# Loops in de Sitter space111To Jacques Bros and Michel Gaudin. In loving
memory

Sergio L. Cacciatori(a,b), Henri Epstein(c) and Ugo Moschella(a,b)

(a)Disat, Università dell’Insubria, Como
(b)INFN, Sezione di Milano, Italia
(c)IHES, Bures-sur-Yvette, France

###### Abstract

We discuss general one and two-loops banana diagrams with arbitrary masses on
the de Sitter spacetime by using direct methods of dS quantum field theory in
the dimensional regularization approach. In the one-loop case we also compute
the effective potential for an $O(N)$ model in $d=4$ dimension as an explicit
function of the cosmological constant $\Lambda$, both exactly and
perturbatively up to order $\Lambda$. For the two-loop case we show that the
calculation is made easy thanks to a remarkable Källén-Lehmann formula that
has been in the literature for a while. We discuss the divergent cases at
$d=3$ using a contiguity formula for generalized hypergeometric functions and
we extract the dominant term at $d=4$ proving a general formula to deal with a
divergent hypergeometric series.

##  1 Introduction

General Relativity and the Standard Model of Particle Physics are the greatest
successes of theoretical physics and are among the highest achievements of
mankind. Starting from the researches of the pre-Socratic philosophers of the
Greek world of the VI century B.C., it took almost three millennia to reach
this level of comprehension of Nature [1].

On its side, General Relativity offers the natural framework for gravity and
cosmology, not only by providing equations that plausibly describe the
evolution of the universe and of the objects that populate it, but also as a
source of ideas and technical tools for improving and sometimes changing the
way we observe and measure the sky; examples are the already mature
gravitational lensing and microlensing techniques and the brand new multi-
messenger astronomy. On the other side, the Standard Model gives a deep
understanding of the microscopic world and there is no need to recall its
great successes here.

The mysterious fact remains that all attempts to construct a single coherent
theory that includes both of them have consistently failed. Nevertheless,
cosmology, astrophysics and the physics of elementary particles are nowadays
inextricably interconnected and even an incomplete understanding of the
history and the dynamics of the universe requires some form of coexistence of
the two theories. At the moment, the best thing that is available is Quantum
Field Theory (QFT) on a curved background, possibly including a quantum
linearized gravitational field and backreaction.

QFT in an inertial frame of flat Minkowski space is quite well understood,
mainly at the perturbative level but non-perturbative methods have also been
developed over the years. The necessity of improving our computational skills
for a deeper understanding of the Standard Model has recently generated a
great flurry of activity; new strategies for computing Feynman-like integrals
in QFT and GR (especially for gravitational waves) are abundant in
contemporary literature. Here are a few examples:

– the proposed reformulation of perturbative QFT in terms of positive
Grassmannian geometry in a complexified momentum space, leading to the notion
of Amplituhedron and its generalizations [25]–[28];

– the traditional method of integration by parts to uncover relations among
different Feynman integrals in order to reduce their calculation to a subset
of Master Integrals; the system of differential equations they have to satisfy
[29]–[30];

– the cohomological techniques based on the interpretation of Feynman
integrals as periods of generators of a suitable twisted-cohomology [31]–[50];

– viewing Feynman integrals as a linear space with the intersection product of
the twisted cohomology used scalar product to determine bases of Master
Integrals, Picard Fuchs-equations and so on [51]–[60];

– other methods that consist in investigating the (algebraic) geometry
underlying Feynman integrals relating them to suitable Calabi-Yau varieties
[61]–[67]. The crucial common ingredient of all the above approaches is the
Fourier momentum-space representation of Feynman integrals. Unfortunately,
such representation is not available on curved backgrounds because translation
invariance, which is the foundation of Fourier analysis, pertains only to flat
space222Even in flat space, a non trivial topology may limit the effectiveness
of the momentum space formulation.; the main road for calculating amplitudes
in curved spacetimes is to do it in position space. This approach is
surprisingly efficient also in Minkowski space: in [2] we presented a study of
one and two-loop diagrams in position space, improving by a margin the
existing literature and also providing new insights into the method of partial
integration.

In this paper we perform the same loop calculations in the de Sitter universe
working in position space at Euclidean times. The status of the de Sitter
spacetime is however rather exceptional. Even in the absence of a linear
momentum space, a complete harmonic analysis has been available for some time
for de Sitter quantum field theory [3, 4, 5, 6] but this possibility has not
yet been fully exploited in applications (see however [7]). We intend to fill
this gap and show how this bunch of rigorous methods and results is also very
effective to compute exact expressions for loop integrals which may in turn
prove to be relevant for cosmology.

In particular, we consider one and two-loop diagrams with no external legs;
calculations are performed without relying on particular choices of
coordinates on the de Sitter manifold; the results are explicit exact formulae
for loop integrals with two and, respectively, three independent scalar
fields; three different masses enter in the loop, no conformal invariance is
supposed, no room for bootstrapping anything.

Our results generalize to the de Sitter spacetime the state of the art of the
knowledge available in flat space [2, 9]. We expect that the methods and the
tools exposed in this papers will be useful to face other calculations
involving loops on the (complex) de Sitter manifold.

The paper is organized as follows: in Sec. 2, after recalling some
generalities about de Sitter QFT, we write multiloop (banana) diagrams with no
external legs in a form suitable for their computation in the rest of the
paper. Sec. 3 is devoted to the one-loop diagram with two independent masses.

The proof of the main formula (3.46) is performed in position space and
crucially relies on the identification of the two-point function as a Legendre
function of the first kind (a Ferrers function of the first kind for the
Euclidean propagator) as opposed to an equivalent but otherwise less specific
hypergeometric function ${}_{2}F_{1}$.

In Sec. 4 we discuss an application: we compute the one-loop effective
potential for the $O(N)$ model on the de Sitter manifold in dimension $d=4$,
exact at one-loop. Computations of effective potentials on the de Sitter
background have sporadically appeared in the literature [8]–[11] but there
exists to date no rigorous deduction based on a formulation sufficiently
general to allow for systematic strategies as is the case in flat spacetime.

In Sec. 6 we compute the flat limit of the effective potential which exactly
reproduces the well-known result in Minkowski space. It is worthwhile to
underline already here that this is indeed a non-trivial fact, as the flat
limit relates quantities that are the outcome of integrals of different
functions over different manifolds and is not just the asymptotic value of
some functions close to a given event. We express the exact one-loop potential
as a function of the cosmological constant and show how to compute it
perturbatively at any desired order, when the cosmological constant is small.

In Sec. 7 we focus on the two-loop diagram with no external legs with three
arbitrary masses; we study the diagram in dimensional regularization for an
arbitrary complex dimension of the de Sitter manifold and produce exact
formulae for it in Eqs. (7.139), (LABEL:ad), (7.158), (7.173) and (7.177). The
crucial ingredient allowing for a full solution of the problem is the Källén-
Lehmann representation of the product of two propagators that was explicitly
constructed a few years ago [7].

In the remaining sections we use our explicit formulae to study the diagram at
integer spacetime dimension. We start with a discussion of odd negative
spacetime-dimensions which also provides a check for our formulae; then we
produce a detailed study of the non-trivial cases $d=2$ and $d=3$.

Extracting the finite part in $d=4$ can be done using the contiguity relations
discussed in Appendix (C) precisely as it is done for the $d=3$ case in
Appendix D. In $d=4$ the situation is complicated by the presence of a double
pole; formulae for the finite part of the diagrams are too long to be
reproduced here. We limit the discussion to the residues that we compute by
using the Erdélyi-Tricomi theorem.

##  2 Banana integrals on the Euclidean sphere

### Geometry

The easiest way to look at either the real or the complex $d$-dimensional de
Sitter manifolds is to visualize them as subsets of the complex Minkowski
spacetime with one spacelike dimension more. This viewpoint allows for a
natural description of the fundamental tubular domains encoding the spectral
condition of de Sitter quantum field theory [3, 4, 5].

Let therefore $M_{d+1}$ be the real $(d+1)$-dimensional Minkowski space-time
and $M_{d+1}^{(c)}$ be its complexification. In a chosen Lorentz frame the
scalar product of two (complex) events is

$z_{1}\cdot
z_{2}=z^{0}_{1}z^{0}_{2}-z^{1}_{1}z^{1}_{2}-\ldots-z^{d}_{1}z^{d}_{2}.$ (2.1)

The future cone $V_{+}$ and the future and past tubes $T_{\pm}$ of the
(complex) Minkowski spacetime are defined as follows:

$\displaystyle V_{+}$ $\displaystyle=$ $\displaystyle\\{x\in M_{d+1}\ :\
x\cdot x>0,\ \ \ x^{0}>0\\},$ (2.2) $\displaystyle T_{\pm}$ $\displaystyle=$
$\displaystyle\\{x+iy\in M_{d+1}^{(c)}\ :\ y\in\pm V_{+}\\}\ .$ (2.3)

The tubes $T_{\pm}$ are the geometrical sets corresponding to the spectral
condition which requires the positivity of the spectrum of the energy operator
in every Lorentz frame [12]. This is the very characteristic property at the
heart of QFT at zero temperature. All the well-known features of QFT, and
above all the Euclidean formulation and renormalization, depend on it.

The real de Sitter universe may be represented as the one-sheeted hyperboloid
immersed in $M_{d+1}$:

$dS_{d}=\\{x\in M_{d+1}\ :\ x\cdot x=-R^{2}=-1\\}\ ;$ (2.4)

the same definition, mutatis mutandis, holds for its complexification:

$dS_{d}^{(c)}=\\{z\in M_{d+1}^{(c)}\ :\ z\cdot z=-R^{2}=-1\\}\ .$ (2.5)

The de Sitter invariant complex variable $\zeta$ is the scalar product in the
ambient spacetime of two complex events $z_{1},z_{2}\in dS_{d}^{(c)}$:

$\zeta=z_{1}\cdot z_{2}.$ (2.6)

Two real events $x_{1}$ and $x_{2}$ in $dS_{d}$ are timelike separated if and
only if

$(x_{1}-x_{2})^{2}=-2-2x_{1}\cdot x_{2}>0.$ (2.7)

The future and past tuboids ${\cal T}_{\pm}$ are the intersections of the
ambient tubes $T_{\pm}$ with the complex de Sitter manifold:

${\cal T}_{\pm}=\\{x+iy\in X_{d}^{(c)}\ :\ y\in\pm V_{+}\\}\ .$ (2.8)

### Harmonic analysis

A natural basis of plane-wave solutions of the de Sitter Klein-Gordon equation

$\Box\psi(z)+m^{2}\psi(z)=0,$ (2.9)

is parameterized by the choice of a lightlike vector $\xi\in C^{+}=\partial
V^{+}$ and a complex number $\lambda$ which is in turn parametrized by the
spacetime dimension and another complex number $\nu$ as follows:

$\psi_{\lambda}(z,\xi)=\left({z\cdot\xi}\right)^{\lambda},\ \ \ \
\lambda=-\frac{d-1}{2}+i\nu.$ (2.10)

The parameters $\lambda$ and $\nu$ are related to the complex mass squared:

$m^{2}=-\lambda(\lambda+d-1)=\frac{(d-1)^{2}}{4}+\nu^{2}.$ (2.11)

Plane waves are well-defined and analytic in each of the tubes $\cal T^{+}$
and $\cal T^{-}$ [3, 4]. Of course their squared mass is real and positive
only when:

a) $\nu$ is real; this correspond in a group-theoretical language to the
principal series of unitary representations of the Lorentz group;

b) $\nu$ is purely imaginary such the $|\nu|<\frac{d-1}{2}$; this correspond
to the complementary series of unitary representations of the Lorentz group.

In de Sitter spacetime there is no global timelike Killing vector. A spectral
condition may however be formulated as the following requirement of normal
analyticity [4]: the two-point distributions are boundary values of functions
analytic in the domain ${\cal T}_{-}\times{\cal T_{+}}$; this condition,
together with de Sitter invariance and the Canonical Commutation Relations,
selects a unique two-point function333It is called - for a strange historical
habit - Bunch-Davis vacuum. Actually, it was W. Thirring the first to find it.
for any de Sitter Klein-Gordon field [4]:

Main result: [3, 4] the normally analytic canonical Wightman function of a de
Sitter Klein-Gordon field in spacetime dimension $d$ whose complex mass
squared $m^{2}=\frac{(d-1)^{2}}{4}+\nu^{2}$ is parametrized by a complex
parameter $\nu$, has the following spectral representation in plane waves and
is holomorphic for $z_{1}\in{\cal T}_{-}$ and $z_{2}\in{\cal T}_{+}$:

$\displaystyle W^{d}_{\nu}(z_{1},z_{2})$
$\displaystyle=\frac{\Gamma\left(\frac{d-1}{2}+i\nu\right)\Gamma\left(\frac{d-1}{2}-i\nu\right)e^{\pi\nu}}{2^{d+1}\pi^{d}}\int_{\gamma}(\xi\cdot
z_{1})^{-\frac{d-1}{2}-i\nu}(\xi\cdot
z_{2})^{-\frac{d-1}{2}+i\nu}\,{\alpha}(\xi)$ (2.12)
$\displaystyle=w^{d}_{\nu}(\zeta)$
$\displaystyle=\frac{\Gamma\left(\frac{d-1}{2}+i\nu\right)\Gamma\left(\frac{d-1}{2}-i\nu\right)}{2(2\pi)^{d/2}}(\zeta^{2}-1)^{-\frac{d-2}{4}}\,P_{-{\frac{1}{2}}+i\nu}^{-\frac{d-2}{2}}(\zeta)$
(2.13)
$\displaystyle=\frac{\Gamma\left(\frac{d-1}{2}+i\nu\right)\Gamma\left(\frac{d-1}{2}-i\nu\right)}{(4\pi)^{d/2}\Gamma\left(\frac{d}{2}\right)}{}_{2}F_{1}\left(\frac{d-1}{2}+i\nu,\frac{d-1}{2}-i\nu;\
{\frac{d}{2}};\ \frac{1-\zeta}{2}\right).$ (2.14)

In standard coordinates, the ($d-1$)-form $\alpha(\xi)$ in Eq. (2.12) is
written

$\alpha(\xi)=(\xi^{0})^{-1}\sum_{j=1}^{d}(-1)^{j+1}\xi^{j}\,d\xi^{1}\ldots\
\widehat{d\xi^{j}}\ldots\ d\xi^{d}\ .$ (2.15)

$\gamma$ denotes any ($d-1$)-cycle in the forward light-cone $C^{+}$. (2.12)
does not depend on the choice of $\gamma$ being the integral of a closed
differential form. In particular we may choose the unit sphere ${S}_{d-1}$
(equipped with its canonical orientation):

$\gamma_{0}={S}_{d-1}=C^{+}\cap\\{\xi\ :\ \xi^{0}=1\\}=\\{\xi\in
C^{+}:{\xi^{1}}^{2}+\ldots+{\xi^{d}}^{2}=1\\}.$ (2.16)

With this choice $\alpha(\xi)$ coincides with the rotation invariant measure
$d\xi$ on ${S}_{d-1}$ normalized as follows:

$\omega_{d}=\int_{\gamma_{0}}d\xi=\frac{2\pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}\right)}.$
(2.17)

By chosing any two points in the respective tubes one can show the validity of
Eq. (2.13). A crucial fact is that the Legendre function [13] of the first
kind in that formula

$P_{\nu}^{\mu}(\zeta)={1\over\Gamma(1-\mu)}\left({\zeta+1\over\zeta-1}\right)^{\mu\over
2}F\left(-\nu,\ 1+\nu\ ;\ 1-\mu\ ;\ {1-\zeta\over 2}\right)\ $ (2.18)

is holomorphic in the cut-plane ${\bf C}\setminus(-\infty,\ 1]$ but the
reduced two-point function $w^{d}_{\nu}(\zeta)$ is holomorphic in the larger
domain444The prefactor $(\zeta^{2}-1)^{-{d-2\over 4}}$ exactly compensates the
singularity of $P_{-{1\over 2}+i\nu}^{-{d-2\over 2}}(\zeta)$ at $\zeta=1$,
making the reduced two-point function regular there.

$\zeta=z_{1}\cdot z_{2}\in\Delta=\\{{\bf C}\setminus(-\infty,\ -1]\\},$ (2.19)

i.e. everywhere except on the causality cut (2.7); this is the maximal
analyticity property.

When $\nu$ is either real or is purely imaginary and such that
$|\nu|<\frac{d-1}{2}$ the corresponding two-point function (2.12) is positive-
definite and admits a direct quantum probabilistic interpretation.

When $\frac{d-1}{2}+i\nu=-n$, where $n$ is zero or a positive integer, a more
involved but yet acceptable quantum interpretation of the above formula is
also possible; on the de Sitter universe there exist tachyonic fields having
no counterpart in flat space [14, 15].

Finally, the Schwinger function (in short: the propagator) is the restriction
of the maximally analytic two-point function to the Euclidean sphere. It can
be obtained as follows: in Eq. (2.12) choose the two points as follows

$\displaystyle z_{1}=\left(\begin{array}[]{c}\mathop{\rm
sh}\nolimits(-i\epsilon)\\\ 0\\\ \vdots\\\ 0\\\ \mathop{\rm
ch}\nolimits(-i\epsilon)\\\ \end{array}\right),\ \ \ \ \
z_{2}(s)=\left(\begin{array}[]{c}\mathop{\rm sh}\nolimits(is)\\\ 0\\\
\vdots\\\ 0\\\ \mathop{\rm ch}\nolimits(is)\end{array}\right),\ \ \ 0<s<\pi,$
(2.30)

so that $z_{1}\cdot z_{2}(s)=-\cos(s-i\epsilon)$; we get the following
expression for the propagator

$\displaystyle G_{\nu}(-\cos
s)=\frac{\Gamma({\frac{d-1}{2}+i\nu})\Gamma({\frac{d-1}{2}-i\nu})}{2(2\pi)^{d/2}}(\sin
s)^{-\frac{d-2}{2}}{\bf P}_{-\frac{1}{2}+i\nu}^{-\frac{d-2}{2}}(-\cos s)$
(2.31)

where ${\bf P}_{\rho}^{\mu}(z)$ is the so called ”Legendre function on the
cut” or Ferrers function of the first kind [13] (see Appendix A). It is
important to keep in mind that Ferrers functions ${\bf P}_{\beta}^{\alpha}(z)$
and Legendre functions $P_{\beta}^{\alpha}(z)$ are holomorphic in different
cut-planes; as regards Ferrers function this is

$\Delta_{2}={\bf C}\setminus\\{(-\infty,-1]\cup[1,\infty)]\\}.$ (2.32)

### Banana integrals

We are now ready to write the $n$-loop banana integrals on the sphere with
$n+1$ propagators:

$\displaystyle I_{n}(\nu_{1},\ldots,\nu_{n+1},d)=\int
G^{d}_{\nu_{1}}(x_{0}\cdot x)G^{d}_{\nu_{2}}(x_{0}\cdot x)\ldots
G^{d}_{\nu_{n+1}}(x_{0}\cdot x)\sqrt{g}\,dx;$ (2.33)

here $x$ varies on the de Sitter sphere, $x_{0}$ is a fixed reference point
over there and $\sqrt{g}\,dx$ is the rotation invariant measure. By
integrating over the angles we get

$\displaystyle I_{n}(\nu_{1},\ldots,\nu_{n+1},d)=$ (2.34)
$\displaystyle=\frac{2\pi^{\frac{d}{2}}}{\Gamma\left({\frac{d}{2}}\right)}\int_{0}^{\pi}G^{d}_{\nu_{1}}(-\cos
s)G^{d}_{\nu_{2}}(-\cos s)\ldots G^{d}_{\nu_{n+1}}(-\cos s)(\sin s)^{d-1}ds.$
(2.35)

The aim of this paper is to actually compute the two and three-lines banana
integrals. Here we add some information on the three lines case: this is
explicitly written as the integral of the product of three Ferrers function on
the interval $(-1,1)$ which is the projection on the plane $\Theta$ of the
Euclidean sphere:

$\displaystyle I_{3}(\nu_{1},\nu_{2},\nu_{3},d)$
$\displaystyle=K_{3}(\nu_{1},\nu_{2},\nu_{3},d)\int_{-1}^{1}{\bf P}_{-{1\over
2}+i\nu_{1}}^{-{d-2\over 2}}(u){\bf P}_{-{1\over 2}+i\nu_{2}}^{-{d-2\over
2}}(u){\bf P}_{-{1\over 2}+i\nu_{3}}^{-{d-2\over
2}}(u){(1-u^{2})^{-\frac{d-2}{4}}}du,$ (2.36)

with

$K_{3}(\nu_{1},\nu_{2},\nu_{3},d)={\prod_{j=1}^{3}\Gamma({d-1\over
2}-i\nu_{j})\Gamma({d-1\over 2}+i\nu_{j}))\over
2^{2+\frac{3d}{2}}\pi^{d}\Gamma\left(\frac{d}{2}\right)}.$ (2.37)

In a previous paper [7] another integral of three Legendre functions has been
computed:

$\displaystyle
h_{d}(\lambda,\nu,\kappa)=\int_{1}^{\infty}{P^{-\frac{d-2}{2}}_{-\frac{1}{2}+i\lambda}(u)P^{-\frac{d-2}{2}}_{-\frac{1}{2}+i\nu}(u)}P^{-\frac{d-2}{2}}_{-\frac{1}{2}+i\kappa}(u)\,{(u^{2}-1)^{-\frac{d-2}{4}}}\
du$ (2.38)
$\displaystyle=\frac{2^{\frac{d}{2}}}{(4\pi)^{\frac{3}{2}}\Gamma\left(\frac{d-1}{2}\right)}\frac{\
\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm
1}\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon\lambda+i\epsilon^{\prime}\nu+i\epsilon^{\prime\prime}\kappa}{2}\right)}{\prod_{\epsilon,\epsilon^{\prime}\epsilon^{\prime\prime}=\pm
1}\Gamma\left(\frac{d-1}{2}+i\epsilon\lambda\right)\Gamma\left(\frac{d-1}{2}+i\epsilon^{\prime}\nu\right)\Gamma\left(\frac{d-1}{2}+i\epsilon^{\prime\prime}\kappa\right)}.$
(2.39)

It was far from obvious and made possible by a mix of geometrical ideas with
analytical and probabilistic tools. The steps involved in the proof-
computation also gave rise to many interesting quantities having possibly
geometrical interpretations that are not yet fully explored in their
mathematical and physical consequences.

