formfactors $F_{1}(\Box)$ and $F_{2}(\Box)$ follow from the RG running of the
relevant constants $\lambda^{(0)}_{2}(\mu)$ and $\lambda^{(2)}_{2}(\mu)$, but
the transition $\lambda^{(0,2)}_{2}(\mu)\to F_{1,2}(\Box)$ is not
straightforward and is mediated by Eqs.(6.23) and (6.27)-(6.28).

##  7 Conclusions

To summarize our notes on conformal anomaly, nonlocal effective action and
running scales let us briefly dwell on possible applications of our results
and related issues.

As it is clear from the above considerations, the conformal anomaly action is
a carrier of the effective rather than fundamental conformal degree of
freedom. Either in the nonlocal or the Wess-Zumino form, it is the difference
of action functionals of two configurations belonging to the orbit of the
conformal group. So unless one of these actions is known the corresponding
physical setup is not complete. In this respect, our approach is very
different from the works which endow the conformal factor $e^{2\sigma}$ the
nature of the fundamental field [77] or, for example, ready to sacrifice the
fundamental nature of the Higgs boson in favor of 36 fundamental zero
dimension scalars $\sigma$ for the sake of a complete eradication of Weyl
anomaly and justification of the primordial cosmological perturbations spectra
[78, 79].

The CFT driven cosmology of Sect.5.1 seems to present such an example where
the physical setup is complete within a certain approximation scheme. This
approximation is associated with the dominance of conformal invariant matter
fields over the loop effects of gravity and other types of matter and
simultaneously puts the model in the subplanckian domain of energies below the
cutoff $M_{P}/\sqrt{\beta}$ when the coefficient of the topological conformal
anomaly $\beta\gg 1$ [43]. To match with the widely accepted bounds on the
energy scale of inflation $\sim 10^{-6}M_{P}$ one needs $\beta\sim 10^{13}$,
which cannot be attained by a contribution of low spin conformal fields
$\beta=(1/360)(N_{0}+11N_{1/2}+62N_{1})$ unless the numbers $N_{s}$ of fields
of spin $s$ are tremendously high.

On the contrary, this bound can be reached by appealing to the idea of
conformal higher spin (CHS) fields [45]. A relatively low tower of higher
spins will be needed, because a partial contribution of spin $s$ to $\beta$
grows as $s^{6}$. These partial contributions $\beta_{s}$ for CHS totally
symmetric tensors and Dirac spin-tensors read in terms of $\nu_{s}$ – their
respective numbers of polarizations (negative for fermions) [80, 81],

$\displaystyle\beta_{s}=\frac{\nu_{s}^{2}(3+14\nu_{s})}{720},\,\nu_{s}=s(s+1),\,s=1,2,3,...\,,$
(7.1)
$\displaystyle\beta_{s}=\frac{\nu_{s}(12+45\nu_{s}+14\nu_{s}^{2})}{1440},$
$\displaystyle\nu_{s}=-2\Big{(}s+\tfrac{1}{2}\Big{)}^{2},\;s=\tfrac{1}{2},\tfrac{3}{2},\tfrac{5}{2},...\,.$
(7.2)

The solution of hierarchy problem thus becomes a playground of $1/N$-expansion
theory for large number $N$ of conformal species. Moreover, with the inclusion
of CHS fields the status of conformal anomaly essentially changes and becomes
similar to that of the chiral anomaly. Chiral anomaly has phenomenological
confirmation within chiral symmetry breaking theory, it also has important
implications in lepton physics, physics of early Universe, its baryon
asymmetry theory, etc. It has a topological nature and is generated in virtue
of Adler-Bardeen theorem only at the one-loop level. Local Weyl anomaly also
has topological (a-type) contribution [13], but for low spins it is
contributed by all orders of loop expansion. CHS spins, however, have their
inverse propagators $\sim\Box^{s}+...$ and, therefore, for high $s$ are UV
finite beyond one loop approximation. So their Weyl anomaly is also exhausted
by the one-loop contribution, and there is a hope that their effect in the CFT
driven cosmology is nonperturbative. As this effect intrinsically, by a
dynamical mechanism of effective equations of motion [27, 43, 44, 46],
provides the upper bound on the energy range of inflation
$M_{P}/\sqrt{\beta}\ll M_{P}$, this also justifies omission of graviton loops
and quantum effects of other (non-conformal) types of matter.

There are, however, serious problems on the road to the realization of this
model. To begin with, CHS fields in curved spacetime are not explicitly known
yet, except conformal gravitino with $s=3/2$ and Weyl graviton with $s=2$.
Recent progress in generalizing these models to arbitrary $s$ on the Einstein-
space background allowed one to compute their 1-loop Weyl anomaly coefficients
(7.1)-(7.2) (by indirect AdS/CFT method in [80] and directly in [81]). This
result, however, leaves the issue of unitarity violation caused by inevitable
higher derivatives in wave operators of these fields. Moreover, these fields
should form a hidden sector not observable at present, which implies the
necessity of their eradication in the course of cosmological expansion. What
might be useful for this purpose is the idea of renormalization group flow
from UV to IR decreasing the value of $\beta$ (the so called $a$-theorem of
[41, 42, 67]) or Weyl symmetry breaking which would generate masses of CHS
fields and thus shorten their massless tower. Finally and most importantly,
the fundamental theory of these interacting CHS fields should necessarily be
organized within a special higher spin symmetry [82, 83]. A complete version
of this theory is still missing, not to say about its constructive extension
to curved spacetime. Thus, the progress here strongly depends on advancing
theory of CHS fields [84, 85, 86, 87, 88].

The issue of RG running constants $\varLambda$ and $G$ has, as it is shown, a
rather unexpected resolution. The manifestation of this UV running actually
takes place in the nonlocal formfactors of the quadratic curvature (dimension
four) terms rather than in the sector of low dimension operators. This
metamorphosis originates from establishing a rather nontrivial scaling
operator (6.11) embracing all powers of the curvature expansion and exploiting
a conventional RG assumption that the renormalized theory does not depend on
the choice of normalization (or subtraction) point. Then simple, though
somewhat tedious, dimensionality considerations lead to this result. Dimension
zero and dimension two cosmological and Einstein terms do not run themselves
but still contribute to the running of the dimension four terms which can be
considered as UV partners of $\varLambda$ and $G$. This metamorphosis of RG
running couplings into the formfactors of the curvature squared terms sounds
important, because it is the quadratic term in the effective action that
mainly determines either the asymptotic freedom of the model or its cutoff
beyond which effective field theory breaks down.

In the IR domain these partners, due to the presence of mass scale $M$, also
start from the quadratic order in the curvature, but they have essential
nonlocality – of the type $M^{4}\int d^{4}x\,\sqrt{g}(\tfrac{1}{\Box}\Re)^{2}$
coming from cosmological constant sector [32] and of the form $M^{2}\int
d^{4}x\,\sqrt{g}\Re\tfrac{1}{\Box}\Re$ originating from the gravitational
constant one. While the latter is well defined in IR limit due to the
subtraction from the IR divergent bulk Einstein action of the Gibbons-Hawking
surface term [51, 52], for the IR cosmological partner [32] the situation is
trickier – in view of IR divergences it requires additional subtraction
procedure. Perhaps even more radical changes will be needed to circumvent this
problem like the curvature expansion on top of the homogeneous (dS or AdS)
background with nonzero curvature.

Of course, there can be other choices of the running scale $D$ different from
(6.11. Nothing prevents from replacing it, say, with
$(\sum_{N}(-\Box_{N})^{k})^{1/2k}$ or other combinations of contracted
derivatives. However, for curvature squared terms of the action all such
choices (satisfying the homogeneity property with respect to derivative
rescalings) lead to one and the same operator $\sim(-\Box)^{1/2}$ because for
the second order of the curvature expansion all d’Alembertians reduce to the
single one, $\Box_{1}=\Box_{2}$, in view of integration by parts (6.25). The
only ambiguity is the choice of the d’Alembertian itself, but it is fixed by
the requirement of general covariance. Alterations in the choice of $D$
certainly affect higher orders in the curvature, but the curvature squared
part, which is most important for UV asymptotic freedom or determination of
the effective field theory cutoff, stays uniquely defined.

