The maxcut of the sunrise with different masses in the continuous Minkoskean
dimensional regularisation
Filippo Calecaa,b 111e-mail<EMAIL_ADDRESS>and Ettore Remiddia,
222e-mail<EMAIL_ADDRESS>
a DIFA, Università di Bologna and INFN, Sezione di Bologna, I-40126 Bologna,
Italy

b Univ Lyon, Ens de Lyon, CNRS, Laboratoire de Physique, F-69342 Lyon, France

We evaluate the maxcut of the two loops sunrise amplitude with three different
masses by using the Minkoskean (as opposed to the usual Euclidean) continuous
dimension regularisation, obtaining in that way six related but different
functions expressed in the form of onedimensional finite integrals. We then
consider the $4$th order homogeneous equation valid for the maxcut, and show
that for arbitrary dimension $d$ the six functions do satisfy the equation. We
further discuss the $d=2,3,4$ cases, verifying that only four of them are
linearly independent. The equal mass limit is also shortly discussed.

Key words: Sunrise, Differential equations, Elliptic Integrals, Elliptic
Polylogarithms

##  1 Introduction

As known, one can derive an inhomogeneous differential equation for any
Feynman amplitude. The solution of that equation can then be obtained by the
Euler’s variation of the constants method, which relies on the knowledge of
the solutions of the associated homogeneous equation. The so-called maxcut
amplitude (it i.e. the loop integral in which all the propagators are replaced
by the corresponding $\delta$-functions) provides always with a solution of
the homogeneous equation. (For the introduction and use of the maxcut, see [1]
).

But it can happen, within the usual continuous dimensional regularisation
approah, that the information so obtained is not sufficient (if the
differential equation is of higher order) to find all the solutions, or even
that the evaluation of the maxcut gives a vanishing result, because the
constraints of the $\delta$-functions lead to a vanishing integration range
for the loop variables.

To overcome the problem, one can give [2] a Minkowskean metric to the
regularising components, as opposed to the Euclidean metric which id always
implicitly used. (Let us recall, in this context, that the regularising loop
components do not mix with the components of the physical external momenta, so
that Wick rotations can be safely carried out). It is found that, when the
Minkowskean metric is used, the evaluation of the maxcut gives a full
information on all the solutions of the homogeneous differential equation.

We consider in this paper the case of the sunrise with three different masses,
and we show that the maxcut evaluated within the Minkowskean metric approach
does in fact provide with a full information on all the solutions of the 4th
order homogeneous equation. Some preliminary work [2] indicates that the new
Minkowskean approach could apply, more in general, to three and four point
amplitudes etc. as well.

In the considered case of the sunrise with three different masses and in
$d$-continuous dimensions, the proposed Minkowskean approach evaluates the
maxcut, for each value of the external physical momentum, as the sum of 6
functions. We verify that each of them separately satisfies the homogeneous
4th order equation equation, and for $d=2,3,4$ we verify also that four of
them are linearly independent.

The mathematical tools which we use are almost elementary (such as integration
by parts identities), but many of the formulas are in general too cumbersome
to be reported explcitly. Correspondingly, for the sake of clarity, our
treatment will be occasionally somewhat pedagogical.

The plan of the paper is as follows.
Section 2 recalls the 4th order equation for the maxcut of the Sunrise
amplitude.
Section 3 describes the evaluation of the maxcut.
Section 4 verifies that the functions introduced in the previous Section
satisfy the 4th order homogeneous equation.
Section 5 deals with the $d=2$ case.
Section 6 deals with the $d=3$ case.
Section 7 deals with the $d=4$ case.
Section 8 discusses the equal mass Sunrise amplitude.
Appendix A describes the explicit evaluation of the maxcut.
Appendix B gives the explicit expression of the 4th order homogeneous
equation.
Appendix C discusses the relations between the functions of the $d=2$ case
with the elliptic functions.

##  2 The Equations for the Sunrise amplitude and its Maxcut

Let us write the Sunrise aplitude with different masses in the continuous
dimensional regularisation as

$S(d,s)=\int\mathcal{D}^{d}k\int\mathcal{D}^{d}q\ \frac{1}{D_{1}-i\epsilon}\
\frac{1}{D_{2}-i\epsilon}\ \frac{1}{D_{3}-i\epsilon}\ ,$ (2.1)

where:
$k$ and $q$ are the integration loop momenta;

$\displaystyle D_{1}$ $\displaystyle=q^{2}+m_{1}^{2}\ ,$ $\displaystyle D_{2}$
$\displaystyle=(k-q)+m_{2}^{2}\ ,$ $\displaystyle D_{3}$
$\displaystyle=(p-k)^{2}+m_{3}^{2}\ ;$ (2.2)

$(m_{1},m_{2},m_{3})$ are the three (in principle different) masses of the
three propagators;
$p$ is the external momentum, and

$s=-p^{2}=p_{0}^{2}-\vec{p}^{2}\ .$ (2.3)

Well established methods and computer programs are available for deriving a
differential equation in $s$ for the amplitude $S(d,s)$; by using LiteRed [3]
we obtained a 4th order inhomogneous linear differential equation for
$S(d,s)$, which reads

$D_{4}(d,s)\ S(d,s)=T(d,s)\ ,$ (2.4)

where $T(d,s)$ is the imhogeneous term (a sum of tedpoles) which will be
ignored in the rest of this paper, and $D_{4}(d,s)$ is a 4th order
differential operator which can be written as

$\displaystyle D_{4}(d,s)$
$\displaystyle=C_{4}(d,s)\frac{d^{4}}{ds^{4}}+C_{3}(d,s)\frac{d^{3}}{ds^{3}}$
$\displaystyle+C_{2}(d,s)\frac{d^{2}}{ds^{2}}+C_{1}(d,s)\frac{d}{ds}+C_{0}(d,s)\
,$ (2.5)

the 5 coefficients $C_{i}(d,s6),i=0,1,..,4$ being polynomials in $d,s$ and in
the three masses $m_{k},k=1,2,3$. The explicit expressions of those
polynomials, which are rather cumbersome, are explicitly written in Appendix
B.

Let us remind here that the Euler’s variation of constants method, commonly
used for the solution of an inhomogeneous 4th order linear differential
equation as Eq.(2.4), requires the knowledge of 4 linearly independent
solutions of the associated 4th order homogeneous equation, which in our case
is

$D_{4}(d,s)\ \phi(d,s)=0\ .$ (2.6)

We consider now the maxcut of the Sunrise amplitude, which in the notation of
Eq.(2.1) is

$M(d,s)=\int\mathcal{D}^{d}k\int\mathcal{D}^{d}q\
\delta(D_{1})\delta(D_{2})\delta(D_{3})\ .$ (2.7)

By repeating the procedure already used for obtaining the inhomogeneous
differential equation for the Sunrise amplitude, one finds that the maxcut
$M(d,s)$ satisfies the above homogeneous equation

$D_{4}(d,s)\ M(d,s)=0\ .$ (2.8)

Indeed, the inhomogeneous part of the equation for $S(d,s)$ is due to the
presence, in the integration by parts identities used in the derivation of the
inhomogeneous equation, of terms of the form

$D_{i}\times\frac{1}{D_{i}-i\epsilon}=1\ ;$ (2.9)

the same derivation (with the same integration by parts identities) can be
used for deriving the equation for the maxcut $M(d,s)$, in which all the
propagators $1/(D_{i}-i\epsilon)$ are replaced by the corresponding
$\delta$-function $\delta(D_{i})$, so that Eq.(2.9) becomes

$D_{i}\times\delta(D_{i}-i\epsilon)=0\ ,$ (2.10)

and inhomogeneous terms drop out.

We will show in the next Section how to evaluate the maxcut $M(d,s)$, and in
the subsequent Sections we will discuss the solutions of Eq.(2.8) and their
properties.

##  3 The Maxcut

We now evaluate the Maxcut $M(d,s)$ defined by Eq.(2.7), with
$D_{1},D_{2},D_{3}$ given by Eq.s(2.2).

We write the components of the 4 dimensional vector $p$, in full generality,
as

$p=(p_{0},p_{x},0,0),$ (3.1)

i.e.setting $p_{y}=p_{z}=0$. As $p$ does not span the $y$ and $z$ directions,
we write the components of the continuous $d$-dimensional loop integration
momenta as

$\displaystyle k$ $\displaystyle=(k_{0},k_{x},K_{n}),{\hskip
14.22636pt}n=1,2,..,d-2\ ,$ $\displaystyle q$
$\displaystyle=(q_{0},q_{x},Q_{n}),{\hskip 14.22636pt}n=1,2,..,d-2\ ,$ (3.2)

where we have merged the $y$ and $z$ components of $k$ and $q$ and their
$(n-4)$ regularising components into the $(d-2)$-dimensional vectors
$(K_{n},Q_{n})$, so that one has

$\mathcal{D}^{d}k=dk_{0}\ dk_{x}\ d^{d-2}K\ ,$ (3.3) $\mathcal{D}^{d}q=dq_{0}\
dq_{x}\ d^{d-2}Q\ .$ (3.4)

In the usual dimensional regularisation, the regularising components are
assumed to have an Euclidean metric; in this paper, however, we give to
$(\vec{K},\vec{Q})$ a Minkowskean metric, so that

$\displaystyle k^{2}$ $\displaystyle=-k_{0}^{2}+k_{x}^{2}-K^{2}\ ,$
$\displaystyle q^{2}$ $\displaystyle=-q_{0}^{2}+q_{x}^{2}-Q^{2}\ ,$
$\displaystyle(k\cdot q)$
$\displaystyle=-k_{0}q_{0}+k_{x}q_{x}-(\vec{K}\cdot\vec{Q})\ .$ (3.5)

A few comments on that choice. As the external physical vector $p$ spans only
the time and $x$ dimensions, scalar products of $p$ and the components of the
regularising vectors $(\vec{K},\vec{Q})$ never occur, and, quite in general,
the metric of the components of the loop momenta corresponding to the
dimensions not spanned by the physical vectors does not matter and can be
changed at will from Euclidean to Minkowskean (Wick rotation).

As a further comment, let us recall that a propagator is equal to a Principal
value plus a $\delta$-function

$\frac{1}{D-i\epsilon}=\frac{\mathcal{P}}{D}+i\pi\delta(D)\ .$ (3.6)

In the amplitude with full propagators $1/(D-i\epsilon)$, as $S(d,s)$ of
Eq.(2.1), the above change of the metric does not affect the value of the
amplitude, but the change does affect the maxcut Eq.(2.7), which involves only
the $\delta$-functions (and not the complete propagators).

At any rate, in the rest of this paper we will be concerned only with the
maxcut, defined by Eq.(2.7) and evaluated with regularising loop momenta with
Minkoskean metric, independently of possible Wick rotations of the complete
amplitude Eq.(2.1). We will therefore limit ourselves to the study of the
solutions of Eq.(2.8), where however $D_{4}(d,s,m_{i}^{2})$ is exactly the
differential operator appearing in the inhomogeneous differential equation
valid for the amplitude Eq.(2.4).

After those preliminary considerations, let us start the actual evaluation of
the maxcut by defining

$B(d,-k^{2})=\int\mathcal{D}^{d}q\ \delta(D_{1})\delta(D_{2})\ .$ (3.7)

Eq.(2.7) then becomes

$M(d,s)=\int\mathcal{D}^{d}k\ B(d,-k^{2})\ \delta(D_{3})\ ;$

by further introducing a factor

$\int db\ \delta(b+k^{2})=1\ ,$

(it is understood that the integration in $b$ ranges from $-\infty$ to
$+\infty$) and suitably interchanging the orders of integration one has

$\displaystyle M(d,s)$ $\displaystyle=\int\mathcal{D}^{d}k\int db\
B(d,-k^{2})\ \delta(b+k^{2})\ \delta(D_{3})$ $\displaystyle=\int db\ B(d,b)\
C(d,-p^{2},b)\ ,$ (3.8)

where we have introduced

$C(d,-p^{2},b)=\int\mathcal{D}^{d}k\ \delta(b+k^{2})\ \delta(D_{3})\ .$ (3.9)

For simplicity, in this paper we discuss only the case of timelike $p$ above
threshold, i.e.with

$-p^{2}=s=E^{2},{\hskip 14.22636pt}E>(m_{1}+m_{2}+m_{3}).$ (3.10)

Referrinng to Appendix A for more details of the integrations (which are
essentially elementary), we report here the explicit expression of $M(d,s)$ as
an integral in $b$, in the case of different masses, ordered as

$m_{1}\geq m_{2}\geq m_{3}.$ (3.11)

To that aim, and for the subsequent discussions of the paper, we introduce the
6 constants (with respect to $b$)

$\displaystyle b_{-1}$ $\displaystyle=-\infty\ ,$ $\displaystyle b_{0}$
$\displaystyle=0\ ,$ $\displaystyle b_{1}$ $\displaystyle=(m_{1}-m_{2})^{2}\
,$ $\displaystyle b_{2}$ $\displaystyle=(m_{1}+m_{2})^{2}\ ,$ $\displaystyle
b_{3}$ $\displaystyle=(E-m_{3})^{2}\ ,$ $\displaystyle b_{4}$
$\displaystyle=(E+m_{3})^{2}\ ,$ $\displaystyle b_{5}$ $\displaystyle=+\infty\
,$ (3.12)

which for $E$ above threshold, Eq.(3.10), are in increasing order

$b_{-1}<b_{0}<b_{1}<b_{2}<b_{3}<b_{4}<b_{5}\ ,$ (3.13)

and the the 4th order polynomial

$R_{4}(b)=(b-b_{1})(b-b_{2})(b-b_{3})(b-b_{4})\ .$ (3.14)

The result of our calculation of $M(d,s)$ can then be written as the sum of 6
terms

$M(d,s)=\sum_{i=0,5}C_{i}(d)\ M_{i}(d,s)\ ,$ (3.15)

where:
the explicit expressions of the factors $C_{i}(d)$, which depend only on $d$
(albeit not needed for the discussions which will follow) are given in
Eq.s(A.15) ;
the $M_{i}(d,s)$ are the following 6 quantities, involving 6 one dimensional
integrals

$M_{i}(d,s)=s^{\frac{2-d}{2}}\int_{b_{i-1}}^{b_{i}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\ .$ (3.16)

The 6 integrals span the whole integration range of the integration variable
$b$ from $-\infty$ to $+\infty$. The convergence of the separate integrals
depends on the behaviour of the integrands at the end points of integration,
which in turn depend on $d$. In particular:
the convergence of $M_{0}(d,s)$, whose integration range is $-\infty<b<0$,
requires $d<8/3$ at $b=b_{-1}=-\infty$ and $d<4$ at $b=b_{0}=0$ ;
$M_{1}(d,s)$ requires $d<4$ at $b_{0}=0$ and $d>1$ at $b=b_{1}$;
$M_{2}(d,s),M_{3}(d,s)$ and $M_{4}(d,s)$ require $d>1$;
$M_{5}(d,s)$ (range $b_{4}<b<b_{5}=+\infty$) requires $d>1$ at $b=b_{4}$ and
$d<8/3$ at $b=b_{5}$;
Therefore all the 6 integrals and their sum $M(d,s)$, Eq.(3.15), converge for
$d$ within the range $1<d<8/3$. The value of $M(d,s)$ as given by Eq.(3.15),
is therefore well defined, and the maxcut $M(d,s)$ of Eq.(2.7) is a solution
of the 4th order homogenous equation Eq.(2.8).

Let us recall here that the imaginary part of $S(d,s)$ (or, more precisely,
its discontinuity in $s$) in the usual Euclidean regularisation scheme is
equal, up to a constant ($s$ independent) factor, to

$\displaystyle{\rm Im}S(d,s)$
$\displaystyle=\theta\left(+E-(m_{1}+m_{2}+m_{3})\right)\times\int\mathcal{D}^{d}k\int\mathcal{D}^{d}q\
\ \theta(+q_{0})\delta(D_{1})\ \theta(k_{0}-q_{0})\delta(D_{2})\
\theta(p_{0}-q_{0})\delta(D_{3})$
$\displaystyle+\theta\left(-E-(m_{1}+m_{2}+m_{3})\right)\times\int\mathcal{D}^{d}k\int\mathcal{D}^{d}q\
\ \theta(-q_{0})\delta(D_{1})\ \theta(q_{0}-k_{0})\delta(D_{2})\
\theta(q_{0}-p_{0})\delta(D_{3})\ ,$ (3.17)

which, when evaluated, is found to be proportional to the single integral
$M_{3}(d,s)$ defined above, see Eq.(3.16), whose integration range in $b$ is
$(m_{1}+m_{2})^{2}<b<(E-m_{3})^{2}\ .$

On the other hand, ${\rm Im}S(d,s)$ is a solution of the homogeneous equation
Eq.(2.6), as the inhomogeneous term $T(d,s)$ of the complete equation Eq.(2.4)
has vanishing discontinuity. Therefore, $M_{3}(d,s)$ alone, independenty of
the other $M_{i}(d,s)$, is also a solution of the homogeneous equation.

Further, again in the conventional Euclidean regularization, (at variance with
the Minkowskean regularisation used in this paper) one finds (we omit details
for obvious space problems) that the whole maxcut is given by a linear
combination, similar to Eq.(3.15), of the 4 integrals $M_{0}(d,s)$,
$M_{1}(d,s)$, $M_{3}(d,s)$ and $M_{5}(d,s)$ only [8].

One is then naturally led to guess that any of the 6 terms $M_{i}(d,s),$
separately taken, is a solution of that 4th order equation; on the other hand,
it is known that a 4th order equation can have only 4 independent solutions,
so that the 6 terms cannot be all linearly independent. Those issues will be
discussed in the rest of this paper.

##  4 Derivatives and differential equations of the $b$ integrals for
arbitrary $d$

In this section we show how to evaluate the 4 $s$-derivatives of each of the 6
functions $M_{i}(d,s)$, (3.16), and check that each of the 6 functions does
separately satisfy the 4th order differential equation Eq.(2.6). The 6
equations for each of the 6 $M_{i}(d,s)$ can be seen as an identity involving
5 functions, namely $M_{i}(d,s)$ and its first 4 derivatives. By means of
suitable integration by parts identities in $b$, we will then express those 5
functions in terms of a set of 4 auxiliary functions only, we will substitute
those expressions in the 4th order equation, which becomes a combination of
those 4 auxiliary functions, and then we will verify that after the
substitutions each of the six 4th order equations does in fact vanish.

To marginally simplify notation and formulas, we introduce the 6 new
functions, for $i=0,1,..,5$,

$N_{i}(d,s)=\int_{b_{i-1}}^{b_{i}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\ ,$ (4.1)

so that, obviously, Eq.(3.16) becomes

$M_{i}(d,s)=s^{\frac{2-d}{2}}N_{i}(d,s)\ ,$ (4.2)

and we will discuss the derivatives of the $N_{i}(d,s)$ instead of the
derivatives of $M_{i}(d,s)$.

As the $b$ integral $N_{i}(d,s)$ depend on the 7 parameters $b_{j}$,
$j=-1,0,..,5$, and therefore depend on $s=E^{2}$ through the presence of $s$
in $b_{3},b_{4}$, see Eq.s(3.12), we have quite in general

$\frac{d}{ds}N_{i}(d,s)=\left(\frac{db_{3}}{ds}\frac{d}{db_{3}}+\frac{db_{4}}{ds}\frac{d}{db_{4}}\right)N_{i}(d,s)\
;$ (4.3)

we will therefore look for derivatives of the integrals $N_{i}(d,s)$ with
respect to $b_{3}$ and $b_{4}$. The approach could be easily extended also to
$b_{1}$ and $b_{2}$, allowing the evaluation of the derivatives with respect
to the masses, which however we will not consider in this paper.

We discuss first the derivatives of the 3 integrals

$N_{n}(d,s),\hskip 28.45274ptn=2,3,4$ (4.4)

whose integration ranges are $b_{1}<b<b_{2}$, $b_{2}<b<b_{3}$ and
$b_{3}<b<b_{4}$.

If $b_{3}$ or $b_{4}$ is an end-point of the integration, a first contribution
to the derivatives of those integrals can arise, in principle, from the
corresponding end-point value of the integrands. As already observed in
Section 3, the 3 integrals listed in Eq.(4.4) converge for $d>1$; therefore,
we can consider the $d>3$ case, so that the contributions from the end-point
values of the derivatives in $b_{3}$ and $b_{4}$ of those integrals vanish
(and by analitic continuation we can extend the final results also to $1<d\leq
3$). We then obtain for $k=3,4$

$\frac{d}{db_{k}}N_{n}(d,s)=\int_{b_{n-1}}^{b_{n}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\left(-\frac{1}{b-b_{k}}\right)\ ,$ (4.5)

which converges for $d>3$. But we need also higher order derivatives, up to
the 4th; by simply taking repeatedly the $d/db_{k}$ derivatives of Eq.(4.5)
(with higher vales of $d$, for convergence sake), we would get up to 4 powers
of the denominator $1/(b-b_{k})$, which including $N_{n}(d,s)$ gives a total
of 5 apparently unrelated functions, while our purpose is to express the
$N_{n}(d,s)$ and their 4 derivatives in terms of 4 independent functions only.

To overcome that problem, for each of the 3 integrals $N_{n}(d,s)$ of Eq.(4.4)
we consider a set of 4 auxiliary integrals

$N_{n}^{(j)}(d,s)=\int_{b_{n-1}}^{b_{n}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times b^{j}\ ,\ ,$ (4.6)

where $j=-1,0,1,2$, and $N_{n}^{(0)}(d,s)=N_{n}(d,s)$.

We then write the following 4 integration by part identities in $b$

$\int_{b_{n-1}}^{b_{n}}db\left(\delta(b-b_{n})-\delta(b-b_{n-1})-\frac{d}{db}\right)\
|b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\left(\frac{|R_{4}(b)|}{b-b_{k}}\right)=0\ .$
(4.7)

For $d>1$ all the end-point contributions vanish; considering for definiteness
the case $b_{k}=b_{3}$ and by evaluating explicitly the $b$ derivative, we can
derive from Eq.(4.7), after some algebra, the relation

$\displaystyle\int_{b_{n-1}}^{b_{n}}db$ $\displaystyle\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\frac{1}{b-b_{3}}=\frac{2}{(d-3)(b_{3}-b_{1})(b_{3}-b_{2})(b_{4}-b_{3})}\int_{b_{n-1}}^{b_{n}}db\
|b|^{\frac{2-d}{2}}\ |R_{4}(b)|^{\frac{d-3}{2}}$
$\displaystyle\times\Biggl{[}$ $\displaystyle+\left(\frac{3}{2}d-2\right)\
\times b^{2}$
$\displaystyle+\left(\frac{d-3}{2}b_{3}-\left(d-\frac{3}{2}\right)(b_{1}+b_{2}+b_{4})\right)\times
b$
$\displaystyle+\left(-\frac{d-3}{2}b_{3}(b_{1}+b_{2}-b_{3}+b_{4})+\frac{d-2}{2}(b_{1}b_{2}+b_{1}b_{4}+b_{2}b_{4})\right)\times
1$ $\displaystyle+\frac{d-2}{2}b_{1}b_{2}b_{4}\times\frac{1}{b}\Biggr{]}\ ,$
(4.8)

which can be used to rewrite the $b_{3}$ derivative of each of the 3 functions
$N_{n}(d,s)=N_{n}^{(0)}(d,s),n=2,3,4$, see Eq.(4.5) for $b_{k}=b_{3}$, in
terms of the 4 auxiliary integrals $N_{n}^{(j)}(d,s),\ j=-1,0,1,2,$ Eq.(4.6).

