImEquation1010"><![CDATA[$x_1$]]></tex-math></inline-formula> and <inline-formula><tex-math notation="LaTeX" id="ImEquation1011"><![CDATA[$x_2$]]></tex-math></inline-formula> have imaginary parts of opposite signs.</p></fn>
<fn id="FN15"><p><sup>15</sup> Contrary to example <inline-formula><tex-math notation="LaTeX" id="ImEquation1012"><![CDATA[$3$]]></tex-math></inline-formula> in Sect. <xref ref-type="sec" rid="SEC2.2">2.2</xref>, we choose to set <inline-formula><tex-math notation="LaTeX" id="ImEquation1013"><![CDATA[$m_3^2 = 0$]]></tex-math></inline-formula> because, for the purpose of this appendix, a non-vanishing mass does not bring anything new with respect to the previous case.</p></fn>
<fn id="FN16"><p><sup>16</sup> We follow the convention of Appendix C of Ref. [<xref ref-type="bibr" rid="B2">2</xref>].</p></fn>
<fn id="FN17"><p><sup>17</sup> One could be tempted here to recover the real mass case results by setting <inline-formula><tex-math notation="LaTeX" id="ImEquation1014"><![CDATA[$m^2_{\rm I} = - \lambda$]]></tex-math></inline-formula>. Doing that could lead to wrong formulae because, when deriving the complex mass case, we have already assumed that <inline-formula><tex-math notation="LaTeX" id="ImEquation1015"><![CDATA[$|m^2_{\rm I}| \gg \lambda$]]></tex-math></inline-formula> and so dropped some <inline-formula><tex-math notation="LaTeX" id="ImEquation1016"><![CDATA[$i \, \lambda$]]></tex-math></inline-formula> terms.</p></fn>
<fn id="FN18"><p><sup>18</sup> The subtlety discussed in Ref. [<xref ref-type="bibr" rid="B1">1</xref>] does not appear in this case.</p></fn>
<fn id="FN19"><p><sup>19</sup> We keep track of this multiplicative change to control various normalizations in the reasoning so as to check the independence of the conclusion with respect to any assumption of &#x201C;hierarchised lambdalogy.&#x201D;</p></fn>
</fn-group>
<ref-list id="ref1">
<title>References</title>
<ref id="B1"><label>[1]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Guillet</surname> <given-names>J. Ph.</given-names></string-name>, <string-name name-style="western"><surname>Pilon</surname> <given-names>E.</given-names></string-name>, <string-name name-style="western"><surname>Shimizu</surname> <given-names>Y.</given-names></string-name>, and <string-name name-style="western"><surname>Zidi</surname> <given-names>M. S.</given-names></string-name></person-group>, <source>Prog. Theor. Exp. Phys.</source> <volume>2019</volume>, <fpage>113B05</fpage> (<year>2019</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1093/ptep/ptz160">http://dx.doi.org/10.1093/ptep/ptz160</ext-link></comment>)</mixed-citation></ref>
<ref id="B2"><label>[2]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Guillet</surname> <given-names>J. Ph.</given-names></string-name>, <string-name name-style="western"><surname>Pilon</surname> <given-names>E.</given-names></string-name>, <string-name name-style="western"><surname>Shimizu</surname> <given-names>Y.</given-names></string-name>, and <string-name name-style="western"><surname>Zidi</surname> <given-names>M. S.</given-names></string-name></person-group>, <source>Prog. Theor. Exp. Phys.</source> <volume>2020</volume>, <fpage>023B04</fpage> (<year>2020</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1093/ptep/ptz160">http://dx.doi.org/10.1093/ptep/ptz160</ext-link></comment>)</mixed-citation></ref>
<ref id="B3"><label>[3]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Guillet</surname> <given-names>J. Ph.</given-names></string-name>, <string-name name-style="western"><surname>Pilon</surname> <given-names>E.</given-names></string-name>, <string-name name-style="western"><surname>Shimizu</surname> <given-names>Y.</given-names></string-name>, and <string-name name-style="western"><surname>Zidi</surname> <given-names>M. S.</given-names></string-name></person-group>, <ext-link ext-link-type="uri" xlink:href="http://arxiv.org/abs/1905.08115">arXiv:1905.08115</ext-link> [hep-ph] [<ext-link ext-link-type="uri" xlink:href="http://www.inspirehep.net/search?p=find+EPRINT+1905.08115">Search <sc>in</sc>SPIRE</ext-link>].</mixed-citation></ref>
<ref id="B4"><label>[4]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Ellis</surname> <given-names>R. K.</given-names></string-name> and <string-name name-style="western"><surname>Zanderighi</surname> <given-names>G.</given-names></string-name></person-group>, <source>J. High Energy Phys.</source> <volume>0802</volume>, <fpage>002</fpage> (<year>2008</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1088/1126-6708/2008/02/002">http://dx.doi.org/10.1088/1126-6708/2008/02/002</ext-link></comment>)</mixed-citation></ref>