Ambiguity in the choice of $D$ can arise in the class of theories which have a
more or less conventional RG running of the gravitational coupling $G$ –
renormalizable Hořava gravity models [53, 89]. In these Lorentz symmetry
violating models a possible covariant curvature expansion undergoes
$(3+1)$-splitting – the set of basic curvatures includes the extrinsic
curvature $K_{ij}$, $i,j=1,2,3$, of spatial slices of constant time $\tau$.
The Einstein term of general relativity is replaced by the sum of the kinetic
term $\sim(16\pi G)^{-1}\\!\int d^{4}x\,\sqrt{g}K_{ij}^{2}$ and the potential
term built as a polynomial in 3-dimensional curvature and its spatial
derivatives. The RG running of $G$ in the kinetic term proceeds as the
insertion of the form factor $G^{-1}(D)$ between two factors of $K_{ij}(x)$,

$\frac{1}{G}\\!\int d^{4}x\sqrt{g}\,K_{ij}^{2}\to\int
d^{4}x\sqrt{g}\,K_{ij}(x)\frac{1}{G(D)}K^{ij}(x).$ (7.3)

Thus no tadpole problem for the RG running of $G$ takes place here – just like
in Yang-Mills type theories this occurs without forming a total derivative
structure.

However the relevant scaling operator $D$ of a unit anisotropic scaling
dimension, which replaces the spacetime covariant square root of $(-\Box)$,
turns out to be ambiguous. Point is that in Lorentz violating models the
notion of physical scaling dimension is replaced by the anisotropic one which
in $(3+1)$-dimensional Hořava gravity is $-3$ for the time coordinate and $-1$
for spatial coordinates. Correspondingly the dimension $6$ wave operator of
the theory is of the second order in time derivatives and of the sixth order
in spatial derivatives. Therefore,
$D\sim(-\partial_{\tau}^{2}-\Delta^{3}/M^{4})^{1/6}$ where $\Delta$ is the
spatial covariant Laplacian and $M$ is a physical mass scale parameter. This
parameter may be different in various (scalar and transverse-traceless)
sectors of the metric field [89], and this is a source of ambiguity in the
running scale of Hořava models. Modulo this problem RG running in
renormalizable non-projectable Hořava gravity is well defined and in
$(3+1)$-dimensional case has a legitimate interpretation of asymptotic freedom
[89].

## Acknowledgements

Essential part of this paper was inspired during the workshop “Quantum
Effective Field Theory and Black Hole Tests of Einstein Gravity” (IFPU,
Miramare, Trieste, Italy, September 12-16, 2022), and A.O.B. is very grateful
to organizers and participants of this workshop. The authors deeply appreciate
the efforts by J.Donoghue, M.Duff, E.Mottola and H.Osborn of critically
reading our manuscript. A.O.B. is also grateful for fruitful discussions and
correspondence with John Donoghue, Michael Duff, Alexander Kamenshchik, Emil
Mottola, Roberto Percacci, Hugh Osborn, Ilya Shapiro, Kostas Skenderis, Arkady
Tseytlin, Alex Vikman, Richard Woodard and especially to G. A. Vilkovisky for
long term collaboration on covariant perturbation theory for quantum effective
action. This work was supported by the Russian Science Foundation grant No
23-12-00051.

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Mellin–Barnes transformation for two-loop master-diagram

S. E. Derkachev‡ , A. V. Ivanov† , L. A. Shumilov♯

‡†St. Petersburg Department of the Steklov Mathematical Institute of Russian
Academy of Sciences,
Fontanka 27, 191023 St. Petersburg, Russia
♯St. Petersburg Academic University,
Khlopina str. 8/3, St. Petersburg, 194021, Russia

‡E-mail<EMAIL_ADDRESS>
†E-mail<EMAIL_ADDRESS>
♯E-mail<EMAIL_ADDRESS>
###### Abstract

In the paper, we obtain an expression for a two-loop master-diagram by using
the Mellin–-Barnes transformation. In the two-dimensional case we managed to
factorize the answer and write it as a bilinear combination of hypergeometric
functions ${}_{3}F_{2}$.

##  1 Introduction

The method of separation of variables, developed by E.K. Sklyanin [3, 4, 5],
is one of the effective methods in the theory of quantum integrable models
(see reviews [1, 2, 3]). It is based on the transition from the coordinate
representation to the separated variables representation, in which the problem
of studying the original Hamiltonian is drastically simplified. Such a unitary
transformation is carried out by an integral operator, the kernel of which is
the mutual eigenfunctions of a special family of commuting operators. The
quantum Toda chain [4, 6, 7] and the two-dimensional model of the non-compact
spin chain [11] are two meaningful examples of quantum integrable models where
the program of separation of variables was fully implemented.

In the works of L.N. Lipatov [8, 9], L.D. Faddeev, and G.P. Korchemskiy [10],
it was discovered that the two-dimensional model of a non-compact spin chain
describes high-energy processes in quantum chromodynamics. The operator of
transition to the representation of separated variables for this model was
constructed explicitly in [11]. In the simplest case, either a two-dimensional
Fourier transform or a two-dimensional generalization of the Mellin transform
is obtained, so the general case is a far-reaching generalization of the two
classical integral transforms. The transition to separated variables turned
out to be a useful tool for calculating multiloop Feynman diagrams in quantum
field theory. Basso and Dixon’s work "Gluing Ladder Feynman Diagrams into
Fishnets" [14] proposed a remarkable determinant representation of a special
class of Feynman diagrams, which play a dominant role in the four-dimensional
Fishnet CFT model [12] as a hypothesis. Then, in [15], using an integral
transformation to separated variables, an analogue of determinant Basso–Dixon
representation was derived for the diagrams in two-dimensional Fishnet CFT
[13], and in [16] a four-dimensional generalization was obtained. Thus, the
Basso–Dixon conjecture was proved, and the main tool used was the
transformation to separated variables. The necessary generalization of the
basic formulae [11] from two-dimensional to four-dimensional case was
developed in [16].

An important open question is the generalization to an arbitrary number of
dimensions, but in the simplest case the necessary formulae are quite obvious
and are known [13, 17]. By analogy with the two-dimensional case, we will call
the simplest integral transform the Mellin–Barnes one. In this work the
Mellin–Barnes transform is used to calculate the two-loop master diagram [32,
33].

The article is organized as follows. In the second section a general
calculation scheme is formulated. In the third section the scheme is applied
to the calculation of the two-loop diagram in the two-dimensional case. The
last section is devoted to the case of an arbitrary number of dimensions. In
the Appendix we give a proof of the unitarity of the Mellin–Barnes transform
for an arbitrary number of dimensions.

The two-dimensional case is highlighted for two reasons. First, in the case,
all calculations can be carried out for a diagram with arbitrary tensor lines,
not being limited only to scalar propagators. Secondly a remarkable
factorization occurs: a two-fold sum is factored into the product of two one-
fold sums.

##  2 Formularion of the problem

In this section we will formulate the problem schematically. Suppose
$\mathcal{N}$ is an abstract set of indices. Elements of the set
$\alpha\in\mathcal{N}$ can have a complex structure, including continuous and
discrete parameters. Let $\mathcal{V}$ be the region of integration and let
the variables $z$ be functions $\mathcal{V}\to\mathbb{C}$. We will notate the
integration measure as $\mathcal{D}z$. Let us consider the family of functions

$D^{\alpha}(z_{1},z_{2}):\,\mathcal{V}\times\mathcal{V}\to\mathbb{\overline{C}},\,\,\,\alpha\in\mathcal{N}.$
(1)

Such functions are the basic elements of diagram technique and is denoted by a
directed line from point $z_{1}$ to point $z_{2}$ with index $\alpha$, as
shown in Fig. 1. We denote the integration over $\mathcal{V}$ with measure
$\mathcal{D}z$ by a point.

Fig. 1: The <<line>>

The subject of this work is shown in Fig. 2. Let us note that such integral is
understood in the sense of analytic continuation from the chosen region of
convergence to the complex plane as a function of external parameters
$\alpha_{1},\ldots,\alpha_{5}$.

Fig. 2: Two-loop diagram.

As a rule, in the theory of integrable models the functions $D^{\alpha}$ have
additional properties that simplify the problem and let us write out an
explicit answer. We highlight the three main ones:

  1. 1.

Let $\alpha,\beta\in\mathcal{N}$, then

$D^{\alpha}(z_{1},z_{2})D^{\beta}(z_{1},z_{2})=D^{c}(z_{1},z_{2}),$ (2)

where $c\in\mathcal{N}$ is a function of parameters $\alpha$ and $\beta$;

  2. 2.

Let $\alpha,\beta\in\mathcal{N}$, then the chain rule is shown in Fig. 3,
where $\hat{c}\in\mathcal{N}$ is a function of $\alpha$ and $\beta$;

  3. 3.