An output of the above result is an explicit Källén–Lehmann representation of
the product of two Wightman two-point functions

$\displaystyle
w^{d}_{\lambda}(\zeta)w^{d}_{\nu}(\zeta)=\int_{-\infty}^{\infty}\rho_{d}({\lambda},\nu,\kappa)w^{d}_{\kappa}(\zeta)\kappa
d\kappa,$ (2.40)

where

$\displaystyle\rho_{d}({\lambda},\nu,\kappa)=\frac{1}{2^{d}{\pi^{{d-1\over
2}}}\kappa\Gamma\left(\frac{d-1}{2}\right)}\frac{\
\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm
1}\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon\lambda+i\epsilon^{\prime}\nu+i\epsilon^{\prime\prime}\kappa}{2}\right)}{\prod_{\epsilon=\pm
1}\Gamma\left(\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{1}{2}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d+1}{4}+\frac{i\epsilon\kappa}{2}\right)}.$
(2.41) (2.42)

Contrary to what happens in flat space, the Källén–Lehmann weight is here a
meromorphic function of the three mass variables; this fact in particular
implies that particle decays that in flat space are forbidden by mass
subadditivity, can take place in the de Sitter universe [7].

It turns out that the direct evaluation of (2.36) is extremely difficult; the
geometry of the complex de Sitter universe is less helpful here than it was in
the evaluation of the previous integral (2.38). The knowledge of the
Källén–Lehmann representation (2.40), which is deeply rooted in the harmonic
analysis that we have recalled above, offers to us one opportunity to solve
the problem; we are going to describe that construction later, in section 7.

##  3 1-loop: the bubble

Let us start with the already nontrivial two-line case, i.e. the bubble:

$\displaystyle I(\lambda,\nu,d)$ $\displaystyle=$
$\displaystyle\frac{2\pi^{\frac{d}{2}}}{\Gamma\left({\frac{d}{2}}\right)}\int_{0}^{\pi}G_{\lambda}(-\cos
s)G_{\nu}(-\cos s)(\sin s)^{d-1}ds$ (3.43) $\displaystyle=$ $\displaystyle
K(\lambda,\nu,d)\int_{-1}^{1}{\bf P}_{-{1\over 2}+i\lambda}^{-{d-2\over
2}}(u){\bf P}_{-{1\over 2}+i\nu}^{-{d-2\over 2}}(u)du,$ (3.44) $\displaystyle
K_{d}(\lambda,\nu)$ $\displaystyle=$ $\displaystyle\Gamma({{d-1\over
2}-i\lambda})\Gamma({{d-1\over 2}+i\lambda})\Gamma({{d-1\over
2}-i\nu)\Gamma({{d-1\over 2}+i\nu})}\over
2\left(2\sqrt{\pi}\right)^{d}\Gamma\left({\frac{d}{2}}\right).$ (3.45)

To the best of our knowledge, as simple as it it may look, the integral at the
rhs of Eq. (3.44) is not listed anywhere in the literature. We will compute it
in three different ways.

### Using the Wronskian

The method that gives the cleanest result is based on the properties of the
Wronskian of two Legendre functions. We summarize this result in the following

###### Formula 1

$I(\lambda,\nu,d)=\frac{\Gamma\left(1-\frac{d}{2}\right)}{2^{d}\pi^{\frac{d}{2}}(\lambda^{2}-\nu^{2})}\
\left(\frac{\Gamma\left(\frac{d-1}{2}-i\nu\right)\Gamma\left(\frac{d-1}{2}+i\nu\right)}{\Gamma\left(\frac{1}{2}-i\nu\right)\Gamma\left(\frac{1}{2}+i\nu\right)}-\frac{\Gamma\left(\frac{d-1}{2}-i\lambda\right)\Gamma\left(\frac{d-1}{2}+i\lambda\right)}{\Gamma\left(\frac{1}{2}-i\lambda\right)\Gamma\left(\frac{1}{2}+i\lambda\right)}\right).\
$ (3.46)

To give a proof of the above formula let us consider two solutions
$u_{\nu}^{\mu}$ and $v_{\sigma}^{\mu}$ of the Legendre equation, with
$u_{\nu}^{\mu}(z)$ standing for either ${\bf P}_{\nu}^{\mu}(z)$ or ${\bf
Q}_{\nu}^{\mu}(z)$ and, independently, $v_{\sigma}^{\mu}(z)$ standing for
either ${\bf P}_{\sigma}^{\mu}(z)$ or ${\bf Q}_{\sigma}^{\mu}(z)$. Then

$\displaystyle\int_{a}^{b}u_{\nu}^{\mu}(z)v_{\sigma}^{\mu}(z)(\nu-\sigma)(\sigma+\nu+1)dz=\Big{[}(1-z^{2})^{\scriptstyle{1\over
2}}(\sigma+\mu)(\sigma-\mu+1)u_{\nu}^{\mu}(z)v_{\sigma}^{\mu-1}(z)$ (3.47)
$\displaystyle-(1-z^{2})^{\scriptstyle{1\over
2}}(\nu+\mu)(\nu-\mu+1)u_{\nu}^{\mu-1}(z)v_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}\
.$ (3.48)

This is the main Wronskian equation. Its derivation and more details are given
in Appendix A (see also [13, 3.12 (1) p. 169]). If $a=-1$ and $b=1$, the
integral on the l.h.s. converges when $-1<\mathop{\rm Re}\nolimits\mu<1$.

In this section we will examine the case when $u_{\nu}^{\mu}={\bf
P}_{\nu}^{\mu}$, $v_{\sigma}^{\mu}={\bf P}_{\sigma}^{\mu}$ which is relevant
to the subject of this paper. The cases when $u_{\nu}^{\mu}={\bf
P}_{\nu}^{\mu}$, $v_{\sigma}^{\mu}={\bf Q}_{\sigma}^{\mu}$, and when
$u_{\nu}^{\mu}={\bf Q}_{\nu}^{\mu}$, $v_{\sigma}^{\mu}={\bf Q}_{\sigma}^{\mu}$
are discussed in Appendix A.

Let us thus substitute $u_{\nu}^{\mu}={\bf P}_{\nu}^{\mu}$ and
$v_{\sigma}^{\mu}={\bf P}_{\sigma}^{\mu}$ in (3.48):

$\displaystyle(\nu-\sigma)(\nu+\sigma+1)\int_{a}^{b}{\bf P}_{\nu}^{\mu}(z){\bf
P}_{\sigma}^{\mu}(z)\,dz=\Big{[}(\sigma+\mu)(\sigma-\mu+1)(1-z^{2})^{\scriptstyle{1\over
2}}{\bf P}_{\nu}^{\mu}(z){\bf P}_{\sigma}^{\mu-1}(z)$ (3.49)
$\displaystyle-(\nu+\mu)(\nu-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu-1}(z){\bf P}_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}$ (3.50)

and evaluate the r.h.s. of this equation under the conditions $a=-1$, $b=1$
and $0<\mu<1$. To do this it suffices to evaluate the first term in the r.h.s.
of (3.50), since the second term is obtained by exchanging $\nu$ and $\sigma$
and by a global change of sign.

Since $0<\mu<1$, as $z\in(0,\ 1)$, $z\rightarrow 1$, Eq. (A.299) shows that

$(\sigma+\mu)(\sigma-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu}(z){\bf P}_{\sigma}^{\mu-1}(z)\sim C(1-z)^{{\scriptstyle{1\over
2}}-{\mu\over 2}-{\mu\over 2}+{\scriptstyle{1\over
2}}}=C(1-z)^{1-\mu}\rightarrow 0\ .$ (3.51)

As $z\in(-1,\ 1)$, $z\rightarrow-1$ we use Eqs. (A.302) and (A.303) to get

$\displaystyle(\sigma+\mu)(\sigma-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu}(z){\bf
P}_{\sigma}^{\mu-1}(z)\rightarrow{2\pi^{-1}\sin(\pi\nu)\Gamma(\mu)\Gamma(1-\mu)\over\Gamma(1+\sigma-\mu)\Gamma(-\sigma-\mu)}\
.$ (3.52)

Putting together the contributions of the two terms in the rhs of (3.50) gives

$\int_{-1}^{1}{\bf P}_{\nu}^{\mu}(z){\bf
P}_{\sigma}^{\mu}(z)\,dz=2\pi^{-1}\Gamma(\mu)\Gamma(1-\mu)\Phi_{{\bf
P}}(\nu,\sigma,\mu),$ (3.53)

where

$\Phi_{{\bf
P}}(\nu,\sigma,\mu)={{\sin(\pi\sigma)\over\Gamma(1+\nu-\mu)\Gamma(-\nu-\mu)}-{\sin(\pi\nu)\over\Gamma(1+\sigma-\mu)\Gamma(-\sigma-\mu)}\over(\nu+\sigma+1)(\nu-\sigma)}.$
(3.54)

$\Phi_{{\bf P}}(\nu,\sigma,\mu)$ extends to an entire function of all its
arguments, it is symmetric in $\nu$ and $\sigma$, invariant under the
involution $\nu\rightarrow-\nu-1$ and vanishes at $\mu=0$. Equation (3.53)
remains valid, by analytic continuation, for $-1<\mathop{\rm
Re}\nolimits\mu<1$. Combining (3.53-3.54) with (3.45) we obtain the beautiful
formula 3.46 for the bubble.

Here are a few consequences of the above result:

  1. 1.

The bubble is regular at $d=2$:

$I(\lambda,\nu,2)=\frac{\left(\psi\left(\frac{1}{2}-i\lambda\right)+\psi\left(\frac{1}{2}+i\lambda\right)-\psi\left(\frac{1}{2}-i\nu\right)-\psi\left(\frac{1}{2}+i\nu\right)\right)\
}{4\pi\left(\lambda^{2}-\nu^{2}\right)}.$ (3.55)

  2. 2.

In odd spacetime dimension the formula becomes very simple; for instance at
$d=3$

$I(\lambda,\nu,3)=\frac{\lambda\coth(\pi\lambda)-\nu\coth(\pi\nu)}{4\pi\lambda^{2}-4\pi\nu^{2}}.$
(3.56)

  3. 3.

At $d=4$ we encounter the first divergence. The Laurent expansion of the
formula near $d=4$ gives

$\displaystyle
I(\lambda,\nu,4)=-\frac{1}{8\pi^{2}(d-4)}+\frac{1-\gamma+\log(4\pi)}{16\pi^{2}}$
(3.57)
$\displaystyle-\frac{\left(4\lambda^{2}+1\right)\left(\psi\left(\frac{3}{2}-i\lambda\right)+\psi\left(\frac{3}{2}+i\lambda\right)\right)-\left(4\nu^{2}+1\right)\left(\psi\left(\frac{3}{2}-i\nu\right)+\psi\left(\frac{3}{2}+i\nu\right)\right)}{64\pi^{2}\left(\lambda^{2}-\nu^{2}\right)}$
(3.58) $\displaystyle+\ {\rm O}(d-4).$ (3.59)

##  4 The $O(N)$ model: a summary

Let $E\cong\mathbb{R}^{N}$ be the $N$ dimensional Euclidean vector space with
the standard scalar product denoted by $\langle\cdot\mid\cdot\rangle$ and
$O(N)$ the corresponding orthogonal group. Let us consider a scalar multi-
component field

$\displaystyle\phi:S_{d}\longrightarrow E,$ (4.60)

where $S_{d}$ is the (Euclidean) de Sitter sphere of radius $R$. The radius
$R$ and the cosmological constant

$\Lambda=\frac{(d-1)(d-2)}{2R^{2}}$ (4.61)

will reappear when necessary (see. Sect. (6)). For the moment we take $R=1$.

The action for $\phi$ is the quartic $O(N)$-invariant action

$\displaystyle
S[\phi]=\int_{S_{d}}\Big{[}\Lambda_{0}+\frac{1}{2}\langle\partial_{\mu}\phi\mid\partial^{\mu}\phi\rangle+\frac{m_{0}^{2}}{2}\langle\phi\mid\phi\rangle+\frac{c_{0}}{4}\langle\phi\mid\phi\rangle^{2}\Big{]}\sqrt{g}\,d^{d}x,$
(4.62)

where $m_{0}$ and $c_{0}$ are respectively the bare mass and bare self-
coupling constant; $x^{\mu}$ are local coordinates on the sphere. We included
also an extra cosmological constant $\Lambda_{0}$ for convenience when
renormalizing.

We will compute the effective potential at $d=4$ for the constant
configuration

$\displaystyle\bar{\phi}\equiv\varphi\,e_{0},$ (4.63)

where $\varphi$ is a real constant and $e_{0}$ is a given vector of norm $1$
in $E$. Of course a nonzero expectation value of $\phi$ breaks the symmetry
down to $O(N-1)$. After choosing any orthonormal basis
$\\{e_{j}\\}_{j=0}^{N-1}$ of $E$ whose first element is $e_{0}$, we may write

$\displaystyle\phi=(\varphi+\psi_{0})e_{0}+\sum_{j=1}^{N-1}\psi_{j}e_{j},$
(4.64)

so that

$\displaystyle S[\phi]=$
$\displaystyle\int_{S_{d}}\Big{[}\Lambda_{0}+\frac{m_{0}^{2}}{2}\varphi^{2}+\frac{c_{0}}{4}\varphi^{4}\Big{]}\sqrt{g}\,d^{d}x+\sum_{a=1}^{4}S_{a}[\psi;\varphi],$
(4.65)

where

$\displaystyle S_{1}[\psi;\varphi]=$
$\displaystyle\int_{S_{d}}\Big{[}{m_{0}^{2}}\varphi\psi_{0}+{c_{0}}\varphi^{3}\psi_{0}\Big{]}\sqrt{g}\,d^{d}x,$
(4.66) $\displaystyle S_{2}[\psi;\varphi]=$
$\displaystyle\frac{1}{2}\int_{S_{d}}\Big{[}\sum_{j=0}^{N-1}\partial_{\mu}\psi_{j}\partial^{\mu}\psi_{j}+(m_{0}^{2}+3c_{0}\varphi^{2})\psi_{0}^{2}+(m_{0}^{2}+c_{0}\varphi^{2})\sum_{j=1}^{N-1}\psi_{j}^{2}\Big{]}\sqrt{g}\,d^{d}x,$
(4.67) $\displaystyle S_{3}[\psi;\varphi]=$
$\displaystyle\int_{S_{d}}\Big{[}c_{0}\varphi\psi_{0}\sum_{j=0}^{N-1}\psi_{j}^{2}\Big{]}\,d^{d}x,$
(4.68) $\displaystyle S_{4}[\psi;\varphi]=$
$\displaystyle\frac{c_{0}}{4}\int_{S_{d}}\sqrt{g}\Big{[}\sum_{j=0}^{N-1}\psi_{j}^{2}\Big{]}^{2}\sqrt{g}\,d^{d}x.$
(4.69)

The effective potential ${\cal V}(\varphi)$ is defined as follows

$\displaystyle\exp(-\Omega_{d}{\cal
V}(\varphi))=\int[\prod_{j}D\psi_{j}]\exp(-S[\phi]);$ (4.70)

here $\Omega_{d}$ is the volume of the Euclidean de Sitter spacetime (which is
finite) and $[\prod_{j}D\psi_{j}]$ is the formal path integral measure. By
construction, $S_{1}$ does not contribute to the effective potential; at
1-loop, we get

$\displaystyle{\cal V}(\varphi)=$
$\displaystyle\Lambda_{0}+\frac{m_{0}^{2}}{2}\varphi^{2}+\frac{c_{0}}{4}\varphi^{4}-\frac{1}{\Omega_{d}}\log\int[\prod_{j}D\psi_{j}]\exp(-S_{2}[\phi]),$
(4.71)

which we rewrite in the standard form:

$\displaystyle{\cal V}(\varphi)=$
$\displaystyle\Lambda_{0}+\frac{m_{0}^{2}}{2}\varphi^{2}+\frac{c_{0}}{4}\varphi^{4}+{\cal
V}_{0}(\varphi)+(N-1){\cal V}_{1}(\varphi),$ (4.72)

where, for $a=0,1$,

$\displaystyle{\cal
V}_{a}(\varphi)=-\frac{1}{\Omega_{d}}\log\int[D\Phi]\exp\left(-\frac{1}{2}\int_{M_{d}}\sqrt{g}\Big{[}\partial_{\mu}\Phi\partial^{\mu}\Phi+M_{a}^{2}(\varphi)\Phi^{2}\Big{]}d^{d}x\right),$
(4.73) $\displaystyle M_{0}^{2}(\varphi)=m_{0}^{2}+3c_{0}\varphi^{2},$ (4.74)
$\displaystyle M_{1}^{2}(\varphi)=m_{0}^{2}+c_{0}\varphi^{2}.$ (4.75)

Several techniques can be used to compute ${\cal V}_{a}(\varphi)$. Instead of
looking at it as a function of $\varphi$, it is useful to see it as a function
of $M_{a}^{2}\equiv z_{a}$. By taking the first derivative w.r.t. $z$ we get
the tadpole integral which leads to the Lee-Sciaccaluga equation for
determining the effective potential [16].

When we differentiate twice, we get the one mass bubble integral; this is the
strategy we want to adopt, since it will allow us to develop methods that will
be helpful to compute the two-loop contribution (a research that will be
considered elsewhere). More precisely, if $J(m^{2}_{1},m^{2}_{2})$ is the two-
masses bubble, one has

$\displaystyle\frac{\partial^{2}}{\partial z^{2}}{\cal
V}(\varphi)=-\frac{1}{2}J(z,z),$ (4.76)

from which we can recover the effective potential. As we will review in Sec.
6, in the flat case and $d=4-2\epsilon$, we have

$\displaystyle
J(z,z)=\frac{1}{16\pi^{2}}\left(\frac{1}{\epsilon}-\log\frac{z}{4\pi
e^{\gamma}}\right)+O(\epsilon).$ (4.77)

After integrating twice in $z$ and renormalising, we get the standard result
[17, 9]

$\displaystyle{\cal V}(\varphi)=$
$\displaystyle\frac{m^{2}}{2}\varphi^{2}+\frac{c_{g}}{4}\varphi^{4}+\frac{(m^{2}+3c_{g}\varphi^{2})^{2}}{64\pi^{2}}\log\frac{m^{2}+3c_{g}\varphi^{2}}{\mu^{2}}$
(4.78)
$\displaystyle+(N-1)\frac{(m^{2}+c_{g}\varphi^{2})^{2}}{64\pi^{2}}\log\frac{m^{2}+c_{g}\varphi^{2}}{\mu^{2}},$
(4.79)

where $m$ and $c_{g}$ are the renormalised mass and coupling constant, and
$\mu^{2}$ is a scale related to the precise renormalisation conditions. We
will now extend this calculation to the de Sitter case.