Similar formulas can also be easily derived from Eq.(4.7) for the other values
of $b_{k}$, namely $(b_{1},b_{2},b_{4})$, and can be used for the derivatives
of the $N_{n}^{(0)}(d,s)$ with respect to those values of $b_{k}$. It is to be
noted that, although those formulas for the derivatives, in which the
$1/(b-b_{k})$ factors are no longer present, were established for $d>3$, they
are are actually valid also for $d>1$.

But the same happens also for all the $b_{k}$ derivatives of all the other
$N_{n}^{(j)}(d,s)$, as it is immediate to check. Indeed, one has, for instance

$\displaystyle\frac{d}{db_{k}}N_{n}^{(1)}(d,s)$
$\displaystyle=\int_{b_{n-1}}^{b_{n}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\left(-1-\frac{b_{k}}{b-b_{k}}\right)\ ,$
$\displaystyle\frac{d}{db_{k}}N_{n}^{(-1)}(d,s)$
$\displaystyle=\int_{b_{n-1}}^{b_{n}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\frac{1}{b_{k}}\times\left(\frac{1}{b}-\frac{1}{b-b_{k}}\right)\
,$

with a similar formula holding for $N_{n}^{(2)}(d,s)$ as well.

Summarising, by replacing systematically the integrals containing the factor
$1/(b-b_{k})$ with the proper analog of Eq.(4.8), all the $b_{k}$ derivatives
of all the integrals $N_{n}^{(j)}(d,s)$ are then completely expressed in terms
of the same $N_{n}^{(j)}(d,s)$ only. By suitably iterating the procedure, one
can express all the higher order derivatives of the 3 functions $N_{n}(d,s)$
of Eq.(4.4) with respect to all the 4 parameters $b_{k}$ in terms of the 4
$N_{n}^{(j)}(d,s)$ only.

By recalling Eq.s(4.2),(4.3) on can then obtain all the $s$ derivatives of the
$M_{i}(d,s)$, Eq.(3.16), for $i=2,3,4$, in terms of the corresponding 4
auxiliary integrals Eq.(4.6). The explicit expressions are rather clumsy, and
involve too many terms (from 189 of the 1st $s$ derivative up to almost 17
thousand of the 4th) to be written here; but by using those expressions one
can check that indeed the 3 integrals $M_{2}(d,s),M_{3}(d,s),M_{4}(d,s)$ do
satisfy, separately, the 4th order differential equation Eq.(2.6).

We consider next the derivatives with respect to $b_{3}$ and $b_{4}$ of

$N_{1}(d,s)=\int_{b_{0}=0}^{b_{1}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\ ,$ (4.9)

which due to the end-points $(0,b_{1})$ converges in the range $1<d<2$. For
this case we take as auxiliary integrals, instead of the Eq.s(4.6), the 4
integrals, for $j=1,..,4$,

$N_{1}^{(j)}(d,s)=\int_{0}^{b_{1}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\left(1\ ;\ \frac{1}{b_{2}-b}\ ;\
\frac{1}{b_{3}-b}\ ;\ \frac{1}{b_{4}-b}\right)\ ,$ (4.10)

with $N_{1}^{(1)}(d,s)=N_{1}(d,s)$, and correspondingly the following 4
integration by parts identities

$\displaystyle\int_{0}^{b_{1}}db\left(\delta(b-b_{1})-\delta(b)-\frac{d}{db}\right)$
$\displaystyle|b|^{\frac{2-d}{2}}\ |R_{4}(b)|^{\frac{d-1}{2}}$
$\displaystyle\times\left[(b_{1}-b)\left(1\ ;\ \frac{1}{b_{2}-b}\ ;\
\frac{1}{b_{3}-b}\ ;\ \frac{1}{b_{4}-b}\right)\right]=0\ .$ (4.11)

By following the same procedure as for the previous case, one finds that,
thanks to the identities Eq.s(4.11), all the $b_{3}$ and $b_{4}$ derivatives
of the auxiliary integrals Eq.(4.10), hence also the $s$ derivative of
$M_{1}(d,s)$, can be expressed in terms of those 4 auxiliary integrals only –
and then one can also check that $M_{1}(d,s)$ does satisfy the 4th order
equation Eq.(2.6).

Similarly, the case of

$N_{5}(d,s)=\int_{b_{4}}^{b_{5}=\infty}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\ ,$ (4.12)

can be treated by introducing the 4 auxiliary integrals, for $j=1,..,4$,

$N_{5}^{(j)}(d,s)=\int_{b_{4}}^{\infty}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\times\left(1\ ;\ \frac{1}{b-b_{1}}\
;\frac{1}{b-b_{2}}\ ;\ \frac{1}{b-b_{3}}\ ;\right)\ ,$ (4.13)

with $N^{(1)}_{5}(d,s)=N_{5}(d,s)[B$, and correspondingly the following 4
integration by parts identities

$\int_{b_{4}}^{\infty}db\ \frac{d}{db}\left[\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-1}{2}}\times\ (b-b_{4})\left(1\ ;\ \frac{1}{b-b_{1}}\ ;\
\frac{1}{b-b_{2}}\ ;\ \frac{1}{b-b_{3}}\right)\right]=0\ ,$ (4.14)

where the vanishing end-point contributions were dropped for ease of typing.

By following the above described procedure, the $s$-derivatives of
$M_{5}(d,s)$, can be expressed in terms of the 4 auxiliary integrals of
Eq.(4.13) – and then one cal also check that $M_{5}(d,s)$ does satisfy the 4th
order equation Eq.(2.6).

The last case to consider is

$N_{0}(d,s)=\int_{b_{-1}=-\infty}^{0}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}=\int_{0}^{\infty}dc\ c^{\frac{2-d}{2}}\
|R_{4}(-c)|^{\frac{d-3}{2}}\ .$ (4.15)

We proceed as in the previous cases, with the auxiliary integrals, for
$j=1,..,4$,

$N_{0}^{(j)}(d,s)=\int_{0}^{\infty}dc\ c^{\frac{2-d}{2}}\
|R_{4}(-c)|^{\frac{d-3}{2}}\times\left(1\ ;\ \frac{1}{c+b_{1}}\
;\frac{1}{c+b_{2}}\ ;\ \frac{1}{c+b_{3}}\ ;\right)\ ,$ (4.16)

with $N_{0}^{(1)}(d,s)=N_{0}(d,s)$, and the integration by parts identities

$\int_{0}^{\infty}dc\ \frac{d}{dc}\left[c^{\frac{2-d}{2}}\
|R_{4}(-c)|^{\frac{d-3}{2}}\times\left(1\ ;\ c\ ;\ \frac{1}{c+b_{3}}\ ;\
\frac{1}{c+b_{4}}\right)\right]=0\ .$ (4.17)

Thanks to the identities, we are able to express the $s$-derivatives of
Eq.(4.15) and of $M_{0}(d,s)$, Eq.(3.16) in terms of the master integrals of
Eq.(4.16) – and to check finally that also $M_{0}(d,s)$ does satisfy the 4th
order equation Eq.(2.6).

Summarising, we have shown that all the 6 functions $M_{i}(d,s),i=0,1,..,5$
defined in Eq.s(4.2,4.1) are solutions of the homogeneous equation Eq.(2.6),

$D_{4}(d,s)M_{i}(d,s)=0\ ,\hskip 28.45274pti=0,1,..,5\ .$ (4.18)

But the differential operator $D_{4}(d,s)$ appearing in the equation is a 4th
order operator, hence the differential equation has only 4 linearly
independent solutions. We were however not able to show, so far, that for
arbitrary $d$, 4 of those 6 solutions are indeed independent. In the following
sections we will consider the cases of $d=2,3,4$, and we will show that in
those cases 4 of the solutions are indeed linearly independent, as expected.
We further, again for $d=2,3,4,$ we show that the knowlegde of the solutions
allows to factorise the 4th order operator $D_{4}(d,s)$.

##  5 Derivatives and differential equation of the $b$ integrals for $d=2$

We study in this section the solutions of Eq.(4.18) at $d=2$. As the six
solutions $M_{i}(d,s)$, discussed in the above Section, are well defined for
$d$ in the range $1<d<8/3$, and $d=2$ is in that range, we can write at once

$D_{4}(2,s)\ M_{i}(2,s)=0\ ,\hskip 22.76219pti=0,1,..,5\ .$ (5.1)

where the functions $M_{i}(2,s)$ correspond to the $d=2$ value of
$M_{i}(d,s)$, see Eq.s(4.2,4.1).

As a first comment, at $d=2$ the factor $s^{\frac{2-d}{2}}$ in front of the
$b$ integral in Eq.(4.2) and the factor $|b|^{\frac{2-d}{2}}$ within the
integrand in Eq.(4.1) are both equal to 1, with an obvious simplification of
the resulting expressions, which become

$\displaystyle M_{i}(2,s)$ $\displaystyle=N_{i}(2,s)\ ,$ (5.2) $\displaystyle
N_{i}(2,s)$ $\displaystyle=\int_{b_{i-1}}^{b_{i}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{b_{i-1}}^{b_{i}}db\
\frac{1}{\sqrt{|(b-b_{1})(b-b_{2})(b-b_{3})(b-b_{4})|}}\ .$ (5.3)

The $N_{i}(2,s)$ are elliptic integrals; their expression in terms of the
complete and incomplete elliptic intregrals by means of the Legendre changes
of variable are discussed in Appendix C. In this Section, in agreement with
the rest of the paper, we discuss their properties by using directly the
$b$-integral representation. For better exploiting the absence of the
$|b|^{\frac{2-d}{2}}$ and of the related absence of $b=0$ as a sensible point,
we will not use the $d=2$ limit of the results obtained in the previous
Section, but will look for simpler identities and formulas.

To start with, we want to evaluate the $s$-derivatives of the six
$b$-integrals, beginning from the integrals whose integration limits
correspond to two vanishing points of $R_{4}(b)$, Eq.(3.14). As a first case,
we consider

$N_{3}(2,s)=\int_{b_{2}}^{b_{3}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{b_{2}}^{b_{3}}db\
\frac{1}{\sqrt{(b-b_{1})(b-b_{2})(b_{3}-b)(b_{4}-b)}}\ .$ (5.4)

For reasons which will be apparent later, we introduce the partner integral

$N^{(1)}_{3}(2,s)=\int_{b_{2}}^{b_{3}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b-b_{1}}\ ,$ (5.5)

and the two integration by parts identities

$\displaystyle\int_{b_{2}}^{b_{3}}db\
\frac{d}{db}\left[\frac{\sqrt{(b-b_{2})(b_{3}-b)}}{\sqrt{(b-b_{1})(b_{4}-b)}}\right]=0\
,$ $\displaystyle\int_{b_{2}}^{b_{3}}db\
\frac{d}{db}\left[\frac{\sqrt{(b-b_{2})(b_{3}-b)}}{\sqrt{(b-b_{1})(b_{4}-b)}}\times\
\frac{1}{b-b_{1}}\right]=0\ ,$ (5.6)

where the vanishing end-point contributions were dropped.

Working out the above expressions is elementary; without writing explicitly
the results for simplicity, let us just say that they establish two identities
between the four integrals

$\int_{b_{2}}^{b_{3}}db\ \frac{1}{\sqrt{|R_{4}(b)|}}\times\left[1\ ;\
\frac{1}{b-b_{1}}\ ;\ \frac{1}{b_{4}-b}\ ;\ \frac{1}{(b-b_{1})^{2}}\right]\ ,$
(5.7)

and the two identities can be used, when needed, to express the last two
integrals in terms of the first two.

As already observed in the previous Section, the $s$-derivatives can be
obtained from the $b_{3}$ and $b_{4}$ derivatives, see Eq.s(3.12) for the
definitions of the $b_{i}$. For $N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$ taking the
$b_{4}$ derivative is straighforward (it gives just a factor $1/(b_{4}-b)$, as
$b_{4}$ lies outside of the integration range), but that is not the case for
the $b_{3}$ derivative; indeed, given the presence of $b_{3}$ as an end point,
the naïve differentiation of the integrand is not allowed here (at variance
with the previous section, in which $d>3$ was assumed; see however the
comments following Eq.(4.8)).

To overcome the probem let us introduce the function $\mbox{\rm
atan}{\sqrt{(b-b_{2})/(b_{3}-b)}}$, whose derivatives in $b$ and $b_{3}$ are

$\displaystyle\frac{d}{db}\left(\mbox{\rm
atan}{\sqrt{\frac{b-b_{2}}{b_{3}-b}}}\right)$ $\displaystyle=+\frac{1}{2}\
\frac{1}{\sqrt{(b-b_{2})(b_{3}-b)}}\ ,$ (5.8)
$\displaystyle\frac{d}{db_{3}}\left(\mbox{\rm
atan}{\sqrt{\frac{b-b_{2}}{b_{3}-b}}}\right)$ $\displaystyle=-\frac{1}{2}\
\frac{b-b_{2}}{b_{3}-b_{2}}\ \frac{1}{\sqrt{(b-b_{2})(b_{3}-b)}}\ .$ (5.9)

By using Eq.(5.8) we can integrate by parts the factor
$1/\sqrt{(b-b_{2})(b_{4}-b)}$ in Eq.s(5.4,5.5), obtaining

$\displaystyle N_{3}(2,s)=$ $\displaystyle\
\frac{\pi}{\sqrt{(b_{4}-b_{3})(b_{3}-b_{1})}}$ $\displaystyle+$
$\displaystyle\int_{b_{2}}^{b_{3}}db\ \left(\mbox{\rm
atan}{\sqrt{\frac{b-b_{2}}{b_{3}-b}}}\right)\
\frac{1}{\sqrt{(b-b_{1})(b_{4}-b)}}\left(\frac{1}{b-b_{1}}-\frac{1}{b_{4}-b}\right)\
,$ (5.10) $\displaystyle N^{(1)}_{3}(2,s)=$ $\displaystyle\
\frac{1}{b_{3}-b_{1}}\frac{\pi}{\sqrt{(b_{4}-b_{3})(b_{3}-b_{1})}}$
$\displaystyle+$ $\displaystyle\int_{b_{2}}^{b_{3}}db\ \left(\mbox{\rm
atan}{\sqrt{\frac{b-b_{2}}{b_{3}-b}}}\right)\
\frac{1}{\sqrt{(b-b_{1})(b_{4}-b)}}\times\left[\frac{3}{(b-b_{1})^{2}}-\frac{1}{b_{4}-b_{1}}\left(\frac{1}{b-b_{1}}+\frac{1}{b_{4}-b}\right)\right]\
.$ (5.11)

The naïve differentiation in $b_{3}$ is now allowed, and gives

$\displaystyle\frac{d}{db_{3}}N_{3}(2,s)=$ $\displaystyle\
\frac{1}{b_{3}-b_{2}}\int_{b_{2}}^{b_{3}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\biggl{[}-1+\frac{1}{2}(b_{2}-b_{1})\times\frac{1}{b-b_{1}}+\frac{1}{2}(b_{4}-b_{2})\times\frac{1}{b_{4}-b}\biggr{]}\
,$ $\displaystyle\frac{d}{db_{3}}N^{(1)}_{3}(2,s)=$
$\displaystyle\frac{1}{b_{3}-b_{2}}\int_{b_{2}}^{b_{3}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\biggl{[}-\biggl{(}\frac{3}{2}\ +\frac{1}{2}\
\frac{b_{2}-b_{1}}{b_{4}-b_{1}}\biggr{)}\times\frac{1}{b-b_{1}}$
$\displaystyle\hskip 91.04881pt+\frac{1}{2}\
\frac{b_{4}-b_{2}}{b_{4}-b_{1}}\times\frac{1}{b_{4}-b}\
+\frac{3}{2}(b_{2}-b_{1})\times\frac{1}{(b-b_{1})^{2}}\biggr{]}\ .$ (5.12)

The two expressions on the r.h.s. of Eq.(5.12) depend on the 4 integrals
listed in Eq.(5.7); as already remarked, by using the Eq.s(5.6), the integrals
with the factors $1/(b_{4}-b),1/(b-b_{1})^{2}$ can be expressed in terms of
only the two integrals $N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$, defined by
Eq.s(5.4,5.5).

As already observed, the allowed naïve differentiation with respect to $b_{4}$
of $N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$ generates only an extra factor
$1/(b_{4}-b)$, so that, again, also the $b_{4}$ derivatives of $N_{3}(2,s)$
and $N^{(1)}_{3}(2,s)$ can be expressed in terms of the two same functions
$N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$ only.

Recalling Eq.(4.3) and collecting results, one can then express the first
order $s$-derivatives of $N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$ in terms of
$N_{3}(2,s)$ and $N_{3}^{(1)}(2,s)$ only, obtaining two equations of the form

$\displaystyle\frac{d}{ds}N_{3}(2,s)$ $\displaystyle=A(s)\ N_{3}(2,s)+B(s)\
N_{3}^{(1)}(2,s)\ ,$ $\displaystyle\frac{d}{ds}N_{3}^{(1)}(2,s)$
$\displaystyle=C(s)\ N_{3}(2,s)+D(s)\ N_{3}^{(1)}(2,s)\ ,\ ,$ (5.13)

where the quantities $A(s),B(s),C(s),D(s)$ are rational expressions in $s$
which we don’t write for space reasons. By iterating the procedure, one can
express also all the higher order $s$-derivatives of $N_{3}(2,s)$ in terms of
$N_{3}(2,s)$ and $N^{(1)}_{3}(2,s)$ only; finally, one can also verify by
brute (algebraic) force that $M_{3}(2,s)=N_{3}(2,s)$, see Eq.(5.2), and its
first 4 $s$-derivatives do satisfy the 4th order equation Eq.(5.1), as
expected.

But we can also rewrite the first of Eq.s(5.13) as

$N_{3}^{(1)}(2,s)=-\frac{A(s)}{B(s)}\ N_{3}(2,s)+\frac{1}{B(s)}\
\frac{d}{ds}N_{3}(2,s)\ ,$ (5.14)

expressing $N_{3}^{(1)}(2,s)$ in terms of $N_{3}(2,s)$ and $dN_{3}(2,s)/ds$.

We have already seen that all the $s$-derivatives of $N_{3}(2,s)$ can be
expressed in terms of $N_{3}(2,s)$ and $N_{3}^{(1)}(2,s)$ only; by using
Eq.(5.14) for replacing $N_{3}^{(1)}(2,s)$ with $dN_{3}(2,s)/ds$, we have that
the 2nd and higher $s$-derivatives of $N_{3}(2,s)$ can be expressed in terms
of $N_{3}(2,s)$ and $dN_{3}(2,s)/ds$ only. In particular, the expression of
the 2nd order derivative $d^{2}N_{3}(2,s)/ds^{2}$ in terms of $N_{3}(2,s)$ and
its 1st derivative $dN_{3}(2,s)/ds$ is nothing but a 2nd order differential
equation in $s$ for $N_{3}(2,s)=M_{3}(2,s)$, which can be written as

$D_{2}(2,s)\ N_{3}(2,s)=0\ ,$ (5.15)

where $D_{2}(2,s)$ is a 2nd order differential operator, whose explicit
expression is given in Eq.(B.12).

A similar procedure can be followed for

$N_{2}(2,s)=\int_{b_{1}}^{b_{2}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{b_{1}}^{b_{2}}db\
\frac{1}{\sqrt{(b-b_{1})(b_{2}-b)(b_{3}-b)(b_{4}-b)}}\ ,$ (5.16)

using as partner integral

$N^{(1)}_{2}(2,s)=\int_{b_{1}}^{b_{2}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b_{4}-b}\ ,$ (5.17)

and as integration by parts identities the relations

$\int_{b_{1}}^{b_{2}}db\
\frac{d}{db}\left[\frac{\sqrt{(b-b_{1})(b_{2}-b)}}{\sqrt{(b_{3}-b)(b_{4}-b)}}\times\left(1\
;\frac{1}{b_{4}-b}\right)\right]=0\ .$

The direct, naïve differentiation (i.e. without a preliminary integration by
parts) of the integrals Eq.(5.16,5.17) with respect to $b_{3}$ and $b_{4}$ is
allowed, and, as in the case of $M_{3}(2,s)$, one finds that also
$M_{2}(2,s)=N_{2}(2,s)$ satisfies, besides the 4th order equation (5.1), also
the 2nd order equation

$D_{2}(2,s)\ M_{2}(2,s)=0\ ,$ (5.18)

where $D_{2}(2,s,m_{i}^{2})$ is the same as in Eq.(5.15).

The above approach can also be extended to

$N_{4}(2,s)=\int_{b_{3}}^{b_{4}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{b_{3}}^{b_{4}}db\
\frac{1}{\sqrt{(b-b_{1})(b_{2}-b)(b-b_{3})(b_{4}-b)}}\ ,$ (5.19)

using as partner integral

$N^{(1)}_{4}(2,s)=\int_{b_{3}}^{b_{4}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b-b_{1}}\ ,$ (5.20)

and as integration by parts identities the relations

$\int_{b_{3}}^{b_{4}}db\
\frac{d}{db}\left[\frac{\sqrt{(b-b_{3})(b_{4}-b)}}{\sqrt{(b-b_{1})(b-b_{2})}}\times\left(1\
;\frac{1}{b-b_{1}}\right)\right]=0\ .$ (5.21)

Similarly to the case of $N_{2}(2,s)$, for carrying out the $b_{3}$ and
$b_{4}$ derivatives a preliminary integration by parts, involving this time
the function $\mbox{\rm atan}{\sqrt{(b-b_{3})/(b_{4}-b)}}$, is needed; as in
the case of $M_{3}(2,s)$ and $M_{2}(2,s)$, one finds that
$M_{4}(2,s)=N_{4}(2,s)$ satisfies both the 4th order equation (5.1) and the
2nd order equation

$D_{2}(2,s)\ M_{4}(2,s)=0\ ,$ (5.22)

where $D_{2}(2,s)$ is again the same differential operator as in Eq.s(5.15)
and (5.18).

To our knowledge, Eq.s(5.15,5.18,5.22) were first obtained in the paper [4].