Let $\alpha\in\mathcal{N}$, then the Mellin–Barnes representation is shown in
Fig. 4, where $\mathcal{\tilde{D}}s$ is measure on the set $\mathcal{N}$,
$z_{0}$ is an arbitrary parameter, $\tilde{c}_{1}\in\mathcal{N}$ and
$\tilde{c}_{2}\in\mathcal{N}$ are functions of the parameters $\alpha$ and
$s$.

Fig. 3: The chain rule.

Fig. 4: Mellin–Barnes representation.

In what follows we assume that all of these properties are satisfied. They
allow us to easily get the answer for the diagram. Indeed, applying
Mellin–Barnes transform to the diagonal with index $\alpha_{5}$ shown in Fig.
2, choosing arbitrary parameter $z_{0}$ equals to $z_{1}$ (or $z_{2}$) and
applying property (2) twice and then the chain rule, we get the answer in the
following form

$D^{b}(z_{1},z_{2})=\int\tilde{\mathcal{D}}s\,C(\alpha_{5},s)B(b_{1}(s),\alpha_{4})B(b_{2}(s),\alpha_{3}),$
(3)

where $b_{1}(s)=c(\alpha_{1},\tilde{c}_{1}(\alpha_{5},s))$,
$b_{2}(s)=c(\alpha_{2},\tilde{c}_{2}(\alpha_{5},s))$, and
$b=c(\hat{c}(b_{1},\alpha_{4}),\hat{c}(b_{2},\alpha_{3}))$.

In the rest of the work we present two special cases (two-dimensional and
multidimensional). In the first case properties 1) –3) are already known and
the main result is the calculation of integral (3) and further factorization
of the answer. The second case is devoted to the study of the above properties
2) and 3), as well as the calculation of the resulting integral by residues.

##  3 Two-dimensional case

In this case the set of indices $\mathcal{N}$ is $\mathbb{C}$. Moreover, it is
convenient to define <<conjugate>> index $\overline{\alpha}$ for each
$\alpha\in\mathbb{C}$ by the rule
$\alpha-\overline{\alpha}=n_{\alpha}\in\mathbb{Z}$. For convenience, let us
also define auxiliary functions

$a(\alpha)=\frac{\Gamma(1-\overline{\alpha})}{\Gamma(\alpha)}\,\,\,\,\,\,\mbox{and}\,\,\,\,\,\,a(\overline{\alpha})=\frac{\Gamma(1-\alpha)}{\Gamma(\overline{\alpha})},$
(4)

where $\Gamma$ is Euler’s Gamma function. For brevity the product of such
objects we denote by $a(\alpha_{1},\ldots,\alpha_{n})=a(\alpha_{1})\cdots
a(\alpha_{n})$. One can verify that such coefficients have the properties

$\displaystyle
a(\alpha)a(1-\overline{\alpha})=1,\,\,\,\,\,\,a(\alpha)a(1-\alpha)=(-1)^{n_{\alpha}},$

and have simple poles, distributed according to the following lemma.

######  Lemma 3.1.

Let $n\in\mathbb{N}$, $m\in\mathbb{Z}$, $x,y\in\mathbb{R}$,
$\nu\in\mathbb{C}$, $s=\frac{n}{2}+i\nu$, $\overline{s}=-\frac{n}{2}+i\nu$,
$\overline{\alpha}=x+iy$, and $\alpha=x+m+iy$, then function $a(1-\alpha\pm
s)$ has simple poles with respect to the variable $\nu$ at the points

$\nu=\mp i\left(\overline{\alpha}+\frac{m+\mid n\mp
m\mid}{2}+k\right),\,k\in\mathbb{N}\cup\\{0\\}.$

Proof: It is sufficient to consider the case with the plus sign. Let us
introduce auxiliary variables

$n=n^{\prime}+m,\,\nu=ip+y,\,b=x+\frac{m}{2},$

then we can rewrite the function in the following form:

$a(1-\alpha+s)=\frac{\Gamma(\overline{\alpha}-\overline{s})}{\Gamma(1-\alpha+s)}=\frac{\Gamma(x+iy+\frac{n}{2}-i\nu)}{\Gamma(1-m-x-
iy+\frac{n}{2}+i\nu)}=\frac{\Gamma(b+\frac{n^{\prime}}{2}+p)}{\Gamma(1-b+\frac{n^{\prime}}{2}-p)}.$

Taking into account the poles of the upper Gamma function and the zeros of the
lower one, we can conclude that singularities exist at the points
$p=-b-\frac{\mid n^{\prime}\mid}{2}-k$, $k\in\mathbb{N}\cup\\{0\\}$ with
respect to the variable $p$. The transition to the original variables proves
the lemma. $\blacksquare$

The integration region $\mathcal{V}$ coincides with $\mathbb{C}$, the
variables are complex numbers, and the integration measure has the form
$\mathcal{D}z=d\,\mathrm{Re}(z)d\,\mathrm{Im}(z)$.

Let $\nu\in\mathbb{R}$, $n\in\mathbb{Z}$, $s=\frac{n}{2}+i\nu$, and
$\overline{s}=-\frac{n}{2}+i\nu$, then, taking into account the definitions

$\int\limits\tilde{\mathcal{D}}s=\sum\limits_{n\in\mathbb{Z}}\int_{\mathbb{R}}d\nu\,\,\,\,\,\,\mbox{and}\,\,\,\,\,\,D^{\alpha}(z_{1},z_{2})=(z_{2}-z_{1})^{\alpha}(\overline{z}_{2}-\overline{z}_{1})^{\overline{\alpha}},$

the remaining coefficients and functions are written as:

$c(\alpha,\beta)=\alpha+\beta,$ (5)
$B(\alpha,\beta)=(-1)^{n_{\alpha}+n_{\beta}}\pi
a(\alpha,\beta,2-\alpha-\beta),\,\,\hat{c}(\alpha,\beta)=\alpha+\beta-1,$ (6)
$C(\alpha,s)=\frac{(-1)^{n_{\alpha}+n_{s}}}{2\pi}a(\alpha,1-s,1+s-\alpha),\,\,\tilde{c}_{1}(\alpha,s)=s,\,\,\tilde{c}_{2}(\alpha,s)=\alpha-s.$
(7)

Then formula (3) has the form

$\frac{\pi}{2}D^{\alpha_{12345}-2}(z_{1},z_{2})a(\alpha_{3},\alpha_{4},\alpha_{5})(-1)^{n_{\alpha_{1234}}}\\\
\int_{\gamma}\tilde{\mathcal{D}}s\,a(1-s,\alpha_{25}-s,2-\alpha_{14}-s)\,a(1+s-\alpha_{5},\alpha_{1}+s,2-\alpha_{235}+s)(-1)^{n},$
(8)

where the contour $\gamma$ coincides with $\mathbb{R}$ except for the point
$\nu=0$, where it goes around from below. For simplicity we use the following
short notation for the sums
$\alpha_{ik}=\alpha_{i}+\alpha_{k}\,,n_{\alpha_{ik}}=n_{\alpha_{i}}+n_{\alpha_{k}}$,
$\alpha_{ikj}=\alpha_{i}+\alpha_{k}+\alpha_{j}\,,n_{\alpha_{ikj}}=n_{\alpha_{i}}+n_{\alpha_{k}}+n_{\alpha_{j}}$
etc.

Obviously, the integrand has six series of poles in the complex plane, three
of which (with the argument $−s$) go to infinity in the upper half-plane, and
the rest – in the lower one.

For convenience, we introduce a set of auxiliary notations:

$\beta_{2}=1-\alpha_{25},\,\,\,\,\,\,\beta_{3}=-1+\alpha_{14},$
$\beta_{4}=\alpha_{5},\,\,\,\,\,\,\beta_{5}=1-\alpha_{1},\,\,\,\,\,\,\beta_{6}=-1+\alpha_{235},$

then, taking into account Lemma 3.1, we can formulate conditions, when the
contour $\mathbb{R}$ separates the <<upper>> series from the <<lower>> ones,
as follows:

$\mathrm{Re}(\overline{\beta}_{i})+n_{\beta_{i}}/2>0,\,\,\,\,\,\,i=2,3,\,\,\,\,\,\,\mbox{and}\,\,\,\,\,\,\mathrm{Re}(\overline{\beta}_{j})+n_{\beta_{j}}/2>0,\,\,\,\,\,\,j=4,5,6.$
(9)

Remark: We will assume that conditions (9) are satisfied and the contour along
the real axis separates the series of poles. All other cases we understand in
the sense of analytic continuation.

######  Theorem 3.2.

Taking into account the above assumptions and conditions (9), integral (8) is
equal to

$\pi^{2}D^{\alpha_{12345}-2}(z_{2},z_{1})(I_{1}+I_{2}+I_{3}),$

where the explicit form for $I_{i}$, $i=1,2,3$, is given after relation (12).

