# General One-loop Generating Function by IBP relations

Bo Fengab111Emails<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>Chang Hucd, Jiyuan Shen cde,
Yaobo Zhangf
aBeijing Computational Science Research Center, Beijing 100193, China
bPeng Huanwu Center for Fundamental Theory, Hefei, Anhui, 230026, China
cSchool of Fundamental Physics and Mathematical Sciences, Hangzhou Institute
for Advanced Study, UCAS, Hangzhou 310024, China
dUniversity of Chinese Academy of Sciences, Beijing 100049, China
eInstitute of Theoretical Physics, Chinese Academy of Sciences, Beijing
100190, China
fSchool of Physics, Ningxia University, Yinchuan, 750002, China

###### Abstract:

In this paper we have studied the most general generating function of
reduction for one loop integrals with arbitrary tensor structure in numerator
and arbitrary power distribution of propagators in denominator. Using IBP
relations, we have established the partial differential equations for these
generating functions and solved them analytically. These results provide
useful guidance for applying generating function method to reductions of
higher loop integrals.

Generating function, Reduction, IBP relations, One-loop integrals

##  1 Motivation

As a connection between the theoretical framework and experimental data, the
scattering amplitude is a central concept of quantum field theory, thus its
efficient computation is extremely important. For tree-level amplitudes, the
on-shell recursion relation [1, 2] has marked a significant advancement in
both new efficient computational tools and a fresh on-shell perspective for
studying quantum field theory222For the introduction of these new
developments, please see following books [3, 4, 5]. Compared to tree-level
amplitudes, computation of loop amplitudes is much more difficult. Many
efforts have been devoted to this subject (see the books [6, 7]).

A fundamental idea of computing loop amplitudes is the reduction of loop
integrals, which states that any loop integral can be written as the linear
combination of some basic integrals (called the master integrals) with
coefficients (called the reduction coefficients) as the rational function of
external momenta, polarization vectors, masses and spacetime dimension. Based
on this observation, loop integrals have been divided into two relatively
independent parts, i.e., the integration of master integrals and the
abstraction of reduction coefficients. Progress in either part will increase
the efficiency of computing loop amplitudes. A nice introduction to recent
huge achievements can be found in the book [8].

In this paper, we will focus on the analytic abstraction of reduction
coefficients. When we talk about the reduction, there are, in fact, two
different layers. The first layer is the algebraic decomposition of loop
integrands, which we will call the ”integrand level reduction”. For one-loop
integrals, the highly influential OPP-method [9] is an example of the method,
which revolutionized the computation of one-loop amplitudes. After realizing
the central mathematical foundation of OPP-method, integrand level reduction
for higher loop integrals has been systematically solved using the
computational algebraic geometry [10, 11, 12]. Although it is highly
systematic, one drawback is the presence of large amounts of spurious terms,
which contribute nothing physically but are necessary for calculation in this
framework. To avoid this inefficiency, we need go to the second layer, i.e.,
the ”integral level reduction”, where only reduction coefficients of physical
contributions are computed.

Historically, the first realization of reduction is done at the integral
level, which was proposed by Passarino and Veltman in [13] and is called the
PV-reduction method. Since then, other alternative integral level reduction
methods have been proposed. Some widely mentioned methods include the
Integration-by-Part (IBP) method [14, 15, 16, 17, 18, 19, 20], the unitarity
cut method [21, 22, 23, 24, 25, 26, 27, 28, 29, 30], the intersection theory
method [31, 32, 33, 34, 35, 36]. It is worth mentioning that different
reduction methods utilize different properties of loop integrals. The PV-
reduction method has used the tensor structure of loop integrals. The IBP
method has used the translation and scaling properties of loop integrals under
dimensional regularization. The unitarity cut method has used the analytic
property (i.e., the imaginary part) of loop integrals. The intersection theory
method has used the deep mathematical insight of twist cohomology and dual
twist cohomology structure of loop integrals [37, 38, 39]. It is interesting
to see if we could explore other properties of loop integrals or combine these
different methods to get new reduction methods.

No matter which reduction method one uses, a notable common feature in the
reduction procedure is the appearance of an iterative structure. For example,
by IBP relation, one can write the integral of higher rank tensor to the
linear combination of integrals of lower rank tensors. Wisely utilizing these
iterative structures without repeatedly computing them is an important step in
enhancing the efficiency of reduction333About the related discussions, see
some recent papers [40, 41, 42]. When encountering an iterative structure, a
very useful approach is to consider the corresponding generating function. In
many examples, it is much easier to find the generating function rather than
to find expansion coefficients of each order. For example, in [43], the
generating function for an operator insertion on an $l$-leg vertex was
introduced, which facilitates the computation of operator matrix elements of
higher loop contributions. Similarly in [44] generating functions for a
specific tensor structure of some two-loop integrals have been introduced and
solved, which provide explicit reductions of that tensor structure with
arbitrary power.

