# Calculation of the one loop box integral at Finite Temperature and Density

A. S. Khvorostukhin<EMAIL_ADDRESS>Joint Institute for Nuclear
Research, 141980 Dubna, Russia  Institute of Applied Physics, MD-2028
Kishineu, Moldova

###### Abstract

Calculation of hadronization, decay or scattering processes at non-zero
temperatures and densities within the Nambu-Jona-Lasinio-like models requires
some techniques for computation of Feynmann diagrams. Decomposition of Feynman
diagrams at the one loop level leads to the appearance of elementary integrals
with one, two, three, and four fermion lines. For example, evaluation of the
$\pi\pi$ scattering amplitude requires calculating a box diagram with four
fermion lines. In this work, the real and imaginary parts of the box integral
at the one loop level are provided in the form suitable for numerical
evaluation. The obtained expressions are applicable to any value of
temperature, particle mass, particle momentum, and chemical potential. Among
the expressions for the box integral, the general formulas for the integral
with an arbitrary number of lines were derived for the case with zero or
collinear fermion momenta.

NJL model, low energy $\pi\pi$ scattering.

###### pacs:

12.38.Mh,12.39.-x,05.90.+m,05.70.Ce

##  I Introduction

Due to the complexity of the QCD Lagrangian and the nonperturbativity of QCD
at energies relevant for the deconfinement/chiral phase transition, effective
theories of strongly interacting matter at non-zero temperatures/densities are
under intensive study, see, e.g. HatsudaKunihiroChiral ; FuLiu2009 ; DBK2013 ;
Moeller2013  and references cited therein.

Extension of QCD Lagrangian-inspired effective models such as the Nambu-Jona-
Lasinio (NJL) one NJLoriginal ; KlevanskyNJL1992 ; Huefner1995 ; Zhuang1995 ;
Huefner1994 ; Rehberg1995  to finite temperatures/densities is performed by
the imaginary time Matsubara formalism Matsubara . The description of the
hadronisation processes starts with Feynman diagrams, which may contain two (a
polarization loop), three or four (a box diagram) fermion lines, and the
obtained amplitudes at the one loop level are the integrals with
$\prod(q_{i}^{2}-m_{i}^{2})^{-1}$. For calculations at zero temperature, it
was shown PassarinoVeltman  that all occurring transition amplitudes at the
one loop level can be decomposed into a number of elementary integrals, which
in turn can be classified by the number of particle lines they contain. This
decomposition technique can be generalized to the Matsubara formalism
Klevansky . The authors of the mentioned paper gave a solution of this task
for diagrams with one, two, and three fermion lines, explicitly displaying
their complex nature.

However, the study of meson scattering processes (for example, $\pi\pi$
scattering) requires the calculation of a box diagram with four fermion lines
FuLiu2009 ; JRS2002 . In the previous works, the amplitude for $\pi\pi$
scattering was calculated only in special kinematics with $s=4M_{\pi}^{2}$ and
$t=u=0$ due to the complexity of calculations in the general case. The aim of
this paper is to solve technical problems associated with such integrals
following the techniques developed in Klevansky  and to give researchers, who
are active in this field, the knowledge sufficient to calculate such
functions. Expressions for both real and imaginary parts will be given in the
form suitable for numerical evaluation. Some problems of applied techniques
will be discussed. The important point of the calculations should be
underlined. Though the integrals with three and more fermion lines are
convergent and do not have to be regularized, the analysis was performed in a
three momentum cutoff regularization scheme with $|\mathbf{p}|<\Lambda$, see
the explanation in Klevansky ; JRS2002 .

The paper is organized as follows. Section II shows how the original integral
can be represented as a sum of elementary integrals. Section III is devoted to
simplification of the result Klevansky  for the three line function; the
problem of integrals with three and greater lines is mentioned. After that,
the results for four line integral are given depending on configurations of
particle momenta. The simplest cases when all three momenta are collinear or
equal to zero are discussed in Sec. IV. More complicated “planar”
configurations are considered in Sec. V. The main original result of the paper
is formulated in Sec. VI where the expression for the most general case of
independent momenta is given. Finally, results are summarized in Conclusions.
Some technical details are eliminated to Appendices.

##  II Decomposition to elementary integrals

In the general case, the integral with $L+1$ fermion lines has the form
Klevansky :

$\displaystyle
Z_{0}^{(L)}\left(\mathcal{X}_{1}^{+},\ldots,\mathcal{X}_{L}^{+};\mathcal{T}\right)=\frac{16\pi^{2}}{\beta}\lim_{\eta\rightarrow
0}\sum_{n}e^{i\omega_{n}\eta}\int\frac{\mathrm{d}\mathbf{p}}{(2\pi)^{3}}\frac{1}{(i\omega_{n}+\mu)^{2}-E^{2}}\prod_{l=1}^{L}\frac{1}{(i\omega_{n}+\mu-\lambda_{l})^{2}-\widetilde{E}_{l}^{2}(\mathbf{k}_{l})}\,,$
(1)

where the arguments were written as a collection of ordered sets (vectors)

$\displaystyle\mathcal{X}_{l}^{\pm}$
$\displaystyle=(\pm\lambda_{l},\mathbf{k}_{l},m_{l}),\quad\mathcal{T}=(T,\mu,m).$
(2)

Here $T$ is the temperature, $\beta=1/T$, $m$ and $m_{l}$ denote the masses of
fermions, $\mu$ and $\mu_{l}$ are the fermion chemical potentials,
$\lambda_{l}$ is defined as $\lambda_{l}=\mu-\mu_{l}+i\nu_{j_{l}}$, and
$\mathbf{k}_{l}$ are the 3-momenta. All considered integrals are integrated
for $p=|\mathbf{p}|\leq\Lambda$, i.e. $\Lambda$ is the 3-momentum cutoff, and

$\displaystyle E$
$\displaystyle=\sqrt{p^{2}+m^{2}},\quad\widetilde{E}_{l}(\mathbf{k})=\sqrt{(\mathbf{p}-\mathbf{k})^{2}+m_{l}^{2}}\,.$
(3)

The Matsubara summation is carried out over the bosonic $i\nu_{j_{l}}=2\pi
j_{l}T$ and fermionic $\omega_{n}=(2n+1)\,\pi\,T$ frequencies. After the
Matsubara summation over $n$ is carried out, the complex bosonic frequencies
are analytically continued to their values on the real axis and become the
zeroth components associated with the corresponding 3-momentum,
$i\nu_{j_{l}}\rightarrow k_{l}^{0}$. The limit $\eta\rightarrow 0$ has to be
taken after the Matsubara summation.

The Matsubara summation in Eq. (1) can be easily performed Klevansky ; MatsSum
. There are $2(L+1)$ poles in Eq. (1)

$\displaystyle i\omega_{n}$ $\displaystyle=-\mu\pm E,\quad
i\omega_{n}=\lambda_{l}-\mu\pm\widetilde{E}_{l}(\mathbf{k}),$ (4)

and after the summation over $n$, which is performed according to the rule

$T\sum_{n}\frac{1}{i\omega_{n}\pm z}=\frac{1}{e^{\pm\beta z}+1}=f(\pm z)$ (5)

with the Fermi-Dirac distribution function $f(\pm z)$, one obtains

$\displaystyle
Z_{0}^{(L)}\left(\mathcal{X}_{1}^{+},\ldots,\mathcal{X}_{L}^{+};\mathcal{T}\right)=16\pi^{2}$
$\displaystyle\left[\int\frac{\mathrm{d}\mathbf{p}}{(2\pi)^{3}}\frac{f(E-\mu)}{2E}\prod_{l=1}^{L}\frac{1}{(E-\lambda_{l})^{2}-\widetilde{E}_{l}^{2}(\mathbf{k}_{l})}\right.$
$\displaystyle-\int\frac{\mathrm{d}\mathbf{p}}{(2\pi)^{3}}\frac{f(-E-\mu)}{2E}\prod_{l=1}^{L}\frac{1}{(E+\lambda_{l})^{2}-\widetilde{E}_{l}^{2}(\mathbf{k}_{l})}$
$\displaystyle+\sum_{s=1}^{L}\int\frac{\mathrm{d}\mathbf{p}}{(2\pi)^{3}}\frac{f(E_{s}-\mu_{s})}{2E_{s}}\frac{1}{(E_{s}+\lambda_{s})^{2}-\widetilde{E}^{2}(\mathbf{k}_{s})}\prod_{l=1,l\neq
s}^{L}\frac{1}{(E_{s}+\lambda_{sl})^{2}-\widetilde{E}_{l}^{2}(\mathbf{k}_{s}-\mathbf{k}_{l})}$
$\displaystyle\left.-\sum_{s=1}^{L}\int\frac{\mathrm{d}\mathbf{p}}{(2\pi)^{3}}\frac{f(-E_{s}-\mu_{s})}{2E_{s}}\frac{1}{(E_{s}-\lambda_{s})^{2}-\widetilde{E}^{2}(\mathbf{k}_{s})}\prod_{l=1,l\neq
s}^{L}\frac{1}{(E_{s}-\lambda_{sl})^{2}-\widetilde{E}_{l}^{2}(\mathbf{k}_{s}-\mathbf{k}_{l})}\right].$
(6)

Above, the substitution $\mathbf{p}\rightarrow\mathbf{k}_{s}-\mathbf{p}$ was
made in third and forth lines, and111Here and everywhere below $l\neq s$ if we
have two indices in the expression.

$\displaystyle\lambda_{ls}$
$\displaystyle=\lambda_{l}-\lambda_{s}=\mu_{s}-\mu_{l}+i\nu_{j_{l}}-i\nu_{j_{s}},\quad
l\neq s,$ (7) $\displaystyle\widetilde{E}(\mathbf{k})$
$\displaystyle=\sqrt{(\mathbf{p}-\mathbf{k})^{2}+m^{2}},\quad
E_{l}=\sqrt{p^{2}+m_{l}^{2}}.$ (8)

So we see that the original integral can be written as a decomposition

$\displaystyle
Z_{0}^{(L)}\left(\mathcal{X}_{1}^{+},\ldots,\mathcal{X}_{L}^{+};\mathcal{T}\right)$
$\displaystyle=\mathcal{Z}^{+}_{L}\left(\mathcal{X}^{-}_{1},\ldots,\mathcal{X}_{L}^{-};\mathcal{T}\right)-\mathcal{Z}^{-}_{L}\left(\mathcal{X}^{+}_{1},\ldots,\mathcal{X}_{L}^{+};\mathcal{T}\right)$
$\displaystyle+\sum_{s=1}^{L}\left[\mathcal{Z}^{+}_{L}\left(\mathcal{X}^{+}_{s1},\ldots,\mathcal{X}^{+}_{sL};\mathcal{T}_{s}\right)-\mathcal{Z}^{-}_{L}\left(\mathcal{X}^{-}_{s1},\ldots,\mathcal{X}^{-}_{sL};\mathcal{T}_{s}\right)\right]$
(9)

with

$\displaystyle\mathcal{T}_{s}$
$\displaystyle=(T,\mu_{s},m_{s}),\quad\mathcal{X}_{ss}^{\pm}=(\pm\lambda_{s},\mathbf{k}_{s},m),\quad\mathcal{X}_{sl}^{\pm}=(\pm\lambda_{sl},\mathbf{k}_{s}-\mathbf{k}_{l},m_{l}).$
(10)

In the spherical coordinates, the elementary integral is given by

$\displaystyle\mathcal{Z}^{\pm}_{L}\left(\mathcal{X}^{+}_{1},\ldots,\mathcal{X}_{L}^{+};\mathcal{T}\right)$
$\displaystyle=\frac{1}{2^{L}\pi}\lim_{\epsilon\rightarrow
0}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,p\,f(\pm
E-\mu)\int\limits_{-1}^{1}{\mathrm{d}}x\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\prod_{l=1}^{L}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+\mathbf{p}\mathbf{k}_{l}}$
$\displaystyle={\rm Re}\mathcal{Z}^{\pm}_{L}+i\,{\rm
Im}\mathcal{Z}^{\pm}_{L},$ (11)

where
$\mathbf{p}=p\,\left(\cos\phi\sqrt{1-x^{2}},\sin\phi\sqrt{1-x^{2}},x\right)$,

$\displaystyle\varepsilon_{l}$
$\displaystyle\equiv\varepsilon(\mathcal{X}_{l}^{+},m)=\varepsilon(\mathcal{X}_{l}^{-},m)=\frac{k_{l}^{2}+m_{l}^{2}-\lambda_{l}^{2}-m^{2}}{2},\quad
k_{l}=|\mathbf{k}_{l}|,$ (12) $\displaystyle E_{0l}$ $\displaystyle\equiv
E_{0}(\varepsilon_{l},\lambda_{l})=\frac{\varepsilon_{l}}{\lambda_{l}}\,,$
(13)

and

$\Lambda_{E}=\sqrt{\Lambda^{2}+m^{2}}.$

So we have reduced the task to evaluating functions of the same kind . For
further consideration, we need to write the components of $\mathbf{k}_{l}$:

$\displaystyle\mathbf{k}_{l}=k_{l}(\sin\delta_{l}\cos\phi_{l},\sin\delta_{l}\sin\phi_{l},\cos\delta_{l}).$
(14)

Let us also introduce

$\displaystyle\cos\psi_{ls}$
$\displaystyle=\frac{\mathbf{k}_{l}\mathbf{k}_{s}}{k_{l}k_{s}},\quad\sin\psi_{ls}=\left|\frac{\mathbf{k}_{l}\times\mathbf{k}_{s}}{k_{l}k_{s}}\right|,$
(15)

where $\mathbf{a}\times\mathbf{b}$ and $\mathbf{a}\mathbf{b}$ denote the
vector and scalar products of two 3-vectors, respectively. As one can easily
see,

$\displaystyle\cos\psi_{ls}$
$\displaystyle=\cos\delta_{l}\cos\delta_{s}+\sin\delta_{l}\sin\delta_{s}\cos(\phi_{l}-\phi_{s}).$
(16)

One should keep in mind that $\delta_{l},\psi_{ls}\in[0,\pi]$ and so we always
have $\sin\delta_{l},\,\sin\psi_{ls}\geq 0$. Also, without loss of generality,
one can choose

$\displaystyle\mathbf{k}_{1}=k_{1}(0,0,1),\quad\mathbf{k}_{2}=k_{2}(\sin\delta_{2},0,\cos\delta_{2}).$
(17)

At the end of this section, let us discuss how Eq. (II) can be converted to
integrals of real functions. The method is based on applying the famous
formula

$\displaystyle\lim_{\epsilon\rightarrow
0}\frac{1}{x-i\epsilon}=\mathcal{P}\frac{1}{x}+i\pi\delta(x)$

which can be generalized as

$\displaystyle\lim_{\epsilon\rightarrow
0}\frac{1}{x-i\lambda\epsilon}=\mathcal{P}\frac{1}{x}+i\pi\,{\rm
sgn}(\lambda)\,\delta(x),\quad\lambda\neq 0.$ (18)

Writing the product under the integral (II) as

$\displaystyle\prod_{l=1}^{L}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+\mathbf{p}\mathbf{k}_{l}}$
$\displaystyle=\prod_{l=1}^{L}\left[\mathcal{P}\frac{1}{\lambda_{l}(E-E_{0l})+\mathbf{p}\mathbf{k}_{l}}+i\pi\,{\rm
sgn}\lambda_{l}\,\delta\left(\lambda_{l}(E-E_{0l})+\mathbf{p}\mathbf{k}_{l}\right)\right]$
(19)

and expanding it, we obtain separate real and imaginary parts.

##  III Correction and generalization of the results Klevansky

Before calculating $\mathcal{Z}^{\pm}_{3}$, we would like make some remarks
related to the results for $L=2$ obtained in Klevansky , see Eq. (5.37). The
authors derived the formulae

$\displaystyle\mathcal{P}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z+x\cos\psi)\,\Theta\\!\left(\Delta(x;z,\psi)\right)}{(x+y)\sqrt{\Delta(x;z,\psi)}}$
$\displaystyle=\frac{\Xi(y,z,\psi)\,\Theta\\!\left(\Delta_{0}(y,z,\psi)\right)+\mathscr{G}(y,z,\psi)\,\Theta\\!\left(-\Delta_{0}(y,z,\psi)\right)}{\sqrt{|\Delta_{0}(y,z,\psi)|}}\,,\quad\sin\psi>0,$
(20)

where

$\displaystyle\Delta(y,z,\psi)$
$\displaystyle=y^{2}+z^{2}+2\,yz\cos\psi-\sin^{2}\psi$
$\displaystyle=(y+z\cos\psi)^{2}-\sin^{2}\psi(1-z^{2})=(z+y\cos\psi)^{2}-\sin^{2}\psi(1-y^{2})$
$\displaystyle=(\cos\psi+yz)^{2}-(1-y^{2})(1-z^{2}),$ (21)

$\Delta_{0}(y,z,\psi)=\Delta(-y;z,\psi)$, and

$\displaystyle\mathscr{G}(y,z,\psi)$ $\displaystyle\triangleq{\rm
sgn}\left(z+\cos\psi\right)\arccos\frac{(1+y)(y-z\cos\psi)-\Delta_{0}(y,z,\psi)}{(1+y)\sin\psi\sqrt{1-z^{2}}}$
$\displaystyle\quad-{\rm
sgn}\left(z-\cos\psi\right)\arccos\frac{-(1-y)(y-z\cos\psi)-\Delta_{0}(y,z,\psi)}{(1-y)\sin\psi\sqrt{1-z^{2}}},\quad\Delta_{0}(y,z,\psi)<0,$
(22) $\displaystyle\Xi(y,z,\psi)$ $\displaystyle\triangleq{\rm
sgn}\left(z+\cos\psi\right)F(1;y,z,\psi)-{\rm
sgn}\left(z-\cos\psi\right)F(-1;y,z,\psi)$
$\displaystyle+\Theta\\!\left(1-z^{2}\right)\,\Big{[}{\rm
sgn}\left(z-\cos\psi\right)F(x^{-};y,z,\psi)-{\rm
sgn}\left(z+\cos\psi\right)F(x^{+};y,z,\psi)\Big{]}$ (23)

with222The function $F(x;y,z,\psi)$ is defined up to an additive constant.

$\displaystyle F(x;y,z,\psi)$
$\displaystyle=\pm\ln\left|\frac{(x+y)(y-z\cos\psi)-\Delta_{0}(y,z,\psi)\pm\sqrt{\Delta_{0}(y,z,\psi)\Delta(x;z,\psi)}}{x+y}\right|,\quad\Delta_{0}(y,z,\psi)>0.$
(24)

Above $x^{\pm}$ are the roots of the quadratic trinomial $\Delta(x;z,\psi)$,

$\displaystyle x^{\pm}$ $\displaystyle\equiv
x^{\pm}(z,\psi)=-z\,\cos\psi\pm\sin\psi\sqrt{1-z^{2}},\quad z^{2}\leq 1.$ (25)

Since further we will often meet the functions $\Xi(y,z,\psi)$ and
$\mathscr{G}(y,z,\psi)$, let us consider them in more detail. The main
disadvantage of Eqs. (III) and (III) is that they are not explicitly symmetric
with respect to the transposal $y\leftrightarrow z$. The mentioned symmetry
follows from Eq. (II), see Klevansky  and below. The original expressions for
both functions are also too complicated. Fortunately, we can simplify both
functions by restoring explicit symmetry. First of all, mentioning that

${\rm sgn}(z\pm\cos\psi)={\rm
sgn}(z)\Theta\\!\left(z^{2}-\cos^{2}\psi\right)\pm{\rm
sgn}(\cos\psi)\Theta\\!\left(\cos^{2}\psi-z^{2}\right),$

and using Eqs. (119), one can show

$\displaystyle\mathscr{G}(y,z,\psi)$ $\displaystyle={\rm sgn}(\cos\psi-
yz)\arccos\frac{\left|\Delta_{0}(y,z,\psi)\right|-(\cos\psi-
yz)^{2}}{(1-y^{2})(1-z^{2})}=2\arctan\frac{\cos\psi-
yz}{\sqrt{\left|\Delta_{0}(y,z,\psi)\right|}}\,.$ (26)

To simplify the $\Xi$-function, it is more convenient to choose

$\displaystyle F(x;y,z,\psi)$
$\displaystyle=\frac{1}{2}\,\ln\left|\frac{\Delta_{0}(y,z,\psi)-(y-z\cos\psi)(x+y)-\sqrt{\Delta_{0}(y,z,\psi)\Delta(x;z,\psi)}}{\Delta_{0}(y,z,\psi)-(y-z\cos\psi)(x+y)+\sqrt{\Delta_{0}(y,z,\psi)\Delta(x;z,\psi)}}\right|.$
(27)

Using $F(x^{\pm};y,z,\psi)=0$ after some transformations, one can prove

$\displaystyle\Xi(y,z,\psi)$ $\displaystyle=\ln\left|\frac{\cos\psi-
yz-\sqrt{\Delta_{0}(y,z,\psi)}}{\cos\psi-
yz+\sqrt{\Delta_{0}(y,z,\psi)}}\right|\,.$ (28)

Application of Eq. (20) to calculate ${\rm Re}\,\mathcal{Z}^{\pm}_{2}$ when
the spherical coordinates are chosen as (17) and all $k_{l}>0$, immediately
gives

$\displaystyle\mathcal{P}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{2}+x\cos\delta_{2})\,\Theta\\!\left(\Delta_{2}(x)\right)}{(x+z_{1})\sqrt{\Delta_{2}(x)}}$
$\displaystyle=\frac{\Xi(z_{1},z_{2},\delta_{2})\,\Theta\\!\left(\Delta_{0}(z_{1},z_{2},\delta_{2})\right)+\mathscr{G}(z_{1},z_{2},\delta_{2})\,\Theta\\!\left(-\Delta_{0}(z_{1},z_{2},\delta_{2})\right)}{\sqrt{|\Delta_{0}(z_{1},z_{2},\delta_{2})|}}\,,$

where $\Delta_{l}(x)=\Delta(x;z_{l},\delta_{l})$ and

$\displaystyle z_{l}$
$\displaystyle=\frac{\lambda_{l}E-\varepsilon_{l}}{p\,k_{l}}\,.$ (29)

For further application, it is better to rewrite this results as

$\displaystyle\mathcal{P}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{s}+x\cos\delta_{s})\,\Theta\\!\left(\Delta_{s}(x)\right)}{(x+z_{l})\sqrt{\Delta_{s}(x)}}$
$\displaystyle=\frac{\Xi_{ls}(E)\,\Theta\\!\left(\Delta_{ls}\right)+\mathscr{G}_{ls}(E)\,\Theta\\!\left(-\Delta_{ls}\right)}{\sqrt{|\Delta_{ls}|}}\,,$
(30)

where
$\Delta_{ls}=\Delta_{0}(z_{l},z_{s},\psi_{ls})=\Delta_{l}(-z_{s})=\Delta_{s}(-z_{l})$,
$\Xi_{ls}(E)=\Xi(z_{l},z_{s},\psi_{ls})$, and
$\mathscr{G}_{ls}(E)=\mathscr{G}(z_{l},z_{s},\psi_{ls})$.

Equation (30) seems like a grateful expression. However, to calculate the box
integral, we need to rewrite it for any coordinate system and an unwaited
problem appears. Let us show it using, as an example, the calculation of ${\rm
Re}\,\mathcal{Z}^{\pm}_{2}$ for two choices of spherical coordinates. We
remind that generally we start with the double integral

$\displaystyle\frac{1}{2\pi}\,\mathcal{P}\int\limits_{-1}^{1}{\mathrm{d}}x\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\,\frac{1}{z_{1}+x\cos\delta_{1}+\sqrt{1-x^{2}}\sin\delta_{1}(\sin\phi\sin\phi_{1}+\cos\phi\cos\phi_{1})}$
$\displaystyle\quad\times\frac{1}{z_{2}+x\cos\delta_{2}+\sqrt{1-x^{2}}\sin\delta_{2}(\sin\phi\sin\phi_{2}+\cos\phi\cos\phi_{2})}\,,$
(31)

where $k_{1,2}>0$. If we choose the spherical coordinates (17), i.e.
$\delta_{1}=0$ and $\phi_{2}=0$, after integration over $\phi$ with the help
of Klevansky

$\displaystyle\int\limits_{0}^{2\pi}\,\frac{{\mathrm{d}}\phi}{a+b\cos\phi}$
$\displaystyle=\frac{2\pi}{\sqrt{|a^{2}-b^{2}|}}\,{\rm
sgn}(a)\,\Theta\\!\left(a^{2}-b^{2}\right)\,,$ (32)

we obtain l.h.s. of Eq. (20) with $z=z_{2}$, $y=z_{1}$, and
$\psi=\delta_{2}=\psi_{12}$.