As $D_{2}(2,s)$ is a 2nd order differential operator, it admits only two
linearly independent solutions, so a relation between the three solutions
$N_{2}(2,s)=M_{2}(2,s)$, $N_{3}(2,s)=M_{3}(2,s)$ and $N_{4}(2,s)=M_{4}(2,s)$
must exist. To find it, consider the contour integral

$\oint_{\mathfrak{C}}db\frac{1}{\sqrt{R_{4}(b)}}$

in the complex $b$ plane, where $R_{4}(b)$ is the usual 4th order polynomial
of Eq.(3.14) (note the presence of $\sqrt{R_{4}(b)}$ and not of
$\sqrt{|R_{4}(b)|}$ as in all the other so far considered integrals) and the
contour $\mathfrak{C}$ is a (big) circle around the origin, of radius $C$ with
$C\gg(b_{i},i=1,..,4)$; as for large $b$ the function $1/\sqrt{R_{4}(b)}$
behaves as $1/b^{2}$, the above contour integral vanishes

$\oint_{\mathcal{C}}db\frac{1}{\sqrt{R_{4}(b)}}=0\ .$ (5.23)

Let us specify the analytic function ${1}/{\sqrt{R_{4}(b)}}$ by giving it the
real (and positive) value ${1}/{\sqrt{|R_{4}(b)}|}$ in the real interval
$b_{2}<b<b_{3}$, two cuts along the two real intervals $b_{1}<b<b_{2}$ and
$b_{3}<b<b_{4}$, and the real negative value
$\left(-{1}/{\sqrt{|R_{4}(b)}|}\right)$ along the remaining parts of the real
$b$ axis, $-\infty<b<b_{1}$ and $b_{4}<b<+\infty$. By shrinking the big circle
into two closed contours around the cuts, one arrives (up to an overall
constant) at the equation

$N_{2}(2,s)-N_{4}(2,s)=0\ ,$ (5.24)

which shows that two of the solutions are not linearly independent. (Strictly
speaking, one should still show that two of them, say $N_{2}(2,s)$ and
$N_{3}(2,s)$, are indeed independent, see below).

We then consider the integrals whose integration limits do not correspond to
the zeroes of $R_{4}(b)$, starting from

$N_{5}(2,s)=\int_{b_{4}}^{\infty}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{b_{4}}^{\infty}db\
\frac{1}{\sqrt{(b-b_{1})(b-b_{2})(b-b_{3})(b-b_{4})}}\ ,$ (5.25)

(recall $b_{5}=\infty$), using as partner integral

$N^{(1)}_{5}(2,s)=\int_{b_{4}}^{\infty}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b-b_{1}}\ ,$ (5.26)

and as integration by parts identities the relations

$\displaystyle\int_{b_{4}}^{\infty}db\
\frac{d}{db}\left[\frac{\sqrt{(b-b_{3})(b-b_{4})}}{\sqrt{(b-b_{1})(b-b_{2})}}\times\left(1\
;\ \frac{1}{b-b_{1}}\right)\right]$
$\displaystyle=\lim_{b\to\infty}\left[\frac{\sqrt{(b-b_{3})(b-b_{4})}}{\sqrt{(b-b_{1})(b-b_{2})}}\times\left(1\
;\ \frac{1}{b-b_{1}}\right)\right]$ $\displaystyle=\left[\ 1\ ;\ 0\right]\ .$
(5.27)

Note that, at variance with the previous cases, see for instance Eq.(5.21),
the right hand side of the identities (5.27) is not zero, as the integration
limits do not correspond to the zeroes of the integrand. We can however follow
closely the above described procedures, expressing in that way all the
$s$-derivatives of $N_{5}(2,s)$ in terms of the two $b$-integrals $N_{5}(2,s)$
and $N^{(1)}_{5}(2,s)$ plus a bunch of additional algebraic terms, originated
by the non vanishing right hand side of Eq.(5.27).

The procedure followed for $N_{5}(2,s)$ can be used also for the two $b$
integrals

$\displaystyle N_{0}(2,s)$ $\displaystyle=\int_{-\infty}^{0}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{-\infty}^{0}db\
\frac{1}{\sqrt{(b_{1}-b)(b_{2}-b)(b_{3}-b)(b_{4}-b)}}\ ,$ $\displaystyle
N_{1}(2,s)$ $\displaystyle=\int_{0}^{b_{1}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}=\int_{0}^{b_{1}}db\
\frac{1}{\sqrt{(b_{1}-b)(b_{2}-b)(b_{3}-b)(b_{4}-b)}}\ ,$ (5.28)

introducing as partner integrals

$\displaystyle N^{(1)}_{0}(2,s)$ $\displaystyle=\int_{-\infty}^{0}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b_{4}-b}\ ,$ $\displaystyle
N^{(1)}_{1}(2,s)$ $\displaystyle=\int_{0}^{b_{1}}db\
\frac{1}{\sqrt{|R_{4}(b)|}}\times\frac{1}{b_{4}-b}\ .$ (5.29)

Skipping details, one finds that, as in the case of $N_{5}(2,s)$, all the
$s$-derivatives of $N_{0}(2,s)$ and $N_{1}(2,s)$ can be expressed in terms of
the pairs of functions $N_{0}(2,s),N^{(1)}_{0}(2,s)$ and
$N_{1}(2,s),N^{(1)}_{1}(2,s)$, plus a bunch of additional algebraic terms.

By evaluating all the $s$-derivatives in that way, one finds as expected that
the three functions $M_{0}(2,s)=N_{0}(2,s)\ $, $M_{1}(2,s)=N_{1}(2,s)\ $ and
$M_{5}(2,s)=N_{5}(2,s)\ $ satisfy the 4th order equation (5.1), but do not
satisfy the 2nd order equation of Eq.(5.15), satisfied by
$M_{2}(2,s),M_{3}(2,s),4_{4}(2,s)$.

Indeed, an explicit calculation gives

$\displaystyle D_{2}(2,s)N_{0}(2,s)$ $\displaystyle=R_{0}(s,m_{i}^{2})\ ,$
$\displaystyle D_{2}(2,s)N_{1}(2,s)$ $\displaystyle=R_{1}(s,m_{i}^{2})\ ,$
$\displaystyle D_{2}(2,s)N_{5}(2,s)$ $\displaystyle=R_{5}(s,m_{i}^{2})\ ,$
(5.30)

where $R_{0}(s,m_{i}^{2}),R_{1}(s,m_{i}^{2}),R_{5}(s,m_{i}^{2})$, generated by
the additional algebraic terms present in the expressions of the
$s$-derivatives, are polynomials in $s$ and in the masses, and are found to
satisfy the relation

$R_{0}(s,m_{i}^{2})+R_{1}(s,m_{i}^{2})+R_{5}(s,m_{i}^{2})=0\ .$ (5.31)

For definiteness, the explicit expressions of $R_{0}(s,m_{i}^{2})$ and
$R_{1}(s,m_{i}^{2})$ are

$\displaystyle R_{0}(s,m_{i}^{2})$
$\displaystyle=(2m_{2}^{2}+2m_{3}^{2}-4m_{1}^{2})\ s^{3}$
$\displaystyle+(12m_{1}^{4}-6m_{2}^{4}-6m_{3}^{4}-14m_{1}^{2}m_{2}^{2}-14m_{1}^{2}m_{3}^{2}+28m_{2}^{2}m_{3}^{2})\
s^{2}$
$\displaystyle+(6m_{2}^{6}+6m_{3}^{6}-12m_{1}^{6}-16m_{1}^{2}m_{2}^{4}-16m_{1}^{2}m_{3}^{4}+22m_{1}^{4}m_{2}^{2}+22m_{1}^{4}m_{3}^{2}$
$\displaystyle\hskip 36.98857pt-6m_{2}^{2}m_{3}^{4}-6m_{2}^{4}m_{3}^{2})\ s$
$\displaystyle-2m_{1}^{2}m_{2}^{2}m_{3}^{4}-2m_{1}^{2}m_{2}^{4}m_{3}^{2}+2m_{1}^{2}m_{2}^{6}+2m_{1}^{2}m_{3}^{6}+4m_{1}^{4}m_{2}^{2}m_{3}^{2}+6m_{1}^{4}m_{2}^{4}+6m_{1}^{4}m_{3}^{4}$
$\displaystyle-10m_{1}^{6}m_{2}^{2}-10m_{1}^{6}m_{3}^{2}+4m_{1}^{8}-2m_{2}^{8}-2m_{3}^{8}+8m_{2}^{2}m_{3}^{6}-12m_{2}^{4}m_{3}^{4}+8m_{2}^{6}m_{3}^{2}\
,$ (5.32) $\displaystyle R_{1}(s,m_{i}^{2})$
$\displaystyle=3(m_{1}^{2}-m_{2}^{2})\
s^{3}+3(3m_{2}^{4}-3m_{1}^{4}+7m_{1}^{2}m_{3}^{2}-7m_{2}^{2}m_{3}^{2})\ s^{2}$
$\displaystyle+(9m_{1}^{6}-9m_{2}^{6}+5m_{1}^{2}m_{3}^{4}-5m_{2}^{2}m_{3}^{4}+14m_{2}^{4}m_{3}^{2}-14m_{1}^{4}m_{3}^{2}+19m_{1}^{2}m_{2}^{4}-19m_{2}^{2}m_{1}^{4})\
s$
$\displaystyle+3(m_{2}^{8}-m_{1}^{8}+m_{1}^{2}m_{3}^{6}-m_{2}^{2}m_{3}^{6}+2m_{1}^{6}m_{2}^{2}-2m_{2}^{6}m_{1}^{2}+3m_{2}^{4}m_{3}^{4}-3m_{1}^{4}m_{3}^{4}$
$\displaystyle\hskip
11.38109pt+3m_{1}^{6}m_{3}^{2}-3m_{2}^{6}m_{3}^{2}+m_{1}^{2}m_{2}^{4}m_{3}^{2}-m_{1}^{4}m_{2}^{2}m_{3}^{2})\
.$ (5.33)

The polynomial $R_{1}(s,m_{i}^{2})$ satisfies (obviously) a 1st order
differential equation, which can be written as

$D_{1a}(2,s)R_{1}(s,m_{i}^{2})=0\ ,$ (5.34)

where $D_{1a}(2,s)$ is the 1st order differential operator

$D_{1a}(2,s)=R_{1}(s,m_{i}^{2})\frac{d}{ds}-\left(\frac{dR_{1}(s,m_{i}^{2})}{ds}\right)\
,$ (5.35)

The polynomial $R_{0}(s,m_{i}^{2})$, which is different from
$R_{1}(s,m_{i}^{2})$, cannot satisfy the same homogeneous 1st order equation
(5.34) as $R_{1}(s,m_{i}^{2})$; indeed, an explicit calculation gives

$D_{1a}(2,s)R_{0}(s,m_{i}^{2})=R_{0a}(s,m_{i}^{2})\ ,$ (5.36)

where $R_{0a}(s,m_{i}^{2})$, whose expression is given by

$R_{0a}(s,m_{i}^{2})=R_{1}(s,m_{i}^{2})\frac{dR_{0}(s,m_{i}^{2})}{ds}-\frac{dR_{1}(s,m_{i}^{2})}{ds}R_{0}(s,m_{i}^{2})\
,$ (5.37)

is again a polynomial in $s$ and the masses (which we don’t list here to save
space).

The polynomial $R_{0a}(s,m_{i}^{2})$ satisfies (obviously) another 1st order
differential equation, which can be written as

$D_{1b}(2,s)R_{0a}(s,m_{i}^{2})=0\ ,$ (5.38)

where $D_{1b}(2,s)$ is the 1st order differential operator

$D_{1b}(2,s)=R_{0a}(s,m_{i}^{2})\frac{d}{ds}-\left(\frac{dR_{0a}(s,m_{i}^{2})}{ds}\right)\
,$ (5.39)

As it is easily verified,

$D_{4}(2,s)=\mathcal{K}\ D_{1b}(2,s)\ D_{1a}(2,s)\ D_{2}(2,s)\ ,$ (5.40)

where $D_{4}(2,s)$ is the same operator as in Eq.(5.1),

$\mathcal{K}=-{8s^{2}}\times\frac{5s^{2}+2(m_{1}^{2}+m_{2}^{2}+m_{3}^{2})\
s-7R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})}{3s^{2}-2(m_{1}^{2}+m_{2}^{2}+m_{3}^{2})\
s-R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})}\times\frac{1}{R_{0a}(s,m_{i}^{2})}\
\frac{1}{R_{1}(s,m_{i}^{2})}$ (5.41)

and

$R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})=m_{1}^{4}+m_{2}^{4}+m_{3}^{4}-2m_{1}^{2}m_{2}^{2}-2m_{2}^{2}m_{3}^{2}-2m_{3}^{2}m_{1}^{2}\
.$ (5.42)

Eq.(5.41) completes the factorisation of $D_{4}(2,s)$ in the product of lower
order differential operators with rational coefficients.

It is to be noted that $D_{4}(2,s)$ and $D_{2}(2,s)$, see Eq.s(B.2) and
Eq.(B.12) for their explicit expessions, are symmetrical under the exchange of
the three masses, but that is not the case for $D_{1a}(2,s)$ and
$D_{1b}(2,s)$.

Let us further observe that the factorisation of a differential operator of a
given order into the product of differential operators of lower order is in
general not unique, as shown for instance by the following elementary example

$\left(\frac{d}{dx}+\frac{1}{x}\right)\left(\frac{d}{dx}-\frac{1}{x}\right)=\frac{d^{2}}{dx^{2}}\
.$

Indeed, an equivalent but different factorization of $D_{4}(2,s)$ can be
obtained by exchanging the order of $R_{0}(s,m_{i}^{2})$, $R_{1}(s,m_{i}^{2})$
and $R_{5}(s,m_{i}^{2})$ in the previous derivation.

We will now look for a relationship of linear dependence between the three
integrals $N_{0}(2,s),N_{1}(2,s)$ and $N_{5}(2,s)$, which do not satisfy the
equation with the 2nd order operator $D_{2}(2,s)$, and the three integrals
$N_{2}(2,s)$, $N_{3}(2,s)$ and $N_{4}(2,s)$, which do satisfy that equation.
In analogy with Eq.(5.23) we can write

$\oint_{\mathfrak{C}}db\frac{1}{\sqrt{-R_{4}(b)}}=0\ ,$ (5.43)

where ${\mathfrak{C}}$ is again the (big) circle around the origin, but this
time we specify the analytic function $1/\sqrt{-R_{4}(b)}$ by assigning it
real and positive value in the real interval $b_{1}<b<b_{2}$, real (negative)
value in the interval $b_{3}<b<b_{4}$ and three cuts along the three real
intervals $-\infty<b<b_{1}$, $b_{2}<b<b_{3}$ and $b_{4}<b<+\infty$. By
shrinking the big circle into three closed contours around the cuts, and
observing that the interval $-\infty<b<b_{1}$ contributes to $N_{0}(2,s)$ plus
$N_{1}(2,s)$, the interval $b_{2}<b<b_{3}$ to $-N_{3}(2,s)$ and
$b_{4}<b<b_{5}=+\infty$ to $N_{5}(2,s)$, one arrives (up to an overall factor)
at the equation

$N_{0}(2,s)+N_{1}(2,s)-N_{3}(2,s)+N_{5}(2,s)=0\ ,$ (5.44)

which gives the expected linear relation between the various $b$-integrals.

As a last argument for the discussion of the actual linear independence of 4
of the 6 $b$-integrals, we evaluate them in the limit of large $s=E^{2}$,
keeping a few leading terms in the $1/E$ expansion. Since
$b_{3}=(E-m_{3})^{2}$ and $b_{4}=(b+m_{3})^{2}$, Eq.(3.12), the expansion in
$1/E$ is equivalent to the expansion in $1/b_{3},1/b_{4}$ with $b_{3},b_{4}$
larger than $b_{1},b_{2}$.

In $N_{1}(2,s)$, see Eq.(5.28), the integration range is $0<b<b_{1}$, the
integration variable $b$ is always smaller than $b_{3},b_{4}$, so that we can
expand $\sqrt{b_{3}-b}$ in $b/b_{3}$, $\sqrt{b_{4}-b}$ in $b/b_{4}$ and then
integrate analytically the resulting terms which contain only the two square
roots $\sqrt{b_{1}-b},\sqrt{b_{2}-b}$; the same argument applies to
$N_{2}(2,s)$, where $b_{1}<b<b_{2}$.

In $N_{4}(2,s)$ the integration range is $b_{3}<b<b_{4}$, so that $b$ is
always larger than $b_{1},b_{2}$; we can therefore expand $\sqrt{b-b_{1}}$ in
$b_{1}/b$, $\sqrt{b-b_{2}}$ in $b_{2}/b$ and then integrate analytically the
resulting terms, which contain only the two square roots
$\sqrt{b-b_{3}},\sqrt{b_{4}-b}$; the same argument applies to $N_{5}(2,s)$,
where $b_{4}<b<+\infty$.

The case of $N_{3}(2,s)$ is slightly more complicated, as the integration
range, $b_{2}<b<b_{3}$, includes both small and large values of the variable
$b$. For that reason, as a first step we split the range in two parts, the
small $b$ range $b_{2}<b<B$ and the large $b$ range $B<b<b_{3}$, with
$b_{2}<{\kern-9.0pt}<B<{\kern-9.0pt}<b_{3}$, and we treat separately the two
ranges.

In the small $b$ range we expand $\sqrt{b_{3}-b}$ and $\sqrt{b_{4}-b}$ in
$b/b_{3}$ and $b/b_{4}$, we then integrate analytically from $b=b_{2}$ up to
$b=B$ the terms so obtained, which contain only the two square roots
$\sqrt{b-b_{1}},\sqrt{b-b_{2}}$, and finally we expand the analytical result
in $E$ and $B$ (up to $1/B$, for the sake of later checks). In the large $b$
range we expand $\sqrt{b-b_{1}},\sqrt{b-b_{2}}$ in $b_{1}/b,\ b_{2}/b$ we then
integrate analytically from $b=B$ up to $b=b_{3}$ the terms so obtained, which
contain only the two square roots $\sqrt{b_{3}-b},\sqrt{b_{4}-b}$, and finally
we expand the analytical result in $E$ and $B$

As a concluding step we sum the contributions from the two $b$ ranges,
checking the cancellation of the dependence on $B$ (as the result corresponds
to the term of zeroth order in $B$, for a stronger check we verify the
disappearance of the $B$-dependence up to the order $1/B$ included, although
the terms in $1/B$ are anyhow irrelevant in the limit of large $B$).

Finally, the expansion of the remaining integral $N_{0}(2,s)$ can be obtained
by following the same procedure used for $N_{3}(2,s)$.

Let us indicate by $N_{k}^{(a)}(2,s),k=0,..,5$, the first 5 terms of the
asymptotic expansions in $1/E^{2}$, for large $s=E^{2}$, of the 6 $b$-integral
$N_{k}(2,s)$, starting from the leading term $1/E^{2}$ up to the term
$1/E^{10}$ included. The results of the above described calculations can be
written as

$\displaystyle N_{0}^{(a)}(2,s)$
$\displaystyle=B_{0}(E,m_{i})\ln\left(\frac{E^{2}}{m_{1}^{2}}\right)+B_{1}(E,m_{i})\
,$ $\displaystyle N_{1}^{(a)}(2,s)$
$\displaystyle=B_{0}(E,m_{i})\ln\left(\frac{m_{1}}{m_{2}}\right)+B_{2}(E,m_{i})\
,$ $\displaystyle N_{2}^{(a)}(2,s)$ $\displaystyle=B_{0}(E,m_{i})\pi\ ,$
$\displaystyle N_{3}^{(a)}(2,s)$
$\displaystyle=B_{0}(E,m_{i})\ln\left(\frac{E^{3}}{m_{1}m_{2}m_{3}}\right)+B_{3}(E,m_{i})\
,$ $\displaystyle N_{4}^{(a)}(2,s)$ $\displaystyle=B_{0}(E,m_{i})\pi\ ,$
$\displaystyle N_{5}^{(a)}(2,s)$
$\displaystyle=B_{0}(E,m_{i})\ln\left(\frac{E}{m_{3}}\right)+B_{4}(E,m_{i})\
,$ (5.45)