Proof: Using definition (4), we represent the integrand as a product of Gamma
functions. Next, we denote the product of factors containing the variable $s$
by $g(s)$, and the product with the variable $\overline{s}$ by
$f(\overline{s})$.

Note that we will close the contour in the upper half-plane. Therefore, it is
convenient to extract the singular parts of the functions and represent
$f(\overline{s})$ in the form

$f(\overline{s})=\Gamma(\overline{s})f_{1}(\overline{s})=\Gamma(\overline{\beta}_{2}+\overline{s})f_{2}(\overline{s})=\Gamma(\overline{\beta}_{3}+\overline{s})f_{3}(\overline{s}).$
(10)

Then the integrand can be represented in three ways

$(-1)^{n}\frac{\Gamma(\overline{s})f_{1}(\overline{s})}{g(s)}=(-1)^{n}\frac{\Gamma(\overline{\beta}_{2}+\overline{s})f_{2}(\overline{s})}{g(s)}=(-1)^{n}\frac{\Gamma(\overline{\beta}_{3}+\overline{s})f_{3}(\overline{s})}{g(s)}.$
(11)

Using Lemma 3.1, we write down the series of poles with respect to variable
$\nu$:

$\left\\{i\left(\frac{\mid
n_{s}\mid}{2}+k\right),\,k\in\mathbb{N}\cup\\{0\\}\right\\},$
$\left\\{i\left(\overline{\beta}_{2}+\frac{n_{\beta_{2}}+\mid
n_{s}+n_{\beta_{2}}\mid}{2}+k\right),\,k\in\mathbb{N}\cup\\{0\\}\right\\},$
$\left\\{i\left(\overline{\beta}_{3}+\frac{n_{\beta_{3}}+\mid
n_{s}+n_{\beta_{3}}\mid}{2}+k\right),\,k\in\mathbb{N}\cup\\{0\\}\right\\}.$

Calculating the residues, we get three terms, each of which contains a double
sum over the set $\mathbb{Z}\times\mathbb{N}$. It turns out that we can
transform the result, using the following statement

######  Lemma 3.3.

Let $k\in\mathbb{N}$, $n,m\in\mathbb{Z}$, and $h_{1},h_{2}\in
C(\mathbb{Z},\mathbb{C})$ are functions that decrease sufficiently rapidly at
infinity, then the following relation holds

$\sum\limits_{n\in\mathbb{Z}}\sum\limits_{k=0}^{\infty}(-1)^{n}h_{1}\left(\frac{m+n+\mid
n+m\mid}{2}+k\right)h_{2}\left(\frac{n-m-\mid n+m\mid}{2}-k\right)=\\\
=\left(\sum\limits_{k=0}^{\infty}(-1)^{k}h_{1}(k)\right)\left(\sum\limits_{n=0}^{\infty}(-1)^{m+n}h_{2}(-m-n)\right).$

Proof: Let us divide the summation over the variable $n\in\mathbb{Z}$ into two
parts: $n+m\geqslant 0$ and $n+m<0$.

We make the change of variables $n\to-m-k-n$, and then use the discrete
Tonelli–Fubini theorem

$\sum\limits_{k=0}^{\infty}\sum\limits_{n=k}^{\infty}\to\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{n}.$

In the second part of the sum we also make the change of variables $n\to-
m-k+n$, then rename the variables $n\leftrightarrow k$, which gives the sum in
the form

$\sum\limits_{n=0}^{\infty}\sum\limits_{k=n+1}^{\infty}.$

Taking into account the fact that in both cases the summand function has the
same form, the two expressions can be combined into one, which completes the
proof of the Lemma. $\blacksquare$

Using Lemma 3.3, we transform the result for each series of the poles:

$2\pi\sum\limits_{n_{s}\in\mathbb{Z}}\sum\limits_{k=0}^{\infty}\frac{(-1)^{\left(\frac{n_{s}+\mid
n_{s}\mid}{2}+k+n_{s}\right)}}{\left(\frac{n_{s}+\mid
n_{s}\mid}{2}+k\right)!}\frac{f_{1}\left(-\frac{n_{s}+\mid
n_{s}\mid}{2}-k\right)}{g\left(\frac{n_{s}-\mid
n_{s}\mid}{2}-k\right)}=2\pi\left(\sum\limits_{k=0}^{\infty}\frac{f_{1}(-k)}{k!}\right)\left(\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}}{g(-n)}\right).$

Similarly, we get the result for the two remaining series:

$2\pi\left(\sum\limits_{k=0}^{\infty}\frac{f_{2}(-k-\overline{\beta}_{2})}{k!}\right)\left(\sum\limits_{n=0}^{\infty}\frac{(-1)^{n+n_{\beta_{2}}}}{g(-n-\beta_{2})}\right),\,\,\,2\pi\left(\sum\limits_{k=0}^{\infty}\frac{f_{3}(-k-\overline{\beta}_{3})}{k!}\right)\left(\sum\limits_{n=0}^{\infty}\frac{(-1)^{n+n_{\beta_{3}}}}{g(-n-\beta_{3})}\right).$

The last sums can be transformed into hypergeometric functions, using the
expansions

${}_{3}F_{2}(a_{1},a_{2},a_{3};1-a_{4},1-a_{5};x)\prod\limits_{i=1}^{5}\Gamma(a_{i})=\\\
=\sum\limits_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+1)}\Gamma(a_{1}+k)\Gamma(a_{2}+k)\Gamma(a_{3}+k)\Gamma(a_{4}-k)\Gamma(a_{5}-k),$
${}_{3}F_{2}(1-a_{1},1-a_{2},1-a_{3};a_{4},a_{5};x)=\\\
=\prod\limits_{i=1}^{5}\Gamma(a_{i})\sum\limits_{k=0}^{\infty}\frac{(-x)^{k}}{\Gamma(k+1)\Gamma(a_{1}-k)\Gamma(a_{2}-k)\Gamma(a_{3}-k)\Gamma(a_{4}+k)\Gamma(a_{5}+k)}.$

Then, taking into account the coefficient in front of the integral (3) and the
property $D^{\alpha}(z_{1},z_{2})=(-1)^{n_{\alpha}}D^{\alpha}(z_{2},z_{1})$,
we can write the final answer in the form of three terms:

$\pi^{2}D^{\alpha_{12345}-2}(z_{2},z_{1})(I_{1}+I_{2}+I_{3}),$ (12)

where

$\displaystyle
I_{1}=a(\alpha_{1},\alpha_{3},\alpha_{4},2-\alpha_{14},\alpha_{25},2-\alpha_{235})$
$\displaystyle{}_{3}F_{2}(\alpha_{5},1-\alpha_{1},-1+\alpha_{235},2-\alpha_{14},\alpha_{25};1)\,{}_{3}F_{2}(\overline{\alpha_{5}},1-\overline{\alpha_{1}},-1+\overline{\alpha_{235}},2-\overline{\alpha_{14}},\overline{\alpha_{25}};1)\,,$
$\displaystyle
I_{2}=a(\alpha_{2},\alpha_{4},\alpha_{5},-1+\alpha_{125},2-\alpha_{25},3-\alpha_{1245})(-1)^{n_{\alpha_{23}}}$
$\displaystyle{}_{3}F_{2}(\alpha_{3},2-\alpha_{125},1-\alpha_{2},2-\alpha_{25},3-\alpha_{1245};1)\,{}_{3}F_{2}(\overline{\alpha_{3}},2-\overline{\alpha_{125}},1-\overline{\alpha_{2}},2-\overline{\alpha_{25}},3-\overline{\alpha_{1245}};1),$
$\displaystyle
I_{3}=a(\alpha_{3},\alpha_{5},3-\alpha_{12345},2-\alpha_{145},\alpha_{14},-1+\alpha_{1245})(-1)^{n_{\alpha_{15}}}$
$\displaystyle{}_{3}F_{2}(\alpha_{4},-2+\alpha_{12345},-1+\alpha_{145},\alpha_{14},-1+\alpha_{1245};1){}_{3}F_{2}(\overline{\alpha_{4}},-2+\overline{\alpha_{12345}},-1+\overline{\alpha_{145}},\overline{\alpha_{14}},-1+\overline{\alpha_{1245}};1)\,,$

which completes the proof of Theorem 3.2. $\blacksquare$

##  4 Multidimensional case

In this case family of functions (1) has the form

$D^{\alpha}(z_{1},z_{2})=\dfrac{(z_{1}-z_{2})^{\mu_{1}\ldots\mu_{n}}}{(z_{1}-z_{2})^{2\left(d/4+n/2+i\nu\right)}},$
(13)

where $\nu\in\mathbb{R}\ \,,\ n=0,1,2,\ldots$, and $z_{1}\,,z_{2}$ are vectors
in $d$-dimensional space ($z_{1}\,,z_{2}\in\mathbb{R}^{d}$). The symbol
$(z_{1}-z_{2})^{\mu_{1}\ldots\mu_{n}}$ means the traceless part of the tensor
$(z_{1}-z_{2})^{\mu_{1}}\cdots(z_{1}-z_{2})^{\mu_{n}}$, for example