<ref id="B5"><label>[5]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Beenakker</surname> <given-names>W.</given-names></string-name> and <string-name name-style="western"><surname>Denner</surname> <given-names>A.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>338</volume>, <fpage>349</fpage> (<year>1990</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/0550-3213(90)90636-R">http://dx.doi.org/10.1016/0550-3213(90)90636-R</ext-link></comment>)</mixed-citation></ref>
<ref id="B6"><label>[6]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Denner</surname> <given-names>A.</given-names></string-name> and <string-name name-style="western"><surname>Dittmaier</surname> <given-names>S.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>844</volume>, <fpage>199</fpage> (<year>2011</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/j.nuclphysb.2010.11.002">http://dx.doi.org/10.1016/j.nuclphysb.2010.11.002</ext-link></comment>)</mixed-citation></ref>
<ref id="B7"><label>[7]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Binoth</surname> <given-names>T.</given-names></string-name></person-group>, J.-Ph. Guillet, G. Heinrich, E. Pilon, and C. Schubert, <source>J. High Energy Phys.</source> <volume>0510</volume>, <fpage>015</fpage> (<year>2005</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1088/1126-6708/2005/10/015">http://dx.doi.org/10.1088/1126-6708/2005/10/015</ext-link></comment>)</mixed-citation></ref>
<ref id="B8"><label>[8]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Binoth</surname> <given-names>T.</given-names></string-name>, <string-name name-style="western"><surname>Guillet</surname> <given-names>J. Ph.</given-names></string-name>, and <string-name name-style="western"><surname>Heinrich</surname> <given-names>G.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>572</volume>, <fpage>361</fpage> (<year>2000</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/S0550-3213(00)00040-7">http://dx.doi.org/10.1016/S0550-3213(00)00040-7</ext-link></comment>)</mixed-citation></ref>
<ref id="B9"><label>[9]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>van Neerven</surname> <given-names>W. L.</given-names></string-name> and <string-name name-style="western"><surname>Vermaseren</surname> <given-names>J. A. M.</given-names></string-name></person-group>, <source>Phys. Lett. B</source> <volume>137</volume>, <fpage>241</fpage> (<year>1984</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/0370-2693(84)90237-5">http://dx.doi.org/10.1016/0370-2693(84)90237-5</ext-link></comment>)</mixed-citation></ref>
<ref id="B10"><label>[10]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Kotikov</surname> <given-names>A. V.</given-names></string-name></person-group>, <source>Phys. Lett. B</source> <volume>267</volume>, <fpage>123</fpage> (<year>1991</year>); <volume>295</volume>, <fpage>409</fpage> (<year>1992</year>) [erratum]. (<comment><ext-link ext-link-type="doi" xlink:href="https://doi.org/10.1016/0370-2693(91)90536-Y">https://doi.org/10.1016/0370-2693(91)90536-Y</ext-link>; <ext-link ext-link-type="doi" xlink:href="https://doi.org/10.1016/0370-2693(92)91582-T">https://doi.org/10.1016/0370-2693(92)91582-T</ext-link></comment>)</mixed-citation></ref>
<ref id="B11"><label>[11]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Bern</surname> <given-names>Z.</given-names></string-name>, <string-name name-style="western"><surname>Dixon</surname> <given-names>L.</given-names></string-name>, and <string-name name-style="western"><surname>Kosower</surname> <given-names>D. A.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>412</volume>, <fpage>751</fpage> (<year>1994</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/0550-3213(94)90398-0">http://dx.doi.org/10.1016/0550-3213(94)90398-0</ext-link></comment>)</mixed-citation></ref>
<ref id="B12"><label>[12]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Tarasov</surname> <given-names>O. V.</given-names></string-name></person-group>, <source>Phys. Rev. D</source> <volume>54</volume>, <fpage>6479</fpage> (<year>1996</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1103/PhysRevD.54.6479">http://dx.doi.org/10.1103/PhysRevD.54.6479</ext-link></comment>)</mixed-citation></ref>