with

$\displaystyle B_{0}(E,m_{i})$
$\displaystyle=1/E^{2}+\bigl{(}m_{1}^{2}+m_{2}^{2}+m_{3}^{2}\bigr{)}/E^{4}$
$\displaystyle+\bigl{(}4m_{1}^{2}m_{2}^{2}+4m_{1}^{2}m_{3}^{2}+4m_{2}^{2}m_{3}^{2}+m_{1}^{4}+m_{2}^{4}+m_{3}^{4}\bigr{)}/E^{6}$
$\displaystyle+\bigl{(}36m_{1}^{2}m_{2}^{2}m_{3}^{2}+9m_{1}^{2}m_{2}^{4}+9m_{1}^{2}m_{3}^{4}+9m_{1}^{4}m_{2}^{2}+9m_{1}^{4}m_{3}^{2}+9m_{2}^{2}m_{3}^{4}+9m_{2}^{4}m_{3}^{2}$
$\displaystyle\hskip 14.22636pt+m_{1}^{6}+m_{2}^{6}+m_{3}^{6}\bigr{)}/E^{8}$
$\displaystyle+\bigl{(}144m_{1}^{2}m_{2}^{2}m_{3}^{4}+144m_{1}^{2}m_{2}^{4}m_{3}^{2}+144m_{1}^{4}m_{2}^{2}m_{3}^{2}+36m_{1}^{4}m_{2}^{4}+36m_{2}^{4}m_{3}^{4}+36m_{1}^{4}m_{3}^{4}$
$\displaystyle\hskip
14.22636pt+16m_{1}^{2}m_{2}^{6}+16m_{1}^{2}m_{3}^{6}+16m_{1}^{6}m_{2}^{2}+16m_{1}^{6}m_{3}^{2}+16m_{2}^{2}m_{3}^{6}+16m_{2}^{6}m_{3}^{2}$
$\displaystyle\hskip 14.22636pt+m_{1}^{8}+m_{2}^{8}+m_{3}^{8}\bigr{)}/E^{10}\
,$ (5.46) $\displaystyle B_{1}(E,m_{i})$
$\displaystyle=\bigl{(}-2m_{2}^{2}-2m_{3}^{2}\bigr{)}/E^{4}+\bigl{(}-4m_{1}^{2}m_{2}^{2}-4m_{1}^{2}m_{3}^{2}-12m_{2}^{2}m_{3}^{2}-3m_{2}^{4}-3m_{3}^{4}\bigr{)}/E^{6}$
$\displaystyle+\bigl{(}-60m_{1}^{2}m_{2}^{2}m_{3}^{2}-15m_{1}^{2}m_{2}^{4}-15m_{1}^{2}m_{3}^{4}-6m_{1}^{4}m_{2}^{2}-6m_{1}^{4}m_{3}^{2}-33m_{2}^{2}m_{3}^{4}$
$\displaystyle\hskip
14.22636pt-33m_{2}^{4}m_{3}^{2}-11m_{2}^{6}/3-11m_{3}^{6}/3\bigr{)}/E^{8}$
$\displaystyle+\bigl{(}-312m_{1}^{2}m_{2}^{2}m_{3}^{4}-312m_{1}^{2}m_{2}^{4}m_{3}^{2}-104m_{1}^{2}m_{2}^{6}/3-104m_{1}^{2}m_{3}^{6}/3-168m_{1}^{4}m_{2}^{2}m_{3}^{2}$
$\displaystyle\hskip
14.22636pt-42m_{1}^{4}m_{2}^{4}-42m_{1}^{4}m_{3}^{4}-8m_{1}^{6}m_{2}^{2}-8m_{1}^{6}m_{3}^{2}-200m_{2}^{2}m_{3}^{6}/3-200m_{2}^{6}m_{3}^{2}/3$
$\displaystyle\hskip
14.22636pt-150m_{2}^{4}m_{3}^{4}-25m_{2}^{8}/6-25m_{3}^{8}/6\bigr{)}/E^{10}\
,$ (5.47) $\displaystyle B_{2}(E,m_{i})$
$\displaystyle=\bigl{(}-m_{1}^{2}+m_{2}^{2}\bigr{)}/E^{4}+\bigl{(}-4m_{1}^{2}m_{3}^{2}+4m_{2}^{2}m_{3}^{2}-3m_{1}^{4}/2+3m_{2}^{4}/2\bigr{)}/E^{6}$
$\displaystyle+\bigl{(}9m_{1}^{2}m_{2}^{4}/2-9m_{1}^{2}m_{3}^{4}-9m_{1}^{4}m_{2}^{2}/2-27m_{1}^{4}m_{3}^{2}/2-11m_{1}^{6}/6+9m_{2}^{2}m_{3}^{4}$
$\displaystyle\hskip
14.22636pt+27m_{2}^{4}m_{3}^{2}/2+11m_{2}^{6}/6\bigr{)}/E^{8}$
$\displaystyle+\bigl{(}+72m_{1}^{2}m_{2}^{4}m_{3}^{2}+40m_{1}^{2}m_{2}^{6}/3-16m_{1}^{2}m_{3}^{6}-72m_{1}^{4}m_{2}^{2}m_{3}^{2}-54m_{1}^{4}m_{3}^{4}-40m_{1}^{6}m_{2}^{2}/3$
$\displaystyle\hskip
14.22636pt-88m_{1}^{6}m_{3}^{2}/3-25m_{1}^{8}/12+16m_{2}^{2}m_{3}^{6}+54m_{2}^{4}m_{3}^{4}+88m_{2}^{6}m_{3}^{2}/3+25m_{2}^{8}/12\bigr{)}/E^{10}\
,$ (5.48) $\displaystyle B_{3}(E,m_{i})$
$\displaystyle=\bigl{(}-2m_{1}^{2}-2m_{2}^{2}-2m_{3}^{2}\bigr{)}/E^{4}$
$\displaystyle+\bigl{(}-10m_{1}^{2}m_{2}^{2}-10m_{1}^{2}m_{3}^{2}m_{1}^{4}-10m_{2}^{2}m_{3}^{2}-3m_{1}^{4}-3m_{2}^{4}-3m_{3}^{4}\bigr{)}/E^{6}$
$\displaystyle+\bigl{(}-90m_{1}^{2}m_{2}^{2}m_{3}^{2}-27m_{1}^{2}m_{2}^{4}-27m_{1}^{2}m_{3}^{4}-27m_{1}^{4}m_{2}^{2}-27m_{1}^{4}m_{3}^{2}-27m_{2}^{2}m_{3}^{4}-27m_{2}^{4}m_{3}^{2}$
$\displaystyle\hskip
14.22636pt-11m_{1}^{6}/3-11m_{2}^{6}/3-11m_{3}^{6}/3\bigr{)}/E^{8}$
$\displaystyle+\bigl{(}-396m_{1}^{2}m_{2}^{2}m_{3}^{4}-396m_{1}^{2}m_{2}^{4}m_{3}^{2}-396m_{1}^{4}m_{2}^{2}m_{3}^{2}-164/3m_{1}^{2}m_{2}^{6}-164/3m_{1}^{2}m_{3}^{6}$
$\displaystyle\hskip
14.22636pt-164/3m_{1}^{6}m_{2}^{2}-164/3m_{1}^{6}m_{3}^{2}-164/3m_{2}^{2}m_{3}^{6}-164/3m_{2}^{6}m_{3}^{2}-117m_{1}^{4}m_{2}^{4}-117m_{1}^{4}m_{3}^{4}$
$\displaystyle\hskip
14.22636pt-117m_{2}^{4}m_{3}^{4}-25m_{1}^{8}/6-25m_{2}^{8}/6-25m_{3}^{8}/6\bigr{)}/E^{10}\
,$ (5.49) $\displaystyle B_{4}(E,m_{i})$
$\displaystyle=\bigl{(}-m_{1}^{2}-m_{2}^{2}\bigr{)}/E^{4}$
$\displaystyle+\bigl{(}-10m_{1}^{2}m_{2}^{2}-10m_{1}^{2}m_{3}^{2}-10m_{2}^{2}m_{3}^{2}-3m_{1}^{4}-3m_{2}^{4}-3m_{3}^{4}\bigr{)}/E^{6}$
$\displaystyle+\bigl{(}-30m_{1}^{2}m_{2}^{2}m_{3}^{2}-33m_{1}^{2}m_{2}^{4}/2-33m_{1}^{4}m_{2}^{2}/2-3m_{1}^{2}m_{3}^{4}-3m_{2}^{2}m_{3}^{4}$
$\displaystyle\hskip
14.22636pt-15m_{1}^{4}m_{3}^{2}/2-15m_{2}^{4}m_{3}^{2}/2-11m_{1}^{6}/6-11m_{2}^{6}/6\bigr{)}/E^{8}$
$\displaystyle+\bigl{(}-84m_{1}^{2}m_{2}^{2}m_{3}^{4}-156m_{1}^{2}m_{2}^{4}m_{3}^{2}-156m_{1}^{4}m_{2}^{2}m_{3}^{2}-100m_{1}^{2}m_{2}^{6}/3-100m_{1}^{6}m_{2}^{2}/3$
$\displaystyle\hskip
14.22636pt-4m_{1}^{2}m_{3}^{6}-4m_{2}^{2}m_{3}^{6}-75m_{1}^{4}m_{2}^{4}-21m_{1}^{4}m_{3}^{4}-21m_{2}^{4}m_{3}^{4}-52m_{1}^{6}m_{3}^{2}/3-52m_{2}^{6}m_{3}^{2}/3$
$\displaystyle\hskip 14.22636pt-25m_{1}^{8}/12-25m_{2}^{8}/12\bigr{)}/E^{10}\
,$ (5.50)

As the functions $N_{k}(2,s)$ are solutions of the 4th differential equation
(5.1), the expansions $N_{k}^{(a)}(2,s)$ can be seen as (actually redundant)
boundary conditions at large $s=E^{2}$ which identify completely the
corresponding solutions $N_{k}(2,s)$.

It is immediate to verify that they satisfy the relations

$\displaystyle N_{4}^{(a)}(2,s)$ $\displaystyle=N_{2}^{(a)}(2,s)\ ,$
$\displaystyle N_{5}^{(a)}(2,s)$
$\displaystyle=-N_{0}^{(a)}(2,s)-N_{1}^{(a)}(2,s)+N_{3}^{(a)}(2,s)\ ;$ (5.51)

as any homogeneous relations valid for the boundary conditions
$N_{k}^{(a)}(2,s)$ holds also for the corresponding solutions $N_{k}(2,s)$.
Eq.s(5.51) imply the already obtained relations Eq.(5.24) end Eq.(5.44).

From the explicit epressions of Eq.s(5.47–5.50) one can also check that the 4
expansions $N_{k}^{(a)}(2,s),k=0,..,3$ are indeed linearly independent from
each other, so that the corresponding 4 functions $N_{k}(2,s),i=0,..,3$ are
also linearly independent of each other.

Let us now briefly discuss the symmetry of the solutions under permutations of
the three masses. To that aim, for evidentiating explicitly the dependence on
the masses, we introduce a new family of 4 functions
$N_{k}(2,s,m_{1},m_{2},m_{3}),k=0,..,3$ defined by

$N_{k}(2,s,m_{1},m_{2},m_{3})=N_{k}(2,s);\hskip 11.38109ptk=0,..,3$ (5.52)

in other words, the $N_{k}(2,s)$, with the already considered asymptotic
expansions $N_{i}^{(a)}(2,s)$, correspond to the permutation
$(m_{1},m_{2},m_{3})$ of the masses, while $N_{k}(2,s,m_{2},m_{3},m_{1})$, for
instance, correspond to the function whose asymptotic expansion is obtained by
the permutation $(m_{2},m_{3},m_{1})$ of the masses in Eq.s(5.45) and
(5.47–5.50).

As already observed, the 4th order differential operator $D_{4}(d,s)$
appearing in Eq.(2.8) is fully symmetrical in the 3 masses, and the same is
obviously true for $D_{4}(2,s)$, its value at $d=2$; therefore, if
$N_{k}(2,s,m_{1},m_{2},m_{3}),k=0,..,3$ is a solution of the equation

$D_{4}(2,s)\ N_{k}(2,s,m_{1},m_{2},m_{3})=0\ ,$ (5.53)

each of the 6 functions $N_{k}(2,s,m_{p_{1}},m_{p_{2}},m_{p_{3}})$, where
$(m_{p_{1}},m_{p_{2}},m_{p_{3}})$ is any of the 6 permutations of the masses,
is also a solution of that equation. We obtain in that way a total of $4\times
6=24$ solutions; but there are only 4 linearly indendent solutions of the
homogeneous equation fo the 4th order differential operator $D_{4}(2,s)$, so
that all the new functions introduced with the permutations of the masses must
be linear combinations of the 4 solutions $N_{k}(2,s),i=0,..,3$.

That is easily verified by the explicit form of the asymptotic expansions
Eq.s(5.45) and (5.47–5.50). One finds immediately that
$N_{2}^{(a)}(2,s,m_{1},m_{2},m_{3})$ and $N_{3}^{(a)}(2,s,m_{1},m_{2},m_{3})$
are fully symmetrical in the masses, so that they remain identical to
$N_{2}^{(a)}(2,s)$ and $N_{3}^{(a)}(2,s)$ under the mass permutations. One
further finds

$\displaystyle N_{0}^{(a)}(2,s,m_{1},m_{3},m_{2})$
$\displaystyle=N_{0}^{(a)}(2,s)\ ,$ $\displaystyle
N_{0}^{(a)}(2,s,m_{3},m_{1},m_{2})$
$\displaystyle=-2N_{0}^{(a)}(2,s)-2N_{1}^{(a)}(2,s)+2N_{3}^{(a)}(2,s)\ ,$
$\displaystyle N_{0}^{(a)}(2,s,m_{2},m_{3},m_{1})$
$\displaystyle=N_{0}^{(a)}(2,s)+2N_{1}^{(a)}(2,s)\ ,$ (5.54)

and

$\displaystyle N_{1}^{(a)}(2,s,m_{2},m_{1},m_{3})$
$\displaystyle=N_{1}^{(a)}(2,s)\ ,$ $\displaystyle
N_{1}^{(a)}(2,s,m_{3},m_{1},m_{2})$
$\displaystyle=\frac{3}{2}N_{0}^{(a)}(2,s)+N_{1}^{(a)}(2,s)-N_{3}^{(a)}(2,s)\
,$ $\displaystyle N_{1}^{(a)}(2,s,m_{3},m_{2},m_{1})$
$\displaystyle=\frac{3}{2}N_{0}^{(a)}(2,s)+2N_{1}^{(a)}(2,s)-N_{3}^{(a)}(2,s)\
.$ (5.55)

showing, as expected, that all the new expansions introduced with the
permutations of the masses are indeed linear combinations of the 4 functions
$N_{k}^{(a)}(2,s),k=0,..,3$, and the same holds for all the functions
$N_{k}(2,s,m_{p_{1}},m_{p_{2}},m_{p_{3}})$, which are therefore linear
combinations of the 4 solutions $N_{k}(2,s),i=0,..,3$.

A last argument to discuss is the equal mass limit $m_{1}=m_{2}=m_{3}=m$. An
elementary calulation shows that the equal mass limit of the 4th order
differential operator $D_{4}(2,s)$, of Eq.(5.1), let us call it $D_{4}(2,s,m)$
(without writing it explicitly for the sake of brevity), is still a 4th order
operator, which therefore admits 4 independent homogeneous solutions.

By calling $N_{i}^{(a)}(2,s,m)$ and $B_{k}(E,m)$ the the equal mass limit of
$N_{i}^{(a)}(2,s)$ and $B_{k}(E,m_{i})$ given by Eq.(5.45), one finds

$\displaystyle N_{0}^{(a)}(2,s,m)$
$\displaystyle=\frac{2}{3}N_{3}^{(a)}(2,s,m)\ ,$ $\displaystyle
N_{1}^{(a)}(2,s,m)$ $\displaystyle=0\ ,$ $\displaystyle N_{2}^{(a)}(2,s,m)$
$\displaystyle=B_{0}(E,m)\pi\ ,$ $\displaystyle N_{3}^{(a)}(2,s,m)$
$\displaystyle=B_{0}(E,m)\ln\left(\frac{E^{3}}{m^{3}}\right)+B_{3}(E,m)\ ,$
(5.56)

showing that only two of the 4 linearly independent asymptotic expansions of
the solutions, say $N_{2}^{(a)}(2,s,m)$ and $N_{3}^{(a)}(2,s,m)$, are left,
corresponding to two solutions only of the 4th order homogeneous equation of
$D_{4}(2,s,m)$, the equal mass limit of $D_{4}(2,s)$. Two solutions are
apparently missing.

But by looking again at Eq.s(5.45) and (5.47–5.50), one finds that the large
$E$ expansions of the two missing solutions are provided by

$\displaystyle\tilde{N}_{0}^{(a)}(2,s,m)$ $\displaystyle=\lim_{m_{i}\to
m}\frac{m_{1}}{m_{3}-m_{1}}\left[\frac{2m_{1}}{m_{2}-m_{1}}N_{1}^{(a)}(2,s)+\frac{m_{1}}{m_{3}-m_{1}}\left(3N_{0}^{(a)}(2,s)-2N_{3}^{(a)}(2,s)\right)\right]\
$
$\displaystyle=-\frac{1}{E^{2}}-5\frac{m^{2}}{E^{4}}-9\frac{m^{4}}{E^{6}}-13\frac{m^{6}}{E^{8}}-17\frac{m^{8}}{E^{10}}\
,$ $\displaystyle\tilde{N}_{1}^{(a)}(2,s,m)$ $\displaystyle=\lim_{m_{i}\to
m}\frac{m_{1}}{m_{2}-m_{1}}N_{1}^{(a)}(2,s)$
$\displaystyle=-\frac{1}{E^{2}}-\frac{m^{2}}{E^{4}}-\frac{m^{4}}{E^{6}}-\frac{m^{6}}{E^{8}}-\frac{m^{8}}{E^{10}}\
,$ (5.57)

where the equal mass limit implies the L’Hôpital’s rule. The two missing
solutions correspond then to two homogeneous solutions of Eq.(5.1) at equal
masses with the asymptotic expansions provided by Eq.s(5.57).

##  6  The differential equation at $d=3$ and its solutions

At $d=3$, Eq.(4.18) becomes

$D_{4}(3,s)\ M_{i}(3,s)=0\ ,\hskip 14.22636pti=0,1,..,5\ .$ (6.1)

where $D_{4}(3,s)$, according to Eq(2.5), is given by

$\displaystyle D_{4}(3,s)$
$\displaystyle=C_{4}(3,s)\frac{d^{4}}{ds^{4}}+C_{3}(3,s)\frac{d^{3}}{ds^{3}}$
$\displaystyle+C_{2}(3,s)\frac{d^{2}}{ds^{2}}+C_{1}(3,s)\frac{d}{ds}+C_{0}(3,s)\
.$ (6.2)

In particular one finds

$\displaystyle C_{4}(3,s)$
$\displaystyle=32s^{3}\mathcal{D}(s)\bigl{[}s^{2}-R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})\bigr{]}\
,$ $\displaystyle C_{0}(3,s)$ $\displaystyle=0\ ,$ (6.3)

see Eq.(B.6) for $\mathcal{D}(s)$ and Eq.(5.42) for
$R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})$.

The naïve $d=3$ limit of the functions $M_{i}(3,s)$ appearing in Eq.(6.1),
according to Eq.s(4.2,4.1), is particularly simple and reads

$M_{i}(3,s)=\frac{1}{E}N_{i}(3,s)\ ,$ (6.4)

with $s=E^{2}$ as usual, and

$N_{i}(3,s)=\int_{b_{i}-1}^{b_{i}}db\ \frac{1}{\sqrt{|b|}}\ .$ (6.5)

But for $i=0$ and $i=5$ the naïve $d=3$ limit cannot be used, because it would
give the non convergent integrals

$\displaystyle N_{0}(3,s)$ $\displaystyle=\int_{-\infty}^{0}db\
\frac{1}{\sqrt{|b|}}\ ,$ $\displaystyle N_{5}(3,s)$
$\displaystyle=\int_{b_{4}}^{+\infty}db\ \frac{1}{\sqrt{|b|}}\ .$ (6.6)

To overcome the inconvenience, one must go back to the definitions of
$N_{0}(d,s),N_{5}(d,s)$, which converge for $d<8/3$, as already remarked in
Section 3, and continue them analytically to $d>3$.

According to Eq.(4.1), the definition of $N_{0}(d,s)$ is

$N_{0}(d,s)=\int_{-\infty}^{0}db\
|b|^{\frac{2-d}{2}}|R_{4}(b)|^{\frac{d-3}{2}}\ ,$ (6.7)

which can be rewritten as

$N_{0}(d,s)=\int_{-\infty}^{0}db\
|b|^{\frac{3}{2}d-5}\left[\left|\frac{R_{4}(b)}{b^{4}}\right|^{\frac{d-3}{2}}\right]$

By integrating by parts the factor $|b|^{\frac{3}{2}d-5}$ one obtains

$N_{0}(d,s)=-\frac{2}{3d-8}\int_{-\infty}^{0}db\ |b|^{\frac{3}{2}d-4}\
\frac{d}{db}\left[\left|\frac{R_{4}(b)}{b^{4}}\right|^{\frac{d-3}{2}}\right]\
,$ (6.8)

an integral representation which converges for $d<10/3$, hence valid at $d=3$.

By a similar argument, for $i=5$ one can arrive at the analytical continuation

$N_{5}(d,s)=-\frac{2}{3d-8}\int_{b_{4}}^{\infty}db\
(b-b_{4})^{\frac{3}{2}d-4}\ \
\frac{d}{db}\left[\left(\frac{b}{b-b_{4}}\right)^{-\frac{d-2}{2}}\left(\frac{R_{4}(b)}{(b-b_{4})^{4}}\right)^{\frac{d-3}{2}}\right]\
,$ (6.9)

which is again converging for $d=3<10/3$.

The naïve $d=3$ limit of Eq.s(6.8,6.9) is now allowed, and the
$b$-integrations can be carried out for all the $N_{i}(3,s)$ integrals. The
results are, with $s=E^{2}$,

$\displaystyle N_{0}(3,s)$ $\displaystyle=0\ ,$ $\displaystyle N_{1}(3,s)$
$\displaystyle=2(m_{1}-m_{2})\ ,$ $\displaystyle N_{2}(3,s)$
$\displaystyle=4m_{2}\ ,$ $\displaystyle N_{3}(3,s)$
$\displaystyle=2(E-m_{1}-m_{2}-m_{3})\ ,$ $\displaystyle N_{4}(3,s)$
$\displaystyle=4m_{3}\ ,$ $\displaystyle N_{5}(3,s)$
$\displaystyle=-2(E+m_{3})\ ,$ (6.10)

It is apparent that only two of the above functions are linearly independent.
Considering the corresponding solutions of Eq.(6.1) $M_{i}(3,s)=N_{i}(3,s)/E$,
see Eq.(6.4), one finds that those solutions are combinations of just two
linearly independent solutions, say, for definiteness

$\displaystyle F_{1}(3,s)$ $\displaystyle=1\ ,$ $\displaystyle F_{2}(3,s)$
$\displaystyle=\frac{1}{E}\ .$ (6.11)

Let us note here, in passing, that inserting $F_{2}(3,s)$ and its derivatives
in the equation

$D_{4}(3,s)F_{2}(3,s)=0\ ,$ (6.12)

and by keeping $D_{4}(3,s)$ in the form of Eq.(6.2), one immediately obtains
the relation

$C_{1}(3,s)-\frac{3}{2s}C_{2}(3,s)+\frac{15}{4s^{2}}C_{3}(3,s)-\frac{105}{8s^{3}}C_{4}(3,s)=0\
.$ (6.13)

As often observed, the 4th order equation Eq.(6.1) is expected to have 4
linearly indepent solutions, while Eq.(6.11) gives only two. To complete the
list of the solutions, let us go back from Eq.(6.1), valid at $d=3$, to the
equation for generic $d$, which we rewrite here as

$D_{4}(d,s)\ G(d,s)=0\ ,$ (6.14)

and expand everything around $d=3$ up to first order in $(d-3)$

$\displaystyle D_{4}(d,s)$ $\displaystyle=D_{4}(3,s)+(d-3)D_{4}(3,1,s)+...\ ,$
$\displaystyle G(d,s)$ $\displaystyle=G(3,s)+(d-3)G(3,1,s)+...\ .$ (6.15)

Correspondingly, Eq.(6.14) splits into an equation at $d=3$ and an equation at
1st order in $(d-3)$

$\displaystyle D_{4}(3,s)\ G(3,s)=0\ ,$ (6.16) $\displaystyle D_{4}(3,s)\
G(3,1,s)+D_{4}(3,1,s)\ G(3,s)=0\ .$ (6.17)

If

$G(3,s)=0\ ,$ (6.18)

Eq.(6.17) becomes

$D_{4}(3,s)\ G(3,1,s)\ =0\ ,$ (6.19)

i.e. if $G(3,s)=0$, then $G(3,1,s)$ is a solution of the equation with the
differential operator $D_{4}(3,s)$.

We will then look for linear combinations $G_{k}(d,s)$ of the functions
$M_{i}(d,s)$ whose $d=3$ limits vanish, $G_{k}(3,s)=0$, so that the first
order terms in $(d-3)$ of their expansions around $d=3$, $G_{k}(3,1,s)$, are
solutions of Eq.(6.36).