However, we can use other spherical coordinates demanding that
$\phi_{1}=\phi_{2}$ while $\sin\delta_{1,2}>0$. Then

$\displaystyle\frac{1}{2\pi}\,\mathcal{P}\int\limits_{-1}^{1}{\mathrm{d}}x\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\,\frac{1}{z_{1}+x\cos\delta_{1}+\sqrt{1-x^{2}}\sin\delta_{1}(\sin\phi\sin\phi_{1}+\cos\phi\cos\phi_{1})}$
$\displaystyle\quad\times\frac{1}{z_{2}+x\cos\delta_{2}+\sqrt{1-x^{2}}\sin\delta_{2}(\sin\phi\sin\phi_{1}+\cos\phi\cos\phi_{1})}$
$\displaystyle=\sin\delta_{1}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{1}+x\cos\delta_{1})\,\Theta\\!\left(\Delta_{1}(x)\right)}{(\beta_{12}+\alpha_{12}x)\sqrt{\Delta_{1}(x)}}-\sin\delta_{2}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{2}+x\cos\delta_{2})\,\Theta\\!\left(\Delta_{2}(x)\right)}{(\beta_{12}+\alpha_{12}x)\sqrt{\Delta_{2}(x)}}\,.$
(33)

Above we denote333We give general definition for any choice of spherical
coordinates.

$\displaystyle\alpha_{ls}$
$\displaystyle=\cos\delta_{s}\sin\delta_{l}-\cos\delta_{l}\sin\delta_{s}\cos(\phi_{l}-\phi_{s}),$
(34) $\displaystyle\beta_{ls}$
$\displaystyle=z_{s}\sin\delta_{l}-z_{l}\sin\delta_{s}\cos(\phi_{l}-\phi_{s}),$
(35)

and use the formula

$\displaystyle\int\limits_{0}^{2\pi}\,\frac{{\mathrm{d}}\phi}{a+b\sin\phi+c\cos\phi}$
$\displaystyle=\frac{2\pi}{\sqrt{|a^{2}-b^{2}-c^{2}|}}\,{\rm
sgn}(a)\,\Theta\\!\left(a^{2}-b^{2}-c^{2}\right)$ (36)

which is a generalization of Eq. (32). The proof can be obtained with the help
of, e.g., Gradshteyn  in the same way as for Eq. (32).

Applying Eq. (20) to every term separately and mentioning that
$\alpha_{12}^{2}=\sin^{2}\psi_{12}$,

$\Delta_{1}\left(-\frac{\beta_{12}}{\alpha_{12}}\right)=\frac{\sin^{2}\delta_{1}}{\sin^{2}\psi_{12}}\,\Delta_{12},\quad\Delta_{2}\left(-\frac{\beta_{12}}{\alpha_{12}}\right)=\frac{\sin^{2}\delta_{2}}{\sin^{2}\psi_{12}}\,\Delta_{12},$

we obtain

$\displaystyle\sin\delta_{1}$
$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{1}+x\cos\delta_{1})\,\Theta\\!\left(\Delta_{1}(x)\right)}{(\beta_{12}+\alpha_{12}x)\sqrt{\Delta_{1}(x)}}-\sin\delta_{2}\int\limits_{-1}^{1}{\mathrm{d}}x\frac{{\rm
sgn}(z_{2}+x\cos\delta_{2})\,\Theta\\!\left(\Delta_{2}(x)\right)}{(\beta_{12}+\alpha_{12}x)\sqrt{\Delta_{2}(x)}}$
(37) $\displaystyle=$
$\displaystyle\frac{\sin\psi_{12}}{\alpha_{12}}\Bigg{[}\frac{\Xi(\beta_{12}/\alpha_{12},z_{1},\delta_{1})\,\Theta\\!\left(\Delta_{12}\right)+\mathscr{G}(\beta_{12}/\alpha_{12},z_{1},\delta_{1})\,\Theta\\!\left(-\Delta_{12}\right)}{\sqrt{|\Delta_{12}|}}$
$\displaystyle-\frac{\Xi(\beta_{12}/\alpha_{12},z_{2},\delta_{2})\,\Theta\\!\left(\Delta_{12}\right)+\mathscr{G}(\beta_{12}/\alpha_{12},z_{2},\delta_{2})\,\Theta\\!\left(-\Delta_{12}\right)}{\sqrt{|\Delta_{12}|}}\Bigg{]}$
$\displaystyle=$
$\displaystyle\frac{\Xi_{12}(E)\,\Theta\\!\left(\Delta_{12}\right)+[\mathscr{G}_{12}(E)-\pi]\,\Theta\\!\left(-\Delta_{12}\right)}{\sqrt{|\Delta_{12}|}}$

which contains the additional $\pi$-term.

The mentioned inconsistency is kept, as we will see below, also for the box
integral.

##  IV The $Z_{3}^{*}$ for collinear momenta

Let us now start the detailed evaluation of $\mathcal{Z}^{\pm}_{3}$ for all
possible orientations of momenta $\mathbf{k}_{1,2,3}$. As one can easily
check, up to permutation of $\mathbf{k}_{1,2,3}$ we have to consider seven
opportunities:

  1. 1.

$k_{1}=k_{2}=k_{3}=0$;

  2. 2.

$k_{1}=k_{2}=0$, $k_{3}>0$;

  3. 3.

$k_{l}>0$, $\mathbf{k}_{l}\times\mathbf{k}_{s}=0$;

  4. 4.

$k_{1,2}>0$, $k_{3}=0$, $\mathbf{k}_{1}\times\mathbf{k}_{2}=0$;

  5. 5.

$k_{1,2}>0$, $k_{3}=0$, $\mathbf{k}_{1}\times\mathbf{k}_{2}\neq 0$;

  6. 6.

$k_{l}>0$, $\mathbf{k}_{l}\times\mathbf{k}_{s}\neq 0$,
$\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3})=0$;

  7. 7.

$k_{l}>0$, $\mathbf{k}_{l}\times\mathbf{k}_{s}\neq 0$,
$\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3})\neq 0$.

The first four cases corresponding to “dot” and “line” configurations are very
simple. For this reason, it is more convenient to derive a general expression
for $\mathcal{Z}_{L}^{\pm}$ and then to apply it to $L=3$. These cases will be
considered below in this Section. Planar configurations (items 5 and 6) when
all three vectors $\mathbf{k}_{1}$, $\mathbf{k}_{2}$, and $\mathbf{k}_{3}$ lie
in the same plane or, more precisely, when
$\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3})=0$ are considered in the
next section. The most complicated situation for three independent vectors is
analyzed separately in Sec. VI.

###  IV.1 Calculation of $\mathcal{Z}_{L}^{\pm}$ when all $k_{l}=0$

The simplest situation is when all $k_{l}$ are equal to zero. Equation (II) is
strongly simplified and takes the form

$\displaystyle\mathcal{Z}^{\pm}_{L}(\mathcal{X}_{1}^{+},\ldots,\mathcal{X}_{L}^{+};\mathcal{T})$
$\displaystyle=\frac{1}{\rho_{L}}\lim_{\epsilon\rightarrow
0}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,G^{\pm}(E;\mathcal{T})\prod_{l=1}^{L}\frac{1}{E-E_{0l}-i\epsilon}$
(38)

with

$\displaystyle\rho_{L}$ $\displaystyle=2^{L-2}\prod_{l=1}^{L}\lambda_{l}$ (39)

and the functions

$\displaystyle G^{\pm}(E;\mathcal{T})$
$\displaystyle=p\,F^{\pm}(E,\mathcal{T}),\quad F^{\pm}(E;\mathcal{T})=f(\pm
E-\mu;T)\,\Theta\\!\left(E-m\right)\,\Theta\\!\left(\Lambda_{E}-E\right).$
(40)

To calculate the integral, we need to translate the product into a sum. Let us
here remind the useful relation Gradshteyn

$\displaystyle\frac{\varphi(x)}{f(x)}$
$\displaystyle=\sum_{l=1}^{L}\frac{c^{\prime}_{l}}{x-x_{l}},\quad
c^{\prime}_{l}=\frac{\varphi(x_{l})}{f^{\prime}(x_{l})},$ (41)

where we assume that the function $f(x)$ has only simple roots $x_{l}$. A
consequence of this wonderful expression is the relation

$\displaystyle\prod_{l=1}^{L}\frac{1}{\alpha_{l}y-y_{l}}$
$\displaystyle=\sum_{l=1}^{L}\frac{c_{l}^{(L)}}{\alpha_{l}y-y_{l}},$ (42)

where $y_{s}\neq y_{l}$ if $s\neq l$, and

$\displaystyle c_{1}^{(1)}=1,\quad c_{l}^{(L)}=\prod_{s=1,s\neq
l}^{L}\frac{\alpha_{l}}{y_{l}\alpha_{s}-y_{s}\alpha_{l}},\quad L>1.$ (43)

Using Eq. (42), the integral (38) can be decomposed as a sum of simple
integrals, and one obtains

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{1}{\rho_{L}}\sum_{l=1}^{L}a_{l}^{(L)}\,I_{G}^{\pm}(E_{0l};\mathcal{T}),$
(44) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{\pi}{\rho_{L}}\sum_{l=1}^{L}a_{l}^{(L)}\,G^{\pm}(E_{0l};\mathcal{T}).$
(45)

The assumption that for every pair $E_{0s}\neq E_{0l}$ if $s\neq l$ is used.
Here

$\displaystyle a_{1}^{(1)}$ $\displaystyle=1,\quad
a_{l}^{(L)}=\prod_{s=1,s\neq l}^{L}\frac{1}{E_{0l}-E_{0s}},\quad L>1,$ (46)

and

$\displaystyle I_{G}^{\pm}(E_{1};\mathcal{T})$
$\displaystyle=\mathcal{P}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\,\frac{G^{\pm}(E;\mathcal{T})}{E-E_{1}}.$
(47)

For $L=3$, one can find

$\displaystyle a_{1}^{(3)}$
$\displaystyle=\frac{1}{(E_{02}-E_{01})(E_{03}-E_{01})},\quad
a_{2}^{(3)}=-\frac{1}{(E_{02}-E_{01})(E_{03}-E_{02})},\quad
a_{3}^{(3)}=\frac{1}{(E_{03}-E_{01})(E_{03}-E_{02})}$

which are connected by

$\displaystyle a_{1}^{(3)}+a_{2}^{(3)}+a_{3}^{(3)}=0.$

###  IV.2 Calculation of $\mathcal{Z}_{L}^{\pm}$ for
$\mathbf{k}_{l}\times\mathbf{k}_{s}=0$

The condition $\mathbf{k}_{l}\times\mathbf{k}_{s}=0$ for some $l,s$ means that
$\mathbf{k}_{l}$ and $\mathbf{k}_{s}$ are collinear if $k_{l}k_{s}\neq 0$. In
this case, we have

$\displaystyle\cos\psi_{ls}$ $\displaystyle\equiv\eta_{ls}=\pm
1,\quad\sin\psi_{ls}=0,$ $\displaystyle\sin\delta_{l}$
$\displaystyle=\sin\delta_{s},\quad\cos\delta_{l}=\eta_{ls}\cos\delta_{s},$
(48) $\displaystyle\sin(\phi_{l}-\phi_{s})=0,\quad\cos(\phi_{l}-\phi_{s})$
$\displaystyle=\eta_{ls},\quad\sin\phi_{l}=\eta_{ls}\sin\phi_{s},\quad\cos\phi_{l}=\eta_{ls}\cos\phi_{s}.$

Another possibility is that $k_{l}=0$ or/and $k_{s}=0$.

Below we assume $L>1$ since we need two or more vectors to speak about
collinearity444The solution for the case $L=1$ can be found in Klevansky ..
Equation (12) can be written as

$\displaystyle\mathcal{Z}^{\pm}_{L}(\mathcal{X}_{1}^{+},\ldots,\mathcal{X}_{L}^{+};\mathcal{T})=\frac{1}{2^{L}\pi}\lim_{\epsilon\rightarrow
0}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,G^{\pm}(E;\mathcal{T})\prod_{s=L-n+1}^{L}\frac{1}{\lambda_{s}(E-E_{0s}-i\epsilon)}$
$\displaystyle\times\int\limits_{-1}^{1}{\mathrm{d}}x\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\prod_{l=1}^{L-n}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x}$

where $n=\overline{0,L-1}$. If $n=0$, the $E$-product is absent.