##  5 The effective potential

To compute the effective potential at $d=4$, we set $d=4-2\epsilon$. Since
$d\nu^{2}=dm^{2}$, we have to integrate twice $-I(\nu,\nu,4)/2$ in
$d\nu^{2}=2\nu d\nu$. From (3.59), we have

$\displaystyle I(\nu,\nu,4-2\epsilon)=$
$\displaystyle\frac{1}{16\pi^{2}\epsilon}+\frac{1-\gamma+\log(4\pi)}{16\pi^{2}}$
(5.80)
$\displaystyle-\frac{1}{128\pi^{2}\nu}\frac{d}{d\nu}\left[\left(4\nu^{2}+1\right)\left(\psi\left(\frac{3}{2}-i\nu\right)+\psi\left(\frac{3}{2}+i\nu\right)\right)\right]$
(5.81) $\displaystyle+\ {\rm O}(\epsilon).$ (5.82)

The nontrivial part comes from the second line. A first integration gives:

$\displaystyle\int\frac{1}{8\nu}\frac{d}{d\nu}\left[\left(4\nu^{2}+1\right)\left(\psi\left(\frac{3}{2}-i\nu\right)+\psi\left(\frac{3}{2}+i\nu\right)\right)\right]2\nu
d\nu=$ (5.83)
$\displaystyle=\left(\nu^{2}+\frac{1}{4}\right)\left(\psi\left(-\frac{1}{2}-i\nu\right)+\psi\left(-\frac{1}{2}+i\nu\right)\right)+C,$
(5.84)

which, integrated once again becomes

$\displaystyle
C\nu^{2}+\int\left(\nu^{2}+\frac{1}{4}\right)\left(\psi\left(-\frac{1}{2}-i\nu\right)+\psi\left(-\frac{1}{2}+i\nu\right)\right)d\nu^{2}.$
(5.85)

An integration by parts then gives

$\displaystyle\int\left(\nu^{2}+\frac{1}{4}\right)$
$\displaystyle\left(\psi\left(-\frac{1}{2}-i\nu\right)+\psi\left(-\frac{1}{2}+i\nu\right)\right)d\nu^{2}=$
(5.86)
$\displaystyle=\frac{1}{2}\left(\nu^{2}+\frac{1}{4}\right)^{2}\left(\psi\left(-\frac{1}{2}-i\nu\right)+\psi\left(-\frac{1}{2}+i\nu\right)\right)$
(5.87) $\displaystyle-\frac{i}{2}\int
d\nu\left(\nu^{2}+\frac{1}{4}\right)^{2}\left(\psi^{\prime}\left(-\frac{1}{2}-i\nu\right)-\psi^{\prime}\left(-\frac{1}{2}+i\nu\right)\right).$
(5.88)

Notice that

$\displaystyle\psi^{\prime}(z)=\zeta(2,z)=\sum_{n=0}^{\infty}\frac{1}{(z+n)^{2}},$
(5.89)

is a Hurwitz zeta function [21]. After setting $y:=\nu^{2}+\frac{1}{4}$, we
can introduce the function

$\displaystyle B(y)\equiv$ $\displaystyle\frac{y^{2}}{4}-\frac{i}{2}\int
d\nu\left(\nu^{2}+\frac{1}{4}\right)^{2}\left(\psi^{\prime}\left(-\frac{1}{2}-i\nu\right)-\psi^{\prime}\left(-\frac{1}{2}+i\nu\right)\right)$
(5.90) $\displaystyle=$ $\displaystyle-\frac{1}{2}\int dy\
y^{2}\left(\sum_{n=0}^{\infty}\frac{2n-1}{(n(n-1)+y)^{2}}-\frac{1}{y}\right),$
(5.91)

where in the second line we have used the series expansion (5.89). By using
the standard Abel-Plana’s formula [22] we can write

$\displaystyle\sum_{n=0}^{\infty}\frac{2n-1}{(n(n-1)+y)^{2}}=$
$\displaystyle-\frac{1}{2y^{2}}-4\int_{0}^{\infty}\frac{t}{e^{2\pi
t}-1}\frac{(t^{2}-y)^{2}-y}{[(t^{2}-y)^{2}+t^{2}]^{2}},$ (5.92)

so that

$\displaystyle B(y)=$
$\displaystyle\frac{y}{3}+\int_{0}^{\infty}\frac{2t^{4}+t^{2}-1+y(4-2t^{2})}{(t^{2}-y)^{2}+t^{2}}\frac{t^{3}dt}{e^{2\pi
t}-1}+6\int_{0}^{\infty}\arctan\left(t-\frac{y}{t}\right)\frac{t^{2}dt}{e^{2\pi
t}-1}$ (5.93)
$\displaystyle+2\int_{0}^{\infty}\left(t^{2}-\frac{1}{2}\right)\log[t^{2}+(t^{2}-y)^{2}]\frac{tdt}{e^{2\pi
t}-1},$ (5.94)

where we have conveniently set to zero an integration constant. With these
notations, the (renormalized) effective potential is

$\displaystyle{\cal V}_{R}=$
$\displaystyle\frac{m_{R}^{2}}{2}\varphi_{R}^{2}+\frac{c_{g}}{4}\varphi_{R}^{4}+\frac{1}{64\pi^{2}}(m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2)^{2}\left(\psi\left(-\frac{1}{2}-i\nu_{0}\right)+\psi\left(-\frac{1}{2}+i\nu_{0}\right)\right)$
(5.95)
$\displaystyle+\frac{1}{32\pi^{2}}B({m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2})+\frac{N-1}{32\pi^{2}}B({m_{R}^{2}+c_{g}\varphi_{R}^{2}-2})$
(5.96)
$\displaystyle+\frac{N-1}{64\pi^{2}}(m_{R}^{2}+c_{g}\varphi_{R}^{2}-2)^{2}\left(\psi\left(-\frac{1}{2}-i\nu_{1}\right)+\psi\left(-\frac{1}{2}+i\nu_{1}\right)\right)$
(5.97)
$\displaystyle-\frac{1}{64\pi^{2}}\left[(m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2)^{2}+(N-1)(m_{R}^{2}+c_{g}\varphi_{R}^{2}-2)^{2}\right]\log{\mu_{R}^{2}}.$
(5.98)

The last line is just the introduction of a reference energy scale, related to
the renormalization, while

$\displaystyle\nu_{0}=$
$\displaystyle\sqrt{m_{R}^{2}+3c_{g}\varphi_{R}^{2}-\frac{9}{4}},$ (5.99)
$\displaystyle\nu_{1}=$
$\displaystyle\sqrt{m_{R}^{2}+c_{g}\varphi_{R}^{2}-\frac{9}{4}}.$ (5.100)

$R$ refers to the de Sitter radius and appears through the formula $m_{R}=mR$,
$\mu_{R}=\mu R$, and $\varphi_{R}=R\varphi$, $m_{R}$, $\mu_{R}$ and
$\varphi_{R}$ being adimensional quantities.
We can find an alternative expression for the function $B$ by directly
integrating (5.91):

$\displaystyle B(y)=-\sum_{n=1}^{\infty}\Big{[}$
$\displaystyle(2n+1)y+\frac{n(n+1)(2n+1)y}{n(n+1)+y}$ (5.101)
$\displaystyle-2(2n+1)(n+1)n\log\Big{(}1+\frac{y}{n(n+1)}\Big{)}\Big{]}+\frac{y^{2}}{4}+D,$
(5.102)

where $D=B(0)$ is a constant that can be determined from (5.94) as

$\displaystyle
D=-\frac{1}{72}+\frac{\gamma-\log(2\pi)}{60}-\frac{3\zeta(3)}{4\pi^{2}}+4\log(A_{3}),$
(5.103)

where $A_{3}$ is the third generalized Glaisher-Kinkelin constant (see
Appendix B).
This formula allows us to write the effective potential at any finite value of
$\nu$, but it is not suitable for taking the flat limit, for which it is
convenient to use (5.94).

##  6 Flat limit

Before discussing the flat limit let us recall that in flat space space the
Schwinger function is proportional to a MacDonald function:

$\displaystyle
G_{m}(x)=\frac{1}{(2\pi)^{d}}\int\frac{e^{-ipx}}{{p^{2}+m^{2}}}dp=\frac{1}{(2\pi)^{\frac{d}{2}}}\left(\frac{r}{m}\right)^{1-\frac{d}{2}}K_{\frac{d}{2}-1}\left(mr\right),\
\ \ r=\sqrt{x^{2}}.$ (6.104)

An easy computation in $x$-space [2] gives the textbook answer for the bubble:

$\displaystyle\int
G_{m_{1}}(x)G_{m_{2}}(x)dx=\frac{\Gamma\left(1-{\frac{d}{2}}\right)}{(4\pi)^{{\frac{d}{2}}}}\frac{m_{2}^{{d}-2}-m_{1}^{{d}-2}}{m_{1}^{2}-m_{2}^{2}}.$
(6.105)

which, when $d=2$, reduces to

$I(m_{1},m_{2},2)=\frac{\log(m_{1})-\log(m_{2})}{2\pi(m_{1}^{2}-m_{2}^{2})}.$
(6.106)

To study the flat limit of Eq. (3.46), we have to restore a generic de Sitter
radius $R$ and rescale the masses accordingly. When $m>0$ and
$R\rightarrow+\infty$, by using the Stirling formula

$\left|\frac{\Gamma\left(\frac{d-1}{2}+iRm\right)}{\Gamma\left({\frac{1}{2}}+iRm\right)}\right|^{2}\sim(Rm)^{d-2}\
.$ (6.107)

Hence,

$I(Rm_{1},Rm_{2},d)\sim-R^{d-4}\frac{\Gamma\left(1-{\frac{d}{2}}\right)(m_{1}^{d-2}-m_{2}^{d-2})}{2^{d}\pi^{\frac{d}{2}}(m_{2}^{2}-m_{1}^{2})}\
\ \ (R\rightarrow+\infty)\ .$ (6.108)

This fits exactly with (6.105).

Once more, we would like to draw the attention of the reader on the non
triviality of this result as it expresses the flat limit of an integrated
quantity as opposed to the asymptotic properties of the correlation functions
which are merely local properties.

Even more interestingly, the same nontrivial result holds true for the
effective potential. To show it, we need to compute the behaviour of $B(\nu)$
for $\nu\to\infty$. This is obtained from (5.94) by noticing that the integral
in the first line goes to zero and using that

$\displaystyle\int_{0}^{\infty}\frac{t^{k}dt}{e^{2\pi
t}-1}=\frac{k!}{(2\pi)^{k+1}}\zeta(k+1).$ (6.109)

Therefore,

$\displaystyle
B\left(\nu^{2}+\frac{1}{4}\right)=\frac{1}{3}\nu^{2}-\frac{1}{15}\log\left(\nu^{2}+\frac{1}{4}\right)+\frac{1}{12}-\frac{3\zeta(3)}{4\pi^{3}}+\ldots,$
(6.110)

where the dots stay for terms vanishing for $\nu\to\infty$.
On the other hand, for $|\arg(x)|\leq\pi-\delta$, $\delta>0$, one has

$\displaystyle\psi(x)=\log
x-\frac{1}{2x}-\sum_{j=1}^{m}\frac{B_{2j}}{2j}\frac{1}{x^{2j}}+O\left(\frac{1}{x^{2m+2}}\right),$
(6.111)

where $B_{j}$ are the Bernoulli numbers. Since $B_{2}=\frac{1}{6}$ and
$B_{4}=-\frac{1}{30}$, it follows that

$\displaystyle\frac{1}{2}\left(\nu^{2}+\frac{1}{4}\right)^{2}\left(\psi\left(-\frac{1}{2}-i\nu\right)+\psi\left(-\frac{1}{2}+i\nu\right)\right)=$
$\displaystyle\frac{1}{2}\left(\nu^{2}+\frac{1}{4}\right)^{2}\log\left(\nu^{2}+\frac{1}{4}\right)$
(6.112) $\displaystyle+\frac{1}{3}\nu^{2}+\frac{1}{20}+\ldots.$ (6.113)

In this limit we have

$\displaystyle{\cal V}_{R}=$
$\displaystyle\frac{m_{R}^{2}}{2}\varphi_{R}^{2}+\frac{c_{g}}{4}\varphi_{R}^{4}+\frac{1}{32\pi^{2}}\Bigg{[}\frac{(m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2)^{2}}{2}\log\frac{m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2}{\mu_{R}^{2}}$
(6.114)
$\displaystyle+\frac{2}{3}(m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2)-\frac{1}{15}\log(m_{R}^{2}+3c_{g}\varphi_{R}^{2}-2)-\frac{1}{30}-\frac{3\zeta(3)}{4\pi^{3}}+\ldots\Bigg{]}$
(6.115)
$\displaystyle+\frac{N-1}{32\pi^{2}}\Bigg{[}\frac{(m_{R}^{2}+c_{g}\varphi_{R}^{2}-2)^{2}}{2}\log\frac{m_{R}^{2}+c_{g}\varphi_{R}^{2}-2}{\mu_{R}^{2}}$
(6.116)
$\displaystyle+\frac{2}{3}(m_{R}^{2}+c_{g}\varphi_{R}^{2}-2)-\frac{1}{15}\log(m_{R}^{2}+c_{g}\varphi_{R}^{2}-2)-\frac{1}{30}-\frac{3\zeta(3)}{4\pi^{3}}+\ldots\Bigg{]}.$
(6.117)

The potential in the flat limit is defined by ${\cal
V}=\lim_{R\to\infty}R^{-4}{\cal V}_{R}$ and gives exactly (4.79). For very
large but finite $R$, we can write

$\displaystyle{\cal V}_{\Lambda}=$
$\displaystyle\frac{m^{2}}{2}\varphi^{2}+\frac{c_{g}}{4}\varphi^{4}+\frac{1}{64\pi^{2}}\Bigg{[}(m^{2}+3c_{g}\varphi^{2})^{2}\log\frac{m^{2}+3c_{g}\varphi^{2}}{\mu^{2}}$
(6.118)
$\displaystyle+(N-1)(m^{2}+c_{g}\varphi^{2})^{2}\log\frac{m^{2}+c_{g}\varphi^{2}}{\mu^{2}}\Bigg{]}$
(6.119)
$\displaystyle-\frac{\Lambda}{48\pi^{2}}\Bigg{[}(m^{2}+3c_{g}\varphi^{2})\Bigg{(}\log\frac{m^{2}+3c_{g}\varphi^{2}}{\mu^{2}}+\frac{1}{6}\Bigg{)}$
(6.120) $\displaystyle\phantom{-\frac{\Lambda}{48\pi^{2}}\Bigg{[}}\
+(N-1)(m^{2}+c_{g}\varphi^{2})\Bigg{(}\log\frac{m^{2}+c_{g}\varphi^{2}}{\mu^{2}}+\frac{1}{6}\Bigg{)}\Bigg{]}$
(6.121)
$\displaystyle+\frac{\Lambda^{2}}{144\pi^{2}}\Bigg{[}-\frac{1}{30}\log\frac{3(m^{2}+3c_{g}\varphi^{2})}{\Lambda}+\log\frac{m^{2}+3c_{g}\varphi^{2}}{\mu^{2}}+\frac{49}{60}-\frac{3\zeta(3)}{4\pi^{3}}$
(6.122)
$\displaystyle+(N-1)\Bigg{(}-\frac{1}{30}\log\frac{3(m^{2}+c_{g}\varphi^{2})}{\Lambda}+\log\frac{m^{2}+c_{g}\varphi^{2}}{\mu^{2}}+\frac{49}{60}-\frac{3\zeta(3)}{4\pi^{3}}\Bigg{)}\Bigg{]}$
(6.123) $\displaystyle+O(\Lambda^{3}),$ (6.124)

where $\Lambda=3R^{-2}$ is the cosmological constant. The first two lines are
the standard Coleman-Weinberg effective potential [17, 9], while the remaining
terms provide the perturbative corrections in the cosmological constant, up to
order two. The exact expression is given by (5.98), which in terms of the full
dimensional quantities is

$\displaystyle{\cal V}_{\Lambda}=$
$\displaystyle\frac{m^{2}}{2}\varphi^{2}+\frac{c_{g}}{4}\varphi^{4}+\frac{1}{64\pi^{2}}(m^{2}+3c_{g}\varphi^{2}-\frac{2}{3}\Lambda)^{2}\left(\psi\left(-\frac{1}{2}-i\nu_{0}\right)+\psi\left(-\frac{1}{2}+i\nu_{0}\right)\right)$
(6.125)
$\displaystyle+\frac{\Lambda^{2}}{288\pi^{2}}B\left(3\frac{m^{2}}{\Lambda}+9c_{g}\frac{\varphi^{2}}{\Lambda}-2\right)+\frac{(N-1)\Lambda^{2}}{288\pi^{2}}B\left(3\frac{m^{2}}{\Lambda}+3c_{g}\frac{\varphi^{2}}{\Lambda}-2\right)$
(6.126)
$\displaystyle+\frac{N-1}{64\pi^{2}}(m^{2}+c_{g}\varphi^{2}-\frac{2}{3}\Lambda)^{2}\left(\psi\left(-\frac{1}{2}-i\nu_{1}\right)+\psi\left(-\frac{1}{2}+i\nu_{1}\right)\right)$
(6.127)
$\displaystyle-\frac{1}{64\pi^{2}}\left[(m^{2}+3c_{g}\varphi^{2}-\frac{2}{3}\Lambda)^{2}+(N-1)(m^{2}+c_{g}\varphi^{2}-\frac{2}{3}\Lambda)^{2}\right]\log\frac{3\mu^{2}}{\Lambda},$
(6.128)

with

$\displaystyle\nu_{0}=$ $\displaystyle
3\sqrt{\frac{m^{2}}{3\Lambda}+c_{g}\frac{\varphi^{2}}{\Lambda}-\frac{1}{4}},$
(6.129) $\displaystyle\nu_{1}=$ $\displaystyle
3\sqrt{\frac{m^{2}}{3\Lambda}+c_{g}\frac{\varphi^{2}}{3\Lambda}-\frac{1}{4}}.$
(6.130)

It is interesting to compare this result with the one for anti de Sitter,
[18]. In that case, the perturbative expansion in the cosmological constant
$\Lambda$ of the effective potential contains also terms of order
$\sqrt{|\Lambda|}$. This is strictly related to the invariance of the Legendre
functions of the first kind $P^{\mu}_{-\frac{1}{2}-\nu}$ under the
$\nu\to-\nu$, which is not true for the functions of the second kind.

##  7 2-loop: the watermelon

For the sake of comparison let us recall at first the flat space formula for
the watermelon:

$\displaystyle I_{3}(m_{1},m_{2},m_{3},d)$
$\displaystyle=-2^{1-2d}\pi^{1-d}\Gamma(2-d)(-S(m_{1},m_{2},m_{3}))^{\frac{d-3}{2}}$
(7.131)
$\displaystyle+\frac{(m_{1}m_{2})^{d-4}\left(m_{1}^{2}+m_{2}^{2}-m_{3}^{2}\right)\,_{2}F_{1}\left(1,2-\frac{d}{2};\frac{3}{2};M_{123}^{2}\right)}{4^{d}\pi^{d-2}(\cos(\pi
d)-1)\Gamma\left(\frac{d}{2}-1\right)\Gamma\left(\frac{d}{2}\right)}$ (7.132)
$\displaystyle+\frac{(m_{2}m_{3})^{d-4}\left(-m_{1}^{2}+m_{2}^{2}+m_{3}^{2}\right)\,_{2}F_{1}\left(1,2-\frac{d}{2};\frac{3}{2};M_{231}^{2}\right)}{4^{d}\pi^{d-2}(\cos(\pi
d)-1)\Gamma\left(\frac{d}{2}-1\right)\Gamma\left(\frac{d}{2}\right)}$ (7.133)
$\displaystyle+\frac{(m_{1}m_{3})^{d-4}\left(m_{1}^{2}-m_{2}^{2}+m_{3}^{2}\right)\,_{2}F_{1}\left(1,2-\frac{d}{2};\frac{3}{2};{M_{312}^{2}}\right)}{4^{d}\pi^{d-2}(\cos(\pi
d)-1)\Gamma\left(\frac{d}{2}-1\right)\Gamma\left(\frac{d}{2}\right)}.$ (7.134)
(7.135)

where

$M_{ijk}=\left(\frac{m_{i}^{2}+m_{j}^{2}-m_{k}^{2}}{2m_{i}m_{j}}\right)$
(7.136)

and

$S(m_{1},m_{2},m_{3})=m_{1}^{4}+m_{2}^{4}+m_{3}^{4}-2m_{1}^{2}m_{2}^{2}-2m_{1}^{2}m_{3}^{2}-2m_{2}^{2}m_{3}^{2}$
(7.137)

is the Symanzik polynomial. The above formula is valid when no mass is bigger
than the sum of the other two; this happens if and only if the Symanzik
polynomial is negative. It may be obtained by solving an appropriate
differential equation for the diagram [9] or else by a direct calculation in
position space which is indeed the easier and shorter way to get it [2]. A
similar formula holds for $S>0$ [2].

As regards de Sitter sphere the situation is trickier. At the moment there is
no substitute for the method of differential equations which seems to be well-
adapted only to flat space but useless otherwise. We are left with $x$-space
calculations.

Faced directly, the integral at the rhs of (2.36) is significantly more
challenging than the already difficult (2.39) and at the moment the only way
we found to evaluate it makes indeed use of Eq. (2.39): by singling out one of
the masses, say $\nu_{3}$, we get an integral representation of the two-loop
watermelon as a superposition of one-loop diagrams:

$\displaystyle I_{3}(\nu_{1},\nu_{2},\nu_{3},d)=\int G_{\nu_{1}}(x_{0}\cdot
x)G_{\nu_{2}}(x_{0}\cdot x)G_{\nu_{3}}(x_{0}\cdot x)\sqrt{g}dx=$ (7.138)
$\displaystyle=\int\rho(\nu_{1},\nu_{2},\kappa)I_{2}(\kappa,\nu_{3},d)\kappa
d\kappa={I}_{3}^{(1)}(\nu_{1},\nu_{2},\nu_{3},d)-I_{3}^{(2)}(\nu_{1},\nu_{2},\nu_{3},d).$
(7.139)

where

$\displaystyle{I}_{3}^{(1)}(\nu_{1},\nu_{2},\nu_{3},d)=\frac{\left(\frac{\Gamma\left(\frac{d-1}{2}-i\nu_{3}\right)\Gamma\left(\frac{d-1}{2}+i\nu_{3}\right)}{\Gamma\left(\frac{1}{2}-i\nu_{3}\right)\Gamma\left(\frac{1}{2}+i\nu_{3}\right)}\right)}{2^{d}\pi^{\frac{d-1}{2}}\Gamma\left(\frac{d-1}{2}\right)}\frac{\Gamma\left(1-\frac{d}{2}\right)}{(4\pi)^{\frac{d}{2}}}{\cal
A}(\nu_{1},\nu_{2},\nu_{3},d)$ (7.140) $\displaystyle{\cal
A}(\nu_{1},\nu_{2},\nu_{3},d)=\int\frac{\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}}\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon^{\prime}\nu_{1}+i\epsilon^{\prime}\nu_{2}+i\epsilon^{\prime\prime}\kappa}{2}\right)}{\prod_{\epsilon}\Gamma\left(\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{1}{2}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d+1}{4}+\frac{i\epsilon\kappa}{2}\right)}\frac{d\kappa}{(\kappa^{2}-\nu_{3}^{2})}\
$ (7.141) (7.142)

and

$\displaystyle{I}_{3}^{(2)}(\nu_{1},\nu_{2},\nu_{3},d)=\frac{1}{2^{d}\pi^{\frac{d-1}{2}}\Gamma\left(\frac{d-1}{2}\right)}\frac{\Gamma\left(1-\frac{d}{2}\right)}{(4\pi)^{\frac{d}{2}}}{\cal
B}(\nu_{1},\nu_{2},\nu_{3})$ (7.143) $\displaystyle{\cal
B}(\nu_{1},\nu_{2},\nu_{3})=\int\frac{\frac{{\Gamma\left(\frac{d-1}{2}-i\kappa\right)\Gamma\left(\frac{d-1}{2}+i\kappa\right)}}{{\Gamma\left(\frac{1}{2}-i\kappa\right)\Gamma\left(\frac{1}{2}+i\kappa\right)}}\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}}\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon^{\prime}\nu_{1}+i\epsilon^{\prime}\nu_{2}+i\epsilon^{\prime\prime}\kappa}{2}\right)}{\prod_{\epsilon}\Gamma\left(\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{1}{2}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d-1}{4}+\frac{i\epsilon\kappa}{2}\right)\Gamma\left(\frac{d+1}{4}+\frac{i\epsilon\kappa}{2}\right)}\frac{d\kappa}{(\kappa^{2}-\nu_{3}^{2})}.$
(7.144)

${I}_{3}^{(1)}(\nu_{1},\nu_{2},\nu_{3},d)$ and
${I}_{3}^{(2)}(\nu_{1},\nu_{2},\nu_{3},d)$ are symmetric only in the exchange
of $\nu_{1}$ and $\nu_{2}$. To work with symmetric expressions we might
replace them by their totally symmetric counterparts but this is not really
helpful.