To that aim, we expand also the functions $M_{i}(d,s)$ and $N_{i}(d,s)$ around
$d=3$ up to first order in $(d-3)$

$\displaystyle M_{i}(d,s)$ $\displaystyle=M_{i}(3,s)+(d-3)M_{i}(3,1,s)\ ,$
$\displaystyle N_{i}(d,s)$ $\displaystyle=N_{i}(3,s)+(d-3)N_{i}(3,1,s)\ ,$
(6.20)

which according to Eq.(4.2) gives

$\displaystyle M_{i}(3,s)$ $\displaystyle=\frac{1}{E}N_{i}(3,s)\ ,$ (6.21)
$\displaystyle M_{i}(3,1,s)$
$\displaystyle=\frac{1}{E}\bigg{(}-\ln(E)N_{i}(3,s)+N_{i}(3,1,s)\bigg{)}\ .$
(6.22)

It is then easily found, on account of Eq.s(6.10), that the following 4 linear
combination $G_{k}(d,s),k=1,..,4$ of the 6 functions $M_{i}(d,s)$

$\displaystyle G_{1}(d,s)$
$\displaystyle=M_{1}(d,s)+M_{2}(d,s)+M_{3}(d,s)+M_{4}(d,s)+M_{5}(d,s)\ ,$
$\displaystyle G_{2}(d,s)$
$\displaystyle=2m_{2}M_{1}(d,s)-(m_{1}-m_{2})M_{2}(d,s)\ ,$ $\displaystyle
G_{3}(d,s)$ $\displaystyle=m_{3}M_{2}(d,s)-m_{2}M_{4}(d,s)\ ,$ $\displaystyle
G_{4}(d,s)$
$\displaystyle=2m_{3}M_{3}(d,s)+(m_{1}+m_{2}+2m_{3})M_{4}(d,s)+2m_{3}M_{5}(d,s)\
,$ (6.23)

do satisfy, by construction, the condition

$G_{k}(3,s)=0\ .\hskip 56.9055ptk=1,..,4$ (6.24)

Therefore, according to Eq.s(6.18,6.19) the functions $G_{k}(3,1,s)$ satisfy
Eq.(6.19), namely

$D_{4}(3,s)\ G_{k}(3,1,s)=0\ .\hskip 56.9055ptk=1,..,4\ .$ (6.25)

The evaluation of the functions $G_{k}(3,1,s)$, according to Eq.(6.23) and
Eq.(6.22) requires the evaluation of the functions $N_{i}(3,1,s)$; one finds

$\displaystyle N_{0}(3,1,s)$ $\displaystyle=2\pi(E+m_{1})\ ,$ $\displaystyle
N_{1}(3,1,s)$ $\displaystyle=2\ m_{1}\big{(}\ln(4m_{1})-3\big{)}-2\
m_{2}\big{(}\ln(4m_{2})-3\big{)}$
$\displaystyle+(E+m_{1}-m_{2}+m_{3})\ln(E+m_{1}-m_{2}+m_{3})$
$\displaystyle+(E+m_{1}-m_{2}-m_{3})\ln(E+m_{1}-m_{2}-m_{3})$
$\displaystyle-(E-m_{1}+m_{2}+m_{3})\ln(E-m_{1}+m_{2}+m_{3})$
$\displaystyle-(E-m_{1}+m_{2}-m_{3})\ln(E-m_{1}+m_{2}-m_{3})$ (6.26)
$\displaystyle N_{2}(3,1,s)$ $\displaystyle=4\
m_{2}\big{(}\ln(4m_{2})-3\big{)}$
$\displaystyle+(E+m_{1}+m_{2}+m_{3})\ln(E+m_{1}+m_{2}+m_{3})$
$\displaystyle+(E+m_{1}+m_{2}-m_{3})\ln(E+m_{1}+m_{2}-m_{3})$
$\displaystyle-(E+m_{1}-m_{2}+m_{3})\ln(E+m_{1}-m_{2}+m_{3})$
$\displaystyle-(E+m_{1}-m_{2}-m_{3})\ln(E+m_{1}-m_{2}-m_{3})$
$\displaystyle+(E-m_{1}+m_{2}+m_{3})\ln(E-m_{1}+m_{2}+m_{3})$
$\displaystyle+(E-m_{1}+m_{2}-m_{3})\ln(E-m_{1}+m_{2}-m_{3})$
$\displaystyle-(E-m_{1}-m_{2}+m_{3})\ln(E-m_{1}-m_{2}+m_{3})$
$\displaystyle-(E-m_{1}-m_{2}-m_{3})\ln(E-m_{1}-m_{2}-m_{3})\ ,$ (6.27)
$\displaystyle N_{3}(3,1,s)$ $\displaystyle=2E\
\big{(}\ln(4E)-3\big{)}-2m_{1}\ \big{(}\ln(4m_{1})-3\big{)}$
$\displaystyle-2m_{2}\ \big{(}\ln(4m_{2})-3\big{)}-2m_{3}\
\big{(}\ln(4m_{3})-3\big{)}$ $\displaystyle-$
$\displaystyle(E+m_{1}+m_{2}+m_{3})\ln(E+m_{1}+m_{2}+m_{3})$ $\displaystyle+$
$\displaystyle(E+m_{1}-m_{2}-m_{3})\ln(E+m_{1}-m_{2}-m_{3})$ $\displaystyle+$
$\displaystyle(E-m_{1}+m_{2}-m_{3})\ln(E-m_{1}+m_{2}-m_{3})$ $\displaystyle+$
$\displaystyle(E-m_{1}-m_{2}+m_{3})\ln(E-m_{1}-m_{2}+m_{3})$ $\displaystyle+$
$\displaystyle 2(E-m_{1}-m_{2}-m_{3})\ln(E-m_{1}-m_{2}-m_{3})\ ,$ (6.28)
$\displaystyle N_{4}(3,1,s)=$ $\displaystyle 4\
m_{3}\big{(}\ln(4m_{3})-3\big{)}$ $\displaystyle+$
$\displaystyle(E+m_{1}+m_{2}+m_{3})\ln(E+m_{1}+m_{2}+m_{3})$ $\displaystyle-$
$\displaystyle(E+m_{1}+m_{2}-m_{3})\ln(E+m_{1}+m_{2}-m_{3})$ $\displaystyle+$
$\displaystyle(E+m_{1}-m_{2}+m_{3})\ln(E+m_{1}-m_{2}+m_{3})$ $\displaystyle-$
$\displaystyle(E+m_{1}-m_{2}-m_{3})\ln(E+m_{1}-m_{2}-m_{3})$ $\displaystyle+$
$\displaystyle(E-m_{1}+m_{2}+m_{3})\ln(E-m_{1}+m_{2}+m_{3})$ $\displaystyle-$
$\displaystyle(E-m_{1}+m_{2}-m_{3})\ln(E-m_{1}+m_{2}-m_{3})$ $\displaystyle+$
$\displaystyle(E-m_{1}-m_{2}+m_{3})\ln(E-m_{1}-m_{2}+m_{3})$ $\displaystyle-$
$\displaystyle(E-m_{1}-m_{2}-m_{3})\ln(E-m_{1}-m_{2}-m_{3})\ ,$ (6.29)
$\displaystyle N_{5}(3,1,s)$ $\displaystyle=-2E\ln(4E)-2m_{3}\ln(m_{3})$
$\displaystyle-(E+m_{1}+m_{2}+m_{3})\ln(E+m_{1}+m_{2}+m_{3})$
$\displaystyle-(E+m_{1}-m_{2}+m_{3})\ln(E+m_{1}-m_{2}+m_{3})$
$\displaystyle-(E-m_{1}+m_{2}+m_{3})\ln(E-m_{1}+m_{2}+m_{3})$
$\displaystyle-(E-m_{1}-m_{2}+m_{3})\ln(E-m_{1}-m_{2}+m_{3})\ .$ (6.30)

The explicit values of the $G_{k}(3,1,s)$ are then easily obtained from the
various definitions and the values of the functions $N_{i}(3,1,s)$ listed
above. For writing them down it is convenient to introduce the new function

$\displaystyle F(3,s,m_{1},m_{2},m_{3})$
$\displaystyle=(m_{1}-m_{2})(E+m_{1}+m_{2}+m_{3})\ln(E+m_{1}+m_{2}+m_{3})$
$\displaystyle+(m_{1}-m_{2})(E+m_{1}+m_{2}-m_{3})\ln(E+m_{1}+m_{2}-m_{3})$
$\displaystyle-(m_{1}+m_{2})(E+m_{1}-m_{2}+m_{3})\ln(E+m_{1}-m_{2}+m_{3})$
$\displaystyle-(m_{1}+m_{2})(E+m_{1}-m_{2}-m_{3})\ln(E+m_{1}-m_{2}-m_{3})$
$\displaystyle+(m_{1}+m_{2})(E-m_{1}+m_{2}+m_{3})\ln(E-m_{1}+m_{2}+m_{3})$
$\displaystyle+(m_{1}+m_{2})(E-m_{1}+m_{2}-m_{3})\ln(E-m_{1}+m_{2}-m_{3})$
$\displaystyle-(m_{1}-m_{2})(E-m_{1}-m_{2}+m_{3})\ln(E-m_{1}-m_{2}+m_{3})$
$\displaystyle-(m_{1}-m_{2})(E-m_{1}-m_{2}-m_{3})\ln(E-m_{1}-m_{2}-m_{3})\ ,$
(6.31)

which satisfies under permutation of the masses the following identities

$\displaystyle F(3,s,m_{1},m_{2},m_{3})+F(3,s,m_{2},m_{1},m_{3})=0\ ,$
$\displaystyle F(3,s,m_{2},m_{3},m_{1})+F(3,s,m_{3},m_{2},m_{1})=0\ ,$
$\displaystyle F(3,s,m_{3},m_{1},m_{2})+F(3,s,m_{1},m_{3},m_{2})=0\ ,$
$\displaystyle
m_{3}F(3,s,m_{1},m_{2},m_{3})+m_{1}F(3,s,m_{2},m_{3},m_{1})+m_{2}F(3,s,m_{3},m_{1},m_{2})=0\
;$ (6.32)

according to the above 4 identities 2 of the 6 functions, say for definiteness
$F(3,s,m_{1},m_{2},m_{3})$ and $F(3,s,m_{2},m_{3},m_{1})$, can be taken as
linearly independent.

As $M_{i}(3,1,s)=N_{i}(3,1,s)/E,i=0,1,..,5$, the explicit expressions of the
four functions $G_{k}(3,1,s),k=1,..,4\ ,$ are given by

$G_{1}(3,1,s)=-\frac{6}{E}(E+m_{3})\ ,$ (6.33)
$G_{2}(3,1,s)=\frac{1}{E}\biggl{[}4m_{1}m_{2}\ln\left(\frac{m_{1}}{m_{2}}\right)-F(3,s,m_{1},m_{2},m_{3})\biggr{]}\
,$ (6.34)
$G_{3}(3,1,s)=\frac{1}{E}\biggl{[}4m_{2}m_{3}\ln\left(\frac{m_{2}}{m_{3}}\right)-F(3,s,m_{2},m_{3},m_{1})\biggr{]}\
,$ (6.35) $\displaystyle G_{4}(3,1,s)$
$\displaystyle=\frac{1}{E}\biggl{[}-12m_{3}(E+m_{3})-4m_{1}m_{3}\ln\left(\frac{m_{1}}{m_{3}}\right)-4m_{2}m_{3}\ln\left(\frac{m_{2}}{m_{3}}\right)$
$\displaystyle+F(3,s,m_{2},m_{3},m_{1})-F(3,s,m_{3},m_{1},m_{2})\biggr{]}\ .$
(6.36)

The four functions $G_{k}(3,1,s),k=1,..,4$. see Eq.(6.25), are all solutions
of Eq.(6.19); we have already seen, Eq.(6.11, two of the four linearly
indepent solutions of the equation, it is not difficult, using in particular
Eq.s(6.32), that two other independent solutions are given by

$\displaystyle F_{3}(3,s)$ $\displaystyle=\frac{1}{E}\
F(3,s,m_{1},m_{2},m_{3})\ ,$ $\displaystyle F_{4}(3,s)$
$\displaystyle=\frac{1}{E}\ F(3,s,m_{2},m_{3},m_{1})\ .$ (6.37)

Summarising, we have found that the 4 linearly independent functions
$F_{k}(3,s),k=1,..,4$, defined by Eq.s(6.11,6.37), can be choosen as a set of
4 linearly independent solutions of the equation

$D_{4}(3,s)\ F_{k}(3,s)=0\ .$ (6.38)

Once the four solutions are knwon, we can factorise the 4th order diferential
operator $D_{4}(3,s)$.

To that aim, by following the same procedure followed in the previous Section,
let us begin by introducing the two first order differential operators

$\displaystyle D_{a}(3,s)$ $\displaystyle=\frac{d}{ds}\ ,$ $\displaystyle
D_{b}(3,s)$ $\displaystyle=2s\ \frac{d}{ds}+3\ ,$ (6.39)

such that

$\displaystyle D_{a}(3,s)\ F_{1}(3,s)=0\ ,$ $\displaystyle D_{b}(3,s)\
D_{a}(3,s)\ F_{2}(3,s)=0\ .$ (6.40)

An elementary calculation gives

$D_{b}(3,s)\ D_{a}(3,s)\ F_{3}(3,s)=R_{3c}(s,m_{i}^{2})\ ,$ (6.41)

where $R_{3c}(s,m_{i}^{2})$ is given by the rational expression

$R_{3c}(3,s,m_{i}^{2})=8\frac{m_{1}m_{2}(m_{2}^{2}-m_{1}^{2})}{\sqrt{s}\
\mathcal{D}(s)}\bigl{[}s^{2}-2(m_{1}^{2}+m_{2}^{2}-3m_{3}^{2})s+R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})\bigr{]}\
,$ (6.42)

with $\mathcal{D}(s)$ and $R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})$ given by
Eq.(B.6) and Eq.(5.42).
The rational expression $R_{3c}(3,s,m_{1}^{2})$ satisfies, by construction,
the first order differential equation

$D_{c}(3,s)R_{3c}(3,s,m_{1}^{2})=0\ ,$ (6.43)

where $D_{c}(3,s)$ is the first order differential operator

$D_{c}(3,s)=R_{3c}(3,s,m_{i}^{2})\frac{d}{ds}-\bigg{(}\frac{dR_{3c}(3,s,m_{i}^{2})}{ds}\bigg{)},$
(6.44)

and $dR_{3c}(3,s,m_{i}^{2})/ds$, like $R_{3c}(3,s,m_{i}^{2})$ is also a
rational expression.

Another explicit calculation gives

$D_{c}(3,s)\ D_{b}(3,s)\ D_{a}(3,s)\ F_{4}(3,s)=R_{3d}(s,m_{i}^{2})\ ,$ (6.45)

with

$R_{3d}(3,s,m_{i}^{2})=512\frac{m_{1}m_{2}^{2}m_{3}}{s\mathcal{D}^{2}(s)}(m_{3}^{2}-m_{2}^{2})(m_{3}^{2}-m_{1}^{2})(m_{2}^{2}-m_{1}^{2})\bigl{[}s^{2}-R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})\bigr{]}\
.$ (6.46)

Again, $R_{3d}(3,s,m_{i}^{2})$ satisfies by construction the first order
differential equation

$D_{d}(3,s)R_{3d}(3,s,m_{1}^{2})=0\ ,$ (6.47)

where $D_{d}(3,s)$ is the first order differential operator

$D_{d}(3,s)=R_{3d}(3,s,m_{i}^{2})\frac{d}{ds}-\bigg{(}\frac{dR_{3d}(3,s,m_{i}^{2})}{ds}\bigg{)},$
(6.48)

and $dR_{3d}(3,s,m_{i}^{2})/ds$, like $R_{3d}(3,s,m_{i}^{2})$, is also a
rational expression.

The full factorisation of $D_{4}(3,s)$ is now completed and reads

$D_{4}(3,s)=\frac{1}{\mathcal{C}}\ D_{d}(3,s)D_{c}(3,s)D_{b}(3,s)D_{a}(3,s)\
,$ (6.49)

with

${\mathcal{C}}=256\frac{m_{1}^{2}m_{2}^{3}m_{3}}{s^{3}\sqrt{s}\mathcal{D}^{4}(s)}(m_{2}^{2}-m_{1}^{2})^{2}(m_{3}^{2}-m_{2}^{2})(m_{3}^{2}-m_{1}^{2})\bigl{[}s^{2}-2(m_{1}^{2}+m_{2}^{2}-3m_{3}^{2})s+R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})\bigr{]}\
.$ (6.50)

##  7 The differential equation at d = 4 and its solutions

In this Section we study the d = 4 limit of the 4th order differential
equation Eq(4.18), which becomes

$D_{4}(4,s)\ M_{i}(4,s)=0\ .\hskip 14.22636pti=0,1,..,5\ .$ (7.1)

As in the $d=3$ case of the previous Section, the values at $d=4$ of the five
coefficients $C_{i}(d,s,m_{i}^{2}),i=0,1,..,4$ which appear in $D_{4}(d,s)$
can be immediately obtained from Eq(2.5),

$\displaystyle D_{4}(4,s)$
$\displaystyle=C_{4}(4,s)\frac{d^{4}}{ds^{4}}+C_{3}(4,s)\frac{d^{3}}{ds^{3}}$
$\displaystyle+C_{2}(4,s)\frac{d^{2}}{ds^{2}}+C_{1}(4,s)\frac{d}{ds}+C_{0}(4,s)\
.$ (7.2)

In particular one finds

$\displaystyle C_{4}(4,s)$
$\displaystyle=8s^{3}\mathcal{D}(s)\bigl{[}3s^{2}-2(m_{1}^{2}m_{2}^{2}m_{3}^{2})s-R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})\bigr{]}\
,$ $\displaystyle C_{0}(4,s)$ $\displaystyle=0\ ,$ (7.3)

see Eq,(B.6) for $\mathcal{D}(s)$ and Eq.(5.42) for
$R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})$.

The first step for discussing the solutions of Eq.(7.1) is obviously to look
at the $d=4$ limit of the 6 functions defined in Eq.s(4.2,4.1), which we
recall here for the convenience of the reader

$M_{i}(d,s)=s^{\frac{2-d}{2}}N_{i}(d,s)\ ,$ (7.4)

with

$N_{i}(d,s)=\int_{b_{i-1}}^{b_{i}}db\ |b|^{\frac{2-d}{2}}\
|R_{4}(b)|^{\frac{d-3}{2}}\ ,$ (7.5)

where the $b_{i},i=-1,0,..,5$ specified in Eq.(3.12). The naïve $d=4$ limit of
Eq.(7.5) then reads

$N_{i}(4,s)=\int_{b_{i-1}}^{b_{i}}\frac{db}{|b|}\ \sqrt{|R_{4}(b|)}\ .$ (7.6)

As already observed in Section 2, for $i=0$ and $i=5$ the above definitions of
the $N_{i}(d,s)$ converge at large $|b|$ only if $d<8/3$, further for $i=0,1$
they converge at $b=0$ only if $d<4$; for reaching $d=4$ an analytic
continuation to $d>4$ is needed.

The analytic continuations can be obtained by suitable integrations by parts.
For $i=0$, with the obvious change $b=-c$ of the integration variable, the
analytic continuation can be written as

$\displaystyle N_{0}(d,s)$ $\displaystyle=\frac{-16}{3(d-4)^{2}(3d-8)(3d-10)}$
$\displaystyle\times\int_{0}^{\infty}c^{-\frac{1}{2}(d-4)}\frac{d}{dc}\bigg{\\{}c^{1+2(d-4)}\frac{d}{dc}\bigg{[}c^{2}\frac{d}{dc}\bigg{(}c^{2}\frac{d}{dc}\bigg{(}\frac{R_{4}(-c)}{c^{4}}\bigg{)}^{\frac{1}{2}(d-3)}\bigg{)}\bigg{]}\bigg{\\}}$
(7.7)

with $R_{4}(-c)=(c+b_{1})(c+b_{2})(c+b_{3})(c+b_{4})$, according to Eq.(3.14).
The above equation can be expanded in Laurent series in $(d-4)$

$N_{0}(d,s)=\frac{1}{(d-4)^{2}}N_{0}(4,-2,s)+\frac{1}{(d-4)}N_{0}(4,-1,s)+N_{0}(4,0,s)+....\
;$ (7.8)

an explicit, relatively simple calculation then gives

$\displaystyle N_{0}(4,-2,s)$ $\displaystyle=0\ ,$ $\displaystyle
N_{0}(4,-1,s)$
$\displaystyle=\frac{4}{3}(-2m_{1}^{2}+m_{2}^{2}+m_{3}^{2})E^{2}$
$\displaystyle+\frac{4}{3}(+m_{1}^{2}m_{2}^{2}+m_{1}^{2}m_{3}^{2}-2m_{2}^{2}m_{3}^{2}).$
(7.9)

By expanding also $M_{0}(d,s)=$ around $d=4$,

$M_{0}(d,s)=\frac{1}{(d-4)^{2}}M_{0}(4,-2,s)+\frac{1}{(d-4)}M_{0}(4,-1,s)+....\
,$

Eq.(7.4) gives as leading term of the expansion of $M_{0}(d,s)$

$M_{0}(4,-1,s)=\frac{1}{E^{2}}N_{0}(4,-1,s)\ ,$ (7.10)

with $s=E^{2}$, and correspondingly Eq.(7.1) for $i=0$ becomes, again at
leading order in $(d-4)$

$\frac{1}{(d-4)}\times D_{4}(4,s)M_{0}(4,-1,s)=0\ ,$ (7.11)

which implies, by dropping the overall $s$-independent facor $1/(d-4)$,

$D_{4}(4,s)M_{0}(4,-1,s)=0\ .$ (7.12)

Hence,

$\displaystyle M_{0}(4,-1,s)$ $\displaystyle=\frac{1}{E^{2}}N_{0}(4,-1,s)$
$\displaystyle=\frac{4}{3}(-2m_{1}^{2}+m_{2}^{2}+m_{3}^{2})$
$\displaystyle+\frac{4}{3E^{2}}(+m_{1}^{2}m_{2}^{2}+m_{1}^{2}m_{3}^{2}-2m_{2}^{2}m_{3}^{2})\
,$ (7.13)

is a non vanishing solution of Eq.(7.1).