Let us define

$\displaystyle\omega_{ls}$
$\displaystyle=k_{l}\lambda_{s}-k_{s}\lambda_{l}\eta_{ls},\quad\zeta_{ls}=\frac{k_{l}\varepsilon_{s}-k_{s}\varepsilon_{l}\eta_{ls}}{\omega_{ls}}.$
(49)

One can see that if $k_{s}=0$, $\zeta_{ls}=E_{0s}$. When $k_{l}k_{s}\neq 0$,
due to $\eta_{ls}^{2}=1$, we have two nice properties:

$\displaystyle\omega_{ls}$
$\displaystyle=-\omega_{sl}\eta_{ls},\quad\zeta_{ls}=\zeta_{sl}.$ (50)

Collinearity of $\mathbf{k}_{l}$ allows us to decompose the product over $x$,
according to Eq. (42), as

$\displaystyle\prod_{l=1}^{L^{\prime}}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x}$
$\displaystyle=\sum_{l=1}^{L^{\prime}}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x}\prod_{s=1,s\neq
l}^{L^{\prime}}\frac{k_{l}}{(E-\zeta_{ls}-i\epsilon)\,\omega_{ls}}.$ (51)

Then we again apply Eq. (42) to the two-component $E$-product

$\prod_{s=1,s\neq
l}^{L^{\prime}}\frac{k_{l}}{(E-\zeta_{ls}-i\epsilon)\,\omega_{ls}}\prod_{s=L^{\prime}+1}^{L^{\prime\prime}}\frac{1}{\lambda_{s}(E-\zeta_{ls}-i\epsilon)}=k_{l}\sum_{s=1,s\neq
l}^{L^{\prime\prime}}\frac{b^{(L^{\prime\prime})}_{ls}}{E-\zeta_{ls}-i\epsilon},$
(52)

where

$\displaystyle b^{(L^{\prime\prime})}_{ls}$
$\displaystyle=\frac{d^{(L^{\prime\prime})}_{ls}}{\omega_{ls}},\quad
d^{(2)}_{12}=1,\quad d^{(L^{\prime\prime})}_{ls}=\prod_{t=1,t\neq s,t\neq
l}^{L^{\prime\prime}}\frac{k_{l}}{\omega_{lt}(\zeta_{ls}-\zeta_{lt})}.$

and, by definition, $L^{\prime}<L^{\prime\prime}$. These coefficients have the
following nice properties:

$\displaystyle b^{(L^{\prime\prime})}_{ls}$
$\displaystyle=-b^{(L^{\prime\prime})}_{sl}\eta_{ls},\quad
d^{(L^{\prime\prime})}_{ls}=d^{(L^{\prime\prime})}_{sl}.$ (53)

Combining two above decompositions and using Eq. (19), one finally gets

$\displaystyle
p\prod_{s=L-n+1}^{L}\frac{1}{\lambda_{s}(E-E_{0s}-i\epsilon)}\prod_{l=1}^{L-n}\frac{1}{\lambda_{l}(E-E_{0l}-i\epsilon)+k_{lx}p\cos\phi\sqrt{1-x^{2}}+k_{ly}p\sin\phi\sqrt{1-x^{2}}+k_{lz}p\,x}$
$\displaystyle=\sum_{l=1}^{L-n}\frac{pk_{l}}{\lambda_{l}(E-E_{0l}-i\epsilon)+k_{lx}p\cos\phi\sqrt{1-x^{2}}+k_{ly}p\sin\phi\sqrt{1-x^{2}}+k_{lz}p\,x}\sum_{s=1,s\neq
l}^{L}\frac{b^{(L)}_{ls}}{E-\zeta_{ls}-i\epsilon}$
$\displaystyle=\sum_{l=1}^{L-n}\sum_{s=1,s\neq
l}^{L}b_{ls}^{(L)}\Bigg{[}\frac{pk_{l}}{\lambda_{l}(E-E_{0l})+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x}\frac{1}{E-\zeta_{ls}}$
$\displaystyle-pk_{l}\pi^{2}{\rm
sgn}\lambda_{l}\,\delta\left(E-\zeta_{ls}\right)\,\delta\left(\lambda_{l}(E-E_{0l})+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x\right)$
$\displaystyle+i\pi\frac{pk_{l}\delta(E-\zeta_{ls})}{\lambda_{l}(E-E_{0l})+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x}$
$\displaystyle\left.+i\pi\frac{pk_{l}\,{\rm
sgn}\lambda_{l}}{E-\zeta_{ls}}\,\delta\left(\lambda_{l}(E-E_{0l})+(k_{l})_{x}p\cos\phi\sqrt{1-x^{2}}+(k_{l})_{y}p\sin\phi\sqrt{1-x^{2}}+(k_{l})_{z}p\,x\right)\right].$
(54)

Performing $\phi$-integration with the help of Eq. (36) and

$\displaystyle\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\>f(\phi)\,\delta\left(a+b\cos\phi+c\sin\phi\right)=\frac{f\left(\phi^{+}\right)+f\left(\phi^{-}\right)}{\sqrt{|a^{2}-b^{2}-c^{2}|}}\,\Theta\\!\left(b^{2}+c^{2}-a^{2}\right),$
(55)

where $\phi^{\pm}$ are defined by the relations

$\displaystyle\cos\phi^{\pm}$ $\displaystyle=\frac{-ab\pm
c\,\sqrt{b^{2}+c^{2}-a^{2}}}{b^{2}+c^{2}},\quad\sin\phi^{\pm}=\frac{-ac\mp
b\,\sqrt{b^{2}+c^{2}-a^{2}}}{b^{2}+c^{2}},$

and applying the well-known result

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\,f(x)\,\delta(a+x)$
$\displaystyle=f(-a)\Theta\\!\left(1-a^{2}\right),$ (56)

we obtain

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{1}{2^{L-1}}\sum_{l=1}^{L-n}\sum_{s=1,s\neq
l}^{L}b_{ls}^{(L)}\Big{[}I_{J}^{\pm}(\zeta_{ls},k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})-\pi^{2}{\rm
sgn}(\lambda_{l})H^{\pm}(\zeta_{ls};k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})\Big{]},$
(57) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{\pi}{2^{L-1}}\sum_{l=1}^{L-n}\sum_{s=1,s\neq
l}^{L}b_{ls}^{(L)}\Big{[}J^{\pm}(\zeta_{ls},k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})+{\rm
sgn}(\lambda_{l})\,I_{H}^{\pm}(\zeta_{ls},k_{l},\lambda_{l},\varepsilon_{l};\mathcal{T})\Big{]},$
(58)

where

$\displaystyle H^{\pm}(E;k,\lambda,\varepsilon,\mathcal{T})$
$\displaystyle=F^{\pm}(E;\mathcal{T})\,\Theta\\!\left(\Phi\left(E;k,\lambda,\varepsilon,m\right)\right)\,\Theta\\!\left(k\right),$
(59) $\displaystyle J^{\pm}(E;k,\lambda,\varepsilon,\mathcal{T})$
$\displaystyle=F^{\pm}(E;\mathcal{T})\ln\left|\frac{\lambda
E-\varepsilon+k\,p}{\lambda E-\varepsilon-k\,p}\right|,$ (60)
$I_{H}^{\pm}(E_{1},k,\lambda,\varepsilon,\mathcal{T})=\int\limits_{m}^{\Lambda_{E}}{\mathrm{d}}E\,\frac{H^{\pm}(E;k,\lambda,\varepsilon,\mathcal{T})}{E-E_{1}},\quad
I_{J}^{\pm}(E_{1},k,\lambda,\varepsilon,\mathcal{T})=\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{J^{\pm}(E;k,\lambda,\varepsilon,\mathcal{T})}{E-E_{1}},$
(61)

and the function $\Phi(E;k,\lambda,\varepsilon,m)=p^{2}k^{2}(1-z^{2})$ is
considered in Appendix B. The above expressions, in principle, are sufficient
for numerical calculations but it is useful to make some additional
transformations which allow one to clarify some properties and the structure
of these formulaes.

Using $\zeta_{ls}=\zeta_{sl}$, we can make a resummation of symmetric terms.
Separating the part with $s>L-n$, one finally has

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{1}{2^{L-1}}\sum_{l=1}^{L-n}\sum_{s=l+1}^{L-n}b_{ls}^{(L)}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{f(\pm
E-\mu)}{E-\zeta_{ls}}\ln\left|\frac{\lambda_{l}E-\varepsilon_{l}+k_{l}p}{\lambda_{l}E-\varepsilon_{l}-k_{l}p}\frac{\lambda_{s}E-\varepsilon_{s}-k_{s}\eta_{ls}p}{\lambda_{s}E-\varepsilon_{s}+k_{s}\eta_{ls}p}\right|$
$\displaystyle-\frac{\pi^{2}}{2^{L-1}}\sum_{l=1}^{L-n}{\rm
sgn}(\lambda_{l})\sum_{s=1,s\neq
l}^{L-n}b_{ls}^{(L)}H^{\pm}(\zeta_{ls};k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})$
$\displaystyle+\frac{1}{2^{L-1}}\sum_{l=1}^{L-n}\sum_{s=L-n+1}^{L}b_{ls}^{(L)}\Big{[}I_{J}^{\pm}(E_{0s},k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})-\pi^{2}{\rm
sgn}(\lambda_{l})H^{\pm}(E_{0s};k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})\Big{]},$
(62) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{L}$
$\displaystyle=\frac{\pi}{2^{L-1}}\sum_{l=1}^{L-n}{\rm
sgn}(\lambda_{l})\sum_{s=1,s\neq
l}^{L-n}b_{ls}^{(L)}I_{H}^{\pm}(\zeta_{ls},k_{l},\lambda_{l},\varepsilon_{l};\mathcal{T})$
$\displaystyle+\frac{\pi}{2^{L-1}}\sum_{l=1}^{L-n}\sum_{s=L-n+1}^{L}b_{ls}^{(L)}\Big{[}J^{\pm}(E_{0s},k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})+{\rm
sgn}(\lambda_{l})\,I_{H}^{\pm}(E_{0s},k_{l},\lambda_{l},\varepsilon_{l};\mathcal{T})\Big{]}.$
(63)

To get Eq. (IV.2), we took into account that
$J^{\pm}(\zeta_{ls},k_{l},\lambda_{l},\varepsilon_{l},\mathcal{T})=0$ when
$k_{l}k_{s}\neq 0$. One can also prove that

$\displaystyle\lim_{E\rightarrow\zeta_{ls}}\frac{1}{E-\zeta_{ls}}\ln\left|\frac{k_{l}p+\lambda_{l}E-\varepsilon_{l}}{k_{l}p-(\lambda_{l}E-\varepsilon_{l})}\frac{pk_{s}\cos\psi_{ls}-(\lambda_{s}E-\varepsilon_{s})}{pk_{s}\cos\psi_{ls}+\lambda_{s}E-\varepsilon_{s}}\right|=2\frac{p_{ls}\,\omega_{sl}}{k_{s}k_{l}\left(p_{ls}^{2}-\frac{\varrho_{ls}^{2}}{\omega_{ls}^{2}}\right)}$
(64)

with

$\displaystyle p_{ls}=p_{sl}=\sqrt{\zeta_{ls}^{2}-m^{2}},\quad\varrho_{ls}$
$\displaystyle=\lambda_{s}\,\varepsilon_{l}-\lambda_{l}\varepsilon_{s}=\lambda_{l}\lambda_{s}\left(E_{0l}-E_{0s}\right),$
(65)

and so the first term in Eq. (IV.2) is not singular.

It can be shown that when all $k_{l}\rightarrow 0$ simultaneously, i.e. $n=L$,
Eqs. (57) and (58) converge into Eqs. (44) and (45), respectively.

##  V Planar momenta

###  V.1 The configuration with $k_{1,2}>0$, $k_{3}=0$,
$\mathbf{k}_{1}\times\mathbf{k}_{2}\neq 0$

Below we will work in the coordinates (17). As a result, the condition of
noncollinearity $\mathbf{k}_{1}\times\mathbf{k}_{2}\neq 0$ is equivalent to

$\displaystyle\sin\psi_{12}=\sin\delta_{2}>0$ (66)

and we have

$\displaystyle\mathcal{Z}^{\pm}_{3}(\mathcal{X}_{1}^{+},\mathcal{X}_{2}^{+},\mathcal{X}_{3}^{+};\mathcal{T})$
$\displaystyle=\frac{1}{8\pi\lambda_{3}}\lim_{\epsilon\rightarrow
0}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{G^{\pm}(E;\mathcal{T})}{E-E_{03}-i\epsilon}\int\limits_{-1}^{1}\frac{{\mathrm{d}}x}{\lambda_{1}(E-E_{01}-i\epsilon)+pk_{1}\,x}$
$\displaystyle\quad\quad\times\int\limits_{0}^{2\pi}\frac{{\mathrm{d}}\phi}{\lambda_{2}(E-E_{02}-i\epsilon)+pk_{2}(\sin\delta_{2}\cos\phi\sqrt{1-x^{2}}+x\cos\delta_{2})}\,.$

The $\phi$-integration is made using of Eqs. (36) and (55); meanwhile
integrals over $x$ can be evaluated with the help of Eqs. (30), (56), and

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\frac{\Theta\\!\left(-\Delta(x;z,\psi)\right)}{(x+y)\sqrt{-\Delta(x;z,\psi)}}$
$\displaystyle=\pi\,\frac{\Theta\\!\left(\Delta_{0}(y,z,\psi)\right)\,\Theta\\!\left(1-z^{2}\right)}{\sqrt{\Delta_{0}(y,z,\psi)}}\,{\rm
sgn}(y-z\cos\psi)$ (67)

which was proved in Klevansky .