### First term

Let us consider the first term ${I}_{3}^{(1)}(\nu_{1},\nu_{2},\nu_{3},d)$; the
change of variables555In the following we will use either
$(\nu_{1},\nu_{2},\nu_{3})$ or $(x,y,w)$ and either $d$ or $\delta(d)$
interchangeably.

$s=\frac{i\kappa}{2},\ \ \ \delta=\frac{d-1}{4},\ \ \ x=\frac{i\nu_{1}}{2},\ \
\ y=\frac{i\nu_{2}}{2},\ \ \ \ w=\frac{i\nu_{3}}{2},\ \ \ \ u=x+y$ (7.146)

allows to rewrite the integral as follows:

$\displaystyle{\cal A}(x,y,w,d)=\int_{-i\infty}^{i\infty}\frac{ds}{2i}\
\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm}\frac{{\
\Gamma\left(\epsilon
s+\delta+\epsilon^{\prime}x+i\epsilon^{\prime\prime}y\right)\Gamma\left(\epsilon
s+w\right)}}{\Gamma\left(\epsilon s\right)\Gamma\left(\frac{1}{2}+\epsilon
s\right)\Gamma\left(\delta+\epsilon
s\right)\Gamma\left(\delta+\frac{1}{2}+\epsilon s\right)\Gamma\left(\epsilon
s+w+1\right)}.$ (7.147)

There are poles on the integration path at

$\displaystyle s=-\delta+\frac{\pm i\nu_{1}\pm i\nu_{2}}{2}-n,\ \ \ \ \
s=-\frac{i\nu_{3}}{2}-n,$ (7.148) $\displaystyle s=-\frac{i\nu_{3}}{2}+n,\ \ \
\ \ \ s=\delta+\frac{\pm i\nu\pm i\lambda}{2}+n.$ (7.149)

By integrating along the imaginary axis (with suitable indentations) the
result may be expressed as a linear combination of four hypergeometric
functions

${I}_{3}^{(1)}(x,y,z,d)=\sum_{\epsilon,\epsilon^{\prime}=\pm 1}A_{d}(\epsilon
x,\epsilon^{\prime}y,w)$ (7.150)

where

$\displaystyle A_{d}(x,y,w)=a_{d}(x,y,w)\times$ (7.151)
$\,{}_{9}F_{8}\left(\begin{array}[]{l}{2\delta,u+\frac{1}{2},u+1,\delta+u+1,2\delta+2x,2\delta+2y,2\delta+2u,\delta-w+u,\delta+w+u}\cr{2x+1,2y+1,2u+1,\delta+u,2\delta+u,2\delta+u+\frac{1}{2}},\delta-w+u+1,\delta+w+u+1\end{array};1\right),$
(7.154) $\displaystyle
a_{d}(x,y,w)=-\frac{\scriptstyle{4^{-2\delta-3}\pi^{-4\delta-\frac{1}{2}}\Gamma\left(\frac{1}{2}-2\delta\right)\cos(2\pi
w)\Gamma(2\delta-2w)\Gamma(2w+2\delta)\Gamma(2x+2\delta)\Gamma(2y+2\delta)}}{\scriptstyle{\sin(2\pi
x)\sin(2\pi
y)\Gamma(2x+1)\Gamma(2y+1)\left(w^{2}-(\delta+u)^{2}\right)\Gamma(-2u-2\delta)\Gamma(2u+4\delta)}}.$
(7.156)

The hypergeometric series in (LABEL:ad) converges absolutely for
$3-4\delta=4-d>0$.

It is possible to proceed to a simplification of the above expressions by
observing that in each term of the hypergeometric series (LABEL:ad) the
following product of Pochhammers reduces nicely to a rational function:

$\displaystyle\frac{(u+\delta+1)_{n}(u-w+\delta)_{n}(u+w+\delta)_{n}}{(u+\delta)_{n}(u-w+\delta+1)_{n}(u+w+\delta+1)_{n}}=\frac{((\delta+u)^{2}-w^{2})\left(\frac{1}{\delta+n+u+w}+\frac{1}{\delta+n+u-w}\right)}{2(\delta+u)}.$
(7.157)

We obtain in this way a formula consisting of eight terms ${}_{7}F_{6}$:

${I}_{3}^{(1)}(x,y,z,d)=\sum_{\epsilon,\epsilon^{\prime}=\pm
1}A^{\prime}_{d}(\epsilon x,\epsilon^{\prime}y,\epsilon^{\prime\prime}w)$
(7.158)

where

$\displaystyle A^{\prime}_{d}(x,y,w)=a^{\prime}_{d}(x,y,w)\times$ (7.159)
$\,{}_{7}F_{6}\left(\begin{array}[]{l}2\delta,u+\frac{1}{2},u+1,\delta+u-w,2\delta+2u,2\delta+2x,2\delta+2y\cr
2u+1,2x+1,2y+1,\delta+u-w+1,2\delta+u,2\delta+u+\frac{1}{2}\end{array};1\right)$
(7.162) $\displaystyle a^{\prime}_{d}(x,y,w)=$ (7.164)
$\displaystyle=\frac{2^{-4\delta-7}\pi^{-4\delta-\frac{1}{2}}\Gamma\left(\frac{1}{2}-2\delta\right)\cos(2\pi
w)\Gamma(2\delta-2w)\Gamma(2w+2\delta)\Gamma(2x+2\delta)\Gamma(2y+2\delta)}{\sin(2\pi
x)\sin(2\pi
y)(\delta+u)\Gamma(2x+1)\Gamma(2y+1)\Gamma(-2u-2\delta)\Gamma(2u+4\delta)(\delta+u-w)}.$
(7.165)

This simplification has not worsened the convergence: (LABEL:add) converges
absolutely under the same conditions: $4\delta-3=-d+4>0$.

### Second term

As regards the second term, we have

$\displaystyle{\cal B}(x,y,w,d)$ $\displaystyle=$
$\displaystyle\int\frac{ds}{2i}\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm
1}\frac{2^{d-2}\Gamma\left(\epsilon s+w\right)\Gamma\left(\epsilon
s+\delta+\epsilon^{\prime}x+\epsilon^{\prime\prime}y\right)}{\Gamma\left(\epsilon
s\right)\Gamma\left(\epsilon s+\frac{1}{4}\right)\Gamma\left(\epsilon
s+\frac{1}{2}\right)\Gamma\left(\epsilon
s+\frac{3}{4}\right)\Gamma\left(\epsilon s+w+1\right)}$ (7.167)

Again, the result may be expressed as a linear combination of four
hypergeometric functions:

$\displaystyle I_{3}^{(2)}=\sum_{\epsilon,\epsilon^{\prime}=\pm
1}B_{d}(\epsilon x,\epsilon^{\prime}y,w)$ (7.169)

where

$\displaystyle B_{d}(x,y,w)=b_{d}(x,y,w)$ (7.170)
$\displaystyle\times\,_{7}F_{6}\left(\begin{array}[]{c}2\delta,\delta+u+1,\delta-w+u,\delta+w+u,2\delta+2x,2\delta+2y,2\delta+2u\cr
2x+1,2y+1,2x+2y+1,\delta+u,\delta-w+u+1,\delta+w+u+1\end{array};1\right)$
(7.173) $\displaystyle
b_{d}(x,y,w)=\frac{2^{-8\delta-4u-7}\pi^{1-4\delta}\Gamma\left(\frac{1}{2}-2\delta\right)\csc(2\pi
x)\csc(2\pi y)\csc(2\pi
u)\Gamma(2x+2\delta)\Gamma(2y+2\delta)}{\left((\delta+u)^{2}-w^{2}\right)\Gamma(2x+1)\Gamma(2y+1)\Gamma(2u+1)\Gamma(-4u-4\delta)\Gamma\left(2u+2\delta+\frac{1}{2}\right)}.$
(7.175)

The hypergeometric series in (7.173) converges absolutely for
$4-8\delta=2(3-d)>0.$ Proceeding exactly as before we may find a
simplification of the above formula:

${I}_{3}^{(2)}(x,y,z,d)=\sum_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm
1}B^{\prime}_{d}(\epsilon x,\epsilon^{\prime}y,\epsilon^{\prime\prime}w)$
(7.177)

where

$\displaystyle
B^{\prime}_{d}(x,y,w)=b^{\prime}_{d}(x,y,w)\,\,_{5}F_{4}\left(\begin{array}[]{l}2\delta,\delta+u+w,2\delta+2u,2\delta+2x,2\delta+2y\cr
2u+1,2x+1,2y+1,\delta+u+w+1\end{array};1\right)$ (7.180) $\displaystyle
b^{\prime}_{d}(x,y,w)=-\frac{2^{-6-4u-8\delta}\pi^{1-4\delta}\csc(2\pi
x)\csc(2\pi y)\csc(2\pi
u)\Gamma\left(\frac{1}{2}-2\delta\right)\Gamma(2x+2\delta)\Gamma(2y+2\delta)}{(w+u+\delta)\Gamma(1+2x)\Gamma(1+2y)\Gamma(1+2u)\Gamma(1-4u-4\delta)\Gamma\left(\frac{1}{2}+2u+2\delta\right)}.$
(7.182)

Again, the simplification has not worsened the convergence: (7.180) converges
absolutely for $4-8\delta=2(3-d)>0$.

In conclusion, ${I}_{3}^{(1)}$ is the most regular of the two terms; it starts
diverging at $d=4$. The second term diverges already at $d=3$. This means that
no compensation among the two terms may be expected. The only exception is at
$d=2$: in that case the watermelon is finite; the two terms have to compensate
each other to render the divergence in the coefficient
$\Gamma\left(1-\frac{d}{2}\right)$ harmless.

##  8 Odd dimensions $d\leq 1$

In odd integer dimension $d\leq 1$ the result simplifies a great deal because
the hypergeometric series reduce to finite sums. Let us examine the simplest
case $d=1$; our general formulae give

$\displaystyle I^{(1)}_{3}(\nu_{1},\nu_{2},\nu_{3},1)$ $\displaystyle=$
$\displaystyle\frac{\coth(\pi\nu_{1})\coth(\pi\nu_{3})\left(\nu_{2}^{2}-\nu_{1}^{2}-\nu_{3}^{2}\right)}{4\nu_{1}\nu_{3}(\nu_{1}-\nu_{2}-\nu_{3})(\nu_{1}+\nu_{2}-\nu_{3})(\nu_{1}-\nu_{2}+\nu_{3})(\nu_{1}+\nu_{2}+\nu_{3})}+$
(8.184) $\displaystyle+$
$\displaystyle\frac{\coth(\pi\nu_{2})\coth(\pi\nu_{3})\left(\nu_{1}^{2}-\nu_{2}^{2}-\nu_{3}^{2}\right)}{4\nu_{2}\nu_{3}(\nu_{1}-\nu_{2}-\nu_{3})(\nu_{1}+\nu_{2}-\nu_{3})(\nu_{1}-\nu_{2}+\nu_{3})(\nu_{1}+\nu_{2}+\nu_{3})}$
(8.185)

and

$\displaystyle I_{3}^{(2)}(\nu_{1},\nu_{2},\nu_{3},1)$ $\displaystyle=$
$\displaystyle\frac{-\coth(\pi\nu_{1})\coth(\pi\nu_{2})\left(\nu_{3}^{2}-\nu_{1}^{2}-\nu_{2}^{2}\right)}{4\nu_{1}\nu_{2}(\nu_{1}-\nu_{2}-\nu_{3})(\nu_{1}+\nu_{2}-\nu_{3})(\nu_{1}-\nu_{2}+\nu_{3})(\nu_{1}+\nu_{2}+\nu_{3})}$
(8.187) $\displaystyle-$
$\displaystyle\frac{1}{2(\nu_{1}-\nu_{2}-\nu_{3})(\nu_{1}+\nu_{2}-\nu_{3})(\nu_{1}-\nu_{2}+\nu_{3})(\nu_{1}+\nu_{2}+\nu_{3})}.$
(8.188)

$I_{3}^{(2)}(\nu_{1},\nu_{2},\nu_{3},1)$ contains a term which is totally
symmetric w.r.t. the mass parameters and a term which is symmetric only w.r.t.
the exchange of $\nu_{1}$ and $\nu_{2}$.

Subtracting $I_{3}^{(2)}$ from $I_{3}^{(1)}$ reestablishes the global symmetry
of the diagram w.r.t. the three mass parameters. Since $I_{3}^{(1)}$ do not
include a totally symmetric term it can be fully deduced from the knowledge of
$I_{3}^{(2)}$.

The above situation is generic in odd negative dimension: the first addendum
contains the terms proportional to $\coth\pi\nu_{3}$ and and the second term
contains a totally symmetric contribution $S(\nu_{1},\nu_{2},\nu_{3},d)$ which
is just a product of poles.

Both the symmetric and the non symmetric contributions admit expansions in
partial fractions. The totally symmetric term has indeed a very simple
expression:

$\displaystyle S(\nu_{1},\nu_{2},\nu_{3},d)$ $\displaystyle=$
$\displaystyle\frac{\cos({2\pi
i\delta})\pi^{-4\delta}\Gamma(1-4\delta)}{32}\prod_{\epsilon,\epsilon^{\prime}=\pm}\frac{\Gamma\left(\delta+\frac{i\nu_{1}}{2}+\frac{i\epsilon\nu_{2}}{2}+\frac{i\epsilon^{\prime}\nu_{3}}{2}\right)}{\Gamma\left(1-\delta+\frac{i\nu_{1}}{2}+\frac{i\epsilon\nu_{2}}{2}+\frac{i\epsilon^{\prime}\nu_{3}}{2}\right)};$
(8.190)

it amounts to a product of simple poles. The non-symmetric term is also
proportional to a product of simple poles but the polynomial at the numerator
becomes more and more cumbersome as the dimension grows (negative). A
straightforward derivation of $I_{3}(\nu_{1},\nu_{2},\nu_{3},1)$ by direct
integration of Eq. (7.138) points towards a fully symmetric expression: in
spacetime dimension $d=1$ the Schwinger function is simply

$\displaystyle G^{1}_{\lambda}(-\cos
s)=\frac{{\Gamma(-i\nu)\Gamma(i\nu)}\mathop{\rm
ch}\nolimits\left(\lambda(\pi-s)\right)}{2\pi}$ (8.191)

and therefore

$\displaystyle
I(\nu_{1},\nu_{2},\nu_{3},1)=2\int_{0}^{\pi}G^{1}_{\nu_{1}}(-\cos
s)G^{1}_{\nu_{2}}(-\cos s)G^{1}_{\nu_{3}}(-\cos s)ds$ (8.192)
$\displaystyle=\frac{\frac{\mathop{\rm
sh}\nolimits(\pi(\nu_{1}+\nu_{2}+\nu_{3}))}{\nu_{1}+\nu_{2}+\nu_{3}}+\frac{\mathop{\rm
sh}\nolimits(\pi(\nu_{1}+\nu_{2}-\nu_{3}))}{\nu_{1}+\nu_{2}-\nu_{3}}+\frac{\mathop{\rm
sh}\nolimits(\pi(\nu_{1}-\nu_{2}+\nu_{3}))}{\nu_{1}-\nu_{2}+\nu_{3}}+\frac{\mathop{\rm
sh}\nolimits(\pi(\nu_{1}-\nu_{2}-\nu_{3}))}{\nu_{1}-\nu_{2}-\nu_{3}}}{16\,\nu_{1}\nu_{2}\nu_{3}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}.$ (8.193)

In the special case when all the masses are equal the formula reduces to

$\displaystyle I(\nu,\nu,\nu,1)=\frac{5+\mathop{\rm
ch}\nolimits(2\pi\nu)}{24\nu^{4}\mathop{\rm sh}\nolimits(\pi\nu)^{2}}.$
(8.194)

Studying the case of three particles with the same mass ab initio is not
easier; it only hides the underlying beautiful structure.

Rescaling the masses $\nu\to\nu R$ and taking the limit $R\to\infty$ gives the
correct result in flat space, as obtained by taking $d=1$ in Eq. (7.135):

$I(m_{1},m_{2},m_{3},1)=\frac{1}{4m_{1}m_{2}m_{3}(m_{1}+m_{2}+m_{3})};$
(8.195)

we stress again that this result is non trivial, being the flat limit of an
integrated quantity and not just the asymptotic behaviour of a correlation
function close to a given point.

Also the above symmetric structure in four terms is generic in odd dimension
$d\leq 1$:

$\displaystyle
I(\nu_{1},\nu_{2},\nu_{3},d)=\sum_{\epsilon,\epsilon^{\prime}=\pm}\frac{\mathop{\rm
sh}\nolimits(\pi(\nu_{1}+\epsilon\nu_{2}+\epsilon^{\prime}\nu_{3}))P_{d}(\nu_{1},\epsilon\nu_{2},\epsilon^{\prime}\nu_{3})}{\nu_{1}\nu_{2}\nu_{3}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})Q_{d}(\nu_{1},\epsilon\nu_{2},\epsilon^{\prime}\nu_{3})}$
(8.196)

where $P_{d}$ and $Q_{d}$ are polynomials. The polynomial at the denominator
is always a product of monomials and the result admits a simple partial
fraction expansion. We will not elabotrate further on this point.

##  9 Spacetime dimension $d=2$

In dimension $d=2$ the watermelon is finite. However, as for the 1-loop
diagram (see Eq. (3.55)) the two terms in Eq. (7.139) are multiplied by the
coefficient $\Gamma\left(1-\frac{d}{2}\right)$ that becomes singular at $d=2$
and a limit procedure is needed. Only in this particular two-dimensional case
${I}_{3}^{(1)}$ and ${I}_{3}^{(2)}$ compensate each other.

To extract the finite result we make use of the elementary identity

$\displaystyle I_{3}(x,y,z,2)=\left.\frac{\partial}{\partial
d}\left((d-2)I_{3}(x,y,z,d)\right)\right|_{d=2}=$ (9.197) (9.198)
$\displaystyle=\left.\partial_{d}\left((d-2)\sum_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm}(A^{\prime}_{d}(\epsilon
x,\epsilon^{\prime}y,\epsilon^{\prime\prime}w)-B^{\prime}_{d}(\epsilon
x,\epsilon^{\prime}y,\epsilon^{\prime\prime}w))\right)\right|_{d=2}.$ (9.199)

It is useful to rewrite the typical term in the above expression as follows:

$\displaystyle
A^{\prime}_{d}(x,y,w)-B^{\prime}_{d}(x,y,w)=a^{\prime}_{d}(x,y,w)\left(\,{}_{7}F_{6}\left(\ldots;1\right)-\frac{b^{\prime}_{d}(x,y,w)}{a^{\prime}_{d}(x,y,w)}\,_{5}F_{4}\left(\ldots;1\right)\right).$
(9.200)

The term in parentheses at the r.h.s. tends to zero when $d\to 2$. This is
precisely the compensation we were expecting. Taking the limit we get

$\displaystyle
T(x,y,w)=\left.\partial_{d}\left((d-2)(A^{\prime}_{d}(x,y,w)-B^{\prime}_{d}(x,y,w))\right)\right|_{d=2}$
(9.201) $\displaystyle=c_{2}(x,y,w)\left.\left(\frac{\partial}{\partial
d}\,{}_{7}F_{6}\left(\ldots;1\right)-\frac{\partial}{\partial
d}\,{}_{5}F_{4}\left(\ldots;1\right)-{}_{5}F_{4}\left(\ldots;1\right)\
\frac{\partial}{\partial
d}\frac{b^{\prime}_{d}(x,y,w)}{a^{\prime}_{d}(x,y,w)}\right)\right|_{d=2}$
(9.202)

where $c_{2}(x,y,w)=\lim_{d\to 2}\ (d-2)a^{\prime}_{d}(x,y,w).$ All the terms
together provide the following finite expression:

$\displaystyle
I_{3}(x,y,z,2)=\sum_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm}T(\epsilon
x,\epsilon^{\prime}y,\epsilon^{\prime\prime}w)$ (9.203)

where

$\displaystyle
T(x,y,w)=\frac{\left(\pi\cot(2\pi(x+y))-\psi\left(\frac{1}{2}-2w\right)-\psi\left(\frac{1}{2}+2w\right)+2\psi(1+2x+2y)\right)}{32\sqrt{\pi}\sin(2\pi
x)\sin(2\pi y)\Gamma\left(\frac{1}{2}-2x-2y\right)}$ (9.204)
$\displaystyle\times\frac{\Gamma\left(\frac{1}{2}+2x\right)\Gamma\left(\frac{1}{2}+2y\right)\,_{5}F_{4}\left(\begin{array}[]{l}\frac{1}{2}+2x,\frac{1}{4}+w+x+y,\frac{1}{2}+2y,\frac{1}{2}+2x+2y,\frac{1}{2}\cr
1+2x,\frac{5}{4}+w+x+y,1+2y,1+2x+2y\end{array};1\right)}{(1+4w+4x+4y)\Gamma(1+2x)\Gamma(1+2y)\Gamma(1+2x+2y)}+$
(9.207)
$\displaystyle-\frac{\Gamma(-2x)\Gamma\left(\frac{1}{2}+2x\right)\Gamma(-2y)\Gamma\left(\frac{1}{2}+2y\right)}{32\pi^{5/2}(1+4w+4x+4y)\Gamma\left(\frac{1}{2}-2x-2y\right)\Gamma(1+2x+2y)}\times$
(9.208)
$\displaystyle\times\left(\,{}_{6}\ddot{F}_{5}\left(\begin{array}[]{l}\frac{1}{2}+x+y,\frac{1}{2}+2x,\frac{1}{4}+w+x+y,\frac{1}{2}+2y,\frac{1}{2}+2x+2y,\frac{1}{2}\cr\frac{1}{2}+x+y,1+2x,\frac{5}{4}+w+x+y,1+2y,1+2x+2y\end{array};1\right)+\right.$
(9.211)
$\displaystyle+\left.\,{}_{6}\ddot{F}_{5}\left(\begin{array}[]{l}1+x+y,\frac{1}{2}+2x,\frac{1}{4}+w+x+y,\frac{1}{2}+2y,\frac{1}{2}+2x+2y,\frac{1}{2}\cr
1+x+y,1+2x,\frac{5}{4}+w+x+y,1+2y,1+2x+2y\end{array};1\right)\right)$ (9.214)

where we introduced the notation

$\displaystyle{}_{p}\dot{F}_{q}\left(\begin{array}[]{l}a,\cdots\cr
b,\cdots\end{array};z\right)=\frac{\partial}{\partial
a}{}_{p}F_{q}\left(\begin{array}[]{l}a,\cdots\cr
b,\cdots\end{array};z\right),\ \ \ \
{}_{p}\ddot{F}_{q}\left(\begin{array}[]{l}a,\cdots\cr
b,\cdots\end{array};z\right)=\frac{\partial}{\partial
b}{}_{p}F_{q}\left(\begin{array}[]{l}a,\cdots\cr
b,\cdots\end{array};z\right).$ (9.224)

While the above result is perfectly finite and usable, there should exist an
underlying simplification, as Eqs. (9.203)-(9.214) should be a one-
parameter666The parameter is the de Sitter radius $R$, which in the formula is
set to 1 deformation of the flat space formula which may simply be written in
terms of dilogarithms [2, 70]. Actually, also in flat space a very natural
construction in position space gives a formula containing derivatives of the
hypergeometric function $\ {}_{2}F_{1}\ $ w.r.t. the parameters [2]. In that
case it is been possible to go further and proceed to dilogarithms which, by
the way, are not more explicit the hypergeometrics. Here the task is much more
difficult and at the moment we are happy with Eq. (9.214).