<ref id="B13"><label>[13]</label><mixed-citation publication-type="book"><person-group person-group-type="author"><string-name name-style="western"><surname>Abramowitz</surname> <given-names>M.</given-names></string-name> and <string-name name-style="western"><surname>Stegun</surname> <given-names>I.</given-names></string-name></person-group>, <source>Handbook of Mathematical Functions</source> (<publisher-name>U.S. Government Printing Office</publisher-name>, <publisher-loc>Washington, DC</publisher-loc>, <year>1972</year>), <edition>10th ed</edition>.</mixed-citation></ref>
<ref id="B14"><label>[14]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Duplan&#x010D;i&#x0107;</surname> <given-names>G.</given-names></string-name> and <string-name name-style="western"><surname>Ni&#x017E;i&#x0107;</surname> <given-names>B.</given-names></string-name></person-group>, <source>Eur. Phys. J. C</source> <volume>20</volume>, <fpage>357</fpage> (<year>2001</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1007/s100520100675">http://dx.doi.org/10.1007/s100520100675</ext-link></comment>)</mixed-citation></ref>
<ref id="B15"><label>[15]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Brandhuber</surname> <given-names>A.</given-names></string-name>, <string-name name-style="western"><surname>Spence</surname> <given-names>B.</given-names></string-name>, and <string-name name-style="western"><surname>Travaglini</surname> <given-names>G.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>706</volume>, <fpage>150</fpage> (<year>2005</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/j.nuclphysb.2004.11.023">http://dx.doi.org/10.1016/j.nuclphysb.2004.11.023</ext-link></comment>)</mixed-citation></ref>
<ref id="B16"><label>[16]</label><mixed-citation publication-type="book"><person-group person-group-type="editor"><string-name name-style="western"><surname>Lewin</surname> <given-names>L.</given-names></string-name></person-group>, ed., <source>Structural Properties of Polylogarithms</source> (<publisher-name>American Mathematical Society</publisher-name>, <publisher-loc>Providence, RI</publisher-loc>, <year>1991</year>).</mixed-citation></ref>
<ref id="B17"><label>[17]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>van Oldenborgh</surname> <given-names>G. J.</given-names></string-name> and <string-name name-style="western"><surname>Vermaseren</surname> <given-names>J. A. M.</given-names></string-name></person-group>, <source>Z. Phys. C</source> <volume>46</volume>, <fpage>425</fpage> (<year>1990</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1007/BF01621031">http://dx.doi.org/10.1007/BF01621031</ext-link></comment>)</mixed-citation></ref>
<ref id="B18"><label>[18]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>Beenakker</surname> <given-names>W.</given-names></string-name>, <string-name name-style="western"><surname>Dittmaier</surname> <given-names>S.</given-names></string-name>, <string-name name-style="western"><surname>Kr&#x00E4;mer</surname> <given-names>M.</given-names></string-name>, <string-name name-style="western"><surname>Pl&#x00FC;mper</surname> <given-names>B.</given-names></string-name>, <string-name name-style="western"><surname>Spira</surname> <given-names>M.</given-names></string-name>, and <string-name name-style="western"><surname>Zerwas</surname> <given-names>P. M.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>653</volume>, <fpage>151</fpage> (<year>2003</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/S0550-3213(03)00044-0">http://dx.doi.org/10.1016/S0550-3213(03)00044-0</ext-link></comment>)</mixed-citation></ref>
<ref id="B19"><label>[19]</label><mixed-citation publication-type="journal"><person-group person-group-type="author"><string-name name-style="western"><surname>&#x2019;t Hooft</surname> <given-names>G.</given-names></string-name> and <string-name name-style="western"><surname>Veltman</surname> <given-names>M.</given-names></string-name></person-group>, <source>Nucl. Phys. B</source> <volume>153</volume>, <fpage>365</fpage> (<year>1979</year>). (<comment><ext-link ext-link-type="doi" xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="http://doi.org/10.1016/0550-3213(79)90605-9">http://dx.doi.org/10.1016/0550-3213(79)90605-9</ext-link></comment>)</mixed-citation></ref>
</ref-list>
</back>
</article>