Finally, taking into account the property (123) or (146), we find

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{1}{4\lambda_{3}}\Bigg{\\{}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{F^{\pm}(E;\mathcal{T})}{E-E_{03}}\frac{\Xi_{12}\,\Theta\\!\left(\Phi_{12}\right)+\left[\mathscr{G}_{12}-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2})\right]\Theta\\!\left(-\Phi_{12}\right)}{\sqrt{|\Phi_{12}(E)|}}$
$\displaystyle-\pi^{2}\,F^{\pm}(E_{03};\mathcal{T})\,\frac{\Theta\\!\left(\Phi_{12}\left(E_{03}\right)\right)}{\sqrt{\Phi_{12}\left(E_{03}\right)}}\,\Pi_{12}\left(E_{03}\right)\Bigg{\\}},$
(68) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{\pi}{4\lambda_{3}}\Bigg{\\{}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{F^{\pm}(E;\mathcal{T})\,\Theta\\!\left(\Phi_{12}(E)\right)}{(E-E_{03})\sqrt{\Phi_{12}(E)}}\,\Pi_{12}(E)$
$\displaystyle+F^{\pm}(E_{03};\mathcal{T})\frac{\Xi_{12}(E_{03})\,\Theta\\!\left(\Phi_{12}(E_{03})\right)+\left[\mathscr{G}_{12}(E_{03})-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2})\right]\Theta\\!\left(-\Phi_{12}(E_{03})\right)}{\sqrt{|\Phi_{12}(E_{03})|}}\Bigg{\\}}$
(69)

with

$\Pi_{ls}(E)={\rm sgn}\lambda_{l}\,{\rm
sgn}(z_{s}-z_{l}\cos\psi_{ls})\,\Theta\\!\left(\Phi_{l}(E)\right)+{\rm
sgn}\lambda_{s}\,{\rm
sgn}(z_{l}-z_{s}\cos\psi_{ls})\,\Theta\\!\left(\Phi_{s}(E)\right)$ (70)

defined for shortness. The quadratic trinomials
$\Phi_{l}=\Phi(E;k_{l},\lambda_{l},\varepsilon_{l},m)$ and
$\Phi_{ls}=p^{2}k_{l}^{2}k_{s}^{2}\Delta_{ls}$ are described in details in
Appendix B.

###  V.2 The case when $k_{l}>0$, $\mathbf{k}_{l}\times\mathbf{k}_{s}\neq 0$,
$\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3})=0$

First of all, let us turn the conditions defining this case into a more
appropriate form. Three noncollinearity requirements give

$\displaystyle\sin\delta_{2},\ \sin\delta_{3},\ \sin\psi_{23}>0,$ (71)

meanwhile the complanarity condition means that $\sin\phi_{3}=0$ and

$\displaystyle\cos\phi_{3}\equiv\eta_{23}=\pm 1.$ (72)

As a result, we can write

$\displaystyle\cos\psi_{23}$
$\displaystyle=\cos(\delta_{2}-\eta_{23}\delta_{3})$ (73)
$\displaystyle\alpha_{23}$
$\displaystyle=\sin(\delta_{2}-\eta_{23}\delta_{3}),\quad\alpha_{32}=-\eta_{23}\alpha_{23},\quad\alpha_{23}^{2}=\alpha_{32}^{2}=\sin^{2}\psi_{23}$
(74) $\displaystyle\beta_{ls}$
$\displaystyle=z_{s}\sin\delta_{l}-\eta_{23}z_{l}\sin\delta_{s},\quad\beta_{32}=-\eta_{23}\beta_{23}.$
(75)

The trinomial $\Phi_{123}(E)=p^{2}k_{1}^{2}k_{2}^{2}k_{3}^{2}\Delta_{123}$ is
simplified in this case and can be written as

$\Phi_{123}(E)=\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]^{2},$

see Appenidx B for details. It is useful to remain $\Phi_{123}(E)$ in the
denominators of integrands without canceling to make more transparent
connection with the 3D case.

Evaluation of $\mathcal{Z}^{\pm}_{3}$ for the considered case is very similar
to the calculations at the end of Sec. III. Equations (36) and (55) are used
to perform $\phi$-integration, Eqs. (20), (56), and (67) are sufficient for
taking integrals over $x$.

To simplify the result, we need the following properties:

$\displaystyle\Delta_{2}\left(-\frac{\beta_{23}}{\alpha_{23}}\right)$
$\displaystyle=\frac{\sin^{2}\delta_{2}}{\sin^{2}\psi_{23}}\,\Delta_{23}(E),\quad\Delta_{3}\left(-\frac{\beta_{23}}{\alpha_{23}}\right)=\frac{\sin^{2}\delta_{3}}{\sin^{2}\psi_{23}}\,\Delta_{23}(E),$
$\displaystyle\eta_{23}\Xi\left(\frac{\beta_{32}}{\alpha_{32}},z_{3},\delta_{3}\right)$
$\displaystyle-\Xi\left(\frac{\beta_{23}}{\alpha_{23}},z_{2},\delta_{2}\right)=-\frac{\alpha_{23}}{\sin\psi_{23}}\,\Xi_{23}$
$\displaystyle\eta_{23}\mathscr{G}\left(\frac{\beta_{32}}{\alpha_{32}},z_{3},\delta_{3}\right)$
$\displaystyle-\mathscr{G}\left(\frac{\beta_{23}}{\alpha_{23}},z_{2},\delta_{2}\right)=-\frac{\alpha_{23}}{\sin\psi_{23}}\,\left(\mathscr{G}_{23}-\pi\eta_{23}\right).$

Finally, we obtain

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{1}{4}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})$
$\displaystyle\quad\times\frac{\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}}{\Phi_{123}(E)}\Bigg{\\{}(\mathbf{k}_{3}\times\mathbf{k}_{1})\frac{\Xi_{13}\Theta\\!\left(\Phi_{13}\right)+\left[\mathscr{G}_{13}-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{3})\right]\Theta\\!\left(-\Phi_{13}\right)}{\sqrt{\left|\Phi_{13}\right|}}$
$\displaystyle\quad\quad+(\mathbf{k}_{1}\times\mathbf{k}_{2})\frac{\Xi_{12}\Theta\\!\left(\Phi_{12}\right)+\left[\mathscr{G}_{12}-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2})\right]\Theta\\!\left(-\Phi_{12}\right)}{\sqrt{\left|\Phi_{12}\right|}}$
$\displaystyle\quad\quad+(\mathbf{k}_{2}\times\mathbf{k}_{3})\frac{\Xi_{23}\Theta\\!\left(\Phi_{23}\right)+\left[\mathscr{G}_{23}-\pi\eta_{23}-\pi\,{\rm
sgn}(\lambda_{2}\lambda_{3})\right]\Theta\\!\left(-\Phi_{23}\right)}{\sqrt{\left|\Phi_{23}\right|}}\Bigg{\\}},$
(76) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{\pi}{4}\Bigg{\\{}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{1}\times\mathbf{k}_{2})\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]\Theta\\!\left(\Phi_{12}(E)\right)}{\Phi_{123}(E)\sqrt{\Phi_{12}(E)}}\,\Pi_{12}(E)$
$\displaystyle+\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{3}\times\mathbf{k}_{1})\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]\Theta\\!\left(\Phi_{13}\right)}{\Phi_{123}(E)\sqrt{\Phi_{13}(E)}}\,\Pi_{13}(E)$
$\displaystyle+\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{2}\times\mathbf{k}_{3})\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]\Theta\\!\left(\Phi_{23}\right)}{\Phi_{123}(E)\sqrt{\Phi_{23}(E)}}\,\Pi_{23}(E)$
$\displaystyle-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2}\lambda_{3})\,\left|\mathbf{k}_{1}\times\mathbf{k}_{2}\right|\left.F^{\pm}(E;\mathcal{T})\frac{\Theta\\!\left(-\Phi_{12}\right)}{\left|\mathbf{b}^{(3)}\right|\sqrt{-\Phi_{12}}}\right|_{E=\mathbf{a}^{(3)}\mathbf{b}^{(3)}/[\mathbf{b}^{(3)}]^{2}}\Bigg{\\}}.$
(77)

As we can see, the real part contains the additional $\pi$-term which breaks
symmetry under interchange $\mathbf{k}_{1,2,3}$. As an ansatz, such terms can
be omitted. Then the result can be represented as a sum of functions

$\displaystyle I_{K;ls}^{(3)}$
$\displaystyle=\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{l}\times\mathbf{k}_{s})[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}]}{\Phi_{123}(E)}\frac{\Xi_{ls}(E)}{\sqrt{\Phi_{ls}(E)}}\,\Theta\\!\left(\Phi_{ls}(E)\right),$
(78) $\displaystyle I_{L;ls}^{(3)}$
$\displaystyle=\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{l}\times\mathbf{k}_{s})[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}]}{\Phi_{123}(E)}\frac{\mathscr{G}_{ls}-\pi\,{\rm
sgn}(\lambda_{l}\lambda_{s})}{\sqrt{-\Phi_{ls}(E)}}\,\Theta\\!\left(-\Phi_{ls}(E)\right),$
(79) $\displaystyle I_{M;ls}^{(3)}$
$\displaystyle=\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,F^{\pm}(E;\mathcal{T})\frac{(\mathbf{k}_{l}\times\mathbf{k}_{s})\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]}{\Phi_{123}(E)}\frac{\Pi_{ls}(E)}{\sqrt{\Phi_{ls}(E)}}\,\Theta\\!\left(\Phi_{ls}(E)\right),$
(80)

in the form

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{1}{4}\left[I_{K;12}^{(3)}+I_{K;23}^{(3)}+I_{K;31}^{(3)}+I_{L;12}^{(3)}+I_{L;23}^{(3)}+I_{L;31}^{(3)}\right],$
(81) $\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{\pi}{4}\left[I_{M;12}^{(3)}+I_{M;23}^{(3)}+I_{M;31}^{(3)}-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2}\lambda_{3})\,\frac{\left|\mathbf{k}_{1}\times\mathbf{k}_{2}\right|}{\left|\mathbf{b}^{(3)}\right|}\frac{F^{\pm}(E_{123};\mathcal{T})\Theta\\!\left(-\Phi_{12}(E_{123})\right)}{\sqrt{-\Phi_{12}(E_{123})}}\right].$
(82)

Here $E_{123}=\mathbf{a}^{(3)}\mathbf{b}^{(3)}/[\mathbf{b}^{(3)}]^{2}$ is the
root of $\Phi_{123}(E)$.

##  VI Box integral for three independent momenta

###  VI.1 Two new definite integrals

To calculate the box integral, we need to evaluate

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\,\frac{K\,x+S}{R_{1}(x)\sqrt{\Delta(x;z,\delta)}}\,{\rm
sgn}\left(z+x\cos\delta\right)\Theta\\!\left(\Delta(x;z,\delta)\right),$ (83)

which is a generalization of Eq. (20), and

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\,\frac{K\,x+S}{R_{1}(x)\sqrt{-\Delta(x;z,\delta)}}\,\Theta\\!\left(-\Delta(x;z,\delta)\right)$
(84)

which is an analog of Eq. (67). Above we have introduced

$\displaystyle R_{1}(x)$ $\displaystyle=Cx^{2}+2Bx+A.$ (85)

Both integrals can be evaluated analytically, e.g., using Gradshteyn .