$\displaystyle(z_{1}-z_{2})^{\mu_{1}\mu_{2}}=(z_{1}-z_{2})^{\mu_{1}}(z_{1}-z_{2})^{\mu_{2}}-\dfrac{1}{d}\delta^{\mu_{1}\mu_{2}}(z_{1}-z_{2})^{2}.$

In the general case we have expressions (20) and (21) as explicit formulae.
Thus, $\alpha$ has discrete parameters and one real continuous parameter
$\nu$, as we had in the two-dimensional case. Discrete parameters include the
number $n$, which specifies the rank of the tensor
$(z_{1}-z_{2})^{\mu_{1}\ldots\mu_{n}}$, and the number that specifies
independent components of the symmetric traceless tensor of rank $n$. We have
only two independent components in the two dimensional case, so only $n$
remains. We integrate over region $\mathcal{V}$, which coincides with
$\mathbb{R}^{d}$, and the integration measure has the standard form
$\mathcal{D}z=d^{d}z$. We will omit notion of the integration region
$\mathbb{R}^{d}$ in most situations.

The chain rule, shown in Fig. 3, gives the answer for the integral

$\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\alpha}}\dfrac{(y-x)^{\mu_{n+1}\ldots\mu_{n+m}}}{\left(y-x\right)^{2\beta}},$
(14)

where $n,m\in\mathbb{N}$. But we do not need the general formula to analyze
the considered diagram, therefore, we are going to use only some special
cases.

In particular, if we put $m=0$ in formula (14), we can obtain the following
chain relation [30, 37]

$\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\alpha}}\dfrac{1}{\left(y-x\right)^{2\beta}}=\pi^{d/2}\dfrac{a_{n}\left(\alpha\right)a_{0}\left(\beta\right)}{a_{n}{\left(\alpha+\beta-d/2\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(\alpha+\beta-d/2\right)}},$
(15)

where

$a_{n}\left(\alpha\right)=\dfrac{\Gamma\left(d/2-\alpha+n\right)}{\Gamma\left(\alpha\right)}\,.$

We also need one more relation. Setting $m=n$ in (14), and contracting
$\mu_{i}$ with $\mu_{n+i}$ for $i\in\\{1,\ldots,n\\}$, we get an integral,
that has the following form after calculation

$\displaystyle\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\alpha}}\dfrac{\left(y-x\right)^{\mu_{1}\ldots\mu_{n}}}{\left(y-x\right)^{2\beta}}=(-1)^{n}\dfrac{\pi^{d/2}a_{n}\left(\alpha\right)a_{n}\left(\beta\right)}{a_{0}\left(\alpha+\beta-d/2-n\right)}\dfrac{\Gamma\left(n+d-2\right)\Gamma\left(d/2-1\right)}{2^{n}\Gamma\left(d-2\right)\Gamma\left(n+d/2-1\right)}\dfrac{1}{y^{2\left(\alpha+\beta-d/2-n\right)}},$
(16)

where summation over the repeated indices is implied. Let us note that the
l.h.s of (16) contains integrals of the form (14) by construction, so the
coefficient in the r.h.s is given by the sum of the same $B$-coefficients
shown in Fig. 3. Thus, we do not need distinct $B$-coefficients, but only
their sum, which can be calculated in a closed form.

Now let us proceed to the derivation of the Mellin-Barnes representation,
shown in Fig. 4. Functions of the form (13), which are the eigenfunctions of
the commuting operators $Q\left(u\right)$ form a complete orthogonal system
(see Appendix). The orthogonality and completeness relations have the
following explicit forms

$\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}\dfrac{x^{\nu_{1}\ldots\nu_{m}}}{x^{2\left(d/4+m/2-i\lambda\right)}}=c_{n}\delta_{nm}\delta\left(\nu-\lambda\right)P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}},$
(17) $\sum_{n\geq
0}\dfrac{1}{c_{n}}\int_{\mathbb{R}}d\nu\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2-i\nu\right)}}=\delta^{\left(d\right)}\left(x-y\right),$
(18)

where the constant $c_{n}$ is given by the expression

$c_{n}=\dfrac{\pi^{d/2+1}n!}{2^{n-1}\Gamma\left(d/2+n\right)},$ (19)

and summation over the repeated indices in (18) is implied.

Operator $P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}$ is a projector onto
the traceless symmetric tensors

$x^{\mu_{1}\ldots\mu_{n}}=P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}x^{\nu_{1}}\ldots
x^{\nu_{n}}$ (20)

and given by the formula [36, 37]

$P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}=\sum_{p\geq
0}\dfrac{n!\left(-1\right)^{p}\Gamma\left(n-p+d/2-1\right)}{2^{2p}p!\left(n-2p\right)!\Gamma\left(n+d/2-1\right)}\hat{S}\left(\delta^{\mu_{1}\mu_{2}}\delta_{\nu_{1}\nu_{2}}\ldots\delta^{\mu_{2p-1}\mu_{2p}}\delta_{\nu_{2p-1}\nu_{2p}}\delta^{\mu_{2p+1}}_{\nu_{2p+1}\ldots}\delta^{\mu_{n}}_{\nu_{n}}\right),$
(21)

where $\hat{S}$ denotes the symmetrization over all indices. Using the
completeness relation, we can get the expression for the scalar propagator in
the Mellin-Barnes representation. The simplest way is to expand the
$\delta$-function in

$\dfrac{1}{\left(x-y\right)^{2\alpha}}=\int
d^{d}z\dfrac{1}{\left(x-z\right)^{2\alpha}}\,\delta^{\left(d\right)}\left(z-y\right),$

using (18), which gives

$\displaystyle\dfrac{1}{\left(x-y\right)^{2\alpha}}=\sum_{n\geq
0}\int_{\mathbb{R}}d\nu\dfrac{1}{c_{n}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2-i\nu\right)}}\int
d^{d}z\dfrac{1}{\left(z-y\right)^{2\alpha}}\dfrac{z^{\mu_{1}\ldots\mu_{n}}}{z^{2\left(d/4+n/2+i\nu\right)}}.$

To calculate the integral with respect to $z$ we use chain rule (15), thus
obtaining a representation for the propagator

$\displaystyle\dfrac{1}{\left(x-y\right)^{2\alpha}}=\sum_{n\geq
0}\int_{\mathbb{R}}d\nu\dfrac{\pi^{d/2}}{c_{n}}\dfrac{a_{n}\left(d/4+n/2+i\nu\right)a_{0}\left(\alpha\right)}{a_{n}\left(\alpha-d/4+n/2+i\nu\right)}\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(\alpha-d/4+n/2+i\nu\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2-i\nu\right)}}.$

The last expression is invariant under translation by an arbitrary vector, so
that the general formula for the Mellin-Barnes transform has the following
form

$\displaystyle\dfrac{1}{\left(x-y\right)^{2\alpha}}=\sum_{n\geq
0}\int_{\mathbb{R}}d\nu\dfrac{\pi^{d/2}}{c_{n}}\dfrac{a_{n}\left(d/4+n/2+i\nu\right)a_{0}\left(\alpha\right)}{a_{n}\left(\alpha-d/4+n/2+i\nu\right)}\dfrac{\left(x-z\right)^{\mu_{1}\ldots\mu_{n}}}{\left(x-z\right)^{2\left(\alpha-d/4+n/2+i\nu\right)}}\dfrac{\left(y-z\right)^{\mu_{1}\ldots\mu_{n}}}{\left(y-z\right)^{2\left(d/4+n/2-i\nu\right)}}.$
(22)

To avoid confusion, let us note that summation with respect to the
Mellin–Barnes measure is done over $n$ as well as the tensor indices
$\mu_{i}$, $i\in\\{1,\ldots,n\\}$.

Now let us compute the diagram shown in Fig. 2. We do not have the formula for
arbitrary parameters in (14) and we can not compute the diagram in the most
general form. For this reason, we consider functions (13) only in scalar form,
that is, when $n=0$. It turns out that in this case, precisely due to relation
(16), it becomes possible to calculate the integral (3).