For the $i=1$ and $i=5$ cases of Eq.(7.5) with the same notation for the
analytic continuations we find

$\displaystyle N_{1}(4,-1,s)$
$\displaystyle=(-2m_{1}^{2}+2m_{2}^{2})E^{2}+2m_{1}^{2}m_{3}^{2}-2m_{2}^{2}m_{3}^{2}\
,$ $\displaystyle N_{5}(4,-1,s)$
$\displaystyle=\frac{16}{3}(-m_{1}^{2}-m_{2}^{2}+2m_{3}^{2})E^{2}$
$\displaystyle+\frac{16}{3}(2m_{1}^{2}m_{2}^{2}-m_{1}^{2}m_{3}^{2}-m_{2}^{2}m_{3}^{2})\
,$ (7.14)

corresponding to the solutions

$\displaystyle M_{1}(4,-1,s)$
$\displaystyle=(-2m_{1}^{2}+2m_{2}^{2})+\frac{1}{E^{2}}(2m_{1}^{2}m_{3}^{2}-2m_{2}^{2}m_{3}^{2}\
,$ $\displaystyle M_{5}(4,-1,s)$
$\displaystyle=\frac{16}{3}(-m_{1}^{2}-m_{2}^{2}+2m_{3}^{2})$
$\displaystyle+\frac{16}{3E^{2}}(2m_{1}^{2}m_{2}^{2}-m_{1}^{2}m_{3}^{2}-m_{2}^{2}m_{3}^{2})\
,$ (7.15)

It is easily seen that the three above solutions, $M_{i}(4,-1,s),i=0,1,5$ are
linear comninations of just two linearly independent solutions, say, for
definiteness

$\displaystyle F_{1}(4,s)$ $\displaystyle=1\ ,$ $\displaystyle F_{2}(4,s)$
$\displaystyle=\frac{1}{E^{2}}\ .$ (7.16)

Following the derivation of Eq.(6.13), by inserting $F_{2}(4,s)$ and its
derivatives in the equation

$D_{4}(4,s)\ F_{2}(4,s)=0\ ,$ (7.17)

and by keeping $D_{4}(4,s)$ in the form of Eq.(7.2), one immediately obtains
the relation

$C_{1}(4,s,m_{i}^{2})-\frac{2}{s}C_{2}(4,s,m_{i}^{2})+\frac{6}{s^{2}}C_{3}(4,s,m_{i}^{2})-\frac{24}{s^{3}}C_{4}(4,s,m_{i}^{2})=0$
(7.18)

between the coefficients of Eq.(7.2).

Let us look now for the other two linearly independent solutions of the 4th
order equation Eq.(7.1).

For the evaluation of the $d=4$ limit of the other three integrals
$N_{i}(d,s),i=2,3,4$ no analytic continuation in $d$ is needed, and the
straightforward use of Eq.(7.6) is allowed.

In order to discuss their linear dependence, in strict analogy with the
discussion leading to Eq.(5.24), let us consider the analytic function

$f(b)=\frac{1}{b}\sqrt{R_{4}(b)}\ ,$ (7.19)

with $R_{4}(b)$ defined by Eq.(3.14), which has a pole at $b=0$, a cut for
$b_{1}<b<b_{2}$ and another cut for $b_{3}<b<b_{4}$.

By integrating that function in the complex b plane along the circle $b=C{\rm
e}^{i\phi},0<\phi<2\pi$ in the limit of large $C\gg E^{2},m_{i}^{2}$,
expanding $f(b)$ in powers of $b$ and dropping terms of order $1/C$ in the
result, one finds

$I_{\mathcal{C}}=\oint_{\mathcal{C}}db\
f(b)=2i\pi\bigl{[}(m_{1}^{2}+m_{2}^{2}-2m_{3}^{2})E^{2}-2m_{1}^{2}m_{2}^{2}+m_{1}^{2}m_{3}^{2}+m_{2}^{2}m_{3}^{2}\bigr{]}\
.\ $ (7.20)

On the other hand, by squeezing the circle into a thin, elongated rectangle
with vertices

$-C-i\epsilon,{\hskip 5.69054pt}+C-i\epsilon,{\hskip
5.69054pt}+C+i\epsilon,{\hskip 5.69054pt}-C+i\epsilon,$

one finds, in the $C\epsilon\to 0$ limit

$\displaystyle J_{\mathcal{C}}$ $\displaystyle=\oint_{\leavevmode\hbox
to8.94pt{\vbox to3.4pt{\pgfpicture\makeatletter\hbox{\hskip
0.2pt\lower-0.2pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{}{{}}{}
{}{{}}{}{}{}{}{{}}{}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@lineto{0.0pt}{3.0pt}\pgfsys@lineto{8.53583pt}{3.0pt}\pgfsys@lineto{8.53583pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{8.53583pt}{3.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}dbf(b)=\int_{-C}^{+C}db\biggl{[}f(b-i\epsilon)-f(b+i\epsilon)\biggr{]}$
$\displaystyle=4i\pi(m_{3}^{2}-m_{2}^{2})(E^{2}-m_{1}^{2})+2i\int_{b_{1}}^{b_{2}}db\frac{1}{b}\sqrt{|R_{4}(b)|}-2i\int_{b_{3}}^{b_{4}}db\frac{1}{b}\sqrt{|R_{4}(b)|}\
,$ (7.21)

where the first term comes from the pole at $b=0$, the other two from the two
cuts. The equality of $I_{\mathcal{C}}$ and $J_{\mathcal{C}}$ gives, in the
notation of Eq.s(7.9,7.14), (7.5) and 7.8,

$N_{4}(4,0,s)=\frac{3}{2}\pi N_{0}(4,-1,s)-2\pi N_{1}(4,-1,s)+N_{2}(4,0,s)\ ,$
(7.22)

showing the that $N_{4}(4,0,s)$ is a linear combination of the integrals in
the r.h.s.. The above relation is to be compared with Eq.(5.24), the analog
and simpler equation relating $N_{4}(2,s)$ and $N_{2}(2,s)$ at $d=2$.

In order to complete (and check) the discussion of the linear dependence or
independence of the six integrals $N_{i}(d,s)$ at $d=4$, one can consider,
besides the simple explicit expressions of $N_{0}(4,-1,s)$ and
$N_{1}(4,-1,s)$, also the expansions of $N_{2}(4,s),N_{3}(4,s)$ and
$N_{4}(4,s)$ at large $E$, up to a sufficiently high oder (say up to
$1/E^{10}$ included, redundantly), which can be evaluated by the same
techniques leading to Eq.s(5.45)-(5.50) at $d=2$. Without reporting the
results for the sake of brevity, let us just say that they provide a check of
Eq.(7.22) and show that the four linearly independent solutions of Eq.(7.1)
can be taken to be the following four functions

$\displaystyle F_{1}(4,s)$ $\displaystyle=1\ ,$ $\displaystyle F_{2}(4,s)$
$\displaystyle=\frac{1}{E^{2}}\ ,$ $\displaystyle F_{3}(4,s)$
$\displaystyle=\frac{1}{E^{2}}N_{2}(4,s)\ ,$ $\displaystyle F_{4}(4,s)$
$\displaystyle=\frac{1}{E^{2}}N_{3}(4,s)\ .$ (7.23)

Correspondingly, we can factorise the 4th order differential operator
$D_{4}(4,s)$ in the form

$D_{4}(4,s)=D_{c}(4,s)D_{b}(4,s)D_{a}(4,s)\ ,$ (7.24)

where

$\displaystyle D_{a}(4,s)$ $\displaystyle=\frac{d}{ds}\ ,$ $\displaystyle
D_{b}(4,s)$ $\displaystyle=\frac{d}{ds}+\frac{2}{s}$

are fixed by imposing

$\displaystyle D_{a}(4,s)F_{1}(4,s)$ $\displaystyle=0\ ,$ $\displaystyle
D_{b}(4,s)D_{a}(4,s)F_{2}(4,s)$ $\displaystyle=0\ ,$

while $D_{c}(4,s)$ is a second order differential operator

$D_{c}(4,s)=A_{2}(4,s)\frac{d^{2}}{ds^{2}}+A_{1}(4,s)\frac{d}{ds}+A_{0}(4,s)\
,$ (7.25)

with

$\displaystyle A_{2}(4,s)$ $\displaystyle=C_{4}(4,s)\ ,$ $\displaystyle
A_{1}(4,s)$ $\displaystyle=C_{3}(4,s)-\frac{2}{s}C_{4}(4,s)\ ,$ $\displaystyle
A_{0}(4,s)$
$\displaystyle=C_{2}(4,s)-\frac{2}{s}C_{3}(4,s)+\frac{8}{s^{2}}C_{4}(4,s,m_{i}^{2})\
,$ (7.26)

as immediately found by comparing Eq.s(7.24) and (7.25) with Eq.(7.2).

Note that the above relations do not involve $C_{1}(4,s)$, which indeed,
according to Eq.(7.18), can be expressed in terms of the other coefficients
$C_{2}(4,s),C_{3}(4,s)$ and $C_{4}(4,s)$.

##  8 Equation and solutions for the equal mass case

At the end of Section 5 we have shortly considered the equal mass limit of the
4th order operator $D_{4}(d,s)$, Eq.(2.5), which is still a 4th order
operator. In this Section we want to discuss the equation for the maxcut of
the sunrise amplitude which is obtained by starting from scratch in the equal
mass case $m_{1}=m_{2}=m_{3}=m$. In that case LiteRed [3] gives for the
amplitude of the maxcut, which will be called $M(d,s,m^{2})$ to emphasize the
equality of the three masses, the 2nd order equation

$D_{2}(d,s,m^{2})\ M(d,s,m^{2})=0\ ,$ (8.1)

where $D_{2}(d,s,m^{2})$ is the 2nd order differential operator

$\displaystyle D_{2}(d,s,m^{2})=$ $\displaystyle\
2s(s-m^{2})(s-9m^{2})\frac{d^{2}}{ds^{2}}$ $\displaystyle-$
$\displaystyle\bigl{[}3(d-4)s^{2}-10(d-6)m^{2}s-9d\,m^{4}\bigr{]}\frac{d}{ds}$
$\displaystyle+$ $\displaystyle(d-3)\bigl{[}(d-4)s+(d+4)m^{2}\bigr{]}\ .$
(8.2)

Eq.(8.2) is to be compared with the corresponding equation valid in the
different mass case Eq.(2.8), which we rewrite here for convenience of the
reader

$D_{4}(d,s)M(d,s)=0\ .$

We recall once more that in the different mass case the differential operator
$D_{4}(d,s)$ is of 4th order, so that $D_{2}(d,s,m^{2})$ is not the equal mass
limit of $D_{4}(d,s)$. Note also the (relative) simplicity of
$D_{2}(d,s,m^{2})$ when compared to the explicit expression of $D_{4}(d,s)$,
Eq.s(B.2),(B.5) etc., given in the Appendix B.

On the contrary, the solutions of Eq.(8.1) can be derived from the functions
$M_{i}(d,s)$ of Eq.(3.16) by considering their equal mass limit; we will call
them $M_{i}(d,s,m^{2})$.

In the equal mass limit the parameter $b_{1}=(m_{1}-m_{2})^{2}$ vanishes, so
that $b_{1}=0$ coincides with $b_{0}=0$ and therefore $M_{1}(d,s,m^{2}),$
which corresponds to the $b$ integration in the range from $b_{0}$ to $b_{1}$,
is also vanishing. Further, as the very definition of $R_{4}(b)$, Eq.(3.14) at
$b_{1}=0$ gives

$R_{4}(b)=b(b-b_{2})(b-b_{3})(b-b_{4})\ ,$

we introduce

$R_{3}(b)=(b-b_{2})(b-b_{3})(b-b_{4})\ ,$ (8.3)

so that Eq.(3.16) becomes

$M_{i}(d,s,m^{2})=s^{\frac{2-d}{2}}\int_{b_{i-1}}^{b_{i}}db\frac{1}{\sqrt{|b|}}|R_{3}(b)|^{\frac{d-3}{2}}\
,$ (8.4)

where the index $i$ takes the 5 values (0,2,3,4,5).

As a first step, we want to check that the functions $M_{i}(d,s,m^{2})$ do
satisfy the 2nd order differential equation Eq.(8.1). In the different mass
case, the differential equation is of 4th order, and we needed an expression
of the solutions and its 4 derivatives in terms of a set of just 4 independent
functions, with $M_{i}(d,s)$ belonging of course to the set. That was achieved
by means of suitable integration by parts identities (ibp-id ’s) for the
occurring $b$ integrals, as discussed in the previous Sections. Those ibp-id
’s, typically, involve 5 terms, and can be used for expressing one of the
terms as a combination of the other four. The presence of 5 terms is related
to the presence of 5 factors, namely
$b,(b-b_{1}),(b-b_{2}),(b-b_{3}),(b-b_{4})$ in the integrands occurring in all
the ibp-id ’s.

In the equal mass case, as Eq.(8.1) is of 2nd order, one has to express each
of the $M_{i}(d,s,m^{2})$ and of its first two $s$-derivatives in terms of
just two related but independent functions; but the $b$-integrands involves 4
monomials (due to $b_{1}=0$, $b$ and $(b-b_{1})$ coincide), so that (see the
continuation for details) the typical ibp-id ’s involve 4 terms and can be
used for expressing just one of the four terms as a combination of the other
three, leaving us with a set of three independent functions, instead of just
two independent functions, as desired.

To start with, let us consider the function

$M_{3}(d,s,m^{2})=s^{\frac{2-d}{2}}N_{3}(d,s,m^{2})\ ,$ (8.5)

with

$N_{3}(d,s,m^{2})=\int_{b_{2}}^{b_{3}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}\
.$ (8.6)

As a set of related functions we can introduce the integrals

$N_{3}^{(k)}(d,s,m^{2})=\int_{b_{2}}^{b_{3}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}b^{k}\
,$ (8.7)

where $k$ is a positive integer, (with $N_{3}(d,s,m^{2})$ corresponding to
$k=0$). We then consider the obvious identities

$\int_{b_{2}}^{b_{3}}db\frac{d}{db}\biggl{[}\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-1}{2}}b^{k}\biggr{]}=0\
,$ (8.8)

valid for $d>1$. By working out the elementary algebra, one obtains for each
$k$ of Eq.(8.8) a 4 terms identity, expressing the integral
$N_{3}^{(k+2)}(d,s,m^{2})$ in terms of the three related functions with
indices $(k+1),k$ and $(k-1)$, so that all the integrals
$N_{3}^{(k)}(d,s,m^{2})$ can be expressed in terms of the three integrals
$N_{3}(d,s,m^{2}),N_{3}^{(1)}(d,s,m^{2})$ and $N_{3}^{(2)}(d,s,m^{2})$.

We then consider the three auxiliary integrals

$I_{3}^{(k)}(d,s,m^{2})=\int_{b_{2}}^{b_{3}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-1}{2}}\
b^{k}\ ,\hskip 11.38109ptk=0,1,2$ (8.9)

and their $s$-derivatives

$\frac{d}{ds}I_{3}^{(k)}(d,s,m^{2})=\frac{d}{ds}\int_{b_{2}}^{b_{3}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-1}{2}}\
b^{k}\ ,\hskip 11.38109ptk=0,1,2\ .$ (8.10)

We can evaluate the $s$-derivatives by following two different paths. As a
first path, ignoring the $b$-integrals on the r.h.s. we express all the three
$I^{(k)}(d,s,m^{2})$ of Eq.(8.9) as linear combinations, with coefficients
depending on $s$, of the three related functions
$N_{3}^{(k)}(d,s,m^{2}),k=0,1,2\ $ by using the identities generated by
Eq.s(8.8), and then we carry out the $s$-derivatives of the products of
coefficients and related functions; in that way one obtains linear
combinations of the three $N_{3}^{(k)}(d,s,m^{2})$ and of their
$s$-derivatives. Alternatively (second path) we can perform the $s$
differentiation of the three $b$ integrands in the r.h.s. of Eq.s(8.10) (for
$d>1$ the end-point contributions vanish), and then we express the results in
terms of the three $N_{3}^{(k)}(d,s,m^{2}),k=0,1,2,$ by using again the
identities Eq.s(8.8). As the two paths must obviously give the same results,
by equating the results one obtains three identities which can be solved by
expressing the $s$-derivatives of the three related functions
$N_{3}^{(k)}(d,s,m^{2})$ in terms of the same three functions. The result can
be written as

$\frac{d}{ds}N_{3}^{(k)}(d,s,m^{2})=\sum_{j=0,1,2}C_{kj}(d,s)N_{3}^{(j)}(d,s,m^{2})\
,\hskip 11.38109pti=0,1,2$ (8.11)

with $N_{3}^{(0)}(d,s,m^{2})=N_{3}(d,s,m^{2})$ as above already specified, and

$\displaystyle C_{00}(d,s)$
$\displaystyle=-\frac{1}{3s}+\frac{d-2}{s-m^{2}}-\frac{2}{3(s-9m^{2})}\ ,$
$\displaystyle C_{01}(d,s)$
$\displaystyle=(3d-4)\biggl{[}\frac{1}{4m^{2}s}-\frac{3}{8m^{2}(s-m^{2})}+\frac{1}{8m^{2}(s-9m^{2})}\biggr{]}\
,$ $\displaystyle C_{02}(d,s)$
$\displaystyle=(3d-4)\biggl{[}-\frac{1}{12m^{4}s}+\frac{3}{32m^{4}(s-m^{2})}-\frac{1}{96m^{4}(s-9m^{2})}\biggr{]}\
,$ (8.12) $\displaystyle C_{10}(d,s)$
$\displaystyle=-\frac{m^{2}}{3s}-\frac{8m^{2}}{3(s-9m^{2})}\ ,$ $\displaystyle
C_{11}(d,s)$
$\displaystyle=\frac{3d-4}{4s}-\frac{d}{2(s-m^{2})}+\frac{3d-4}{2(s-9m^{2})}\
,$ $\displaystyle C_{12}(d,s)$
$\displaystyle=(3d-4)\biggl{[}-\frac{1}{12m^{2}s}+\frac{1}{8m^{2}(s-m^{2})}-\frac{1}{24m^{2}(s-9m^{2})}\biggr{]}\
,$ (8.13) $\displaystyle C_{20}(d,s)$
$\displaystyle=-m^{2}-\frac{m^{4}}{3s}-\frac{32m^{4}}{3(s-9m^{2})}\ ,$
$\displaystyle C_{21}(d,s)$
$\displaystyle=\frac{d}{4}+(3d-4)\frac{m^{2}}{4s}-\frac{2dm^{2}}{s-m^{2}}+2\frac{(3d-4)m^{2}}{s-9m^{2}}\
,$ $\displaystyle C_{22}(d,s)$
$\displaystyle=(3d-4)\biggl{[}-\frac{1}{12s}+\frac{1}{2(s-m^{2})}-\frac{1}{6(s-9m^{2})}\biggr{]}\
.$ (8.14)

The second $s$-derivative of $N_{3}(d,s,m^{2})$ can then be obtained by
further differentiating Eq.(8.11) for $k=0$, and then by using again the
Eq.s(8.11),(8.12),(8.13),(8.14).

Once established the expressions of the $s$-derivative of $N_{3}(d,s,m^{2})$,
one can easily obtain the corresponding derivatives of $M_{3}(d,s,m^{2})$,
Eq.(8.5), plug them into the 2nd order equation Eq.(8.1) and check that they
do actually satisfy that equation, as expected.

At first sight, that might look a bit surprising; the fact that
$M_{3}(d,s,m^{2})$ and its first two derivatives satisfy the 2nd order
equation (8.1) tells us that a relation should exist between those three
functions (namely $M_{3}(d,s,m^{2})$ and its two derivatives), but those three
functions were expressed as combinations of the three functions
$N_{3}^{(k)}(d,s,m^{2}),\ k=0,1,2$ for which we were unable to find a relation
of linear dependence.

A closer investigation shows that by introducing a new partner of
$N_{3}(d,s,m^{2})$, let’s call it $\widetilde{N}_{3}(d,s,m^{2}),$ defined as

$\widetilde{N}_{3}(d,s,m^{2})=N_{3}^{(2)}(d,s,m^{2})-(s+3m^{2})N_{3}^{(1)}(d,s,m^{2})\
,$ (8.15)

one can rewrite Eq.(8.11) at $k=0$ as

$\displaystyle\frac{d}{ds}N_{3}(d,s,m^{2})$
$\displaystyle=\left(-\frac{1}{3s}+\frac{d-2}{s-m^{2}}-\frac{2}{3(s-9m^{2})}\right)N_{3}(d,s,m^{2})$
$\displaystyle+\frac{3d-4}{m^{4}}\left(-\frac{1}{12s}+\frac{3}{32(s-m^{2})}-\frac{1}{96(s-9m^{2})}\right)\widetilde{N}_{3}(d,s,m^{2})\
.$ (8.16)

One further finds that the derivative of $\widetilde{N}_{3}(d,s,m^{2})$ can be
written as

$\displaystyle\frac{d}{ds}\widetilde{N}_{3}(d,s,m^{2})$
$\displaystyle=\left(2m^{2}+\frac{2m^{4}}{3s}+\frac{64m^{4}}{3(s-9m^{2})}\right)N_{3}(d,s,m^{2})$
$\displaystyle+(3d-4)\left(\frac{1}{6s}+\frac{1}{3(s-9m^{2})}\right)\widetilde{N}_{3}(d,s,m^{2})\
,$ (8.17)

i.e. can also be expressed in terms of the same two functions
$N_{3}(d,s,m^{2})$ and $\widetilde{N}_{3}(d,s,m^{2})$.

By repeated use of Eq.s(8.16,8.17) one can then express all the
$s$-derivatives of $N_{3}(d,s,m^{2})$ and its partner
$\widetilde{N}_{3}(d,s,m^{2})$ in terms of the same two functions and then
verify that

$M_{3}(d,s,m^{2})=s^{\frac{2-d}{2}}N_{3}(d,s,m^{2})$

satisfies the second order differential equation Eq.(8.1) introduced at the
beginning of this Section.

Besides Eq.(8.15), let us consider the function

$S_{3}(d,s)=N_{3}^{(1)}(d,s,m^{2})-\frac{1}{3}(s+3m^{2})N_{3}^{(0)}(d,s,m^{2})\
;$ (8.18)

remarkably, thanks to Eq.s(8.11), its $s$-derivative is given by the simple
expression

$\frac{d}{ds}S_{3}(d,s)=\frac{d-2}{s-m^{2}}S_{3}(d,s)\ .$ (8.19)

Eq.(8.19) is actually a 1st order differential equation in $s$, which can be
immediately solved giving

$S_{3}(d,s)=X_{3}(d)\ (s-m^{2})^{d-2}\ ,$ (8.20)

where $X_{3}(d)$ is an as yet unspecified normalization constant (later on, we
will give its value for d =2,3 ).

It is easy to check that the arguments so far used for deriving the properties
of $N_{3}(d,s,m^{2})$ of Eq.(8.6) rely on the absence of end-point
contributions in the various ibp-id ’s, and therefore apply as well to

$N_{2}(d,s,m^{2})=\int_{b_{1}=0}^{b_{2}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}\
.$ (8.21)

and

$N_{4}(d,s,m^{2})=\int_{b_{3}}^{b_{4}}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}\
;$ (8.22)

further, they apply also to

$N_{0}(d,s,m^{2})=\int_{0}^{\infty}dc\frac{1}{\sqrt{c}}R_{3}(c)^{\frac{d-3}{2}}$
(8.23)

(see Eq.(4.15) for the notation); indeed, the convergence at $c=0$ is
independent of $d$, so that the convergence at large $c$ can be achieved for
$d$ sufficiently small (in fact negative), and the results can be extended by
analytic continuation also to positive values of $d$.