##  10 Spacetime dimension $d=3$

### Flat space recapitulation

Let us review at first the three-dimensional case in flat space. Only the
first line in formula (7.135) has a pole at $d=3$, namely the simple pole in
the Gamma function. A Laurent expansion near $d\sim 3$ gives

$\displaystyle
I_{3}(m_{1},m_{2},m_{3},d)=-\frac{1}{32\pi^{2}(d-3)}+\frac{2-2\gamma+\log\left(16\pi^{2}\right)}{64\pi^{2}}+$
(10.225) $\displaystyle-\frac{\mathop{\rm
th}\nolimits^{-1}\left(\frac{2m_{1}m_{2}}{m_{1}^{2}+m_{2}^{2}-m_{3}^{2}}\right)+\mathop{\rm
th}\nolimits^{-1}\left(\frac{2m_{1}m_{3}}{m_{1}^{2}-m_{2}^{2}+m_{3}^{2}}\right)+\mathop{\rm
th}\nolimits^{-1}\left(\frac{2m_{2}m_{3}}{-m_{1}^{2}+m_{2}^{2}+m_{3}^{2}}\right)}{32\pi^{2}}+$
(10.226)
$\displaystyle-\frac{\log\left(-m_{1}^{4}-m_{2}^{4}-m_{3}^{4}+2m_{1}^{2}m_{2}^{2}+2m_{1}^{2}m_{3}^{2}+2m_{2}^{2}m_{3}^{2}\right)}{64\pi^{2}}+{\rm
O}(d-3).$ (10.227)

By using the identity $\mathop{\rm
th}\nolimits^{-1}(x)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$ the result
simplifies to

$\displaystyle I_{3}(m_{1},m_{2},m_{3},d)=$
$\displaystyle-\frac{1}{32\pi^{2}(d-3)}-\frac{1}{16\pi^{2}}\log(m_{1}+m_{2}+m_{3})$
(10.228) $\displaystyle+\frac{1-\gamma+\log(4\pi)}{32\pi^{2}}+{\rm O}(d-3).$
(10.229)

Alternatively, we may proceed by an ultraviolet cutoff of the integral
defining the watermelon; in position space we cut a little sphere surrounding
the origin $x=0$:

$\displaystyle I_{3,\Lambda}(m_{1},m_{2},m_{3},3)$
$\displaystyle=\frac{\left({m_{1}m_{2}m_{3}}\right)^{\frac{1}{2}}}{(2\pi)^{\frac{5}{2}}\pi}\int_{\Lambda^{2}}^{\infty}\
r^{\frac{1}{2}}K_{\frac{1}{2}}\left(m_{1}r\right)K_{\frac{1}{2}}\left(m_{2}r\right)K_{\frac{1}{2}}\left(m_{3}r\right)dr$
(10.230)
$\displaystyle=\int_{\Lambda^{2}}^{\infty}\frac{e^{-r(m_{1}+m_{2}+m_{3})}}{16\pi^{2}r}dr=\frac{\Gamma(0,(m_{1}+m_{2}+m_{3})\Lambda^{2})}{16\pi^{2}}.$
(10.231)

Expansion in $\Lambda$ now gives

$\displaystyle I_{3,\Lambda}(m_{1},m_{2},m_{3},3)$
$\displaystyle=-\frac{\log(\Lambda^{4})}{32\pi^{2}}-\frac{1}{16\pi^{2}}{\log(m_{1}+m_{2}+m_{3})}-\frac{\gamma}{16\pi^{2}}.$
(10.232)

We see that the mass dependent finite parts coincide in the two expression. On
the other hand the additive constants depend on the renormalization scheme.

### de Sitter: UV cutoff

At $d=3$ the Schwinger function is elementary:

$G^{3}_{\nu}(-\cos s)=\frac{\mathop{\rm
sh}\nolimits(\nu(\pi-s))}{4\pi\mathop{\rm sh}\nolimits(\pi\nu)\sin s}.$
(10.233)

We may cutoff the integral as in flat space

$\displaystyle
I_{3,K}(\nu_{1},\nu_{2},\nu_{3},d)=4\pi\int_{K}^{\pi}G^{3}_{\nu_{1}}(-\cos
s)G^{3}_{\nu_{2}}(-\cos s)G^{3}_{\nu_{3}}(-\cos s){\sin^{2}s}{ds}=$ (10.234)
$\displaystyle={\int_{K}^{\pi}\frac{\scriptstyle{\mathop{\rm
sh}\nolimits((\pi-s)(\nu_{1}-\nu_{2}-\nu_{3}))-\mathop{\rm
sh}\nolimits((\pi-s)(\nu_{1}+\nu_{2}-\nu_{3}))-\mathop{\rm
sh}\nolimits((\pi-s)(\nu_{1}-\nu_{2}+\nu_{3}))+\mathop{\rm
sh}\nolimits((\pi-s)(\nu_{1}+\nu_{2}+\nu_{3}))}}{{64\pi^{2}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}\sin s}ds}.$ (10.235)

This leads us to consider the following indefinite integral

$\displaystyle F_{a}(s)=\int\frac{\mathop{\rm sh}\nolimits(a(\pi-s))}{\sin
s}{ds}=\frac{1}{2}e^{as-\pi
a+is}\Phi\left(e^{2is},1,\frac{1}{2}-\frac{ia}{2}\right)-\frac{1}{2}\psi\left(\frac{1}{2}-\frac{ia}{2}\right)+$
(10.237) $\displaystyle-\frac{1}{2}e^{-as+\pi
a+is}\Phi\left(e^{2is},1,\frac{1}{2}+\frac{ia}{2}\right)+\frac{1}{2}\psi\left(\frac{1}{2}+\frac{ia}{2}\right).$
(10.238)

Here $\Phi$ is the Lerch transcendent function and we adjusted the primitive
$F_{a}(s)$ so that at the upper bound it vanishes at $s=\pi$. It follows that

$\displaystyle\int_{K}^{\pi}\frac{\mathop{\rm sh}\nolimits(a(\pi-s))}{\sin
s}{ds}$ $\displaystyle=-F_{a}(K)\simeq-\mathop{\rm sh}\nolimits(\pi
a)(\log(2K)+\gamma)+$ (10.239) $\displaystyle-\frac{1}{2}\mathop{\rm
sh}\nolimits(\pi
a)\left(\psi\left(\frac{1}{2}-\frac{ia}{2}\right)+\psi\left(\frac{1}{2}+\frac{ia}{2}\right)\right)+{\rm
O}(K)$ (10.240)

and therefore

$\displaystyle\int_{K}^{\pi}G^{3}_{\nu_{1}}(-\cos s)G^{3}_{\nu_{2}}(-\cos
s)G^{3}_{\nu_{3}}(-\cos s){\sin^{2}s}{ds}=-\frac{\log(2K)+\gamma}{16\pi^{2}}+$
(10.241)
$\displaystyle-\frac{\psi\left(\frac{1}{2}+\frac{i(\nu_{1}-\nu_{2}-\nu_{3})}{2}\right)+\psi\left(\frac{1}{2}-\frac{i(\nu_{1}-\nu_{2}-\nu_{3})}{2}\right)}{128\pi^{2}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}\mathop{\rm
sh}\nolimits(\pi(\nu_{1}-\nu_{2}-\nu_{3}))+$ (10.242)
$\displaystyle-\frac{\psi\left(\frac{1}{2}+\frac{i(\nu_{1}+\nu_{2}+\nu_{3})}{2}\right)+\psi\left(\frac{1}{2}-\frac{i(\nu_{1}+\nu_{2}+\nu_{3})}{2}\right)}{128\pi^{2}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}\mathop{\rm
sh}\nolimits(\pi(\nu_{1}+\nu_{2}+\nu_{3}))+$ (10.243)
$\displaystyle+\frac{\psi\left(\frac{1}{2}+\frac{i(\nu_{1}-\nu_{2}+\nu_{3})}{2}\right)+\psi\left(\frac{1}{2}-\frac{i(\nu_{1}-\nu_{2}+\nu_{3})}{2}\right)}{128\pi^{2}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}\mathop{\rm
sh}\nolimits(\pi(\nu_{1}-\nu_{2}+\nu_{3}))+$ (10.244)
$\displaystyle+\frac{\psi\left(\frac{1}{2}+\frac{i(\nu_{1}+\nu_{2}-\nu_{3})}{2}\right)+\psi\left(\frac{1}{2}-\frac{i(\nu_{1}+\nu_{2}-\nu_{3})}{2}\right)}{128\pi^{2}\mathop{\rm
sh}\nolimits(\pi\nu_{1})\mathop{\rm sh}\nolimits(\pi\nu_{2})\mathop{\rm
sh}\nolimits(\pi\nu_{3})}\mathop{\rm
sh}\nolimits(\pi(\nu_{1}+\nu_{2}-\nu_{3}))+{\rm O}(K).$ (10.245)

### Dimensional regularization

The task is now to extract the finite part of the watermelon at $d=3$ from our
general formulae. We accomplish this task in full detail, as in this example
we explain the algorithm to extract information from our general formula for
integer dimensions. The hypergeometric series at the rhs of Eq. (LABEL:ad)
converges for $d<4$ and therefore the first term $I^{(1)}_{3}$ is finite. A
direct calculation shows that

$\displaystyle A_{3}(x,y,w)=\frac{\cot(2\pi w)(\cot(2\pi x)+\cot(2\pi
y))\left(\psi\left(\frac{1}{2}+w+x+y\right)-\psi\left(\frac{1}{2}-w+x+y\right)\right)}{128\pi^{2}}.$
(10.246)

The full $I_{3}^{(1)}(x,y,w,3)$ follows from (7.150). On the other hand
$I_{3}^{(2)}$ contains the divergent part of the diagram, here a simple pole:

$\displaystyle{I_{3}^{(2)}}(x,y,w,d)\simeq\frac{R(x,y,w,3)}{d-3}+\widetilde{I_{3}}(x,y,w,3).$
(10.247)

Both the residue and the finite term may be computed with the help of the
contiguity relations explained in Appendix C. Details are provided in Appendix
D. In particular the residue of $B_{d}(x,y,w)$ at $d=3$ is given by

$R_{B}(x,y,w)=b_{3}(x,y,w)f_{3}(x,y,w)=\frac{1-\cot(2\pi x)\cot(2\pi
y)}{128\pi^{2}}$ (10.248)

and does not depend on $w$ (see Eqs. (7.173) and (D.437)). In turn, the
residue of $I_{3}^{(2)}$ at $d=3$ is the sum of four terms:

$\displaystyle
R(x,y,w)=R_{B}(x,y,w)+R_{B}(x,-y,w)+R_{B}(-x,y,w)+R_{B}(-x,-y,w)=\frac{1}{32\pi^{2}}.$
(10.249)

We get the same result as in flat space. This fact is indeed nontrivial: the
calculation in flat space and the calculation on the Euclidean sphere have
essentially nothing to do with each other; the latter expresses a global
quantity resulting from integrating on the curved sphere (here the radius of
the sphere has been set to 1, but the result does not depend on the curvature)
the de Sitter propagators which are completely different from the propagators
in flat space in the large; they are similar to each other only in a small
region as compared with the curvature and for lage value of the masses.

As regards the finite term, calculations are cumbersome; some details may
found in Appendix D. In the special case where $w=0$ (i.e. $\nu_{3}=0$), Eqs.
(D.421), (D.437), (D.440) and (D.473) together give

$\displaystyle\widetilde{B}_{3}(x,y,0)=$ (10.250)
$\displaystyle=\frac{(\cot(2\pi x)\cot(2\pi
y)-1)\left(1-\gamma-\pi\cot(4\pi(x+y))+\log(\pi)-2\psi\left(\frac{1}{2}+x+y\right)\right)}{128\pi^{2}}$
(10.251)

which allows for the calculation of $\widetilde{I_{3}}(x,y,0,3)$. We may
actually solve for $\widetilde{I_{3}}(x,y,w,3)$. in full generality by
observing that the functions

$\displaystyle\Delta I_{3}^{(2)}=I_{3}^{(2)}(x,y,w,d)-I_{3}^{(2)}(x,y,0,d),$
(10.253) $\displaystyle\Delta B_{d}=B_{d}(x,y,w)-B_{d}(x,y,0),$ (10.254)

are regular at $d=3$ because the residues (10.248) and (10.249) do not depend
on $w$. In particular, by using Eq. (7.139) and the identity

$\frac{1}{\left(\kappa^{2}-\nu^{2}\right)}-\frac{1}{\kappa^{2}}=\frac{\nu^{2}\Gamma\left(-\frac{i\kappa}{2}\right)\Gamma\left(\frac{i\kappa}{2}\right)}{4\left(\kappa^{2}-\nu^{2}\right)\Gamma\left(1-\frac{i\kappa}{2}\right)\Gamma\left(\frac{i\kappa}{2}+1\right)}$

we get

$\displaystyle\Delta I_{3}^{(2)}$ $\displaystyle=$
$\displaystyle-\frac{w^{2}}{2\pi
i}\int\prod_{\epsilon,\epsilon^{\prime},\epsilon^{\prime\prime}=\pm
1}\frac{2^{d-2}\pi\Gamma\left(\epsilon s+w\right)\Gamma\left(\epsilon
s+\delta+x+y\right)}{\Gamma\left(1+\epsilon s\right)\Gamma\left(\epsilon
s+\frac{1}{4}\right)\Gamma\left(\epsilon
s+\frac{1}{2}\right)\Gamma\left(\epsilon
s+\frac{3}{4}\right)\Gamma\left(\epsilon s+w+1\right)}.$ (10.255)

By expressing $\Delta I_{3}^{(2)}$ in terms of hypergeometrics as in Eq.
(7.173), after some work, we obtain that

$\displaystyle\Delta B_{3}=\lim_{d\to
3}(B_{d}(x,y,w)-B_{d}(x,y,0))=-\frac{w^{2}(\cot(2\pi x)\cot(2\pi
y)-1)}{8\pi^{2}(2x+2y+1)\left(4w^{2}-(2x+2y+1)^{2}\right)}\times$ (10.257)
$\displaystyle\times{}_{4}F_{3}\left(\begin{array}[]{r}1,x+y+\frac{1}{2},w+x+y+\frac{1}{2},-w+x+y+\frac{1}{2}\cr
x+y+\frac{3}{2},w+x+y+\frac{3}{2},-w+x+y+\frac{3}{2}\end{array};1\right)$
(10.260) (10.261) (10.262) $\displaystyle=\frac{(\cot(2\pi x)\cot(2\pi
y)-1)\left(2\psi\left(\frac{1}{2}+x+y\right)\right)-\psi\left(\frac{1}{2}-w+x+y\right)-\psi\left(\frac{1}{2}+w+x+y\right)}{128\pi^{2}}.$
(10.263) (10.264)

Combining Eqs. (LABEL:bbdd0) and (10.264) we get

$\displaystyle\widetilde{B}_{3}(x,y,w)=\frac{(1-\cot(2\pi x)\cot(2\pi
y))\left(\pi\cot(4\pi(x+y))-1+\gamma-\log(\pi)\right)}{128\pi^{2}}$ (10.265)
$\displaystyle+\frac{(1-\cot(2\pi x)\cot(2\pi
y))\left(\psi\left(\frac{1}{2}-w+x+y\right)+\psi\left(\frac{1}{2}+w+x+y\right)\right)}{128\pi^{2}}$
(10.266)

and $\widetilde{I}_{2}(x,y,w,3)$ is computed by (8.188).

Collecting all the contributions the final result at $d\sim 3$ is given by

$\displaystyle{I}(x,y,w,d\sim
3)=-\frac{1}{32\pi^{2}(d-3)}+\frac{1-\gamma+\log(\pi)}{32\pi^{2}}+$ (10.267)
$\displaystyle\sum_{\epsilon,\epsilon^{\prime}=\pm}\frac{\left(\psi\left(\frac{1}{2}-w-\epsilon
x-\epsilon^{\prime}y\right)+\psi\left(\frac{1}{2}+w+\epsilon
x+\epsilon^{\prime}y\right)\right)\sin\left(2\pi\left(w+\epsilon
x+\epsilon^{\prime}y\right)\right)}{128\pi^{2}\sin(2\pi w)\sin(2\pi\epsilon
x)\sin\left(2\pi\epsilon^{\prime}y\right)}$ (10.268)

to be compared with (10.245): the mass-dependent finite part is the same as in
Eq. (10.245).

##  11 Spacetime dimension $d=4$

In the four dimensional case both $I_{3}^{(1)}$ and $I_{3}^{(2)}$ diverge. The
situation is complicated even more by the presence of a double pole. While the
algorithm remains the same, the whole procedure of extracting the residues and
the finite parts from our general formulae through the contiguity scheme is
quite heavy and the formulae cumbersome.

We will refrain to reproduce those formulae here. Instead we proceed to
identify the residues of the poles but yet another method which has a quite
general domain of applicability. We describe the general construction first.

### Convergence, poles and residues

In Sec. 7 we considered integrals of the form:

$\displaystyle{\bf J}=\int_{\eta+i{\bf R}}J((a_{p}),(c_{p}),s)ds\ ,$ (11.269)
$\displaystyle
J((a_{p}),(c_{p}),s)={\prod_{j=1}^{p-1}\Gamma(a_{j}+s)\Gamma(a_{j}-s)\over(a_{p}+s)(a_{p}-s)\prod_{j=1}^{p-1}\Gamma(c_{j}+s)\Gamma(c_{j}-s)}={\prod_{j=1}^{p}\Gamma(a_{j}+s)\Gamma(a_{j}-s)\over\prod_{j=1}^{p}\Gamma(c_{j}+s)\Gamma(c_{j}-s)}\
,$ (11.270)

where $c_{p}=a_{p}+1$. Note that $J((a_{p}),(c_{p}),s)=J((a_{p}),(c_{p}),-s)$.
It is assumed that $\mathop{\rm Re}\nolimits a_{j}>0$ for $1\leq j\leq p-1$,
$a_{p}$ is pure imaginary, $\eta>0$ is small, so that all the poles of ${J}$
(as a function of $s$) on the right of the contour are those located at
$s=a_{j}+n$, $n\geq 0$ integer, $1\leq j\leq p-1$. We also assume that outside
of $\\{|s|\leq K\\}$ the contour is deformed so that it lies on the imaginary
axis $i{\bf R}$. Let

$Z=2\sum_{j=1}^{p}(a_{j}-c_{j})\ .$ (11.272)

Since $|\arg s|\leq{\pi\over 2}$ we may use the Erdélyi-Tricomi Theorem [21,
Eq. 5.11.13], [79, pp. 118 ff] : as $s$ tends to $\pm i\infty$

${\Gamma(a_{j}+s)\over\Gamma(c_{j}+s)}\sim
s^{a_{j}-c_{j}}\sum_{k=0}^{\infty}G_{k}(a_{j},\ c_{j})\,s^{-k}\ $ (11.273)

where the first three terms are given by

$\displaystyle G_{0}=1,\ \ G_{1}(a,b)={\scriptstyle{1\over 2}}(a-b)(a+b-1),$
(11.274) $\displaystyle G_{2}(a,b)={1\over 12}\left({a-b\atop
2}\right)\Big{(}3(a+b-1)^{2}-(a-b+1)\Big{)}\ .$ (11.275)

In general, $G_{n}(a,b)$ is a polynomial with real rational coefficients in
$a$ and $b$.