The two-loop integral

$I\left(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\right)=\int\dfrac{d^{d}xd^{d}y}{x^{2\alpha_{1}}y^{2\alpha_{2}}\left(z-x\right)^{2\alpha_{4}}\left(z-y\right)^{2\alpha_{3}}\left(x-y\right)^{2\alpha_{5}}}$
(23)

is computed, according to the general scheme, outlined in the second section.
At the first step we use formula (22) for the propagator between vertices $x$
and $z$, shifting both vertices by the vector $y$. The integral takes the form

$\displaystyle
I\left(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\right)=$
$\displaystyle\sum_{n\geq
0}\int_{\mathbb{R}}d\nu\dfrac{\pi^{d/2}}{c_{n}}\dfrac{a_{n}\left(d/4+n/2+i\nu\right)a_{0}\left(\alpha_{4}\right)}{a_{n}\left(\alpha_{4}+n/2-d/4+i\nu\right)}$
(24) $\displaystyle\int
d^{d}xd^{d}y\dfrac{\left(x-y\right)^{\mu_{1}\ldots\mu_{n}}\left(z-y\right)^{\mu_{1}\ldots\mu_{n}}}{x^{2\alpha_{1}}y^{2\alpha_{2}}\left(x-y\right)^{2\left(\alpha_{45}-d/4+n/2+i\nu\right)}\left(z-y\right)^{2\left(\alpha_{3}+d/4+n/2-i\nu\right)}},$

so we can compute the remaining integrals over $x$ and $y$ by using the chain
rule.

Indeed, let us use relation (15) and calculate integral over $x$. Then, the
second line of (24) can be rewritten as

$\left(-1\right)^{n}\pi^{d/2}\dfrac{a_{n}\left(\alpha_{45}-d/4+n/2+i\nu\right)a_{0}\left(\alpha_{1}\right)}{a_{n}\left(\alpha_{145}-3d/4+n/2+i\nu\right)}\int
d^{d}y\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{\alpha_{1245}-3d/4+n/2+i\nu}}\dfrac{\left(z-y\right)^{\mu_{1}\ldots\mu_{n}}}{\left(z-y\right)^{2\left(\alpha_{3}+d/4+n/2-i\nu\right)}}.$

Next, according to the scheme from Section 2, we need to apply the chain rule
to every summand in the sum over tensor indices. However, due to the absence
of the needed chain relation, we add summation over tensor indices to the
integration over the variable $y$, and then use relation (16). Then we get
equality

$\displaystyle\int
d^{d}y\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{\alpha_{1245}-3d/4+n/2+i\nu}}\dfrac{\left(z-y\right)^{\mu_{1}\ldots\mu_{n}}}{\left(z-y\right)^{2\left(\alpha_{3}+d/4+n/2-i\nu\right)}}=(-1)^{n}\pi^{d/2}\dfrac{\Gamma\left(n+d-2\right)\Gamma\left(d/2-1\right)}{2^{n}\Gamma\left(d-2\right)\Gamma\left(n+d/2-1\right)}$
$\displaystyle\dfrac{a_{n}\left(\alpha_{1245}-3d/4+n/2+i\nu\right)a_{n}\left(\alpha_{3}+d/4+n/2-i\nu\right)}{a_{0}\left(\alpha_{12345}-d\right)}\dfrac{1}{z^{2\left(\alpha_{12345}-d\right)}}.$

Thus, the result for the diagram has the following general form

$I\left(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\right)=\dfrac{C}{z^{2\left(\alpha_{12345}-d\right)}},$

where

$\displaystyle C=$
$\displaystyle\dfrac{\pi^{d-1}}{2}\dfrac{\Gamma\left(d/2-1\right)}{\Gamma\left(d-2\right)}\dfrac{a_{0}\left(\alpha_{1}\right)a_{0}\left(\alpha_{4}\right)}{a_{0}\left(\alpha_{12345}-d\right)}\sum_{n}\int_{\mathbb{R}}d\nu\,M_{n}\,\dfrac{a_{n}\left(d/4+n/2+i\nu\right)}{a_{n}\left(\alpha_{4}-d/4+n/2+i\nu\right)}$
(25)
$\displaystyle\dfrac{a_{n}\left(\alpha_{45}-d/4+n/2+i\nu\right)}{a_{n}\left(\alpha_{145}-3d/4+n/2+i\nu\right)}a_{n}\left(\alpha_{1245}-3d/4+n/2+i\nu\right)a_{n}\left(\alpha_{3}+d/4+n/2-i\nu\right),$

where we use the convenient notation

$M_{n}=\dfrac{1}{n!}\left(n+d/2-1\right)\Gamma\left(n+d-2\right).$ (26)

We calculate the integral over $\nu$ by residues by analogy with the two-
dimensional case. In order to understand the structure of the poles, we
rewrite the integrand in terms of Gamma functions

$\displaystyle\dfrac{\Gamma\left(d/4+n/2-i\nu\right)}{\Gamma\left(d/4+n/2+i\nu\right)}$
$\displaystyle\dfrac{\Gamma\left(\alpha_{4}-d/4+n/2+i\nu\right)}{\Gamma\left(3d/4-\alpha_{4}+n/2-i\nu\right)}\dfrac{\Gamma\left(3d/4-\alpha_{45}+n/2-i\nu\right)}{\Gamma\left(\alpha_{45}-d/4+n/2+i\nu\right)}\dfrac{\Gamma\left(\alpha_{145}-3d/4+n/2+i\nu\right)}{\Gamma\left(5d/4-\alpha_{145}+n/2-i\nu\right)}$
$\displaystyle\dfrac{\Gamma\left(5d/4-\alpha_{1245}+n/2-i\nu\right)}{\Gamma\left(\alpha_{1245}-3d/4+n/2+i\nu\right)}\dfrac{\Gamma\left(d/4-\alpha_{3}+n/2+i\nu\right)}{\Gamma\left(\alpha_{3}+d/4+n/2-i\nu\right)}.$

This expression has six infinite series of poles, three of them are located in
the upper half plane and three in the lower one. As in the two-dimensional
case, we will close the contour in the upper half-plane, so we proceed to
enumerating the corresponding poles. The gamma function has simple poles at
the non-positive integers.

$\Gamma\left(x\right)=\dfrac{\left(-1\right)^{n}}{n!}\dfrac{1}{x+n}+\ldots,$

so that the poles in the upper half-plane appear in three cases

$\begin{cases}\nu=i\left(k_{1}+\alpha_{4}-d/4+n/2\right);\ \\\
\nu=i\left(k_{2}+\alpha_{1}+\alpha_{4}+\alpha_{5}-3d/4+n/2\right);\ \\\
\nu=i\left(k_{3}+d/4-\alpha_{3}+n/2\right).\end{cases}$

Summing the residues, we get the final answer

$C=\pi^{d}\dfrac{\Gamma\left(d/2-1\right)}{\Gamma\left(d-2\right)}\dfrac{a_{0}\left(\alpha_{1}\right)a_{0}\left(\alpha_{4}\right)}{a_{0}\left(\alpha_{12345}-d\right)}\,\left(I_{1}+I_{2}+I_{3}\right),$
(27)

where

$\displaystyle
I_{1}=\sum_{n=0}^{\infty}M_{n}\sum_{k=0}^{\infty}\dfrac{\left(-1\right)^{k}}{k!}\dfrac{\Gamma\left(\alpha_{4}+n+k\right)}{\Gamma\left(d/2-\alpha_{4}-k\right)}\dfrac{\Gamma\left(d/2-\alpha_{5}+n+k\right)}{\Gamma\left(\alpha_{5}-k\right)}\dfrac{\Gamma\left(\alpha_{15}-d/2-k\right)}{\Gamma\left(d-\alpha_{15}+n+k\right)}$
$\displaystyle\dfrac{\Gamma\left(d-\alpha_{125}+n+k\right)}{\Gamma\left(\alpha_{125}-d/2-k\right)}\dfrac{\Gamma\left(d/2-\alpha_{34}-k\right)}{\Gamma\left(\alpha_{34}+n+k\right)}\dfrac{1}{\Gamma\left(d/2+n+k\right)},$
$\displaystyle
I_{2}=\sum_{n=0}^{\infty}M_{n}\sum_{k=0}^{\infty}\dfrac{\left(-1\right)^{k}}{k!}\dfrac{\Gamma\left(\alpha_{145}-d/2+n+k\right)}{\Gamma\left(d-\alpha_{145}-k\right)}\dfrac{\Gamma\left(d/2-\alpha_{15}-k\right)}{\Gamma\left(\alpha_{15}+n+k\right)}\dfrac{\Gamma\left(\alpha_{1}+n+k\right)}{\Gamma\left(d/2-\alpha_{1}-k\right)}$
$\displaystyle\dfrac{\Gamma\left(d/2-\alpha_{2}+n+k\right)}{\Gamma\left(\alpha_{2}-k\right)}\dfrac{\Gamma\left(d-\alpha_{1345}-k\right)}{\Gamma\left(\alpha_{1345}-d/2+n+k\right)}\dfrac{1}{\Gamma\left(d/2+n+k\right)},$
$\displaystyle
I_{3}=\sum_{n=0}^{\infty}M_{n}\sum_{k=0}^{\infty}\dfrac{\left(-1\right)^{k}}{k!}\dfrac{\Gamma\left(d/2-\alpha_{3}+n+k\right)}{\Gamma\left(\alpha_{3}-k\right)}\dfrac{\Gamma\left(\alpha_{34}-d/2-k\right)}{\Gamma\left(d-\alpha_{34}+n+k\right)}\dfrac{\Gamma\left(d-\alpha_{345}+n+k\right)}{\Gamma\left(\alpha_{345}-d/2-k\right)}$
$\displaystyle\dfrac{\Gamma\left(\alpha_{1345}-d-k\right)}{\Gamma\left(3d/2-\alpha_{1345}+n+k\right)}\dfrac{\Gamma\left(3d/2-\alpha_{12345}+n+k\right)}{\Gamma\left(\alpha_{12345}-d-k\right)}\dfrac{1}{\Gamma\left(d/2+n+k\right)}.$