We can then generalize Eq.s(8.18,8.19,8.20) into

$\displaystyle S_{i}(d,s)$
$\displaystyle=N_{i}^{(1)}(d,s,m^{2})-\frac{1}{3}(s+3m^{2})N_{i}^{(0)}(d,s,m^{2})\
,$ (8.24) $\displaystyle\frac{d}{ds}S_{i}(d,s)$
$\displaystyle=\frac{d-2}{s-m^{2}}S_{i}(d,s)\ ,$ (8.25)

valid for $i=0,2,3,4$. The above equation is easily integrated, giving the
obvious generalization of Eq.(8.20)

$S_{i}(d,s)=X_{i}(d)\ (s-m^{2})^{(d-2)}\ ,$ (8.26)

where the constants $X(d,i)$ are still to be fixed.

Let us also note that, by defining

$T_{i}^{(n)}(d,s)=(s-m^{2})^{n}S_{i}(d,s)\ ,$ (8.27)

Eq.(8.25 ) becomes

$\frac{d}{ds}T_{i}^{(n)}(d,s)=\frac{d-n-2}{s-m^{2}}T_{i}^{(n)}(d,s)\ ,$ (8.28)

giving in particular

$\frac{d}{ds}T_{i}^{(2)}(d,s)=\frac{d-4}{s-m^{2}}T_{i}^{(2)}(d,s)\ .$ (8.29)

The case of

$N_{5}(d,s,m^{2})=\int_{b_{4}}^{\infty}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}\
;$ (8.30)

is different; if the related functions are defined as in Eq.(8.7), the
convergence of the factor $R_{3}(b)^{\frac{d-3}{2}}$ at $b=b_{4}=(E+m)^{2}$
requires $d>1$, which is incompatible with the convergence of all the same
related functions at large $b$.

In this case one can however introduce as related functions the integrals

$N_{5}^{(k)}(d,s,m^{2})=\int_{b_{4}}^{\infty}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-3}{2}}\frac{1}{b^{k}}\
,$ (8.31)

with $k\geq 0$, and as auxiliary integrals

$I_{5}^{(k)}(d,s,m^{2})=\int_{b_{4}}^{\infty}db\frac{1}{\sqrt{b}}R_{3}(b)^{\frac{d-1}{2}}\
\frac{1}{b^{k}}\ ,\hskip 11.38109ptk=0,1,2\ .$ (8.32)

It is then found that by introducing as partner function

$\displaystyle\widetilde{N}_{5}(d,s,m^{2})$
$\displaystyle=3m^{2}(s-m^{2})^{2}(s+3m^{2})N_{5}^{(2)}(d,s,m^{2})$
$\displaystyle+\left(\frac{d-4}{4}(s+3m^{2})^{3}+3m^{2}(s-m^{2})^{2}\right)N_{5}^{(1)}(d,s,m^{2})\
,$ (8.33)

one can verify that all the $s$-derivatives of $N_{5}(d,s,m^{2})$ and
$\widetilde{N}_{5}(d,s,m^{2})$, in particular the 1st and the 2nd derivative
of $N_{5}(d,s,m^{2})$, can be expressed in terms of the two same functions
$N_{5}(d,s,m^{2}),\widetilde{N}_{5}(d,s,m^{2})$, and, finally, that also
$N_{5}(d,s,m^{2})$ does satisfy Eq.(8.1), as expected. We were however unable
to find the analog of Eq.(8.25) for $N_{5}(d,s,m^{2})$.

Coming back to Eq.(8.24), skipping the case of $S_{0}(d,s)$ which would
require an analytical continuation for $d=3,4$, let us observe that for
$i=2,3,4$ it can be written as

$S_{i}(d,s)=\int_{b_{i-1}}^{b_{i}}\frac{db}{\sqrt{b}}|R_{3}(b,m^{2})|^{\frac{d-3}{2}|}\left(b-\frac{1}{3}(s+3m^{2})\right)\
,$ (8.34)

which for $d=2$ Eq.(8.24) becomes

$S_{i}(2,s)=\int_{b_{i-1}}^{b_{i}}\frac{db}{|\sqrt{b(b-4m^{2}}\sqrt{R_{2}(s,b,m^{2})}|}\left(b-\frac{1}{3}(s+3m^{2})\right)\
,$ (8.35)

with $R_{2}(s,b,m^{2})$ given, as usual, from Eq.(5.42).

It can be recalled here the formula (see [5], after fixing a typo)

$\frac{d}{db}\ln{\left(\frac{b(s+3m^{2}-b)+\sqrt{b(b-4m^{2})}\sqrt{R_{2}(s,b,m^{2})}}{b(s+3m^{2}-b)-\sqrt{b(b-4m^{2})}\sqrt{R_{2}(s,b,m^{2})}}\right)}=\frac{s+3m^{2}-3b}{\sqrt{b(b-4m^{2})}\sqrt{R_{2}(s,b,m^{2})}}\
,$ (8.36)

whose integration in $b$ gives at once the formula (exact)

$S_{i}(2,s)=-\frac{1}{3}\ln{\left(\frac{b(s+3m^{2}-b)+\sqrt{b(b-4m^{2})}\sqrt{R_{2}(s,b,m^{2})}}{b(s+3m^{2}-b)-\sqrt{b(b-4m^{2})}\sqrt{R_{2}(s,b,m^{2})}}\right)}\Biggr{|}_{b_{i-1}}^{b_{i}}\
,$ (8.37)

where the proper analytical continuation of the logarithm is understood. The
above equation gives

$\displaystyle S_{2}(2,s)$ $\displaystyle=-\frac{1}{3}\pi\ ,$ $\displaystyle
S_{3}(2,s)$ $\displaystyle=0\ ,$ $\displaystyle S_{4}(2,s)$
$\displaystyle=+\frac{2}{3}\pi\ .$ (8.38)

To our knowledge, for $i=3$ Eq.s(8.35,8.38) reproduce a result contained in
equation (85) of the Sabry paper [6], (see Eq.s(32,67,68) of that paper for
the notations).
For $d=3,5$ the integration of Eq.(8.34) is trivial; one finds

$\displaystyle S_{2}(3,s)$ $\displaystyle=-\frac{4}{3}m(s-m^{2})\ ,$
$\displaystyle S_{3}(3,s)$ $\displaystyle=0\ ,$ $\displaystyle S_{4}(3,s)$
$\displaystyle=+\frac{8}{3}m(s-m^{2})\ ,$ (8.39)

and

$\displaystyle S_{2}(5,s)$ $\displaystyle=-\frac{32}{9}m^{3}(s-m^{2})^{3}\ ,$
$\displaystyle S_{3}(5,s)$ $\displaystyle=0\ ,$ $\displaystyle S_{4}(5,s)$
$\displaystyle=+\frac{64}{9}m^{3}(s-m^{2})^{3}\ .$ (8.40)

For $d=4$ the direct calculation is not possible; we used the large $s$
expansion for $S_{2}(4,s),S_{4}(4,s)$, and an expansion just avove the
threshold $s=(3m)^{2}$ (we recall that our integration formulas are valid for
$(3m)^{2}<s<+\infty$). We find

$\displaystyle S_{2}(4,s)$ $\displaystyle=-\frac{2}{3}\pi\ m^{2}(s-m^{2})^{2}\
,$ $\displaystyle S_{3}(4,s)$ $\displaystyle=0\ ,$ $\displaystyle S_{4}(4,s)$
$\displaystyle=+\frac{4}{3}\pi\ m^{2}(s-m^{2})^{2}\ .$ (8.41)

The above equations are (of course) in agreent with Eq.(8.20).

## Acknowledgements

We are grateful to J. Vermaseren for his assistance in the use of the
algebraic program FORM [7] which was intensively used in all the steps of this
work.

##  Appendix A Evaluation of the Maxcut

We give some details of the evaluation of the integral appearing in Section 3,
starting from $B(d,-k^{2})$ Eq.(3.7),

$B(d,-k^{2})=\int\mathcal{D}^{d}q\ \delta(D_{1})\delta(D_{2})\ .$

As a first step, due to the presence of the factor $\delta(D_{1})$ in the
integrand, one can write

$B(d,-k^{2})=\int\mathcal{D}^{d}q\ \delta(D_{1})\delta(D_{2}-D_{1})\ ,$ (A.1)

and from Eq.s(2.2) we have at once

$D_{2}-D_{1}=k^{2}-2(k\cdot q)-m_{1}^{2}+m_{2}^{2}\ .$

According to Eq.(3.5) (where the Minkoskean regularisation is used) we then
write $(k\cdot q)$ as

$(k\cdot
q)=-k_{0}q_{0}+k_{x}q_{x}-(\vec{K}\cdot\vec{Q})=-k_{0}q_{0}+k_{x}q_{x}-KQ_{1}\
,$

where K is the modulus of $\vec{K}$, while the first component of $\vec{Q}$,
namely $Q_{1}$, is the component of $\vec{Q}$ along the direction of
$\vec{K}$. By performing the change of integration variables
$(q_{0},Q_{1})\to({q_{0}}^{\prime},{Q_{1}}^{\prime})$ (which is in fact a
rotation, with Jacobian equal 1)

$\displaystyle q_{0}$
$\displaystyle=\frac{1}{\sqrt{k_{0}^{2}+K^{2}}}(k_{0}{q_{0}}^{\prime}-K{Q_{1}}^{\prime})\
,$ $\displaystyle Q_{1}$
$\displaystyle=\frac{1}{\sqrt{k_{0}^{2}+K^{2}}}(K{q_{0}}^{\prime}+k_{0}{Q_{1}}^{\prime})\
,$

and then by rewriting again $(q_{0},Q_{1})$ instead of
$({q_{0}}^{\prime},{Q_{1}}^{\prime})$, one finds

$(k\cdot q)=-\sqrt{k_{0}^{2}+K^{2}}\ q_{0}+k_{x}q_{x}\ ,$

where $Q_{1}$ has disappeared. At this point one has

$\displaystyle D_{1}$ $\displaystyle=-q_{0}^{2}+q_{x}^{2}-Q^{2}+m_{1}^{2}\ ,$
(A.2) $\displaystyle D_{2}-D_{1}$ $\displaystyle=2\sqrt{k_{0}^{2}+K^{2}}\
q_{0}-2k_{x}q_{x}+k^{2}-m_{1}^{2}+m_{2}^{2}\ ,$ (A.3)

and, as the components of $\vec{Q}$ do not appear explicitly in the integrand,
we can introduce polar coordinates in $\mathcal{D}^{d}q$, Eq.(3.4),

$\mathcal{D}^{d}q=dq_{0}\ dq_{x}\ d^{d-2}Q=dq_{0}\ dq_{x}\
\Omega(d-2)\int_{0}^{\infty}Q^{d-3}\ dQ\ ,$

where

$\Omega(n)=2\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\ ,$ (A.4)

is the solid angle in $n$ dimensions, so that Eq.(A.1) becomes

$B(d,-k^{2})=\Omega(d-2)\int dq_{0}\ dq_{x}\int_{0}^{\infty}Q^{d-3}\ dQ\
\delta(D_{1})\delta(D_{2}-D_{1})\ .$ (A.5)

Let us consider separately the cases of timelike and spacelike $k$.

If $k$ is timelike, ($-k^{2}=k_{0}^{2}-k_{x}^{2}+K^{2}>0$), we perform the
change of variable (a Lorentz boost, with Jacobian equal 1)

$\displaystyle q_{0}$
$\displaystyle=\frac{1}{\sqrt{-k^{2}}}\left(\sqrt{k_{0}^{2}+K^{2}}{q_{0}}^{\prime}+k_{x}{q_{x}}^{\prime}\right)\
,$ $\displaystyle q_{x}$
$\displaystyle=\frac{1}{\sqrt{-k^{2}}}\left(k_{x}{q_{0}}^{\prime}+\sqrt{k_{0}^{2}+K^{2}}{q_{x}}^{\prime}\right)\
,$

and then we rewrite again $(q_{0},q_{x})$ instead of
$({q_{0}}^{\prime},{q_{x}}^{\prime}).$ Eq.(A.2) does not change, while
Eq.(A.3) becomes

$D_{2}-D_{1}=2\sqrt{-k^{2}}\ q_{0}+k^{2}-m_{1}^{2}+m_{2}^{2}\ .$ (A.6)

By integrating $\delta(D_{2}-D_{1})$ in $q_{0}$ and $\delta(D_{1})$ in $q_{x}$
Eq.(A.5) becomes (for timelike $k$)

$\Theta(-k^{2})B(d,-k^{2})={\Theta}(-k^{2})\
\Omega(d-2)\frac{1}{2\sqrt{-k^{2}}}\int_{0}^{\infty}dQ\frac{Q^{d-3}}{\sqrt{Q^{2}-\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})}}\
,$ (A.7)

where

$R_{2}(a,b,c)=a^{2}+b^{2}+c^{2}-2ab-2ac-2bc\ ,$ (A.8)

in agreement with Eq.(5.42), and it is understood that the argument of the
square root must be positive, i.e. the constraint

$Q^{2}-\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})\geq 0\ $ (A.9)

must hold.

If $R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})$ is positive, as $k^{2}$ is negative,
Eq.(A.9) is always satisfied; one can rescale $Q^{2}\to v$ according to

$Q^{2}=-\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})\ v\ ,$

and the calculation is completed with the integral

$\int_{0}^{\infty}dv\ v^{\frac{d-4}{2}}\
(1+v)^{-\frac{1}{2}}=B\left(\frac{d-2}{2},\frac{3-d}{2}\right)=\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{d-2}{2}\right)\Gamma\left(\frac{3-d}{2}\right)\
,$

where $B(x,y),\Gamma(x)$ are the Euler’s Beta and Gamma functions.

If $R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})$ is negative, as $k^{2}$ is also
negative, Eq.(A.9) gives the condition

$Q^{2}\geq\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})\ ;$

one can then rescale

$Q^{2}=+\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})\ w\ ,$

and the calculation is completed with the integral

$\int_{1}^{\infty}dw\ w^{\frac{d-4}{2}}\
(w-1)^{-\frac{1}{2}}=B\left(\frac{3-d}{2},\frac{1}{2}\right)=\frac{\sqrt{\pi}}{\Gamma\left(\frac{4-d}{2}\right)}\Gamma\left(\frac{d-3}{2}\right)\
.$

If $k$ is spacelike, ($k^{2}=-k_{0}^{2}+k_{x}^{2}-K^{2}>0$), in analogy with
the procedure followed for $k$ timeike, we perform the change of variable
(again a Lorentz boost, with Jacobian equal 1)

$\displaystyle q_{0}$
$\displaystyle=\frac{1}{\sqrt{k^{2}}}\left(k_{x}{q_{0}}^{\prime}+\sqrt{k_{0}^{2}+K^{2}}{q_{x}}^{\prime}\right)\
,$ $\displaystyle q_{x}$
$\displaystyle=\frac{1}{\sqrt{k^{2}}}\left(\sqrt{k_{0}^{2}+K^{2}}{q_{0}}^{\prime}+k_{x}{q_{x}}^{\prime}\right)\
,$

we rewrite again $(q_{0},q_{x})$ instead of
$({q_{0}}^{\prime},{q_{x}}^{\prime}),$ and then we integrate
$\delta(D_{2}-D_{1})$ in $q_{x}$ and $\delta(D_{1})$ in $q_{0}$; Eq.(A.5)
becomes (for spacelike $k$)

$\Theta(k^{2})B(d,-k^{2})={\Theta}(k^{2})\
\Omega(d-2)\frac{1}{2\sqrt{k^{2}}}\int_{0}^{\infty}dQ\frac{Q^{d-3}}{\sqrt{\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})-Q^{2}}}\
,$ (A.10)

and it is understood that the argument of the square root must be positive,
i.e. the constraint

$\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})-Q^{2}\geq 0\ $ (A.11)

must hold. Note that we can write

$R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})=(k^{2}+(m_{1}-m_{2})^{2})(k^{2}+(m_{1}+m_{2})^{2})\
,$

so that, for $k^{2}$ positive ($k$ spacelike)
$R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})$ is always positive; we can then rescale
$Q^{2}\to t$ according to

$Q^{2}=+\frac{1}{4k^{2}}R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})\ t\ ,$

and the calculation is completed with the integral

$\int_{0}^{1}dt\ t^{\frac{d-4}{2}}\
(1-t)^{-\frac{1}{2}}=B\left(\frac{d-2}{2},\frac{1}{2}\right)=\frac{\sqrt{\pi}}{\Gamma\left(\frac{d-1}{2}\right)}\Gamma\left(\frac{d-2}{2}\right)\
.$

By collecting results we have

$\displaystyle B(d,-k^{2})=$ $\displaystyle\
\Theta(-k^{2})\Theta(+R_{2}(-k^{2},m_{1}^{2},m_{2}^{2}))\
|k^{2}|^{\frac{2-d}{2}}\ |R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})|^{\frac{d-3}{2}}$
$\displaystyle{\hskip 8.53581pt}\times 2^{1-d}\Omega(d-2)\
\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{d-2}{2}\right)\Gamma\left(\frac{3-d}{2}\right)$
$\displaystyle+$ $\displaystyle\
\Theta(-k^{2})\Theta(-R_{2}(-k^{2},m_{1}^{2},m_{2}^{2}))\
|k^{2}|^{\frac{2-d}{2}}\ |R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})|^{\frac{d-3}{2}}$
$\displaystyle{\hskip 8.53581pt}\times 2^{1-d}\Omega(d-2)\
\frac{\sqrt{\pi}}{\Gamma\left(\frac{4-d}{2}\right)}\Gamma\left(\frac{3-d}{2}\right)$
$\displaystyle+$ $\displaystyle\
\Theta(+k^{2})\Theta(+R_{2}(-k^{2},m_{1}^{2},m_{2}^{2}))\
|k^{2}|^{\frac{2-d}{2}}\ |R_{2}(-k^{2},m_{1}^{2},m_{2}^{2})|^{\frac{d-3}{2}}$
$\displaystyle{\hskip 8.53581pt}\times 2^{1-d}\Omega(d-2)\
\frac{\sqrt{\pi}}{\Gamma\left(\frac{d-1}{2}\right)}\Gamma\left(\frac{d-2}{2}\right)\
.$ (A.12)

The evaluation of $C(d,b)$, Eq.(3.9) is similar, with $p^{2}$ playing the role
of $k^{2}$ in $B(d,-k^{2})$. For simplicity, we restrict ourself to consider
only the case of timelike $p$, and above the unitarity threshold, i.e.

$s=-p^{2}=E^{2}\ ,{\hskip 14.22636pt}E>(m_{1}+m_{2}+m_{3})\ ,$

we find

$\displaystyle C(d,E^{2},b)=$ $\displaystyle\
\Theta(+R_{2}(E^{2},b,m_{3}^{2}))\ E^{2-d}\
|R_{2}(E^{2},b,m_{3}^{2})|^{\frac{d-3}{2}}$ $\displaystyle{\hskip
8.53581pt}\times 2^{1-d}\Omega(d-2)\
\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{d-2}{2}\right)\Gamma\left(\frac{3-d}{2}\right)$
$\displaystyle+$ $\displaystyle\ \Theta(-R_{2}(E^{2},b,m_{3}^{2}))\ E^{2-d}\
|R_{2}(E^{2},b,m_{3}^{2})|^{\frac{d-3}{2}}$ $\displaystyle{\hskip
8.53581pt}\times 2^{1-d}\Omega(d-2)\
\frac{\sqrt{\pi}}{\Gamma\left(\frac{4-d}{2}\right)}\Gamma\left(\frac{3-d}{2}\right)\
.$ (A.13)

The integrand in $b$ of $M(d,s)$, Eq.(3.8), is the product of $B(d,b)$ and
$C(d,E^{2},b)$, Eq.s(A.12,A.13). From the definition of $R_{2}(a,b,c)$,
Eq.(A.8), one finds

$\displaystyle R_{2}(b,m_{1}^{2},m_{2}^{2})$
$\displaystyle=(b-(m_{1}-m_{2})^{2})(b-(m_{1}+m_{2})^{2})=(b-b_{1})(b-b_{2})\
,$ $\displaystyle R_{2}(E^{2},b,m_{3}^{2})$
$\displaystyle=(b-(E-m_{3})^{2})(b-(E+m_{3})^{2})=(b-b_{3})(b-b_{4})\ ,$

where the $b_{i}$ are defined as in Eq.s(3.12), so that $R_{4}(b)$, Eq.(3.14),
is the product of $R_{2}(b,m_{1}^{2},m_{2}^{2})$ and
$R_{2}(E^{2},b,m_{3}^{2})$,

$R_{4}(b)=R_{2}(b,m_{1}^{2},m_{2}^{2})\ R_{2}(E^{2},b,m_{3}^{2})\ .$

The product of $B(d,b)$ and $C(d,E^{2},b)$ contains also the product of
several $\Theta$-functions whose arguments depend on $b$, see the formulas
above. To work out the constraints implied by those $\Theta$’s, let us
introduce a new $\Theta$-function with 3 arguments, $\Theta(b_{i},b,b_{j})$,
where $b_{i},b_{j}$ are defined as in Eq.s(3.12) and $b_{i}<b_{j}$ (an
essential condition), whose value is 1 for $b$ in the range $b_{i}<b<b_{j}$,
and zero otherwise. We have, obviously

$\displaystyle\Theta(b)=\Theta(b_{0},b,b_{5})\ ,{\hskip
8.53581pt}\Theta(-b)=\Theta(b_{-1},b,b_{0})\ ,$
$\displaystyle\Theta(+R_{2}(b,m_{1}^{2},m_{2}^{2}))=\Theta(b_{-1},b,b_{1})+\Theta(b_{2},b,b_{5})\
,{\hskip
8.53581pt}\Theta(-R_{2}(b,m_{1}^{2},m_{2}^{2}))=\Theta(b_{1},b,b_{2})\ ,$
$\displaystyle\Theta(+R_{2}(E^{2},b,m_{3}^{2}))=\Theta(b_{-1},b,b_{3})+\Theta(b_{4},b,b_{5})\
,{\hskip 8.53581pt}\Theta(-R_{2}(E^{2},b,m_{3}^{2}))=\Theta(b_{3},b,b_{4})\ .$

As a next step, consider separately each term $\Theta(b_{i},b,b_{j})$; if
$b_{i}$ and $b_{j}$ are not adjacent, i.e. if $i\neq j-1$, write it as a sum
of adjacent terms, as in the following examples

$\displaystyle\Theta(b_{-1},b,b_{3})$
$\displaystyle=\Theta(b_{-1},b,b_{0})+\Theta(b_{0},b,b_{1})+\Theta(b_{1},b,b_{2})+\Theta(b_{2},b,b_{3})\
,$ $\displaystyle\Theta(b_{2},b,b_{5})$
$\displaystyle=\Theta(b_{2},b,b_{3})+\Theta(b_{3},b,b_{4})+\Theta(b_{3},b,b_{4})\
.$