Let us start from the consideration of the first one. In the general case,
integration is quite difficult. However, our case is much simpler since
$\Delta(x)$ and $R_{1}(x)$ are connected by the relation

$\displaystyle D$ $\displaystyle=B^{2}-AC=-\frac{k}{C-k}\,D_{0},\quad C>k\geq
0,$ (86)

with

$\displaystyle D_{0}=u^{2}-C(C-k)\,d,\quad u=B-Cb,\quad
d=\sin^{2}\delta(1-z^{2}),\quad b=z\cos\delta.$ (87)

It means that $R_{1}(x)=(C-k)\left(x+b+\frac{u}{C-k}\right)^{2}+k\Delta(x)$.
Then one can prove

$\displaystyle\int\,{\mathrm{d}}x\frac{K\,x+S}{R_{1}(x)\sqrt{\Delta(x)}}$
$\displaystyle=\frac{\mathcal{V}}{\sqrt{|D_{0}|}}\Bigg{\\{}\frac{\Theta\\!\left(D_{0}\right)}{2}\,\,\ln\left|\frac{\sqrt{D_{0}\Delta(x)}+u(x+b)+(C-k)d}{\sqrt{D_{0}\Delta(x)}-[u(x+b)+(C-k)d]}\right|$
$\displaystyle-\Theta\\!\left(-D_{0}\right)\,\arctan\frac{u(x+b)+(C-k)d}{\sqrt{|D_{0}|\Delta(x)}}\Bigg{\\}}+\mathcal{U}\arctan\frac{(C-k)(x+b)+u}{\sqrt{(C-k)k\Delta(x)}}$
(88)

with

$\displaystyle\mathcal{V}$
$\displaystyle=(C-k)\frac{u(S-Kb)-K(C-k)d}{u^{2}-(C-k)^{2}d},\quad\mathcal{U}=\sqrt{\frac{C-k}{k}}\frac{(S-Kb)(C-k)-Ku}{u^{2}-(C-k)^{2}d}\,.$
(89)

Applying Eq. (VI.1) and taking into account that

$\displaystyle\Theta\\!\left(\Delta(x;z,\delta)\right)$
$\displaystyle=\Theta\\!\left(z^{2}-1\right)+\Theta\\!\left(1-z^{2}\right)\left[\Theta\\!\left(x^{-}-x\right)+\Theta\\!\left(x-x^{+}\right)\right],$
(90)

we find

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\,\frac{K\,x+S}{R_{1}(x)\sqrt{\Delta(x;z,\delta)}}\,{\rm
sgn}\left(z+x\cos\delta\right)\Theta\\!\left(\Delta(x;z,\delta)\right)=\mathcal{V}\frac{\mathscr{V}\,\Theta\\!\left(D_{0}\right)+\mathscr{W}\,\Theta\\!\left(-D_{0}\right)}{\sqrt{|D_{0}|}}+\mathcal{U}\mathscr{U},$
(91)

where

$\displaystyle\mathscr{V}$
$\displaystyle=\frac{1}{2}\left[\ln\left|\frac{(z+\cos\delta)\sqrt{D_{0}}+q^{+}}{(z+\cos\delta)\sqrt{D_{0}}-q^{+}}\right|+\ln\left|\frac{(z-\cos\delta)\sqrt{D_{0}}+q^{-}}{(z-\cos\delta)\sqrt{D_{0}}-q^{-}}\right|\right],$
(92) $\displaystyle\mathscr{W}$ $\displaystyle=\left[{\rm
sgn}(z+\cos\delta)\,{\rm sgn}\left(u+(C-k)\sqrt{d}\right)+{\rm
sgn}(z-\cos\delta)\,{\rm sgn}\left(u-(C-k)\sqrt{d}\right)\right]\frac{\pi}{2}$
$\displaystyle-\arctan\frac{q^{+}}{(z+\cos\delta)\sqrt{|D_{0}|}}-\arctan\frac{q^{-}}{(z-\cos\delta)\sqrt{|D_{0}|}},$
(93) $\displaystyle q^{+}$ $\displaystyle=u(1+z\cos\delta)+(C-k)d,\quad
q^{-}=u(1-z\cos\delta)-(C-k)\sin^{2}\delta(1-z^{2}),$ (94)
$\displaystyle\mathscr{U}$
$\displaystyle=\arctan\frac{u+(C-k)(1+z\cos\delta)}{(z+\cos\delta)\sqrt{(C-k)k}}-\arctan\frac{u-(C-k)(1-z\cos\delta)}{(z-\cos\delta)\sqrt{(C-k)k}}$
$\displaystyle+\frac{\pi}{2}\,{\rm sgn}\,u\,\left[{\rm sgn}(z-\cos\delta)-{\rm
sgn}(z+\cos\delta)\right]\Theta\\!\left(d\right).$ (95)

In a similar way, we find

$\displaystyle\int\,{\mathrm{d}}x\frac{K\,x+S}{R_{1}(x)\sqrt{-\Delta(x)}}=\Theta\\!\left(d\right)\Theta\\!\left(D_{0}\right)\,\frac{\mathcal{V}}{\sqrt{|D_{0}|}}\,\arctan\frac{u(x+b)+\left(C-k\right)d}{\sqrt{-D_{0}\Delta(x)}}$
$\displaystyle-\Theta\\!\left(d\right)\Theta\\!\left(D_{0}\right)\frac{\mathcal{U}}{2}\,\,\ln\left|\frac{u+(x+b)\left(C-k\right)-\sqrt{-k(C-k)\Delta(x)}}{u+(x+b)\left(C-k\right)+\sqrt{-k(C-k)\Delta(x)}}\right|$
$\displaystyle+\Theta\\!\left(d\right)\Theta\\!\left(-D_{0}\right)\frac{C\,}{4\sqrt{dD}}\left[\Theta\\!\left(-f^{-}\right)\frac{\tau^{-}\,f^{-}-\tau^{+}}{\sqrt{|f^{-}|}}\,\ln\left|\frac{(x-x^{-})\sqrt{|f^{-}|}-\sqrt{-\Delta(x)}}{(x-x^{-})\sqrt{|f^{-}|}+\sqrt{-\Delta(x)}}\right|\right.$
$\displaystyle\left.-\Theta\\!\left(-f^{+}\right)\frac{\tau^{-}\,f^{+}-\tau^{+}}{\sqrt{|f^{+}|}}\ln\left|\frac{(x-x^{-})\sqrt{|f^{+}|}-\sqrt{-\Delta(x)}}{(x-x^{-})\sqrt{|f^{+}|}+\sqrt{-\Delta(x)}}\right|\right],\quad\Delta(x)<0.$
(96)

Keeping in mind the property

$\Theta\\!\left(-\Delta(x;z,\delta)\right)=\Theta\\!\left(1-z^{2}\right)\Theta\\!\left(x^{+}-x\right)\Theta\\!\left(x-x^{-}\right),$

we see that the second, third, and fourth terms vanish555When $D_{0}>0$, we
have $R_{1}(x)>0$ for any $x$, and so the integrand has no additional poles.
For $D_{0}<0$, one also has to take into account two intermediate poles when
$R_{1}(x)=0$. However, they do not contribute. and thus we do not need the
exact values of the constants $\tau^{\pm}$ and $f^{\pm}$. Observing
that$\left|u\right|\geq\sqrt{C(C-k)d}\geq(C-k)\sqrt{d}$ if $D_{0}>0$, and
$\left|x^{\pm}+b\right|=\sqrt{d}$, we find

$\Theta\\!\left(d\right)\Theta\\!\left(D_{0}\right){\rm
sgn}\left[u(x^{\pm}+b)+(C-k)d\right]=\Theta\\!\left(d\right)\Theta\\!\left(D_{0}\right){\rm
sgn}\left[u(x^{\pm}+b)\right]=\pm\Theta\\!\left(d\right)\Theta\\!\left(D_{0}\right){\rm
sgn}\,u.$ (97)

Combining all these remarks with Eq. (VI.1), one obtains

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\,\frac{K\,x+S}{R_{1}(x)\sqrt{\left|\Delta(x;z,\delta)\right|}}\,\Theta\\!\left(-\Delta(x;z,\delta)\right)=\pi\mathcal{V}\frac{\Theta\\!\left(1-z^{2}\right)\Theta\\!\left(D_{0}\right)}{\sqrt{|D_{0}|}}\,{\rm
sgn}\,u.$ (98)

Let us take one more step to get expressions that can be directly applied in
further evaluations. Namely, we shall deal with integrals like

$\displaystyle\int\,{\mathrm{d}}x\frac{Mx+N}{(x-x_{0})R_{1}(x)\sqrt{\pm\Delta(x;\delta,z)}}=\frac{1}{R_{1}(x_{0})}\left[Q\int\frac{{\mathrm{d}}x}{(x-x_{0})\sqrt{\pm\Delta(x;\delta,z)}}+\int\,{\mathrm{d}}x\frac{S-CQ\,x}{R_{1}(x)\sqrt{\pm\Delta(x;\delta,z)}}\right]$
(99)

with $Q=Mx_{0}+N$, $S=MA-2NB-NCx_{0}$. The integral with
$\sqrt{-\Delta(x;\delta,z)}$ in the denominator is taken using Eqs. (67) and
(98). The result is

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\frac{M\,x+N}{(x-x_{0})R_{1}(x)\sqrt{-\Delta(x;\delta,z)}}\,\Theta\\!\left(-\Delta(x;\delta,z)\right)=\frac{\pi\,\Theta\\!\left(1-z^{2}\right)}{R_{1}(x_{0})}$
$\displaystyle\quad\times\Bigg{[}{\rm
sgn}\left(u\right)\frac{\mathcal{V}^{\prime}}{\sqrt{|D_{0}|}}\,\Theta\\!\left(D_{0}\right)+{\rm
sgn}(-x_{0}-z\cos\delta)\frac{Q}{\sqrt{|\Delta_{0}(\delta,z,x_{0})|}}\,\Theta\\!\left(\Delta_{0}(\delta,z,x_{0})\right)\Bigg{]}.$
(100)

When we deal with $\sqrt{\Delta(x;\delta,z)}$, we have the additional factor
${\rm sgn}\left(z+x\cos\delta\right)$ under integral. The second term is
evaluated with the help of Eq. (91); meanwhile the first term is just Eq.
(20). We obtain

$\displaystyle\int\limits_{-1}^{1}{\mathrm{d}}x\frac{Mx+N}{(x-x_{0})R_{1}(x)\sqrt{\Delta(x;\delta,z)}}\,{\rm
sgn}(z+x\cos\delta)\Theta\\!\left(\Delta(x;\delta,z)\right)$
$\displaystyle=Q\,\frac{\Xi(z,x_{0},\delta)\,\Theta\\!\left(\Delta_{0}(z,x_{0},\delta)\right)+\mathscr{G}(z,x_{0},\delta)\,\Theta\\!\left(-\Delta_{0}(z,x_{0},\delta)\right)}{R_{1}(x_{0})\sqrt{|\Delta_{0}(z,x_{0},\delta)|}}+\mathcal{V}^{\prime}\frac{\mathscr{V}\,\Theta\\!\left(D_{0}\right)+\mathscr{W}\,\Theta\\!\left(-D_{0}\right)}{R_{1}(x_{0})\sqrt{|D_{0}|}},$
(101)

where

$\displaystyle\mathcal{V}^{\prime}$
$\displaystyle=(C-k)\frac{u(S+CQb)+CQ(C-k)d}{u^{2}-(C-k)^{2}d}.$ (102)

###  VI.2 The real and imaginary parts of the integral with four fermion lines

The expressions obtained in the previous paragraph are sufficient to our
purpose. To move further, we need to know $A$, $B$, $C$, $M$, and $N$. As
usual, we start with integration over $\phi$ of the the product under
integral. To do this, we need the following expression:

$\displaystyle\int\limits_{0}^{2\pi}\frac{{\mathrm{d}}\phi}{(a_{1}+b_{1}\cos\phi+c_{1}\sin\phi)(a_{2}+b_{2}\cos\phi+c_{2}\sin\phi)}=\frac{2\pi}{(a_{1}b_{2}-a_{2}b_{1})^{2}-(b_{1}c_{2}-b_{2}c_{1})^{2}+(a_{2}c_{1}-a_{1}c_{2})^{2}}$
$\displaystyle\times\left[\frac{a_{2}(b_{1}^{2}+c_{1}^{2})-a_{1}(b_{1}b_{2}+c_{1}c_{2})}{\sqrt{|a_{1}^{2}-b_{1}^{2}-c_{1}^{2}|}}\,{\rm
sgn}(a_{1})\,\Theta\\!\left(a_{1}^{2}-b_{1}^{2}-c_{1}^{2}\right)+\frac{a_{1}(b_{2}^{2}+c_{2}^{2})-a_{2}(b_{1}b_{2}+c_{1}c_{2})}{\sqrt{|a_{2}^{2}-b_{2}^{2}-c_{2}^{2}|}}\,{\rm
sgn}(a_{2})\,\Theta\\!\left(a_{2}^{2}-b_{2}^{2}-c_{2}^{2}\right)\right].$
(103)

As one can see, Eq. (36) is the limit of Eq. (VI.2).

Using

$\displaystyle\delta\left([g(x)]^{2}-[h(x)]^{2}\right)$
$\displaystyle=\frac{\delta(g(x)-h(x))+\delta(g(x)+h(x))}{2\left|h(x)\right|}$
(104)

together with Eq. (55), we find

$\displaystyle\int\limits_{0}^{2\pi}{\mathrm{d}}\phi\,\delta\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)$
$\displaystyle\quad\times\delta\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right)$
$\displaystyle=2\sin\delta_{2}\sin\delta_{3}\left|\sin\phi_{3}\right|\delta\left(\Psi_{23}(x)\right).$
(105)

Equations (55), (VI.2), and (VI.2) allow us to make integration over $\phi$.
See Appendix C for details.