It follows that for a fixed integer $N\geq 1$, there is a $L>0$ and a bounded
analytic function $h_{N}((a_{p}),(c_{p}),s)$ such that, for $s=it$, $t\geq L$,

$J((a_{p}),(c_{p}),s)=t^{Z}\Big{(}\sum_{k=0}^{N}{u_{k}\over
t^{k}}\Big{)}+t^{Z-N-1}h_{N}((a_{p}),(c_{p}),s)\ .$ (11.276)

Here $u_{0}=1$ and the other $u_{k}$ are polynomial expressions of the
$G_{n}(a_{j},c_{j})$, $0\leq n\leq N$, $1\leq j\leq p$.

Suppose that $\mathop{\rm Re}\nolimits Z<-1$: integrating
$J((a_{p}),(c_{p}),s)$ over the whole integration contour gives

$2i\int_{L}^{\infty}t^{Z}\Big{(}\sum_{k=0}^{N}{u_{k}\over
t^{k}}\Big{)}\,dt=\sum_{k=0}^{N}{-2iu_{k}L^{Z-k+1}\over Z-k+1}\ $ (11.277)

plus a bounded analytic function of the parameters.

The result has a meromorphic continuation for $\mathop{\rm Re}\nolimits Z<N$ ,
with poles given by (11.277). The residue at $Z=k-1$ is $-2iu_{k}$. In
particular the residue at $Z=-1$ is $-2i$ and is independent of the
parameters. Also, $u_{2n+1}=0$, since $J((a_{p}),(c_{p}),s)$ is an even
function of $s$. For $N=2$ we can write

$J((a_{p}),(c_{p}),s)=t^{Z}\left(\Big{(}1+s^{-2}\sum_{j=1}^{p}(2G_{2}(a_{j},c_{j})-G_{1}(a_{j},c_{j})^{2})\Big{)}+O(s^{-3})\right).$
(11.278)

Since $s^{2}=-t^{2}$, we get

$u_{2}=\sum_{j=1}^{p}(G_{1}(a_{j},c_{j})^{2}-2G_{2}(a_{j},c_{j}))=\sum_{j=1}^{p}(f_{2}(a_{j})-f_{2}(c_{j})),$
(11.279)

where

$f_{2}(a)={a(a-1)(2a-1)\over 6}\ .$ (11.280)

### Summary

J extends to a meromorphic function of the parameters having poles at

$Z=-1+2n,\ \ \ n\geq 0\ \ {\rm integer}.$ (11.281)

Near $Z=-1+2n$,

${\bf J}\sim{-2iu_{2n}\over{\zeta}+1-2n}\ .$ (11.282)

### Application at $d=3$ and $d=4$

Let us consider at first the integral (7.147). It fits with the definition
(11.270) with $Z=4\delta-4=d-5$. Here we trade the variable $Z$ for the
spacetime dimension $d$ and conclude that there are poles $d=4+2n$. The
residue of ${\cal A}$ at $d=4$ is thus simply:

${\rm residue\ of\ }{\cal A}\ \ {\rm at\ \ }d=4:\ \ \frac{-2iu_{0}}{2i}=-1\ .$
(11.283)

Also the integral (LABEL:bubble2) fits with the definition (11.270); now
$Z=8\delta-5=2d-7$. There are poles at $d=3+n$ with residues (in the variable
$d$) $-2^{d-2}iu_{2n}/(2i)=-2^{d-3}u_{2n}$; in particular

$\displaystyle\hbox{residue of ${\cal B}$ at $d=3$ : \ \ \ }-u_{0}=-1\ ,$
(11.284) $\displaystyle\hbox{residue of ${\cal B}$ at $d=4$ : \ \ \
}-2u_{2}=2w^{2}-2x^{2}-2y^{2}+\frac{1}{8}\ .$ (11.285)

Multiplying by the right normalizations as in Eqs. (7.140) and (7.143) and
restoring the mass variables we obtain the leading singularities as follows:

$\displaystyle\hbox{ ${I_{3}}$ at $d=3$ : \ \ \ }-\frac{1}{32\pi^{2}(d-3)},\ $
(11.286) $\displaystyle\hbox{ ${I_{3}}$ at $d=4$ : \ \ \
}-\frac{4\nu_{1}^{2}+4\nu_{2}^{2}+4\nu_{3}^{2}+3}{512\pi^{4}(d-4)^{2}}.\ $
(11.287)

Once more, we notice that the dominant divergence in $d=3$ is exactly the same
as in the flat case. (11.287) reproduces the flat dominant divergence in the
limit $R\to\infty$.

##  12 Conclusions

In a recent paper we have shown that calculating loop integrals in position
space may be advantageous also in flat Minkowski space from several
viewpoints. While we believe that that possibility in flat space deserves
attention and is not just a luxury or a mathematical ornament, in curved space
performing loop calculation in position space is compulsory. Here we have
started this program in de Sitter space. Calculations are significantly more
intricate than in flat space but can be performed successfully till the end.
In a companion paper we present a study of the same diagrams in anti de Sitter
space. We have found that in the AdS case things are a little simpler than in
the present de Sitterian study because of the presence of a true spectral
condition. The two papers together open a new way for precision calculations
of QFT in the presence of a cosmological constant.

## Acknowledgments

U. M. and S. L. C. are grateful to the Institut des Hautes Etudes
Scientifiques - Bures-sur-Yvette, for the generous hospitality and support
they received while writing this paper.

##  Appendix A Details about section 3

###  A.1 The functions ${\bf P}_{\nu}^{\mu}$ and ${\bf Q}_{\nu}^{\mu}$

These functions are solutions of the Legendre equation [13, 3.2 (1) p. 121]:

$\displaystyle{d\over dz}(1-z^{2})w^{\prime}(z)+C_{\nu,\mu}(z)w(z)=0\ ,$
(A.288) $\displaystyle C_{\nu,\mu}(z)=\nu(\nu+1)-\mu^{2}(1-z^{2})^{-1}\ .$
(A.289)

They are called ”Legendre functions on the cut” in [13] or Ferrers functions
[21, 14.3.1, 14.3.2]. They are holomorphic in the domain

$\Delta_{2}=\\{z\in{\bf C}\ :\ \mathop{\rm Im}\nolimits z\not=0\ \ {\rm or}\ \
-1<z<1\\}$ (A.290)

and given there by:

$\displaystyle{\bf P}_{\beta}^{\alpha}(z)$
$\displaystyle={1\over\Gamma(1-\alpha)}\left({1+z\over
1-z}\right)^{\alpha\over 2}F\left(-\beta,\ 1+\beta\ ;\ 1-\alpha\ ;\ {1-z\over
2}\right)\ ,$ (A.291)
$\displaystyle={\Gamma(-\alpha)\over\Gamma(1+\beta-\alpha)\Gamma(-\beta-\alpha)}\left({1+z\over
1-z}\right)^{\alpha\over 2}F\left(-\beta,\ 1+\beta\ ;\ 1+\alpha\ ;\ {1+z\over
2}\right)$ (A.292)
$\displaystyle-\pi^{-1}\sin(\pi\beta)\Gamma(\alpha)\left({1-z\over
1+z}\right)^{\alpha\over 2}F\left(-\beta,\ 1+\beta\ ;\ 1-\alpha\ ;\ {1+z\over
2}\right)\ $ (A.293) $\displaystyle=\frac{\mathop{\rm
tg}\nolimits(\pi\alpha)+\mathop{\rm tg}\nolimits(\pi\beta)}{\pi}{\bf
Q}^{\alpha}_{\beta}(z)+\frac{\mathop{\rm tg}\nolimits(\pi\alpha)-\mathop{\rm
tg}\nolimits(\pi\beta)}{\pi}{\bf Q}^{\alpha}_{-\beta-1}(z),$ (A.294)
$\displaystyle{\bf Q}_{\beta}^{\alpha}(z)$
$\displaystyle={\pi\cos(\pi\alpha)\over 2\sin(\pi\alpha)}{\bf
P}_{\beta}^{\alpha}(z)-{\pi\Gamma(\beta+\alpha+1)\over
2\sin(\pi\alpha)\Gamma(\beta-\alpha+1)}{\bf P}_{\beta}^{-\alpha}(z)$ (A.296)
$\displaystyle={\scriptstyle{1\over
2}}\Gamma(\alpha)\cos(\alpha\pi)\left({1+z\over 1-z}\right)^{\alpha\over
2}F\left(-\beta,\ \beta+1\ ;\ 1-\alpha\ ;\ {1-z\over 2}\right)$ (A.297)
$\displaystyle+$ $\displaystyle{\Gamma(1+\beta+\alpha)\Gamma(-\alpha)\over
2\Gamma(1+\beta-\alpha)}\left({1-z\over 1+z}\right)^{\alpha\over
2}F\left(-\beta,\ \beta+1\ ;\ \alpha+1\ ;\ {1-z\over 2}\right).$ (A.298)

(See [13, 3.4 (6) p. 143, 3.2 (15) pp 124-125, 3.4 (13), (10) p. 144]). From
this it follows that:

1) as $z\rightarrow 1$

$\displaystyle{\bf P}_{\beta}^{\alpha}(z)$
$\displaystyle\sim{1\over\Gamma(1-\alpha)}\left({1-z\over
2}\right)^{-{\alpha\over 2}}\ ,$ (A.299) $\displaystyle{\bf
Q}_{\beta}^{\alpha}(z)$ $\displaystyle\sim{\scriptstyle{1\over
2}}\Gamma(\alpha)\cos(\pi\alpha)\left({1-z\over 2}\right)^{-{\alpha\over 2}}\
\ {\rm if}\ \ \mathop{\rm Re}\nolimits\alpha>0\ ,$ (A.300) $\displaystyle{\bf
Q}_{\beta}^{\alpha}(z)$
$\displaystyle\sim{\Gamma(1+\beta+\alpha)\Gamma(-\alpha)\over
2\Gamma(1+\beta-\alpha)}\left({1-z\over 2}\right)^{{\alpha\over 2}}\ \ {\rm
if}\ \ \mathop{\rm Re}\nolimits\alpha<0\ ;$ (A.301)

2) as $z\rightarrow-1$

$\displaystyle{\bf P}_{\beta}^{\alpha}(z)$
$\displaystyle\sim-\pi^{-1}\sin(\pi\beta)\Gamma(\alpha)\left({1+z\over
2}\right)^{-{\alpha\over 2}}\ \ {\rm if}\ \ \mathop{\rm Re}\nolimits\alpha>0\
\ ,$ (A.302) $\displaystyle{\bf P}_{\beta}^{\alpha}(z)$
$\displaystyle\sim{\Gamma(-\alpha)\over\Gamma(1+\beta-\alpha)\Gamma(-\beta-\alpha)}\left({1+z\over
2}\right)^{{\alpha\over 2}}\ \ {\rm if}\ \mathop{\rm Re}\nolimits\ \alpha<0\ \
,$ (A.303) $\displaystyle{\bf Q}_{\beta}^{\alpha}(z)$
$\displaystyle\sim-{\Gamma(\alpha)\cos(\pi\beta)\over 2}\left({1+z\over
2}\right)^{-{\alpha\over 2}}\ \ {\rm if}\ \ \mathop{\rm Re}\nolimits\alpha>0\
,$ (A.304) $\displaystyle{\bf
Q}_{\beta}^{\alpha}(z)\sim{\Gamma(-\alpha)\Gamma(1+\beta+\alpha)\over
2\sin(\pi\alpha)\Gamma(1+\beta-\alpha)}\Big{[}\sin(\pi\beta)-\cos(\pi\alpha)\sin\pi(\alpha+\beta)\Big{]}\left({1+z\over
2}\right)^{{\alpha\over 2}}$ (A.305)
$\displaystyle=-{\Gamma(-\alpha)\Gamma(1+\beta+\alpha)\cos\pi(\alpha+\beta)\over
2\Gamma(1+\beta-\alpha)}\left({1+z\over 2}\right)^{{\alpha\over 2}}\ \ {\rm
if}\ \ \mathop{\rm Re}\nolimits\alpha<0\ .$ (A.306)

###  A.2 Derivation of the main Wronskian equation (3.48)

Let $u_{1}$, $u_{2}$ be solutions of

${d\over dz}(1-z^{2})u^{\prime}_{j}(z)+B_{j}(z)u_{j}(z)=0\ ,\ \ j=1,\ 2,$
(A.307)

where $B_{j}$ is analytic in the domain we consider. Let $D(z)$ be the
determinant

$D(z)=\left|\begin{array}[]{cc}u_{1}(z)&u_{2}(z)\\\
(1-z^{2})u^{\prime}_{1}(z)&(1-z^{2})u^{\prime}_{2}(z)\\\ \end{array}\right|\
.$ (A.308)

Then

${d\over dz}D(z)=\left|\begin{array}[]{cc}u_{1}(z)&u_{2}(z)\\\
-B_{1}(z)u_{1}(z)&-B_{2}(z)u_{2}(z)\\\
\end{array}\right|=u_{1}(z)u_{2}(z)\Big{[}B_{1}(z)-B_{2}(z)\Big{]}\ .$ (A.309)

Thus

$\displaystyle\int_{a}^{b}u_{1}(z)u_{2}(z)$
$\displaystyle\Big{[}B_{1}(z)-B_{2}(z)\Big{]}dz=\Big{[}D(z)\Big{]}_{a}^{b}=$
(A.310)
$\displaystyle=\Big{[}(1-z^{2})[u_{1}(z)u^{\prime}_{2}(z)-u_{2}(z)u^{\prime}_{1}(z)]\Big{]}_{a}^{b}\
.$ (A.311)

We now take $u_{1}=u_{\nu}^{\mu}$, $u_{2}=v_{\sigma}^{\rho}$ with the
understanding that $u_{\nu}^{\mu}$ stands for ${\bf P}_{\nu}^{\mu}$ or ${\bf
Q}_{\nu}^{\mu}$, and independently $v_{\sigma}^{\rho}$ stands for ${\bf
P}_{\sigma}^{\rho}$ or ${\bf Q}_{\sigma}^{\rho}$. Thus
$B_{1}(z)=C_{\nu,\mu}(z)$ and $B_{2}(z)=C_{\sigma,\rho}(z)$, and

$B_{1}(z)-B_{2}(z)=(\nu-\sigma)(\nu+\sigma+1)+(\rho^{2}-\mu^{2})(1-z^{2})^{-1}\
.$ (A.312)

We recall the formulae [13, 3.8 (19),(15) p.161]

$(1-z^{2}){d{\bf P}_{\nu}^{\mu}(z)\over dz}=-\nu z{\bf
P}_{\nu}^{\mu}(z)+(\nu+\mu){\bf P}_{\nu-1}^{\mu}(z)\ ,$ (A.313) ${\bf
P}_{\nu-1}^{\mu}(z)=z{\bf
P}_{\nu}^{\mu}(z)+(\nu-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu-1}(z)\ .$ (A.314)

Hence

$(1-z^{2}){d{\bf P}_{\nu}^{\mu}(z)\over dz}=\mu z{\bf
P}_{\nu}^{\mu}(z)+(\nu+\mu)(\nu-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu-1}(z)]\ .$ (A.315)

As stated in [13] the formulae (A.313-A.315) remain valid when ${\bf P}$ is
replaced by ${\bf Q}$. Therefore, setting $\rho=\mu$, we get:

$\displaystyle\int_{a}^{b}u_{\nu}^{\mu}(z)v_{\sigma}^{\mu}(z)$
$\displaystyle(\nu-\sigma)(\sigma+\nu+1)dz=$ (A.316)
$\displaystyle=\Big{[}(1-z^{2})^{\scriptstyle{1\over
2}}(\sigma+\mu)(\sigma-\mu+1)u_{\nu}^{\mu}(z)v_{\sigma}^{\mu-1}(z)$ (A.317)
$\displaystyle-(1-z^{2})^{\scriptstyle{1\over
2}}(\nu+\mu)(\nu-\mu+1)u_{\nu}^{\mu-1}(z)v_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}\
.$ (A.318)

This is eq. (3.48). Recall that here the integration is over an arc (with
extremities $a$ and $b$) contained in the domain $\Delta_{2}=\\{z\in{\bf C}\
:\ \mathop{\rm Im}\nolimits z\not=0\ \ {\rm or}\ \ |z|<1\\}$, and that
$z\mapsto(1-z^{2})^{\scriptstyle{1\over 2}}$ is the function holomorphic in
this domain and equal to $|1-z^{2}|^{\scriptstyle{1\over 2}}$ when $z\in(-1,\
1)$.

####  A.2.1 Case of the functions $P_{\nu}^{\mu}$ and $Q_{\nu}^{\mu}$

Some details have to be modified if we now suppose that $u_{\nu}^{\mu}$ stands
for $P_{\nu}^{\mu}$ or $Q_{\nu}^{\mu}$, and independently $v_{\sigma}^{\rho}$
stands for $P_{\sigma}^{\rho}$ or $Q_{\sigma}^{\rho}$. These functions777We
use the notations of [13, 3.2 (3),(5) p. 122]. Note that $Q_{\nu}^{\mu}$ is
not defined where $\Gamma(\nu+\mu+1)$ has a pole. will be considered as
holomorphic in the domain $\Delta_{1}=\\{z\in{\bf C}\ :\ \mathop{\rm
Im}\nolimits z\not=0\ \ {\rm or}\ \ z>1\\}$. Then eqs (A.313-A.315) have to be
replaced by

$(1-z^{2}){dP_{\nu}^{\mu}(z)\over dz}=-\nu
zP_{\nu}^{\mu}(z)+(\nu+\mu)P_{\nu-1}^{\mu}(z)\ ,$ (A.319)
$P_{\nu-1}^{\mu}(z)=zP_{\nu}^{\mu}(z)-(\nu-\mu+1)(z^{2}-1)^{\scriptstyle{1\over
2}}P_{\nu}^{\mu-1}(z)\ ,$ (A.320)

(see [13, 3.8 (10), (5) p. 161])

$(1-z^{2}){dP_{\nu}^{\mu}(z)\over dz}=\mu
zP_{\nu}^{\mu}(z)-(\nu+\mu)(\nu-\mu+1)(z^{2}-1)^{\scriptstyle{1\over
2}}P_{\nu}^{\mu-1}(z)]\ .$ (A.321)

Here $z\mapsto(z^{2}-1)^{\scriptstyle{1\over 2}}$ is the function holomorphic
in $\Delta_{1}$ equal to $|z^{2}-1|^{\scriptstyle{1\over 2}}$ when $z>1$.
Again as stated in [13] eqs (A.319-A.321) remain valid if $P$ is replaced by
$Q$. Therefore, setting $\rho=\mu$, we get from (A.311):

$\displaystyle\int_{a}^{b}u_{\nu}^{\mu}(z)v_{\sigma}^{\mu}(z)$
$\displaystyle(\nu-\sigma)(\sigma+\nu+1)dz=$ (A.322)
$\displaystyle=\Big{[}-(z^{2}-1)^{\scriptstyle{1\over
2}}(\sigma+\mu)(\sigma-\mu+1)u_{\nu}^{\mu}(z)v_{\sigma}^{\mu-1}(z)$ (A.323)
$\displaystyle+(z^{2}-1)^{\scriptstyle{1\over
2}}(\nu+\mu)(\nu-\mu+1)u_{\nu}^{\mu-1}(z)v_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}\
.$ (A.324)

Here the integration is over an arc contained in $\Delta_{1}$.

###  A.3 The basic case $u_{\nu}^{\mu}={\bf Q}_{\nu}^{\mu}$,
$v_{\sigma}^{\mu}={\bf Q}_{\sigma}^{\mu}$

This case may be considered as basic since by using (A.294) it is possible to
obtain the other cases from it. The formula (3.48) becomes in this case:

$\displaystyle(\nu-\sigma)(\nu+\sigma+1)\int_{a}^{b}$ $\displaystyle{\bf
Q}_{\nu}^{\mu}(z){\bf Q}_{\sigma}^{\mu}(z)\,dz=$ (A.325)
$\displaystyle=\Big{[}(\sigma+\mu)(\sigma-\mu+1)(1-z^{2})^{\scriptstyle{1\over
2}}{\bf Q}_{\nu}^{\mu}(z){\bf Q}_{\sigma}^{\mu-1}(z)$ (A.326)
$\displaystyle-(\nu+\mu)(\nu-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
Q}_{\nu}^{\mu-1}(z){\bf Q}_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}.$ (A.327)

As in sect. 3 it is sufficient to evaluate the contribution of the first term
in the rhs of (A.327) since the contribution of the second can be obtained by
exchanging $\nu$ and $\sigma$ and a global change of sign.