In so doing one obtains a proliferation of terms, consisting of the products
of several $\Theta$-functions with 3 arguments; as a last step, use the
(obvious) identity

$\Theta(b_{i-1},b,b_{i})\ \Theta(b_{j-1},b,b_{j})=\delta_{i,j}\
\Theta(b_{i-1},b,b_{i})$

(the products of two $\Theta$’s is zero unless the arguments are identical)
and one is left with few terms containing a single $\Theta$ factor. By
observing that

$\int_{-\infty}^{\infty}db\ \Theta(b_{i-1},b,b_{i})\
f(b)=\int_{b_{i-1}}^{b_{i}}db\ f(b)\ ,$

Eq.s(3.15,3.16) are recovered. For completeness, we report here the explicit
values of the coefficients $C_{i}(d)$ appearing in Eq.(3.15) (even if they
don’t play any role in this paper)

$\displaystyle C_{0}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{d-2}{2}\right)\Gamma\left(\frac{3-d}{2}\right)\
,$ $\displaystyle C_{1}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac{d-2}{2}\right)\Gamma^{2}\left(\frac{3-d}{2}\right)\
,$ $\displaystyle C_{2}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\sqrt{\pi}\Gamma\left(\frac{4-d}{2}\right)}\Gamma\left(\frac{d-1}{2}\right)\Gamma^{2}\left(\frac{3-d}{2}\right)\
,$ $\displaystyle C_{3}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac{d-2}{2}\right)\Gamma^{2}\left(\frac{3-d}{2}\right)\
,$ $\displaystyle C_{4}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\sqrt{\pi}\Gamma\left(\frac{4-d}{2}\right)}\Gamma^{2}\left(\frac{3-d}{2}\right)\
,$ $\displaystyle C_{5}(d)$
$\displaystyle=2^{2d-3}\Omega(d-1)\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac{d-2}{2}\right)\Gamma^{2}\left(\frac{3-d}{2}\right)\
.$ (A.15)

##  Appendix B The Differential Equations

The $4$th order differential equation for the maxcut of the Sunrise amplitude
was generated by the Mathematica package LiteRed [3]. The equation was already
written in Eq.(2.8), which we repeat here for the convenience of the reader

$D_{4}(d,s,m_{i}^{2})\ M(d,s)=0\ ,$ (B.1)

where $M(d,s)$ is a generic solution, and $D_{4}(d,s,m_{i}^{2})$ is the 4th
order differential operator, already written in Eq.(B.2) and again rewritten
here

$\displaystyle D_{4}(d,s,m_{i}^{2})$
$\displaystyle=C_{4}(d,s,m_{i}^{2})\frac{d^{4}}{ds^{4}}+C_{3}(d,s,m_{i}^{2})\frac{d^{3}}{ds^{3}}$
$\displaystyle+C_{2}(d,s,m_{i}^{2})\frac{d^{2}}{ds^{2}}+C_{1}(d,s,m_{i}^{2})\frac{d}{ds}+C_{0}(d,s,m_{i}^{2})\
,$ (B.2)

The 5 coefficients $C_{i}(d,s,m_{i}^{2}),i=0,1,..,4$ are polynomials in $d,s$
and the three masses $m_{i},i=1,2,3$. As the equation and all those
polynomials are symmetrical under the permutations of the three masses, it can
be convenient to express them through the three symmetrical combinations of
the three masses, namely

$\displaystyle U=$ $\displaystyle\ m_{1}^{2}+m_{2}^{2}+m_{3}^{2}\ ,$
$\displaystyle D=$ $\displaystyle\
m_{1}^{2}m_{2}^{2}+m_{2}^{2}m_{3}^{2}+m_{3}^{2}m_{1}^{2}\ ,$ $\displaystyle
T=$ $\displaystyle\ m_{1}^{2}m_{2}^{2}m_{3}^{2}\ .$ (B.3)

With the above definition, $R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})$, defined as
in Eq.(A.8), becomes

$R_{2}(m_{1}^{2},m_{2}^{2},m_{3}^{2})=U^{2}-4D\ .$ (B.4)

The explicit expression of the coefficient $C_{4}(d,s,m_{i}^{2})$ is

$C_{4}(d,s,m_{i}^{2})=8s^{3}\ \mathcal{D}(s)\ \bigg{[}(7-d)s^{2}+2(3-d)s\
U+(3d-13)(U^{2}-4D)\bigg{]}$ (B.5)

where

$\displaystyle\mathcal{D}(s)=$ $\displaystyle\
(s-(m_{1}+m_{2}+m_{3})^{2})(s-(m_{1}+m_{2}-m_{3})^{2})$ $\displaystyle\
(s-(m_{1}-m_{2}+m_{3})^{2})(s-(m_{1}-m_{2}-m_{3})^{2})$ $\displaystyle=$
$\displaystyle\
s^{4}-4s^{3}U+s^{2}(6U^{2}-8D)+s(16UD-4U^{3}-64T)+(U^{2}-4D)^{2}\ .$ (B.6)

We further have

$\displaystyle C_{3}(d,s,m_{i}^{2})=$ $\displaystyle\
s^{8}(1008-284d+20d^{2})+s^{7}(-2000+328d-8d^{2})U$ $\displaystyle+$
$\displaystyle\ s^{6}\big{[}\
(4736-2224d+208d^{2})D+(-2080+1372d-132d^{2})U^{2}\ \big{]}$ $\displaystyle+$
$\displaystyle\ s^{5}\big{[}\
(-10752-256d+256d^{2})T+(8160-3088d+208d^{2})U^{3}$ $\displaystyle\hskip
22.76219pt+(-25728+9664d-704d^{2})UD\ \big{]}$ $\displaystyle+$
$\displaystyle\ s^{4}\big{[}\
(-35072+10944d-576d^{2})D^{2}+(-7120+2012d-52d^{2})U^{4}$ $\displaystyle\hskip
22.76219pt+(37248-10784d+352d^{2})U^{2}D+(-12288+2560d+512d^{2})UT\ \big{]}$
$\displaystyle+$ $\displaystyle\ s^{3}\big{[}\
(-133120+17408d+3072d^{2})DT+(2032-184d-72d^{2})U^{5}$ $\displaystyle\hskip
22.76219pt+(-16256+1472d+576d^{2})U^{3}D+(33280-4352d-768d^{2})U^{2}T$
$\displaystyle\hskip 22.76219pt+(32512-2944d-1152d^{2})UD^{2}\ \big{]}$
$\displaystyle+$ $\displaystyle\ s^{2}\big{[}\
(9984d-2304d^{2})D^{3}+(-156d+36d^{2})U^{6}$ $\displaystyle\hskip
22.76219pt+(1872d-432d^{2})U^{4}D+(-7488d+1728d^{2})U^{2}D^{2}\ \big{]}\ ,$
(B.7) $\displaystyle C_{2}(d,s,m_{i}^{2})=$ $\displaystyle\
s^{7}(5376-2448d+366d^{2}-18d^{3})+s^{6}(-4720+652d+124d^{2}-16d^{3})U$
$\displaystyle+$ $\displaystyle\ s^{5}\big{[}\
(44832-24528d+4152d^{2}-216d^{3})D+(-18720+11436d-1998d^{2}+102d^{3})U^{2}\
\big{]}$ $\displaystyle+$ $\displaystyle\ s^{4}\big{[}\
(2016-5664d+1440d^{2}-96d^{3})T+(30720-13896d+1704d^{2}-48d^{3})U^{3}$
$\displaystyle\hskip 22.76219pt+(-110016+50720d-6816d^{2}+256d^{3})UD\
\big{]}$ $\displaystyle+$ $\displaystyle\ s^{3}\big{[}\
(-68224+18432d-224d^{2}-96d^{3})D^{2}+(-13280+3272d+146d^{2}-38d^{3})U^{4}$
$\displaystyle\hskip
22.76219pt+(70176-17696d-528d^{2}+176d^{3})U^{2}D+(-576-6720d+2880d^{2}-192d^{3})UT\
\big{]}$ $\displaystyle+$ $\displaystyle\ s^{2}\big{[}\
(-24960-74112d+23424d^{2}-1152d^{3})DT+(624+828d-228d^{2})U^{5}$
$\displaystyle\hskip
22.76219pt+(-4992-6624d+1824d^{2})U^{3}D+(6240+18528d-5856d^{2}+288d^{3})U^{2}T$
$\displaystyle\hskip 22.76219pt+(9984+13248d-3648d^{2})UD^{2}\ \big{]}$
$\displaystyle+$ $\displaystyle\ s\ \big{[}\
(-9984d+7296d^{2}-1152d^{3})D^{3}+(156d-114d^{2}+18d^{3})U^{6}$
$\displaystyle\hskip
22.76219pt+(-1872d+1368d^{2}-216d^{3})U^{4}D+(7488d-5472d^{2}+864d^{3})U^{2}D^{2}\
\big{]}\ ,$ (B.8) $\displaystyle C_{1}(d,s,m_{i}^{2})$
$\displaystyle=s^{6}(9408-6440d+1610d^{2}-175d^{3}+7d^{4})$
$\displaystyle+s^{5}(1440-3792d+1708d^{2}-270d^{3}+14d^{4})U$
$\displaystyle+s^{4}\big{[}\ (104256-77088d+20536d^{2}-2316d^{3}+92d^{4})D$
$\displaystyle\hskip
22.76219pt+(-36480+27352d-7182d^{2}+767d^{3}-27d^{4})U^{2}\big{]}$
$\displaystyle+s^{3}\big{[}\ (-4032+4384d-1888d^{2}+416d^{3}-32d^{4})T$
$\displaystyle\hskip 22.76219pt+(29760-16688d+2712d^{2}-52d^{3}-12d^{4})U^{3}$
$\displaystyle\hskip
22.76219pt+(-117888+69440d-13024d^{2}+688d^{3}+16d^{4})UD\big{]}$
$\displaystyle+s^{2}\big{[}\ (-29952+10368d+736d^{2}-720d^{3}+80d^{4})D^{2}$
$\displaystyle\hskip 22.76219pt+(-2880-1416d+1342d^{2}-273d^{3}+17d^{4})U^{4}$
$\displaystyle\hskip
22.76219pt+(19008+3072d-5552d^{2}+1272d^{3}-88d^{4})U^{2}D$
$\displaystyle\hskip
22.76219pt+(1152+576d-1472d^{2}+576d^{3}-64d^{4})UT\big{]}$ $\displaystyle+s\
\big{[}\ (49920-23168d-7296d^{2}+3968d^{3}-384d^{4})DT$ $\displaystyle\hskip
22.76219pt+(-1248+1088d-292d^{2}+34d^{3}-2d^{4})U^{5}$ $\displaystyle\hskip
22.76219pt+(9984-8704d+2336d^{2}-272d^{3}+16d^{4})U^{3}D$ $\displaystyle\hskip
22.76219pt+(-12480+5792d+1824d^{2}-992d^{3}+96d^{4})U^{2}T$
$\displaystyle\hskip
22.76219pt+(-19968+17408d-4672d^{2}+544d^{3}-32d^{4})UD^{2}\big{]}$
$\displaystyle+\hskip 8.53581pt\big{[}\
(6656d-6528d^{2}+1984d^{3}-192d^{4})D^{3}$ $\displaystyle\hskip
22.76219pt+(-104d+102d^{2}-31d^{3}+3d^{4})U^{6}$ $\displaystyle\hskip
22.76219pt+(1248d-1224d^{2}+372d^{3}-36d^{4})U^{4}D$ $\displaystyle\hskip
22.76219pt+(-4992d+4896d^{2}-1488d^{3}+144d^{4})U^{2}D^{2}\big{]}\ ,$ (B.9)
$\displaystyle C_{0}(d,s,m_{i}^{2})$ $\displaystyle=(d-3)(d-4)\times\big{\\{}$
$\displaystyle+s^{5}(336-146d+21d^{2}-d^{3})+s^{4}(360-258d+51d^{2}-3d^{3})U$
$\displaystyle+s^{3}\big{[}\
(4752-2280d+344d^{2}-16d^{3})D+(-1280+568d-70d^{2}+2d^{3})U^{2}\ \big{]}$
$\displaystyle+s^{2}\big{[}\
(336-104d-48d^{2}+8d^{3})T+(240+120d-66d^{2}+6d^{3})U^{3}$
$\displaystyle\hskip 22.76219pt+(-1056-400d+248d^{2}-24d^{3})UD\ \big{]}$
$\displaystyle+s\ \big{[}\
(2496-2240d+592d^{2}-48d^{3})D^{2}+(240-182d+33d^{2}-d^{3})U^{4}$
$\displaystyle\hskip
22.76219pt+(-1584+1288d-280d^{2}+16d^{3})U^{2}D+(-96+176d-96d^{2}+16d^{3})UT\
\big{]}$ $\displaystyle+\hskip 8.53581pt\big{[}\
(-4160+3872d-1088d^{2}+96d^{3})DT+(104-102d+31d^{2}-3d^{3})U^{5}$
$\displaystyle\hskip
22.76219pt+(-832+816d-248d^{2}+24d^{3})U^{3}D+(1040-968d+272d^{2}-24d^{3})U^{2}T$
$\displaystyle\hskip 22.76219pt+(1664-1632d+496d^{2}-48d^{3})UD^{2}\
\big{]}\big{\\}}\ ,$ (B.10)

For completeness, we report here the explitcit expressions of some
coefficients appearing in Section 5, which are also symmetrical in the three
masses. According to Eq.(5.40) of Section 5, at $d=2$ the 4th order operator
$D_{4}(d,s,m_{i}^{2})$, Eq.(2.5) or Eq.(B.2) can be factorised as

$D_{4}(2,s,m_{i}^{2})=\mathcal{K}\ D_{1b}(2,s,m_{i}^{2})\
D_{1a}(2,s,m_{i}^{2})\ D_{2}(2,s,m_{i}^{2})\ ,$ (B.11)

with

$D_{2}(2,s,m_{i}^{2})=B_{2}(2,s,m_{i}^{2})\frac{d^{2}}{ds^{2}}+B_{1}(2,s,m_{i}^{2})\frac{d}{ds}+B_{0}(2,s,m_{i}^{2})\
.$ (B.12)

In the notation of the previous formulas, the explicit expressions of the
coefficients $B_{k}(2,s,m_{i}^{2})$ are

$\displaystyle B_{2}(2,s,m_{i}^{2})$ $\displaystyle=s\ \mathcal{D}(s)\
(-3s^{2}+2sU+U^{2}-4D)\ ,$ $\displaystyle B_{1}(2,s,m_{i}^{2})$
$\displaystyle=-9s^{6}+32s^{5}U+s^{4}(-37U^{2}+4D)+s^{3}(32UD+8U^{3})$
$\displaystyle+s^{2}(-128UT-88U^{2}D+13U^{4}+144D^{2})$
$\displaystyle+s(-128UD^{2}-128U^{2}T+64U^{3}D-8U^{5}+512DT)$
$\displaystyle+48U^{2}D^{2}-12U^{4}D+U^{6}-64D^{3}\ ,$ $\displaystyle
B_{0}(2,s,m_{i}^{2})$
$\displaystyle=-2s^{5}+7s^{4}U+s^{3}(-2U^{2}-12D)+s^{2}(32UD-6U^{3}-60T)$
$\displaystyle+s^{2}(32UD-6U^{3}-60T)+s(8UT-28U^{2}D+5U^{4}+32D^{2})$
$\displaystyle-16UD^{2}-12U^{2}T+8U^{3}D-U^{5}+48DT\ .$ (B.13)

##  Appendix C The $b$ integrals at $d=2$ as elliptic functions

According to the definitions Eq.s(4.1,3.14), at $d=2$ the six $b$ integrals
become

$N_{i}(2,s)=\int_{b_{i-1}}^{b_{i}}\frac{db}{\sqrt{|(b-b_{1})(b-b_{2})(b-b_{3})(b-b_{4})|}},\hskip
14.22636pti=0,1,..,5$ (C.1)

where the $b_{i}$’s, given by Eq.s(3.12), are in increasing order

$b_{-1}<b_{0}<b_{1}<b_{2}<b_{3}<b_{4}<b_{5}\ .$ (C.2)

The six integrals can be expressed, see [8] by using the Legendre changes of
integration variable, in terms of elliptic integrals of the first kind,
defined as

$\begin{split}F(x,k)&=\int_{0}^{x}\frac{dt}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}\
,\\\ K(k)&=\int_{0}^{1}\frac{dt}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}\
,\end{split}$ (C.3)

where $F(x,k)$ is the incomplete elliptic integral (with phase $x$) and $K(k)$
is the complete elliptic integral. $K(k)$ is a solution of the 2nd order
differential equation

$\left[k(1-k^{2})\frac{d^{2}}{dk^{2}}+(1-3k^{2})\frac{d}{dk}-k\right]K(k)=0\
,$ (C.4)

i.e.

$k(1-k^{2})K^{\prime\prime}(k)+(1-3k^{2})K^{\prime}(k)-kK(k)=0\ ,$ (C.5)

with a second solution of the same equation provided by $K(k^{\prime})$, with

$k^{\prime 2}=1-k^{2}\ .$ (C.6)

For $N_{0}(2,s)$ and $N_{1}(2,s)$, the change of integration variable is

$t^{2}=\frac{(b_{4}-b_{2})(b_{1}-b)}{(b_{4}-b_{1})(b_{2}-b)}\ ,\hskip
14.22636ptb=\frac{(b_{4}-b_{2})b_{1}-(b_{4}-b_{1})b_{2}t^{2}}{(b_{4}-b_{2})-(b_{4}-b_{1})t^{2}}\
,$ (C.7)

giving

$\displaystyle
N_{0}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\Bigg{[}F\bigg{(}\sqrt{\frac{b_{4}-b_{2}}{b_{4}-b_{1}}},q\bigg{)}-F\bigg{(}\sqrt{\frac{b_{1}(b_{4}-b_{2})}{b_{2}(b_{4}-b_{1})}},q\bigg{)}\bigg{]}\
,$ (C.8) $\displaystyle
N_{1}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\
F\bigg{(}\sqrt{\frac{b_{1}(b_{4}-b_{2})}{b_{2}(b_{4}-b_{1})}},q\bigg{)}\ ,$
(C.9)

with

$q^{2}=\frac{(b_{4}-b_{1})(b_{3}-b_{2})}{(b_{3}-b_{1})(b_{4}-b_{2})}.$ (C.10)

For $N_{2}(2,s)$ the change of variable is

$t^{2}=\frac{(b_{4}-b_{2})(b-b_{1})}{(b_{2}-b_{1})(b_{4}-b)}\ ,$ (C.11)

giving

$N_{2}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\ K(q^{\prime})\ ,$
(C.12)

where

$q^{\prime
2}=1-q^{2}=\frac{(b_{4}-b_{3})(b_{2}-b_{1})}{(b_{3}-b_{1})(b_{4}-b_{2})}\ .$
(C.13)

For $N_{3}(2,s)$ one proceeds in a similar way. By writing

$t^{2}=\frac{(b_{3}-b_{1})(b-b_{2})}{(b_{3}-b_{2})(b-b_{1})}\ ,$ (C.14)

the result is

$N_{3}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\ K(q)\ .$ (C.15)

For $N_{4}(2,s)$ the change of variable is

$t^{2}=\frac{(b_{3}-b_{1})(b_{4}-b)}{(b_{4}-b_{3})(b-b_{1})}\ ,$ (C.16)

which gives

$N_{4}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\ K(q^{\prime})\ ,$
(C.17)

equal to $N_{3}(2,s),$ according to Eq.(5.24).

Finally, for $N_{5}(2,s)$ the change of variable is

$t^{2}=\frac{(b_{3}-b_{1})(b-b_{4})}{(b_{4}-b_{1})(b-b_{3})}\ ,$ (C.18)

and the result is

$N_{5}(2,s)=\frac{2}{\sqrt{(b_{4}-b_{2})(b_{3}-b_{1})}}\
F\bigg{(}\sqrt{\frac{b_{3}-b_{1}}{b_{4}-b_{1}}},q\bigg{)}\ .$ (C.19)

Let us observe that Eq.s(C.12,C.15,C.17) are of the form

$N(s)=f(s)K(g(s))\ ,$ (C.20)

where $f(s),g(s)$ are known and simple functions of $s$ (involving at most
square roots). By using the chain differentiation rule, one can therefore
differentiate repeatedly Eq.(C.20) with respect to $s$, and express in that
way the derivatives of $K(g(s))$ in terms of the derivatives of $N(s)$ times
known coefficients. By inserting those expressions into Eq.(C.5) one obtains a
second order differential equation in $s$ for $N(s)$, i.e. for
$N_{2}(2,s)=N_{4}(2,s)$ and $N_{3}(2,s)$. That equation coincides (of course)
with Eq.(B.12) or, which is the same, Eq.(5.15) of Section 5.

As a further remark, Eq.(5.44) can be written as

$N_{0}(2,s)+N_{1}(2,s)+N_{5}(2,s)=N_{3}(2,s)\ ,$ (C.21)

which in terms of the above elliptic functions reads

$F\bigg{(}\sqrt{\frac{b_{4}-b_{2}}{b_{4}-b_{1}}},q\bigg{)}+F\bigg{(}\sqrt{\frac{b_{3}-b_{1}}{b_{4}-b_{1}}},q\bigg{)}=K(q)\
.$ (C.22)

## References

  * [1] A. Primo and L. Tancredi, Maximal cuts and differential equations for Feynman integrals. An application to the three-loop massive banana graph, Nucl. Phys. B921 (2017) 316–356, [arXiv:1704.05465].
  * [2] E. Remiddi, Maximal Cuts and Wick Rotations, PoS MA2019 (2022) 002.
  * [3] R. N. Lee, Presenting litered: a tool for the loop integrals reduction, 2012\.
  * [4] L. Adams, C. Bogner, and S. Weinzierl, The two-loop sunrise graph in two space-time dimensions with arbitrary masses in terms of elliptic dilogarithms, J.Math.Phys. 55 (2014), no. 10 102301, [arXiv:1405.5640].
  * [5] R. Barbieri and E. Remiddi, Electron and Muon 1/2(g-2) from Vacuum Polarization Insertions, Nucl. Phys. B90 (1975) 233–266.
  * [6] A. Sabry, Fourth order spectral functions for the electron propagator, Nucl. Phys. 33 (1962), no. 17 401–430.
  * [7] J. Vermaseren, New features of FORM, [math-ph/0010025], 2000.
  * [8] F. Caleca, Equazione differenziale omogenea di secondo ordine per il grafico sunrise a 2-loop con masse arbitrarie. University of Bologna, Tesi di Laurea 2020, [http://amslaurea.unibo.it/21035/]