Above $\Psi_{23}(x)$ is the quadratic trinomial playing the role of
$R_{1}(x)$. It can be written as

$\displaystyle\Psi_{ls}(x)$
$\displaystyle\equiv(\beta_{ls}+\alpha_{ls}x)^{2}+\Delta_{l}(x)\sin^{2}\delta_{s}\sin^{2}(\phi_{l}-\phi_{s}).$
(106)

The symmetry $\Psi_{ls}(x)=\Psi_{sl}(x)$ can be easily checked. We also need
to define666Compare with the definition of $\Delta_{0}$.
$\Delta_{123}=\Psi_{23}(-z_{1})$. Also, taking into account the property

$\displaystyle\alpha_{ls}^{2}+\sin^{2}\delta_{s}\sin^{2}(\phi_{l}-\phi_{s})=\sin^{2}\psi_{ls},$
(107)

we find

$\displaystyle C=\sin^{2}\psi_{ls},\quad
D=-\sin^{2}\delta_{l}\sin^{2}\delta_{s}\sin^{2}(\phi_{l}-\phi_{s})\Delta_{ls}.$
(108)

Other quantities in Eqs. (VI.1) and (VI.1), like $k$, $D_{0}$, $u$, are not
$(l\leftrightarrow s)$ symmetric since depend on $b$. For integrals with
$\Delta(x;\delta,z)=\Delta_{l}(x)$, we have

$\displaystyle k_{l}$
$\displaystyle=\sin^{2}\delta_{s}\sin^{2}(\phi_{l}-\phi_{s}),\quad
D_{0l}=\alpha_{ls}^{2}\sin^{2}\delta_{l}\Delta_{ls},\quad
u_{l}=(\beta_{ls}-z_{l}\alpha_{ls}\cos\delta_{l})\alpha_{ls},$ (109)

and

$\displaystyle M_{l}$ $\displaystyle=\alpha_{ls},\quad N_{l}=\beta_{ls}.$
(110)

The expressions allow us to find

$\displaystyle\mathcal{V}_{l}^{\prime}$ $\displaystyle=\alpha_{ls}T_{1},$
(111)

where we used that $u=(N-Mb)M$ and $C-k_{l}=M^{2}$ which result in
$\mathcal{V}^{\prime}=-M(B+Cx_{0})$, and we also introduced the quantities

$T_{l}=z_{l}\sin^{2}\psi_{st}-z_{s}(\cos\psi_{ls}-\cos\psi_{lt}\cos\psi_{st})-z_{t}(\cos\psi_{lt}-\cos\psi_{ls}\cos\psi_{st}),\quad
t\neq l,s.$ (112)

To simplify the result, we shall also use the following relations:

$\alpha_{ls}\sin\delta_{l}=\cos\delta_{s}-\cos\delta_{l}\cos\psi_{ls}$ (113)

and

$\displaystyle\beta_{ls}-z_{l}\alpha_{ls}\cos\delta_{l}$
$\displaystyle=(z_{s}-z_{l}\cos\psi_{ls})\sin\delta_{l}.$ (114)

Taking into account all the mentioned properties and applying Eqs. (56),
(VI.1), and (VI.1), we obtain

$\displaystyle{\rm Re}Z_{3}^{\pm}$
$\displaystyle=\frac{1}{4}\left[I_{K;12}^{(3)}+I_{K;23}^{(3)}+I_{K;31}^{(3)}+I_{L;12}^{(3)}+I_{L;23}^{(3)}+I_{L;31}^{(3)}\right],$
(115)

where we omit the $\pi$-term in $I_{L;23}^{(3)}$ as in the planar case, and

$\displaystyle{\rm Im}Z_{3}^{\pm}$
$\displaystyle=\frac{\pi}{4}\left[I_{M;12}^{(3)}+I_{M;23}^{(3)}+I_{M;31}^{(3)}-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2}\lambda_{3})\,\frac{F^{\pm}(E^{+}_{123};\mathcal{T})+F^{\pm}(E^{-}_{123};\mathcal{T})}{2\sqrt{D^{\prime}_{123}}}\,\Theta\\!\left(D^{\prime}_{123}\right)\right].$
(116)

The separated terms $\mathscr{V}_{2,3}$ and $\mathscr{W}_{2,3}$ were joined
together using the relations

$\displaystyle\Theta\\!\left(\Delta_{ls}\right)\left[{\rm
sgn}(\alpha_{ls})\mathscr{V}_{l}+{\rm sgn}(\alpha_{sl})\mathscr{V}_{s}\right]$
$\displaystyle=\Xi_{ls}\,\Theta\\!\left(\Delta_{ls}\right)$ (117)

and

$\displaystyle\Theta\\!\left(-\Delta_{ls}\right)\left[{\rm
sgn}(\alpha_{ls})\mathscr{W}_{l}+{\rm sgn}(\alpha_{sl})\mathscr{W}_{s}\right]$
$\displaystyle=\left(\mathscr{G}_{ls}+\pi\mathscr{G}^{0}_{ls}\right)\Theta\\!\left(-\Delta_{ls}\right).$
(118)

At last, one should take into account the property

$\displaystyle pk_{1}k_{2}k_{3}k_{s}k_{t}T_{l}$
$\displaystyle=\pm\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right](\mathbf{k}_{s}\times\mathbf{k}_{t})$

to reduce the formulas to the form similar to the “planar” case.

We would like to underline that the origin of the second ($\pi$) term in Eq.
(118) is the same as in Eq. (37) since it depends on the coordinate system
through $\cos\delta_{l}$. One can prove that

$\displaystyle\mathscr{G}^{0}_{ls}$ $\displaystyle=-{\rm
sgn}\left(\cos\psi_{ls}-z_{l}z_{s}\right)\Theta\\!\left([|\Delta_{ls}|(z_{l}-\cos\delta_{l})^{2}-(q_{ls})^{-})^{2}][|\Delta_{ls}|(z_{s}-\cos\delta_{s})^{2}-(q_{sl}^{-})^{2}]\right)$

with

$\displaystyle q^{\pm}_{ls}$
$\displaystyle=(z_{s}\pm\cos\delta_{s})(1-z_{l}^{2})-(z_{l}\pm\cos\delta_{l})\left(\cos\psi_{ls}-z_{l}z_{s}\right)$

obtained from $q^{\pm}$ introduced in Eq. (94).

As one can see, we immediately obtain Eqs. (81) and (82), respectively, when
$\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3})=0$.

##  VII Conclusions

Above we have given the formulas required for a numerical evaluation of the
four fermion line integral at finite temperature/density at any values of its
arguments, which are often needed for theoretical calculations such as within
the NJL model. Both real and imaginary parts of these functions are presented
separately. Generalization to integrals with bosonic lines can be also proceed
in an analogous fashion.

We have also obtained, as a by-product, expressions for integrals with any
number of lines when momenta are in simple ’dot’ or ’line’ configurations.

It is shown that evaluation of integrals with the help of Eq. (19) results in
some inconsistent (ambiguous) expressions for three or greater lines.

Acknowledgements: We are grateful to A. Friesen for the idea of this paper and
permanent help. My work was supported by the RFBR grant, no. 18-02-40137.

##  Appendix A Summation of inverse trigonometric functions

In this Appendix we just cite some beautiful and useful formulas from
Gradshteyn . The sum and difference of two arccosines are

$\displaystyle\arccos X+\arccos Y$
$\displaystyle=2\pi\Theta\\!\left(-X-Y\right)+{\rm
sgn}(X+Y)\arccos\left(XY-\sqrt{1-X^{2}}\sqrt{1-Y^{2}}\right),$ (119)
$\displaystyle\arccos X-\arccos Y$ $\displaystyle={\rm
sgn}(Y-X)\arccos\left(XY+\sqrt{1-X^{2}}\sqrt{1-Y^{2}}\right).$ (120)

For arctangents we have

$\displaystyle\arctan X+\arctan Y$
$\displaystyle=\arctan\frac{X+Y}{1-XY}+\pi\,{\rm
sgn}(X+Y)\Theta\\!\left(XY-1\right),$
$\displaystyle=\Theta\\!\left(|X+Y|\right)\arctan\frac{XY-1}{X+Y}+\frac{\pi}{2}\,{\rm
sgn}(X+Y),$ (121)

where the second line is obtained with the help of

$\displaystyle\arctan X+\arctan\frac{1}{X}$ $\displaystyle=\frac{\pi}{2}\,{\rm
sgn}X.$ (122)

##  Appendix B Some properties of quadratic trinomials arising during the
evaluation of integrals

First of all, let us note that

$\displaystyle\Theta\\!\left(-\Delta_{0}(y,z,\psi)\right)$
$\displaystyle=\Theta\\!\left(-\Delta_{0}(y,z,\psi)\right)\Theta\\!\left(1-y^{2}\right)\Theta\\!\left(1-z^{2}\right).$
(123)

Another point to mention, is the following. When we make integration over
angles, it is convenient to exclude the variables $E$ using $z_{l}$. But after
that, in the final expressions, we should make the inverse transition as a
result of which the following functions arise:

$\displaystyle\Phi_{l}(E)$
$\displaystyle\equiv\Phi(E;k_{l},\lambda_{l},\varepsilon_{l},m)=p^{2}k_{l}^{2}(1-z_{l}^{2}),$
(124) $\displaystyle\Phi_{ls}(E)$
$\displaystyle=p^{2}k_{l}^{2}k_{s}^{2}\Delta_{ls},$ (125)
$\displaystyle\Phi_{123}(E)$
$\displaystyle=p^{2}k_{1}^{2}k_{2}^{2}k_{3}^{2}\Delta_{123}.$ (126)

Recognizing that these functions are just trinomials, let us introduce the
generalized trinomial

$\displaystyle\Omega_{L}(E;\mathcal{X}^{+}_{1},\ldots,\mathcal{X}^{+}_{L},m)$
$\displaystyle=C_{\Omega}(\mathcal{X}^{+}_{1},\ldots,\mathcal{X}^{+}_{L},m)\,E^{2}-2B_{\Omega}(\mathcal{X}^{+}_{1},\ldots,\mathcal{X}^{+}_{L},m)\,E+A_{\Omega}(\mathcal{X}^{+}_{1},\ldots,\mathcal{X}^{+}_{L},m)$
$\displaystyle=\left[\mathbf{b}^{(L)}E-\mathbf{a}^{(L)}\right]^{2}-\left[w^{(L)}\right]^{2}p^{2},$
(127)

where we demand $\left|C_{\Omega}\right|+\left|B_{\Omega}\right|\neq 0$. The
coefficients are

$\displaystyle A_{\Omega}$
$\displaystyle=\left[\mathbf{a}^{(L)}\right]^{2}+m^{2}\left[w^{(L)}\right]^{2},\quad
B_{\Omega}=\mathbf{a}^{(L)}\mathbf{b}^{(L)},\quad
C_{\Omega}=\left[\mathbf{b}^{(L)}\right]^{2}-\left[w^{(L)}\right]^{2}.$ (128)

As we can easily see, the determinant is

$\displaystyle
D_{\Omega}=\left[w^{(L)}\right]^{2}\left\\{A_{\Omega}-m^{2}\left[\mathbf{b}^{(L)}\right]^{2}-\frac{\left[\mathbf{a}^{(L)}\times\mathbf{b}^{(L)}\right]^{2}}{[w^{(L)}]^{2}}\right\\},$
(129)

where we used that

$\displaystyle\mathbf{a}^{2}\mathbf{b}^{2}$
$\displaystyle=\left(\mathbf{a}\times\mathbf{b}\right)^{2}+\left(\mathbf{a}\mathbf{b}\right)^{2}.$
(130)

The vectors $\mathbf{a}^{(L)}$ and $\mathbf{b}^{(L)}$ can be expressed as
linear combinations of a set of vectors $\mathbf{q}_{l}$ for any $L$:

$\displaystyle\mathbf{a}^{(L)}$
$\displaystyle=\sum_{l=1}^{L}\varepsilon_{l}\mathbf{q}_{l},\quad\mathbf{b}^{(L)}=\sum_{l=1}^{L}\lambda_{l}\mathbf{q}_{l},$
(131)

and
$\mathbf{q}_{l}=\mathbf{q}_{l}\left(\mathbf{k}_{1},\ldots,\mathbf{k}_{L}\right)$.

Using the generalized trinomial, one can write

$\displaystyle\Phi_{l}(E)$
$\displaystyle=-\Omega_{1}\left(E;\mathcal{X}_{l}^{+},m\right)$ (132)

with

$\displaystyle\mathbf{a}^{(1)}_{l}$
$\displaystyle=\varepsilon_{l}\frac{\mathbf{k}}{k},\quad\mathbf{b}^{(1)}_{l}=\lambda_{l}\frac{\mathbf{k}}{k},\quad
w^{(1)}_{l}=k_{l},$ (133)

and $\mathbf{k}$ is an arbitrary vector.