We again fix $\mu\in(0,\ 1)$. Using the formulae (A.300) and (A.301) we find
that as $z\rightarrow 1$ the first term in the rhs of (A.327) tends to

${\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\cos(\pi\mu)\over
2\Gamma(1+\sigma-\mu)}\ .$ (A.328)

As $z\rightarrow-1$ (with a minus sign due to dealing with the lower bound
$a$) this first term contributes:

${\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\cos(\pi\nu)\cos\pi(\sigma+\mu)\over
2\Gamma(1+\sigma-\mu)}\ .$ (A.329)

The rhs of (A.327) is thus given by

$\displaystyle{\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)[\cos(\pi\mu)+\cos(\pi\nu)\cos\pi(\sigma+\mu)]\over
2\Gamma(1+\sigma-\mu)}$ (A.330)
$\displaystyle-{\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\nu+\mu)[\cos(\pi\mu)+\cos(\pi\sigma)\cos\pi(\nu+\mu)]\over
2\Gamma(1+\nu-\mu)}.$ (A.331)

In this expression let us set $\nu=-\sigma-1$. The first term becomes

$\displaystyle{\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)[\cos(\pi\mu)-\cos(\pi\sigma)\cos\pi(\sigma+\mu)]\over
2\Gamma(1+\sigma-\mu)}$ (A.332)
$\displaystyle={\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\sin\pi(\sigma+\mu)\sin(\pi\sigma)\over
2\Gamma(1+\sigma-\mu)}$ (A.333)
$\displaystyle=-{\pi\Gamma(\mu)\Gamma(1-\mu)\sin(\pi\sigma)\over
2\Gamma(1+\sigma-\mu)\Gamma(-\sigma-\mu)}\ .$ (A.334)

The second term becomes

$\displaystyle-{\Gamma(\mu)\Gamma(1-\mu)\Gamma(\mu-\sigma)[\cos(\pi\mu)-\cos(\pi\sigma)\cos\pi(\mu-\sigma)]\over
2\Gamma(-\sigma-\mu)}$ (A.335)
$\displaystyle={\Gamma(\mu)\Gamma(1-\mu)\Gamma(\mu-\sigma)\sin\pi(\mu-\sigma)\sin(\pi\sigma)\over
2\Gamma(-\sigma-\mu)}$ (A.336)
$\displaystyle={\pi\Gamma(\mu)\Gamma(1-\mu)\sin(\pi\sigma)\over
2\Gamma(1+\sigma-\mu)\Gamma(-\sigma-\mu)}\ .$ (A.337)

Thus the expression (A.331) vanishes if we set $\nu=-\sigma-1$. It also
obviously vanishes if we set $\sigma=\nu$. We obtain

$\displaystyle\int_{-1}^{1}{\bf Q}_{\nu}^{\mu}(z){\bf
Q}_{\sigma}^{\mu}(z)\,dz=\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\nu+\mu)\Gamma(1+\sigma+\mu)\Phi_{{\bf
Q}{\bf Q}}(\nu,\ \sigma,\ \mu)\ ,$ (A.338) $\displaystyle\Phi_{{\bf Q}{\bf
Q}}(\nu,\ \sigma,\ \mu)={1\over(\nu-\sigma)(\nu+\sigma+1)}\times$ (A.339)
$\displaystyle\qquad\qquad\Bigg{[}{\cos(\pi\mu)+\cos(\pi\nu)\cos\pi(\sigma+\mu)\over
2\Gamma(1+\nu+\mu)\Gamma(1+\sigma-\mu)}-{\cos(\pi\mu)+\cos(\pi\sigma)\cos\pi(\nu+\mu)\over
2\Gamma(1+\sigma+\mu)\Gamma(1+\nu-\mu)}\Bigg{]}\ .$ (A.340)

$\Phi_{{\bf Q}{\bf Q}}(\nu,\ \sigma,\ \mu)$ extends to an entire function of
all its arguments. It is symmetric in $\nu$ and $\sigma$ and vanishes when
$\mu=0$. The equality (A.338) extends by analytic continuation to values of
$\mu$ with $-1<\mathop{\rm Re}\nolimits\mu<1$.

###  A.4 The case $u_{\nu}^{\mu}={\bf P}_{\nu}^{\mu}$, $v_{\sigma}^{\mu}={\bf
Q}_{\sigma}^{\mu}$

This case may be dealt with by two different methods: the first is to use the
Wronskian equation as in the two preceding cases. The second is to use eq.
(A.294) and the result of subsect. A.3. The formula (3.48) becomes in this
case:

$\displaystyle(\nu-\sigma)(\nu+\sigma+1)\int_{a}^{b}$ $\displaystyle{\bf
P}_{\nu}^{\mu}(z){\bf Q}_{\sigma}^{\mu}(z)\,dz=$ (A.341)
$\displaystyle=\Big{[}(\sigma+\mu)(\sigma-\mu+1)(1-z^{2})^{\scriptstyle{1\over
2}}{\bf P}_{\nu}^{\mu}(z){\bf Q}_{\sigma}^{\mu-1}(z)$ (A.342)
$\displaystyle-(\nu+\mu)(\nu-\mu+1)(1-z^{2})^{\scriptstyle{1\over 2}}{\bf
P}_{\nu}^{\mu-1}(z){\bf Q}_{\sigma}^{\mu}(z)\Big{]}_{a}^{b}.$ (A.343)

We fix $\mu\in(0,\ 1)$. To evaluate the rhs of this formula as $z\rightarrow
1$ we use the formulae (A.299-A.301): as $z\rightarrow 1$ the first term in
the rhs of (A.343) tends to

${\Gamma(1+\sigma+\mu)\over\Gamma(1+\sigma-\mu)}\ .$ (A.344)

The second term in the rhs of (A.343) behaves like ${\rm Const.\
}(1-z)^{1-\mu}$ and tends to 0. To evaluate the rhs of the formula (A.343) as
$z\rightarrow-1$ we use the formula (A.302-A.306). In the end we find that,
for $a=-1$ and $b=1$ the rhs of (A.343) is equal to

$\displaystyle{\Gamma(1+\sigma+\mu)\over\Gamma(1+\sigma-\mu)}$ (A.345)
$\displaystyle+{\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\sin(\pi\nu)\cos\pi(\mu+\sigma)\over\pi\Gamma(1+\sigma-\mu)}$
(A.346)
$\displaystyle+{\Gamma(\mu)\Gamma(1-\mu)\cos(\pi\sigma)\over\Gamma(1+\nu-\mu)\Gamma(-\nu-\mu)}\
.$ (A.347)

If we set $\nu=\sigma$ in this expression, it becomes

$\displaystyle{\Gamma(1+\sigma+\mu)\over\Gamma(1+\sigma-\mu)}$ (A.348)
$\displaystyle-{\Gamma(1+\sigma+\mu)\Gamma(\mu)\Gamma(1-\mu)\over\pi\Gamma(1+\sigma-\mu)}\times$
(A.349)
$\displaystyle\Bigg{[}-\sin(\pi\sigma)\cos\pi(\mu+\sigma)-{\pi\cos(\pi\sigma)\over\Gamma(1+\sigma+\mu)\Gamma(-\sigma-\mu)}\Bigg{]}=0\
,$ (A.350)

as it must be. We can therefore subtract from (A.347) the same expression with
$\nu=\sigma$ and we obtain, for $0<\mu<1$,

$\displaystyle\int_{-1}^{1}{\bf P}_{\nu}^{\mu}(z){\bf
Q}_{\sigma}^{\mu}(z)\,dz=\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\Phi_{{\bf
P}{\bf Q}}(\nu,\ \sigma,\ \mu)\ ,$ (A.351) $\displaystyle\Phi_{{\bf P}{\bf
Q}}(\nu,\ \sigma,\ \mu)={1\over(\nu-\sigma)(\nu+\sigma+1)}\times$ (A.352)
$\displaystyle\Bigg{[}{[\sin(\pi\nu)-\sin(\pi\sigma)]\cos\pi(\mu+\sigma)\over\pi\Gamma(1+\sigma-\mu)}$
(A.353)
$\displaystyle+{\cos(\pi\sigma)\over\Gamma(1+\sigma+\mu)}\left[{1\over\Gamma(1+\nu-\mu)\Gamma(-\nu-\mu)}-{1\over\Gamma(1+\sigma-\mu)\Gamma(-\sigma-\mu)}\right]\Bigg{]}\
.$ (A.354)

The function $\Phi_{{\bf P}{\bf Q}}(\nu,\ \sigma,\ \mu)$ is entire in all its
variables. It vanishes at $\mu=0$ and it is invariant under the change
$\nu\rightarrow-1-\nu$. The equality (A.351) extends by analytic continuation
to $-1<\mathop{\rm Re}\nolimits\mu<1$.

The second method is to use (A.294) to write

$\displaystyle\int_{-1}^{1}{\bf P}_{\nu}^{\mu}(z){\bf
Q}_{\sigma}^{\mu}(z)\,dz=$ (A.355) $\displaystyle\frac{\mathop{\rm
tg}\nolimits(\pi\mu)+\mathop{\rm tg}\nolimits(\pi\nu)}{\pi}\int_{-1}^{1}{\bf
Q}_{\nu}^{\mu}(z){\bf Q}_{\sigma}^{\mu}(z)\,dz+\frac{\mathop{\rm
tg}\nolimits(\pi\mu)-\mathop{\rm tg}\nolimits(\pi\nu)}{\pi}\int_{-1}^{1}{\bf
Q}^{\mu}_{-\nu-1}(z){\bf Q}_{\sigma}^{\mu}(z)\,dz=$ (A.356)
$\displaystyle=\frac{1}{(\nu-\sigma)(\nu+\sigma+1)}\left(\frac{\Gamma(\mu+\sigma+1)(\sec(\pi\mu)\cos(\pi\nu)\cos(\pi(\mu+\sigma))+1)}{\Gamma(-\mu+\sigma+1)}\right.+$
(A.357) $\displaystyle\left.+\frac{\pi\nu\sec(\pi\mu)\mathop{\rm
tg}\nolimits(\pi\nu)\cos(\pi\sigma)-\pi\mu\csc(\pi\mu)(\sec(\pi\nu)+\cos(\pi\sigma))}{\Gamma(-\mu-\nu+1)\Gamma(-\mu+\nu+1)}\right).$
(A.358)

###  A.5 Other integrals

Using the formulae [13, 3.4 (14), (15), (17), (18) p. 144]

${\bf P}_{\beta}^{\alpha}(-z)=\cos\pi(\beta+\alpha){\bf
P}_{\beta}^{\alpha}(z)-{2\sin\pi(\beta+\alpha)\over\pi}{\bf
Q}_{\beta}^{\alpha}(z)\ ,$ (A.359) ${\bf
Q}_{\beta}^{\alpha}(-z)=-\cos\pi(\beta+\alpha){\bf
Q}_{\beta}^{\alpha}(z)<-{\sin\pi(\beta+\alpha)\over 2\pi}{\bf
P}_{\beta}^{\alpha}(z)\ ,$ (A.360) $\Gamma(1+\beta+\alpha){\bf
P}_{\beta}^{-\alpha}(z)=\Gamma(1+\beta-\alpha)\Big{[}{\bf
P}_{\beta}^{\alpha}(z)\cos(\pi\alpha)-{2\over\pi}\sin(\pi\alpha){\bf
Q}_{\beta}^{\alpha}(z)\Big{]}\ ,$ (A.361) $\Gamma(1+\beta+\alpha){\bf
Q}_{\beta}^{-\alpha}(z)=\Gamma(1+\beta-\alpha)\Big{[}{\bf
Q}_{\beta}^{\alpha}(z)\cos(\pi\alpha)+{\pi\over 2}\sin(\pi\alpha){\bf
P}_{\beta}^{\alpha}(z)\Big{]}\ ,$ (A.362)

we can immediately obtain $\int_{-1}^{1}{\bf P}_{\nu}^{\pm\mu}(\pm z){\bf
P}_{\sigma}^{\pm\mu}(\pm z)dz$, $\int_{-1}^{1}{\bf P}_{\nu}^{\pm\mu}(\pm
z){\bf Q}_{\sigma}^{\pm\mu}(\pm z)dz$,$\int_{-1}^{1}{\bf Q}_{\nu}^{\pm\mu}(\pm
z){\bf Q}_{\sigma}^{\pm\mu}(\pm z)dz$, for instance :

$\displaystyle\int_{-1}^{1}{\bf P}_{\nu}^{\mu}(z){\bf
P}_{\sigma}^{\mu}(-z)\,dz=2\pi^{-1}\cos\pi(\sigma+\mu)\Gamma(\mu)\Gamma(1-\mu)\Phi_{{\bf
P}{\bf P}}(\nu,\ \sigma,\ \mu)$ (A.363)
$\displaystyle-{2\over\pi}\sin\pi(\sigma+\mu)\Gamma(\mu)\Gamma(1-\mu)\Gamma(1+\sigma+\mu)\Phi_{{\bf
P}{\bf Q}}(\nu,\ \sigma,\ \mu)\ .$ (A.364)

Finally let us use subsubsect. A.2.1 to evaluate the integral

$\int_{1}^{\infty}Q_{\nu}^{\mu}(z)Q_{\sigma}^{\mu}(z)dz\ .$ (A.365)

It follows from [13, 3.9.2 (21), (5), (6) pp 163-164] that:

${\rm as}\ \ z\rightarrow+\infty,\ \ Q_{\nu}^{\mu}(z)\sim{\rm const.\
}z^{-\nu-1}\ ;$ (A.366) ${\rm if\ }\mathop{\rm Re}\nolimits\mu>0,\ \ {\rm as}\
\ z\rightarrow 1,\ \ Q_{\nu}^{\mu}(z)\sim e^{i\pi\mu}2^{{\mu\over
2}-1}\Gamma(\mu)(z-1)^{-{\mu\over 2}}\ ;$ (A.367) ${\rm if\ }\mathop{\rm
Re}\nolimits\mu<0,\ \ {\rm as}\ \ z\rightarrow 1,\ \
Q_{\nu}^{\mu}(z)\sim{e^{i\pi\mu}2^{-{\mu\over
2}-1}\Gamma(-\mu)\Gamma(\nu+\mu+1)(z-1)^{{\mu\over 2}}\over\Gamma(\nu-\mu+1)}\
.$ (A.368)

Note also ([13, 3.2 (2) p. 140])

$e^{i\pi\mu}\Gamma(\nu+\mu+1)Q_{\nu}^{-\mu}(z)=e^{-i\pi\mu}\Gamma(\nu-\mu+1)Q_{\nu}^{\mu}(z)\
.$ (A.369)

It follows that the integral (A.365) converges if $\mathop{\rm
Re}\nolimits(\nu+\sigma)>-1$ and $|\mathop{\rm Re}\nolimits\mu|<1$. For our
evaluation we suppose $\mathop{\rm Re}\nolimits(\nu+\sigma)>-1$ and $0<\mu<1$.
We then set $u_{\nu}^{\mu}(z)=Q_{\nu}^{\mu}(z)$ and
$v_{\sigma}^{\mu}=Q_{\sigma}^{\mu}$ in (A.324) and let $b$ tend to infinity
and $a$ tend to 1. The result is that, under our assumptions,

$\displaystyle\int_{1}^{\infty}Q_{\nu}^{\mu}(z)Q_{\sigma}^{\mu}(z)\,dz=$
(A.370) $\displaystyle{e^{2i\pi\mu}\Gamma(\mu)\Gamma(1-\mu)\over
2(\nu-\sigma)(\sigma+\nu+1)}\Bigg{[}{\Gamma(\nu+\mu+1)\over\Gamma(\nu-\mu+1)}-{\Gamma(\sigma+\mu+1)\over\Gamma(\sigma-\mu+1)}\Bigg{]}\
.$ (A.371)

This equation remains valid, by analytic continuation, when $|\mathop{\rm
Re}\nolimits\mu|<1$ (and $\mathop{\rm Re}\nolimits(\nu+\sigma)>-1$) (note that
the bracket vanishes for $\mu=0$). The formula is compatible with (A.369).
Letting $\nu$ tend to $\sigma$ we find

$\displaystyle\int_{1}^{\infty}$ $\displaystyle
Q_{\sigma}^{\mu}(z)Q_{\sigma}^{\mu}(z)\,dz=$ (A.372)
$\displaystyle={e^{2i\pi\mu}\Gamma(\mu)\Gamma(1-\mu)\Gamma(\sigma+\mu+1)\over
2(2\sigma+1)\Gamma(\sigma-\mu+1)}\Big{[}\psi(\sigma+\mu+1)-\psi(\sigma-\mu+1)\Big{]}\
.$ (A.373)

Again this equation is valid when $|\mathop{\rm Re}\nolimits\mu|<1$ and
$\mathop{\rm Re}\nolimits 2\sigma>-1$, but its r.h.s. can be continued outside
of this region.

##  Appendix B The constant $D$

We have

$\displaystyle D=$ $\displaystyle
B(0)=\int_{0}^{\infty}\frac{2t^{4}+t^{2}-1+}{t^{4}+t^{2}}\frac{t^{3}dt}{e^{2\pi
t}-1}$ (B.374)
$\displaystyle+6\int_{0}^{\infty}\arctan\left(t\right)\frac{t^{2}dt}{e^{2\pi
t}-1}$ (B.375)
$\displaystyle+2\int_{0}^{\infty}\left(t^{2}-\frac{1}{2}\right)\log[t^{2}+t^{4}]\frac{tdt}{e^{2\pi
t}-1}.$ (B.376)

The first line gives

$\displaystyle\int_{0}^{\infty}\Bigg{(}2-\frac{1}{t^{2}}\Bigg{)}\frac{t^{3}dt}{e^{2\pi
t}-1}=2\zeta(4)\frac{\Gamma(4)}{(2\pi)^{4}}-\frac{\zeta(2)}{(2\pi)^{2}}=-\frac{1}{30}.$
(B.377)

For the second line we use

$\displaystyle\int_{0}^{\infty}\arctan\left(t\right)\frac{t^{2}dt}{e^{2\pi
t}-1}=-\frac{2}{9}+\frac{\zeta(3)}{8\pi^{2}}+\frac{1}{4}\log(2\pi)-\log A,$
(B.378)

where $A$ is the Glaisher-Kinkelin constant related to the first derivative of
the Riemann $\zeta$ function by [23]

$\displaystyle\log A=\frac{1}{12}-\zeta^{\prime}(-1).$ (B.379)

For the last line we use

$\displaystyle\int_{0}^{\infty}(2t^{2}-1)\log\left(1+t^{2}\right)\frac{t^{2}dt}{e^{2\pi
t}-1}=$
$\displaystyle\frac{11}{8}-\frac{3}{2}\left(\log(2\pi)+\frac{\zeta(3)}{\pi^{2}}\right)-2\zeta^{\prime}(-3)+5\log
A,$ (B.380) $\displaystyle\int_{0}^{\infty}(2t^{2}-1)\log
t^{2}\frac{t^{2}dt}{e^{2\pi t}-1}=$
$\displaystyle\frac{1}{60}\left(-\frac{19}{6}+\gamma-\log(2\pi)\right)+\frac{3\zeta^{\prime}(4)}{2\pi^{4}}+\log
A.$ (B.381)

The standard functional relation for $\zeta(z)$ gives

$\displaystyle\zeta^{\prime}(-3)=-\frac{4!}{2(2\pi)^{4}}\zeta^{\prime}(4)+\frac{B_{4}}{4}(H_{3}-\gamma-\log(2\pi)).$
(B.382)

On the other hand, we can use the Adamchik formula [24]

$\displaystyle\zeta^{\prime}(-3)=\frac{B_{4}}{4}H_{3}-\log A_{3},$ (B.383)

where $A_{3}$ is the third generalized Glaisher-Kinkelin number (or third
Bendersky’s number) defined by [23]

$\displaystyle\log A_{3}=$
$\displaystyle\lim_{n\to\infty}\Bigg{[}\sum_{k=1}^{n}k^{3}\log
k-\left(\frac{n^{4}}{4}+\frac{n^{3}}{2}+\frac{n^{2}}{4}-\frac{1}{120}\right)\log
n+\frac{n^{4}}{16}-\frac{n^{2}}{12}\Bigg{]}$ (B.384) $\displaystyle\simeq$
$\displaystyle-0.02065635,$ (B.385)

$H_{n}$ are the harmonic numbers, and $B_{n}$ the Bernoulli numbers. Summing
up, we get

$\displaystyle
D=-\frac{1}{72}+\frac{\gamma-\log(2\pi)}{60}-\frac{3\zeta(3)}{4\pi^{2}}+4\log(A_{3})\approx-0.2088707.$
(B.386)

##  Appendix C Poles from contiguity in hypergeometrics

Recall that

${}_{q+1}F_{q}\left(\left.\begin{array}[]{l}a_{1},\ \dots,\ a_{q+1}\\\ b_{1},\
\dots,\ b_{q}\\\
\end{array}\right|z\right)=\sum_{n=0}^{\infty}{(a_{1})_{n}\dots(a_{q+1})_{n}\,z^{n}\over
n!\,(b_{1})_{n}\dots(b_{q})_{n}}\ .$ (C.387)

Define

$Z=a_{1}+\cdots+a_{q+1}-b_{1}-\cdots-b_{q}\ ,$ (C.388)

then

$\begin{array}[]{llll}\hbox{the series is}&\hbox{absolutely convergent
for}&|z|<1&\\\ &\hbox{absolutely convergent for}&|z|=1&{\rm if}\ \ \mathop{\rm
Re}\nolimits Z<0\ ,\\\ &\hbox{conditionally convergent for}&|z|=1,\
z\not=1&{\rm if}\ \ 0\leq\mathop{\rm Re}\nolimits Z<1\ .\end{array}$ (C.389)

A general fact for ${}_{q+1}F_{q}$ is provided by the following formula [PBM3
p. 441 (37)](after a small correction), [Rainville (19) p. 719]

$\displaystyle\left[\sigma+z\sum_{j=1}^{q}(a_{j}-b_{j})\right]{}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\
\sigma\\\ (b_{q})\\\ \end{array}\right|z\right)+$ (C.392)
$\displaystyle+z\sum_{j=1}^{q}{(b_{j}-\sigma)\prod_{k=1}^{q}(b_{j}-a_{k})\over
b_{j}\prod_{k=1\atop
k\not=j}^{q}(b_{j}-b_{k})}{}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\
\sigma\\\ b_{1},\ldots,b_{j-1},b_{j}+1,b_{j+1},\ldots,b_{q}\\\
\end{array}\right|z\right)$ (C.395)
$\displaystyle=\sigma(1-z)\,{}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\
\sigma+1\\\ (b_{q})\\\ \end{array}\right|z\right).$ (C.398)

This requires $b_{j}\not=b_{k}$ for all $j\not=k$. Denote

$Z=\sigma+\sum_{j=1}^{q}(a_{j}-b_{j})\ .$ (C.399)

First take the values of the parameters such that

$\displaystyle\mathop{\rm Re}\nolimits Z<-1\ ,$ (C.400) $\displaystyle
b_{j}-b_{k}\not=0\ \ \forall\ \ j\not=k\ ,\ \ -b_{j}\not\in{\bf Z}_{+}\ .$
(C.401)

In this case it is legitimate to let $z$ tend to 1 everywhere and we get

$\displaystyle(Z)\ {}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\
\sigma\\\ (b_{q})\\\ \end{array}\right|1\right)=$ (C.404)
$\displaystyle-\sum_{j=1}^{q}{(b_{j}-\sigma)\prod_{k=1}^{q}(b_{j}-a_{k})\over
b_{j}\prod_{k=1\atop
k\not=j}^{q}(b_{j}-b_{k})}{}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\
\sigma\\\ b_{1},\ldots,b_{j-1},b_{j}+1,b_{j+1},\ldots,b_{q}\\\
\end{array}\right|1\right)\ .$ (C.407)

The rhs continues to be regular provided

$\displaystyle\mathop{\rm Re}\nolimits(1-Z)=\sum_{j=1}^{q}\mathop{\rm
Re}\nolimits b_{j}-\sum_{j=1}^{q}\mathop{\rm Re}\nolimits a_{j}-\mathop{\rm
Re}\nolimits\sigma+1>0\ ,$ (C.409) $\displaystyle b_{j}-b_{k}\not=0\ \
\forall\ \ j\not=k\ ,\ \ -b_{j}\not\in{\bf Z}_{+}\ .$ (C.410)

so that, in this region,

${}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\ \sigma\\\ (b_{q})\\\
\end{array}\right|1\right)={\Phi_{0}((a_{q}),\ \sigma,\ (b_{q}))\over Z}\ ,$
(C.411)

where $\Phi_{0}((a_{q}),\ \sigma,\ (b_{q}))$ is analytic in (C.410).