##  5 Conclusion

We have covered in detail the application of the simplest integral transform
to the separated variables representation to compute the two-loop master-
diagram. Apparently, the proposed approach is equivalent to the Gegenbauer
polynomial technique [35, 36, 37, 38], so our work is rather methodological.

The next step is to compare the obtained formulae with the numerous already
available results [30, 32, 33, 37]. In particular, an important task is to
calculate a special master diagram, arising in the $1/N$ expansion [30, 31,
34].

The proposed method can be applied to the calculation of ladder diagrams [27,
28, 29]. In the multidimensional case all calculations repeat step by step
those in the two-dimensional [15] and four-dimensional [13] cases. The
corresponding formulae are given in the Appendix.

Let us note interesting connections with other works. The two-dimensional
transformation to separated variables turned out to be a convenient tool in
the study of Racah operators in the representation theory of
$SL(2,\mathbb{C})$, to which the works of R.S. Ismagilov [19, 20] and [21] are
devoted. The works of Yu.A. Neretin [22, 23] explore the Barnes–Ismagilov
integrals and their generalizations, and the work of Yu.A. Neretin and V.F.
Molchanov [24] essentially investigated in detail the second transformation
from the hierarchy of transformations to separated variables.

The work [18] is devoted to the application of the transform to separated
variables to derive a two-dimensional version of the Gustafson integrals. An
interesting open question is the study of the connection between the method of
separation of variables and the Dotsenko–Fateev and Fateev–Litvinov integral
identities in conformal field theory[25, 26].

#### Acknowledgments:

This work was supported by the Russian Science Foundation (project
19-11-00131). Also A.V. Ivanov is a winner of the Young Russian Mathematician
award and would like to thank its sponsors and jury.

##  6 Appendix

###  6.1 Projector

Let us introduce a projector onto the space of symmetric and traceless tensors

$\psi^{\prime}_{\mu_{1}\ldots\mu_{n}}=P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}\,\psi_{\nu_{1}\ldots\nu_{n}}.$

The initial tensor $\psi_{\nu_{1}\ldots\nu_{n}}$ is symmetric and its image
$\psi^{\prime}_{\mu_{1}\ldots\mu_{n}}$ is traceless. Considering a polynomial
of the form

$P\left(u,v\right)=\dfrac{1}{n!}P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}u^{\mu_{1}}\cdots
u^{\mu_{n}}v^{\nu_{1}}\cdots v^{\nu_{n}}$

and rewriting previous equality by using the generating function

$\psi^{\prime}\left(u\right)=\psi^{\prime}_{\mu_{1}\ldots\mu_{n}}u^{\mu_{1}}\ldots
u^{\mu_{n}},\,\,\,\,\psi\left(v\right)=\psi_{\mu_{1}\ldots\mu_{n}}v^{\mu_{1}}\ldots
v^{\mu_{n}},$

we get

$\psi^{\prime}\left(u\right)=P\left(u,\partial_{v}\right)\psi\left(v\right).$

The tracelessness of $\psi^{\prime}_{\mu_{1}\ldots\mu_{n}}$ is equivalent to
$\Delta_{u}P(u,v)=0$. Substituting the general decomposition

$P\left(u,v\right)=\left(uv\right)^{n}+a_{1}\left(uv\right)^{n-2}u^{2}v^{2}+a_{2}\left(uv\right)^{n-4}u^{4}v^{4}+\ldots$

in the equation $\Delta_{u}P(u,v)=0$, we obtain the recurrence relation

$a_{p}=-a_{p-1}\dfrac{\left(n-2p+2\right)\left(n-2p+1\right)}{2p\left(d+2n-2-2p\right)}\,.$
(28)

A solution which satisfies the initial condition $a_{0}=1$ has the form

$a_{p}=\dfrac{n!\left(-1\right)^{p}\Gamma\left(n-p+d/2-1\right)}{2^{2p}p!\left(n-2p\right)!\Gamma\left(n+d/2-1\right)}\,.$
(29)

Thus, we have formula (21), where $\hat{S}$ denotes the symmetrization over
all indices

$\displaystyle\hat{S}\,\psi_{\mu_{1}\cdots\mu_{n}}=\frac{1}{n!}\sum_{p\in
S_{n}}\psi_{\mu_{p(1)}\cdots\mu_{p(n)}}.$ (30)

###  6.2 Orthogonality

The orthogonality relation has the following form

$\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}\dfrac{x^{\nu_{1}\ldots\nu_{m}}}{x^{2\left(d/4+m/2-i\lambda\right)}}=c_{n}\delta_{nm}\delta\left(\nu-\lambda\right)P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}.$
(31)

We obtain $\delta\left(\nu-\lambda\right)$ by integrating over the radial
variable, therefore it is convenient to separate this part by going to the
spherical coordinates

$\int
d^{d}x\dfrac{x^{\mu_{1}\ldots\mu_{n}}\,x^{\nu_{1}\ldots\nu_{m}}}{x^{2\left(d/2+(n+m)/2+i(\nu-\lambda)\right)}}=\pi\delta(\lambda-\nu)\,\int_{\mathbb{S}^{d-1}}d\Omega\,\hat{x}^{\mu_{1}\ldots\mu_{n}}\,\hat{x}^{\nu_{1}\ldots\nu_{m}},$

where $d\Omega$ is the standard measure on the sphere, $\hat{x}^{\mu}$ is a
vector of unit length, and the integral over the radius is done by
substituting $r=e^{p}$

$\int^{\infty}_{0}\dfrac{dr}{r}r^{2i(\lambda-\nu)}=\pi\delta(\lambda-\nu).$

Thus, we need to prove the identity

$\displaystyle\int_{\mathbb{S}^{d-1}}d\Omega\,\hat{x}^{\mu_{1}\ldots\mu_{n}}\,\hat{x}^{\nu_{1}\ldots\nu_{m}}=\dfrac{c_{n}}{\pi}\,\delta_{nm}\,P^{\mu_{1}\ldots\mu_{n}}_{\nu_{1}\ldots\nu_{n}}.$

Consider an equivalent relation by contracting both sides with complex vectors
$u$ and $v$, satisfying the condition $u^{2}=v^{2}=0$,

$\displaystyle\int_{\mathbb{S}^{d-1}}d\Omega\left(\hat{x}u\right)^{n}\,\left(\hat{x}v\right)^{m}=\dfrac{c_{n}}{\pi}\,\delta_{nm}\,\left(uv\right)^{n},$

and prove this relation by using the standard formula for the Gaussian
integral

$\displaystyle\int d^{d}x\,e^{-x^{2}+xa}=\pi^{d/2}e^{\frac{a^{2}}{4}}.$

Substituting $a=tu+sv$ and differentiating the required number of times by $t$
and $s$, we get

$\displaystyle\int
d^{d}xe^{-x^{2}}\,(xu)^{n}\,(xv)^{m}=\pi^{d/2}n!2^{-n}\,\delta_{nm}(uv)^{n}.$

The integral above differs from the one we need by a constant factor which can
be restored by going to the spherical coordinates

$\displaystyle\int
d^{d}x\,e^{-x^{2}}\,(xu)^{n}\,(xv)^{m}=\frac{1}{2}\Gamma\left(\frac{d+n+m}{2}\right)\int_{\mathbb{S}^{d-1}}d\Omega\,(\hat{x}u)^{n}\,(\hat{x}v)^{m},$

so that the constant $c_{n}$ has the form (19).