For $L=2$, we have

$\displaystyle\Phi_{ls}(E)$
$\displaystyle=\Omega_{2}\left(E;\mathcal{X}_{l}^{+},\mathcal{X}_{s}^{+},m\right),$
(134) $\displaystyle\mathbf{a}^{(2)}_{ls}$
$\displaystyle=\varepsilon_{s}\mathbf{k}_{l}-\varepsilon_{l}\mathbf{k}_{s},\quad\mathbf{b}^{(2)}_{ls}=\lambda_{s}\mathbf{k}_{l}-\lambda_{l}\mathbf{k}_{s},\quad
w^{(2)}_{ls}=\left|\mathbf{k}_{l}\times\mathbf{k}_{s}\right|.$ (135)

We also need the coefficients for $L=3$:

$\displaystyle\mathbf{a}^{(3)}$
$\displaystyle=\varepsilon_{1}\mathbf{k}_{2}\times\mathbf{k}_{3}+\varepsilon_{2}\mathbf{k}_{3}\times\mathbf{k}_{1}+\varepsilon_{3}\mathbf{k}_{1}\times\mathbf{k}_{2},$
$\displaystyle\mathbf{b}^{(3)}$
$\displaystyle=\lambda_{1}\mathbf{k}_{2}\times\mathbf{k}_{3}+\lambda_{2}\mathbf{k}_{3}\times\mathbf{k}_{1}+\lambda_{3}\mathbf{k}_{1}\times\mathbf{k}_{2},$
(136) $\displaystyle w^{(3)}$
$\displaystyle=\mathbf{k}_{1}(\mathbf{k}_{2}\times\mathbf{k}_{3}).$ (137)

One can easily prove,

$\displaystyle\mathbf{a}_{ls}^{(2)}\times\mathbf{b}_{ls}^{(2)}$
$\displaystyle=\varrho_{ls}(\mathbf{k}_{l}\times\mathbf{k}_{s})$ (138)
$\displaystyle\mathbf{a}^{(3)}\times\mathbf{b}^{(3)}$
$\displaystyle=w^{(3)}\left(\varrho_{12}\mathbf{k}_{3}-\varrho_{13}\mathbf{k}_{2}+\varrho_{23}\mathbf{k}_{1}\right).$
(139)

Then, using

$\displaystyle\mathbf{a}^{(3)}$
$\displaystyle=\varepsilon_{t}(\mathbf{k}_{l}\times\mathbf{k}_{s})-\mathbf{a}^{(2)}_{ls}\times\mathbf{k}_{t},$
$\displaystyle\mathbf{b}^{(3)}$
$\displaystyle=\lambda_{t}(\mathbf{k}_{l}\times\mathbf{k}_{s})-\mathbf{b}^{(2)}_{ls}\times\mathbf{k}_{t},$
(140)

and Eq. (130), we find

$\displaystyle\Phi_{123}(E)\left(\mathbf{k}_{l}\times\mathbf{k}_{s}\right)^{2}$
$\displaystyle=\left\\{\left[\mathbf{b}^{(3)}E-\mathbf{a}^{(3)}\right]\left(\mathbf{k}_{l}\times\mathbf{k}_{s}\right)\right\\}^{2}+\left[w^{(3)}\right]^{2}\Phi_{ls}(E)$
(141)

form which it follows that

$\displaystyle\Theta\\!\left(D_{123}\right)=\Theta\\!\left(D_{123}\right)\Theta\\!\left(-\Phi_{12}(E_{123}^{\pm})\right)\Theta\\!\left(-\Phi_{13}(E_{123}^{\pm})\right)\Theta\\!\left(-\Phi_{23}(E_{123}^{\pm})\right).$
(142)

Here $E_{123}^{\pm}$ are the zeros of $\Phi_{123}(E)$. Also, Eq. (129) can be
transformed as

$\displaystyle D_{123}=\left[w^{(3)}\right]^{2}D^{\prime}_{123},\quad
D^{\prime}_{123}=A_{\Omega}^{(3)}-m^{2}\left[\mathbf{b}^{(3)}\right]^{2}-\left(\varrho_{12}\mathbf{k}_{3}-\varrho_{13}\mathbf{k}_{2}+\varrho_{23}\mathbf{k}_{1}\right)^{2}.$
(143)

Similar expressions are also valid for $\Phi_{ls}(E)$. We have

$\displaystyle\Phi_{ls}(E)k_{s}^{2}$
$\displaystyle=\left\\{\left[\mathbf{b}^{(2)}E-\mathbf{a}^{(2)}\right]\mathbf{k}_{s}\right\\}^{2}-\left[w^{(2)}\right]^{2}\Phi_{l}(E),$
(144) $\displaystyle D_{ls}$
$\displaystyle=\left[w^{(2)}\right]^{2}\left\\{A_{\Omega}^{(2)}-m^{2}\left[\mathbf{b}^{(2)}\right]^{2}-\varrho_{ls}^{2}\right\\}.$
(145)

Equation (144) leads to one more representation of Eq. (123):

$\displaystyle\Theta\\!\left(D_{ls}\right)=\Theta\\!\left(D_{ls}\right)\Theta\\!\left(\Phi_{l}(E_{ls}^{\pm})\right)\Theta\\!\left(\Phi_{s}(E_{ls}^{\pm})\right).$
(146)

The generalized trinomial $\Omega_{L}$ allows one to deal in the same way with
integrals containing $\Theta\\!\left(\pm\Omega_{L}\right)$ or
$\Omega_{L}^{-1}$ like $I_{H,K,L,M}^{\pm}$. One has to find the roots and, if
they are real, to make integration over the corresponding intervals in the
first case or extracting singularities in the second one.

##  Appendix C The product and $\phi$ integration for $L=3$

The product part of the integrand can be explicitly written as

$\displaystyle\lim_{\epsilon\rightarrow 0}$
$\displaystyle\frac{pk_{1}}{\lambda_{1}(E-E_{01}-i\epsilon)+pk_{1}x}\frac{pk_{2}}{\lambda_{2}(E-E_{02}-i\epsilon)+pk_{2}(\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}+x\cos\delta_{2})}$
(147)
$\displaystyle\quad\times\frac{pk_{3}}{\lambda_{3}(E-E_{03}-i\epsilon)+pk_{3}\big{[}(\cos\phi_{3}\cos\phi+\sin\phi_{3}\sin\phi)\sin\delta_{3}\sqrt{1-x^{2}}+x\cos\delta_{3}\big{]}}$
$\displaystyle=$
$\displaystyle\frac{1}{(z_{1}+x)\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}$
$\displaystyle\quad\times\frac{1}{z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}}$
$\displaystyle-\pi^{2}\,{\rm
sgn}(\lambda_{1}\lambda_{2})\,\frac{\delta\left(z_{1}+x\right)\,\delta\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}{z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}}$
$\displaystyle-\pi^{2}\,{\rm
sgn}(\lambda_{1}\lambda_{3})\frac{\delta\left(z_{1}+x\right)\,\delta\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right)}{z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}}$
$\displaystyle-\pi^{2}\,{\rm
sgn}(\lambda_{2}\lambda_{3})\frac{\delta\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}{z_{1}+x}$
$\displaystyle\quad\times\delta\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right)$
$\displaystyle+i\pi\,{\rm
sgn}\lambda_{1}\,\frac{\delta\left(z_{1}+x\right)}{\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}$
$\displaystyle\quad\times\frac{1}{z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}}$
$\displaystyle+i\pi\,{\rm
sgn}\lambda_{2}\frac{\delta\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}{(z_{1}+x)\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right)}$
$\displaystyle+i\pi\,{\rm
sgn}\lambda_{3}\frac{\delta\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right)}{(z_{1}+x)\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)}$
$\displaystyle-i\pi^{3}\,{\rm
sgn}(\lambda_{1}\lambda_{2}\lambda_{3})\,\delta\left(z_{1}+x\right)\,\delta\left(z_{2}+x\cos\delta_{2}+\cos\phi\sin\delta_{2}\sqrt{1-x^{2}}\right)$
$\displaystyle\quad\times\delta\left(z_{3}+x\cos\delta_{3}+\cos\phi\sin\delta_{3}\cos\phi_{3}\sqrt{1-x^{2}}+\sin\phi\sin\delta_{3}\sin\phi_{3}\sqrt{1-x^{2}}\right).$

After $\phi$ integration using Eqs. (55), (VI.2), and (VI.2), we obtain

$\displaystyle{\rm Re}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{1}{4k_{1}k_{2}k_{3}}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{f(\pm
E-\mu)}{p^{2}}$
$\displaystyle\quad\int\limits_{-1}^{1}{\mathrm{d}}x\Bigg{\\{}\sin\delta_{2}\frac{\beta_{23}+x\alpha_{23}}{(z_{1}+x)\Psi_{23}(x)\sqrt{\Delta_{2}(x)}}\,{\rm
sgn}(z_{2}+x\cos\delta_{2})\Theta\\!\left(\Delta_{2}(x)\right)$
$\displaystyle\quad+\sin\delta_{3}\frac{\beta_{32}+x\alpha_{32}}{(z_{1}+x)\Psi_{23}(x)\sqrt{\Delta_{3}(x)}}\,{\rm
sgn}(z_{3}+x\cos\delta_{3})\Theta\\!\left(\Delta_{3}(x)\right)$
$\displaystyle\quad-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{2})\,\sin\delta_{2}\,\delta\left(z_{1}+x\right)\frac{\beta_{23}+x\alpha_{23}}{\Psi_{23}(x)\sqrt{-\Delta_{2}(x)}}\,\Theta\\!\left(-\Delta_{2}(x)\right)$
$\displaystyle\quad-\pi\,{\rm
sgn}(\lambda_{1}\lambda_{3})\,\sin\delta_{3}\,\delta\left(z_{1}+x\right)\frac{\beta_{32}+x\alpha_{32}}{\Psi_{23}(x)\sqrt{-\Delta_{3}(x)}}\,\Theta\\!\left(-\Delta_{3}(x)\right)$
$\displaystyle\quad-\pi\,{\rm
sgn}(\lambda_{2}\lambda_{3})\frac{\sin\delta_{2}\sin\delta_{3}\left|\sin\phi_{3}\right|\delta\left(\Psi_{23}(x)\right)}{z_{1}+x}\Bigg{\\}}$
(148)

and

$\displaystyle{\rm Im}\mathcal{Z}^{\pm}_{3}$
$\displaystyle=\frac{1}{4k_{1}k_{2}k_{3}}\int\limits_{m}^{\Lambda_{E}}\\!{\mathrm{d}}E\,\frac{f(\pm
E-\mu)}{p^{2}}$
$\displaystyle\times\int\limits_{-1}^{1}{\mathrm{d}}x\Bigg{\\{}\pi\,{\rm
sgn}\lambda_{1}\sin\delta_{2}\,\delta\left(z_{1}+x\right)\frac{\beta_{23}+x\alpha_{23}}{\Psi_{23}(x)\sqrt{\Delta_{2}(x)}}\,{\rm
sgn}(z_{2}+x\cos\delta_{2})\Theta\\!\left(\Delta_{2}(x)\right)$
$\displaystyle\quad+\pi\,{\rm
sgn}\lambda_{1}\sin\delta_{3}\,\delta\left(z_{1}+x\right)\frac{\beta_{32}+x\alpha_{32}}{\Psi_{23}(x)\sqrt{\Delta_{3}(x)}}\,{\rm
sgn}(z_{3}+x\cos\delta_{3})\Theta\\!\left(\Delta_{3}(x)\right)$
$\displaystyle\quad+{\rm
sgn}\lambda_{2}\,\sin\delta_{2}\frac{\beta_{23}+x\alpha_{23}}{(z_{1}+x)\Psi_{23}(x)\sqrt{-\Delta_{2}(x)}}\,\Theta\\!\left(-\Delta_{2}(x)\right)$
$\displaystyle\quad+{\rm
sgn}\lambda_{3}\,\sin\delta_{3}\frac{\beta_{32}+x\alpha_{32}}{(z_{1}+x)\Psi_{23}(x)\sqrt{-\Delta_{3}(x)}}\,\Theta\\!\left(-\Delta_{3}(x)\right)$
$\displaystyle\quad-\pi^{2}\,{\rm
sgn}(\lambda_{1}\lambda_{2}\lambda_{3})\,\delta\left(z_{1}+x\right)\,\sin\delta_{2}\sin\delta_{3}\left|\sin\phi_{3}\right|\delta\left(\Psi_{23}(x)\right)\Bigg{\\}}.$
(149)

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