This process can be iterated, i.e. it can be applied to each of the
hypergeometic functions appearing in the rhs of (LABEL:r.210), and it can be
iterated further, so that, for any integer $n\geq 0$, there is a function
$\Phi_{n}((a_{q}),\ \sigma,\ (b_{q}))$, analytic in

$\displaystyle\mathop{\rm Re}\nolimits(n+1-Z)>0\ ,$ (C.412) $\displaystyle
b_{j}-b_{k}\notin{\bf Z}\ \forall\ \ j\not=k\ ,\ \ -b_{j}\not\in{\bf Z}_{+}\
,$ (C.413)

such that, in this region,

${}_{q+1}F_{q}\left(\left.\begin{array}[]{l}(a_{q}),\ \sigma\\\ (b_{q})\\\
\end{array}\right|1\right)={\Phi_{n}((a_{q}),\ \sigma,\ (b_{q}))\over
Z(Z-1)\ldots(Z-n)}\ .$ (C.414)

Of course some of the poles appearing in this formula may not actually occur,
i.e. their residues might be zero.

###### Remark C.1

Let $f$ be a function meromorphic on a domain $D\subset{\bf C}$ that has at
$a\in D$ a Laurent expansion

$f(z)={f_{-1}\over z-a}+f_{0}+f_{1}(z-a)+\ldots+f_{n}(z-a)^{n}+\ldots$ (C.415)

Then we have

$f_{-1}=(z-a)f(z)\Big{|}_{z=a}\ ,\ \ \ f_{0}=\Big{(}{d\over
dz}\Big{)}(z-a)f(z)\Big{|}_{z=a}\ ,\ \ \ f_{1}={\scriptstyle{1\over
2}}\Big{(}{d\over dz}\Big{)}^{2}(z-a)f(z)\Big{|}_{z=a}\ ,\ \ \ {\rm etc.}$
(C.416)

In case e.g. $f$ is a hypergeometric function that, as a function of $Z$, as
above, has a pole at $Z=a$, the contiguity formula (possibly iterated) may
provide an expression for $(Z-a)f(Z)$, as a sum of convergent hypergeometric
series, allowing a computation of the first terms of the Laurent series of $f$
at $a$. If

$f(z)={f_{-p}\over(z-a)^{p}}+\dots+{f_{-1}\over
z-a}+f_{0}+f_{1}(z-a)+\ldots+f_{n}(z-a)^{n}+\ldots$ (C.417)

then

$f_{n}={1\over(n+p)!}\Big{(}{d\over
dz}\Big{)}^{n+p}(z-a)^{p}f(z)\Big{|}_{z=a}\ ,\ \ \ n\geq-p\ .$ (C.418)

##  Appendix D Details about the derivation of the finite term in the three-
dimensional case

The hypergeometric series in (7.173) converges absolutely for
$4-8\delta=2(3-d)>0.$ The residue and the finite term at $d=3$ are computed as
follows:

$\displaystyle{R_{B}}(x,y,w,3)$ $\displaystyle=\lim_{d\to
3}\left((d-3)B_{d}(x,y,w)\right),$ (D.419)
$\displaystyle\widetilde{B}_{3}(x,y,w)$ $\displaystyle=\lim_{d\to 3}\
\partial_{d}\left((d-3)B_{d}(x,y,w)\right)$ (D.420)
$\displaystyle=b_{3}(x,y,w)F_{3}(x,y,w)+f_{3}(x,y,w)\partial_{d}b_{d}(x,y,w)$
(D.421)

where

$\displaystyle f_{3}(x,y,w)=\lim_{d\to 3}\ \ (d-3)\
{}_{7}F_{6}\left(\begin{array}[]{c}2\delta,\ldots,2\delta+2u\cr
2x+1,\ldots,\delta+w+u+1\end{array};1\right),$ (D.424) $\displaystyle
F_{3}(x,y,w)=\lim_{d\to 3}\ \frac{\partial}{\partial d}\
(d-3)\,{}_{7}F_{6}\left(\begin{array}[]{c}2\delta,\ldots,2\delta+2u\cr
2x+1,\ldots,\delta+w+u+1\end{array};1\right).$ (D.427)

To compute the above quantities the first step is to rewrite the
hypergeometric functions at the rhs of Eq. (7.173) by using the contiguity
relation

$\displaystyle(d-3)\,_{7}F_{6}\left(\begin{array}[]{l}2\delta,1+u+\delta,u-w+\delta,u+w+\delta,2u+2\delta,2x+2\delta,2y+2\delta\cr
1+2x,1+2y,1+2u,u+\delta,1+u-w+\delta,1+u+w+\delta\end{array};1\right)=$
(D.430)
$\displaystyle=\frac{\scriptstyle(2\delta-1)(1+2x-2\delta)(1-2y-2\delta)(1+2x-2y-2\delta)(x-y-\delta)(1-w+x-y-\delta)(1+w+x-y-\delta)\,_{7}F_{6}(2\delta,\ldots;2+2x,\ldots;1)}{8(1+2x)(x-y)y(1+x-y-\delta)(w+x-y-\delta)(w-x+y+\delta)}$
(D.431)
$\displaystyle+\frac{\scriptstyle(2\delta-1)(1+2y-2\delta)(x-y+\delta)(1+w-x+y-\delta)(1-w-x+y-\delta)(1-2x-2\delta)(1-2x+2y-2\delta)\,_{7}F_{6}(2\delta,\ldots;2+2y,\ldots;1)}{8x(x-y)(1+2y)(1-x+y-\delta)(w-x+y-\delta)(w+x-y+\delta)}$
(D.432)
$\displaystyle-\frac{\scriptstyle(2\delta-1)(1+2x-2\delta)(1+2y-2\delta)(1+2x+2y-2\delta)(x+y-\delta)(1-w+x+y-\delta)(1+w+x+y-\delta)\,_{7}F_{6}(2\delta,\ldots;2+2x+2y,\ldots;1)}{8xy(1+2x+2y)(1+x+y-\delta)(w-x-y+\delta)(w+x+y-\delta)}$
(D.433) $\displaystyle+\frac{\scriptstyle
w^{2}(x-y-\delta)(x+y-\delta)(x-y+\delta)\,_{7}F_{6}\left(2\delta,\ldots;1+x+y+\delta,\ldots;1\right)}{2(1+x-y-\delta)(1-x+y-\delta)(1+x+y-\delta)(1-w)(1+w)}$
(D.434)
$\displaystyle-\frac{\scriptstyle(1+2w)(1+w-x-y-\delta)(1+w+x-y-\delta)(1+w-x+y-\delta)(1+w+x+y-\delta)\,_{7}F_{6}(2\delta,\ldots;2+w+x+y+\delta,\ldots;1)}{4(1+w)(w-x-y+\delta)(w+x-y+\delta)(w-x+y+\delta)(1+w+x+y+\delta)}$
(D.435)
$\displaystyle+\frac{\scriptstyle(1-2w)(1-w-x-y-\delta)(1-w+x-y-\delta)(1-w-x+y-\delta)(1-w+x+y-\delta)\,_{7}F_{6}(2\delta,\ldots;2-w+x+y+\delta,\ldots;1)}{4(1-w)(w+x+y-\delta)(w-x+y-\delta)(w+x-y-\delta)(1-w+x+y+\delta)}.$
(D.436)

At the r.h.s., we have only specified the parameters of the hypergeometric
functions which are shifted by 1; the hypergeometric functions at the r.h.s.
converge for $d<7/2$. Eq. (D.436) gives us immediately the residue at $d=3$ of
the hypergeometric function at the l.h.s.:

$\displaystyle f_{3}(x,y,w)=\frac{4w^{2}-(2x+2y+1)^{2}}{8x+8y+4}.$ (D.437)

Calculating the finite part is more difficult. Let us start with the easiest:

$\displaystyle\left.f_{3}(x,y,w)\,\partial_{d}b_{d}(x,y,w)\right|_{d=3}=-\frac{(2x+2y+1)(\cot(2\pi
x)\cot(2\pi y)-1)}{128\pi^{2}\left(4w^{2}-(2x+2y+1)^{2}\right)}+$ (D.438)
$\displaystyle-\frac{(\cot(2\pi x)\cot(2\pi
y)-1)\left(\psi(2x+1)+\psi(2y+1)-\gamma+2+\log\left(4\pi^{2}\right)\right)}{256\pi^{2}}+$
(D.439) $\displaystyle+\frac{(\cot(2\pi x)\cot(2\pi
y)-1)\left(\psi\left(2x+2y+\frac{3}{2}\right)-2\psi(-4x-4y-2)\right)}{256\pi^{2}}.$
(D.440)

On the other hand computing $F_{3}(x,y,w)$ is quite involved. The first three
lines at the r.h.s. of (D.436) may be grouped into a single expression that we
denote $(d-3)R_{123}$; their contribution to $F_{3}(x,y,z)$ is therefore

$\displaystyle R_{123}(x,y,w)=$ (D.441)
$\displaystyle=\frac{\scriptscriptstyle
x(2x-2y-1)\left((1+2x+2y)^{2}-4w^{2}\right)\left((2+4x-4y)\psi(1+2x)-(1-2w+2x-2y)\psi\left(\frac{1}{2}-w+x+y\right)-(1+2w+2x-2y)\psi\left(\frac{1}{2}+w+x+y\right)\right)}{4\left((1+2x)^{2}-4y^{2}\right)\left((1-2x+2y)^{2}-4w^{2}\right)}+$
(D.442) $\displaystyle\frac{\scriptscriptstyle
y(2y-2x-1)\left((1+2x+2y)^{2}-4w^{2}\right)\left((2-4x+4y)\psi(1+2y)-(1-2w-2x+2y)\psi\left(\frac{1}{2}-w+x+y\right)-(1+2w-2x+2y)\psi\left(\frac{1}{2}+w+x+y\right)\right)}{4\left((1+2y)^{2}-4x^{2}\right)\left((1+2x-2y)^{2}-4w^{2}\right)}+$
(D.443)
$\displaystyle\frac{\scriptscriptstyle(x+y)(2x+2y-1)\left((1+2x+2y)^{2}-4w^{2}\right)\left((1-2w+2x+2y)\psi\left(\frac{1}{2}-w+x+y\right)+(1+2w+2x+2y)\psi\left(\frac{1}{2}+w+x+y\right)-2(1+2x+2y)\psi(1+2x+2y)\right)}{4(1+2x+2y)^{2}\left(4w^{2}-(1-2x-2y)^{2}\right)}.$
(D.444)

The other terms are trickier as they involve derivatives of the hypergeometric
functions. Let us consider the fourth line. After some work we get the
following expression:

$\displaystyle R_{4}=\frac{\scriptstyle
w\left((1-2x-2y)\left((1+2x+2y)^{2}-4w^{2}\right)\right)\left(\left(H_{-\frac{1}{2}-w+x+y}\right){}^{2}-\left(H_{-\frac{1}{2}+w+x+y}\right){}^{2}\right)}{32\left(w^{2}-1\right)(1+2x+2y)}$
(D.446) $\displaystyle+\frac{\scriptstyle
w\left(2w^{2}\left(-3+12x^{2}+8xy+12y^{2}\right)+(1+2x+2y)^{2}\left(2-x-8x^{2}+4x^{3}-y-4x^{2}y-8y^{2}-4xy^{2}+4y^{3}\right)\right)\psi\left(\frac{1}{2}+w+x+y\right)}{4\left(-1+w^{2}\right)(1+2x+2y)^{2}\left(-1+4x^{2}-8xy+4y^{2}\right)}$
(D.447) $\displaystyle-\frac{\scriptstyle
w\left(2w^{2}\left(-3+12x^{2}+8xy+12y^{2}\right)+(1+2x+2y)^{2}\left(2+4x^{3}-y-8y^{2}+4y^{3}-4x^{2}(2+y)-x\left(1+4y^{2}\right)\right)\right)\psi\left(\frac{1}{2}-w+x+y\right)}{4\left(-1+w^{2}\right)(1+2x+2y)^{2}\left(-1+4x^{2}-8xy+4y^{2}\right)}$
(D.448) $\displaystyle+\frac{\scriptstyle
2w^{2}\left(-\frac{1}{2}+x+y\right)\left({}_{4}\dot{F}_{3}\left(\left\\{1+2x,1,\frac{1}{2}-w+x+y,\frac{1}{2}+w+x+y\right\\},\left\\{1+2x,\frac{3}{2}-w+x+y,\frac{3}{2}+w+x+y\right\\},1\right)\right)}{4(-1+w)(1+w)\left(\frac{1}{2}+x+y\right)}$
(D.449) $\displaystyle+\frac{\scriptstyle
2w^{2}\left(-\frac{1}{2}+x+y\right)\left({}_{4}\dot{F}_{3}\left(\left\\{1+2y,1,\frac{1}{2}-w+x+y,\frac{1}{2}+w+x+y\right\\},\left\\{1+2y,\frac{3}{2}-w+x+y,\frac{3}{2}+w+x+y\right\\},1\right)\right)}{4(-1+w)(1+w)\left(\frac{1}{2}+x+y\right)}$
(D.450) $\displaystyle+\frac{\scriptstyle
2w^{2}\left(-\frac{1}{2}+x+y\right)\left({}_{4}\dot{F}_{3}\left(\left\\{1+2x+2y,1,\frac{1}{2}-w+x+y,\frac{1}{2}+w+x+y\right\\},\left\\{1+2x+2y,\frac{3}{2}-w+x+y,\frac{3}{2}+w+x+y\right\\},1\right)\right)}{4(-1+w)(1+w)\left(\frac{1}{2}+x+y\right)}.$
(D.451)

As regards the fifth and the sixth line it is useful to compute at first the
following auxiliary expression

${}_{4}F_{3}\left(\begin{array}[]{r
}s,\frac{3}{2}+u,\frac{1}{2}+u-w,\frac{1}{2}+u+w\cr\frac{1}{2}+u,\frac{3}{2}+u-w,\frac{5}{2}+u+w\end{array};1\right)=$
(D.455)
$\displaystyle=\frac{\Gamma(1-s)\left(\frac{(1+2u+2w)(3+2u+2w)\Gamma\left(\frac{3}{2}+u-w\right)}{\Gamma\left(\frac{3}{2}-s+u-w\right)}-\frac{(1+2u-2w)(-1+2u-2w+s(2+4w))\Gamma\left(\frac{5}{2}+u+w\right)}{\Gamma\left(\frac{5}{2}-s+u+w\right)}\right)}{4(1+2u)(1+2w)}$
(D.456)

which allows to compute ${}_{4}F_{3}\left(\begin{array}[]{r
}1,\frac{3}{2}+u,\frac{1}{2}+u-w,\frac{1}{2}+u+w\cr\frac{1}{2}+u,\frac{3}{2}+u-w,\frac{5}{2}+u+w\end{array};1\right)$,
its derivative w.r.t $u$ and ${}_{4}\dot{F}_{3}\left(\begin{array}[]{r
}1,\frac{3}{2}+u,\frac{1}{2}+u-w,\frac{1}{2}+u+w\cr\frac{1}{2}+u,\frac{3}{2}+u-w,\frac{5}{2}+u+w\end{array};1\right)$.
We get in this way an expression for $R_{5}$ where, as before, there remain
three terms expressed as derivatives of hypergeometric functions:

$\displaystyle
R_{5}(x,y,w)=\frac{\scriptstyle(1+2w+2x+2y)\left(4w^{2}+4w^{3}+(1+2x+2y)^{2}(w+2(1+x+y))\right)}{32(1+w)(1+2x+2y)^{2}(3+2w+2x+2y)}(H_{-\frac{1}{2}-w+x+y}-H_{\frac{1}{2}+w+x+y})$
(D.458)
$\displaystyle+\frac{\gamma\left(4w^{2}-(1+2x+2y)^{2}\right)\left(2+4w+(1+2w+2x+2y)\left(H_{\frac{1}{2}+w+x+y}-H_{-\frac{1}{2}-w+x+y}\right)\right)}{64(1+w)(1+2x+2y)}$
(D.459)
$\displaystyle+\frac{(1-2w+2x+2y)(1+2w+2x+2y)^{2}\left(\psi\left(\frac{1}{2}-w+x+y\right)^{2}-\psi\left(\frac{3}{2}+w+x+y\right)^{2}\right)}{128(1+w)(1+2x+2y)}+$
(D.460)
$\displaystyle+\frac{\scriptstyle\left(5+8w-\frac{8(1+w)(1+2w)}{1+2w-2x-2y}-\frac{8(1+w+x)(1+2w+2x)}{1+2w+2x-2y}-\frac{8x(1+2x)}{1+2w-2x+2y}\right)}{16(1+w)(1+2x+2y)(6+4w+4x+4y)}\left((1+2x+2y)^{2}-4w^{2}\right)\times$
(D.461)
$\displaystyle\times\left(\psi\left(\frac{1}{2}-w+x+y\right)+(1+2w)\psi\left(\frac{1}{2}+w+x+y\right)-2(1+w)\psi\left(\frac{3}{2}+w+x+y\right)\right)$
(D.462)
$\displaystyle+\frac{\scriptscriptstyle(1+2w)(1+2w+2x+2y)\left(4w(1+w)+(1+2x+2y)^{2}-(1-2w+2x+2y)(1+2x+2y)(3+2w+2x+2y)\left(H_{\frac{1}{2}+w+x+y}-\gamma\right)\right)}{32(1+w)(1+2x+2y)^{2}(3+2w+2x+2y)}$
(D.463) $\displaystyle-\frac{\scriptscriptstyle(1+2w)(1+2w+2x+2y)\
{}_{5}\dot{F}_{4}\left(\left\\{1+2x,\frac{3}{2}+x+y,\frac{1}{2}-w+x+y,\frac{1}{2}+w+x+y,1\right\\},\left\\{1+2x,\frac{1}{2}+x+y,\frac{3}{2}-w+x+y,\frac{5}{2}+w+x+y\right\\},1\right)}{8(1+w)(3+2w+2x+2y)}$
(D.464) $\displaystyle-\frac{\scriptstyle(1+2w)(1+2w+2x+2y)\
\left({}_{5}\dot{F}_{4}\left(\left\\{1+2y,\ldots\right\\},\left\\{1+2y,\ldots\right\\},1\right)+{}_{5}\dot{F}_{4}\left(\left\\{1+2x+2y,\ldots\right\\},\left\\{1+2x+2y,\ldots\right\\},1\right)\right)}{8(1+w)(3+2w+2x+2y)}.$
(D.465)

Finally, $R_{6}(x,y,w)=R_{5}(x,y,-w)$. Collecting all the above terms we are
able to fully construct $F_{3}(x,y,w)$ determine $B_{3}(x,y,w)$ and finally
$\tilde{I_{2}}(x,y,w)$. The result is however utterly complicated. It
simplifies a lot when $w=0$ by taking into account the following identity:

$\displaystyle{}_{3}\dot{F}_{2}\left(\begin{array}[]{l}a,b,1\cr
a,b+2\end{array};1\right)=\left.\frac{d}{ds}F_{2}\left(\begin{array}[]{l}a+s,b,1\cr
a,b+2\end{array};1\right)\right|_{s=0}=\sum_{n=1}^{\infty}\frac{b(1+b)(\psi(a+n)-\psi(a))}{(b+n)(1+b+n)}$
(D.470)
$\displaystyle=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{b(1+b)}{(b+n)(1+b+n)(k+a)}=\sum_{k=0}^{\infty}\sum_{n=k+1}^{\infty}\frac{b(1+b)}{(b+n)(1+b+n)(k+a)}$
(D.471)
$\displaystyle=\sum_{k=0}^{\infty}\frac{b(1+b)}{(a+k)(1+b+k)}=\frac{b(1+b)(\psi(1+b)-\psi(a))}{1-a+b}.$
(D.472)

Collecting everything we land on an indeed very simple result:

$\displaystyle
F(x,y,0)=\frac{(2x+2y+1)}{8}\left(H_{2x+2y}+H_{2x}+H_{2y}-4H_{x+y-\frac{1}{2}}\right)-\frac{1}{8}.$
(D.473)

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