###  6.3 Completeness

The completeness relation has the form ($x,y\in\mathbb{R}^{d}$)

$\sum_{n\geq
0}\int_{\mathbb{R}}d\nu\dfrac{1}{c_{n}}\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2-i\nu\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2+i\nu\right)}}=\delta^{\left(d\right)}\left(x-y\right).$
(32)

We can expand the delta function in the r.h.s. as a product of the radial and
angular parts

$\delta^{d}\left(x-y\right)=\dfrac{1}{x^{2\left(d/2-1/2\right)}}\delta\left(|x|-|y|\right)\dfrac{1}{J}\delta\left(\hat{x}-\hat{y}\right),$
(33)

where $|x|$ is the length of the vector $x$, $\hat{x}=x/|x|~{}$ is a vector of
unit length, and $J$ is the Jacobian, corresponding to change of variables to
$d$-dimensional spherical coordinates.

The l.h.s of (32) has a similar structure. The integral over $\nu$ depends
only on the length of the vectors, and the sum over $n$ is related to the
angular part. We can integrate over $\nu$ by substituting $|x|=e^{u}$ ,
$|y|=e^{v}$

$\int_{\mathbb{R}}d\nu\dfrac{1}{|x|^{-2i\nu}}\dfrac{1}{|y|^{2i\nu}}=\pi\delta\left(u-v\right).$
(34)

Therefore we need to prove the relation for the angular part only

$\sum_{n\geq
0}\dfrac{\pi}{c_{n}}\,\hat{x}^{\mu_{1}\ldots\mu_{n}}\hat{y}^{\mu_{1}\ldots\mu_{n}}=\dfrac{1}{J}\delta\left(\hat{x}-\hat{y}\right),$
(35)

which we can rewrite as the completeness relation for Gegenbauer polynomials

$\sum_{n\geq
0}\left(2n+d-2\right)C_{n}^{\left(d/2-1\right)}\left(\hat{x}\cdot\hat{y}\right)=\left(d-2\right)S_{d-1}\,\dfrac{1}{J}\delta\left(\hat{x}-\hat{y}\right),$
(36)

where $S_{d-1}=2\pi^{d/2}\Gamma^{-1}(d/2)$ is the surface of the unit sphere
in $d$-dimensional space by using the relation between Gegenbauer polynomials
and traceless tensors [35]

$\hat{x}^{\mu_{1}\ldots\mu_{n}}\hat{y}^{\mu_{1}\ldots\mu_{n}}=\dfrac{n!\Gamma\left(d/2-1\right)}{2^{n}\Gamma\left(n+d/2-1\right)}C_{n}^{\left(d/2-1\right)}\left(\hat{x}\cdot\hat{y}\right),$
(37)

The function

$G\left(x,y\right)=\dfrac{1}{\left(x-y\right)^{2\left(d/2-1\right)}}$

is the fundamental solution of the $d-$ dimensional Laplace equation

$-\operatorname{\Delta}_{x}G\left(x,y\right)=\left(d-2\right)S_{d-1}\delta^{\left(d\right)}\left(x-y\right),$
(38)

and can be considered as a generating function for Gegenbauer polynomials [36]
(see multipole expansion)

$G\left(x,y\right)=\begin{cases}\sum_{n\geq
0}\dfrac{|x|^{n}}{|y|^{n+d-2}}\,C^{\left(d/2-1\right)}_{n}\left(\hat{x}\cdot\hat{y}\right),\
|x|<|y|;\\\ \sum_{n\geq
0}\dfrac{|y|^{n}}{|x|^{n+d-2}}\,C^{\left(d/2-1\right)}_{n}\left(\hat{x}\cdot\hat{y}\right),\
|y|<|x|.\end{cases}$ (39)

Integrating (38) over $r=|x|$ in a neighborhood of the point $|y|$, we get

$-r^{d-1}\dfrac{\partial}{\partial
r}G\left(x,y\right)\bigg{|}_{|y|-0}^{|y|+0}=\left(d-2\right)\dfrac{S_{d-1}}{J}\delta\left(\hat{x}-\hat{y}\right).$

If we substitute decomposition (39) into the l.h.s, we immediately obtain
identity (36).

###  6.4 Operator $Q(u)$ and ladder diagrams

Consider the integral operator $Q\left(u\right)$

$\left[Q\left(u\right)\Phi\right]\left(x\right)=\int
d^{d}y\dfrac{1}{\left(x-y\right)^{2\left(d/2-u\right)}y^{2u}}\Phi\left(y\right).$

The operators $Q\left(u\right)$ form a commutative family. Commutativity
$Q\left(u\right)Q\left(v\right)=Q\left(v\right)Q\left(u\right)$ can be proved
by finding the kernel of the operator $Q\left(u\right)Q\left(v\right)$, using
the star-triangle relation [30, 29], and checking that it is invariant under
permutation $u\rightleftarrows v$.

We can check that the functions

$\Psi^{\mu_{1}\ldots\mu_{n}}\left(x\right)=\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}$

are eigenfunctions of the operator $Q(u)$ explicitly by using chain rule (15)

$\displaystyle\left[Q\left(u\right)\Psi^{\mu_{1}\ldots\mu_{n}}\right]\left(x\right)=\tau\left(u,\nu,n\right)\,\Psi^{\mu_{1}\ldots\mu_{n}}(x),$
$\displaystyle\tau\left(u,\nu,n\right)=\pi^{d/2}\dfrac{a_{n}\left(d/4+n/2+u+i\nu\right)a_{0}\left(d/2-u\right)}{a_{n}\left(d/4+n/2+i\nu\right)}.$

The system of functions $\Psi^{\mu_{1}\ldots\mu_{n}}\left(x\right)$ is
orthogonal and complete, so that we obtain the spectral decomposition of for
the kernel of operator $Q\left(u\right)$ in the form

$\displaystyle
Q_{u}\left(x,y\right)=\dfrac{1}{\left(x-y\right)^{2\left(d/2-u\right)}y^{2u}}=\sum_{n\geq
0}\dfrac{1}{c_{n}}\int_{\mathbb{R}}d\nu~{}\tau\left(u,\nu,n\right)\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2-i\nu\right)}}.$

Since the operators $Q(u)$ form a commutative family and have a common set of
eigenfunctions, it follows that we can extend the previous formula to the
product $Q(u_{1})\cdots Q(u_{L})$

$Q_{u_{1}\ldots u_{L}}\left(x,y\right)=\sum_{n\geq
0}\dfrac{1}{c_{n}}\int_{\mathbb{R}}d\nu\dfrac{x^{\mu_{1}\ldots\mu_{n}}}{x^{2\left(d/4+n/2+i\nu\right)}}\dfrac{y^{\mu_{1}\ldots\mu_{n}}}{y^{2\left(d/4+n/2-i\nu\right)}}\prod\limits_{k=1}^{L}\tau\left(u_{k},\nu,n\right).$
(40)

Now if we recall that the kernel of the product $Q(u_{1})\cdots Q(u_{L})$ is
given by the convolution of the kernels of $Q(u)$, we get

$Q_{u_{1}\ldots u_{L}}\left(x,y\right)=\prod\limits_{k=1}^{L-1}\int
d^{d}x_{k}\dfrac{1}{\left(x_{k-1}-x_{k}\right)^{2\left(d/2-u_{k}\right)}}\dfrac{1}{x_{k}^{2u_{k}}}\times\dfrac{1}{\left(x_{L}-y\right)^{2\left(d/2-u_{L}\right)}}\dfrac{1}{y^{2u_{L}}},$

where $x_{0}=x$. This integral corresponds to the so-called Feynman ladder
diagram. Formula (40) gives a closed form expression, containing only one
integration over $\nu$ and summation over $n$.

Also, if we recall relation (37), we can rewrite formula (40) using the
Gegenbauer polynomials

$Q_{u_{1}\ldots u_{L}}\left(x,y\right)=\Gamma\left(d/2-1\right)\sum_{n\geq
0}\left(d/2-1+n\right)\int_{\mathbb{R}}d\nu\dfrac{C_{n}^{\left(d/2-1\right)}\left(\hat{x}\cdot\hat{y}\right)}{x^{2\left(d/4+i\nu\right)}y^{2\left(d/4-i\nu\right)}}\prod\limits_{k=1}^{L}\tau\left(u_{k},\nu,n\right).$

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