Our aim is to solve the reduction of general loop integrals, thus we need to
find the general generating function. When we talk about the general loop
integrals, we mean following two things: the general polynomial of loop
momenta in the numerator and the general power distribution of propagators in
the denominator. A general numerator implies a general tensor structure, which
is the starting point of the PV-reduction method. Recently, to simplify the
manipulation of tensor structure, an auxiliary vector $R$ has been introduced
in [45, 46, 47, 48, 49, 50, 40, 41, 42]. Taking advantage of this method, the
generating function for tensor reduction of one-loop integrals has been
constructed and solved in [51, 52]. For the general denominator it is pointed
out in [53] that masses could play the role of fugacity variable in generating
functions.

Finding the proper organization to define the generating function is only the
first step. What is more crucial is to establish its differential equations
and then solve them analytically. In [51], differential equations have been
established using the PV-reduction method. When using the same idea to treat
two-loop integrals, we could not get enough differential equations to
completely determine the generating function. It is not so surprise, since
compared to other methods the PV-reduction method utilizes the least
information about loop integrals, i.e., only the tensor structure. Thus when
encountering irreducible scalar products in higher loop integrals, the PV-
reduction method faces a serious challenge as demonstrated in [48], where the
analytic reduction results of sub one loop integrals must be used to solve the
two-loop integral reduction. In fact, the formulation of the differential
equation for generating function in [52] is grounded on the observation of
such sub one loop reduction. This result was initially conjectured in [48] and
subsequently proved in [40] and [50] employing either module intersection
method or the projective space language associated to their Feynman
parametrization.

In contrast to the PV-reduction method, the IBP method is considered to be a
complete method, i.e., it can reduce any integral to basis regardless of the
number of loops or the complexity of integrands. Therefore we believe that
using the IBP method we can get sufficient differential equations for the
generating function for any loop level. To verify the belief, we will restudy
generating functions of one loop integrals in this paper. We will demonstrate
that using the IBP method, we can derive a much simpler differential equation
for the generating function of tensor reduction compared to the results in
[51]. Furthermore, the differential equation for the generating function of
general denominators could also easily be written down and solved. These
results will provide a valuable guidance when we explore generating functions
for higher loop integrals.

The structure of the paper is as follows. In the section two, we set up the
general framework for generating functions. In the section three, we establish
corresponding differential equations using the IBP method. In the section
four, we resolve the generating function for tensor reduction, using either
the form of series expansion or the form of hypergeometric function. In the
section five, we solve the matrix ${\cal H}$, which represents the reduction
of scalar integrals with only single propagator with power two. In the section
six, we address the generating function for an arbitrary denominator. Finally,
a brief conclusion and discussion are presented in the section seven. For
self-sufficiency, some technical details are presented in the Appendix.

##  2 The setup

In this paper, we will discuss the generating function for the most general
one-loop integrals, i.e., with arbitrary numerator and denominator. In this
section, we will set up the foundation of our discussion.

We start from the generating function of tensor reduction of $(n+1)$-gon,
which is defined as

$I_{n+1}^{R}\equiv\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot
R}\over\prod_{i=0}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}=\bm{J}_{n+1}\cdot\vec{{\alpha}}^{gen}(R)~{},~{}~{}$ (1)

where the $\bm{J}_{n+1}$ is the row vector of the scalar basis and the
$\vec{{\alpha}}^{gen}$ is the column vector of generating functions of tensor
reduction. Since we have assumed the general $D$-dimensional spacetime, there
are no kinematic relations between external momenta $K_{i}$ for general
momentum configurations, thus each scalar $m$-gon is a master integral and we
can arrange the order of $\bm{J}_{n+1}$ according to the topology. Let us
define

$J_{n+1;\widehat{S}}=\int{d^{D}\ell\over i\pi^{D/2}}{1\over\prod_{i=0,i\not\in
S}^{n}((\ell-K_{i})^{2}-M_{i}^{2})},~{}~{}~{}~{}~{}~{}$ (2)

where $\widehat{S}$ is the list of removed propagators, then the row vector
$\bm{J}_{n+1}$ will be ordered as

${\bm{J}}=\\{J_{n+1};J_{n+1;\widehat{0}},J_{n+1;\widehat{1}},...,J_{n+1;\widehat{n}};J_{n+1;\widehat{01}},J_{n+1;\widehat{02}},...,J_{n+1;\widehat{(n-1)n}};...;J_{n+1;\widehat{01...(n-1)}},..\\}~{}.~{}~{}~{}$
(3)

Similarly we define the generating functions

$I_{n+1;\widehat{S}}^{R}\equiv\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot
R}\over\prod_{i=0,i\not\in S}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}=\bm{J}_{n+1}\cdot\vec{{\alpha}}^{gen}_{n+1;\widehat{S}}~{}.~{}~{}$
(4)

There are two remarks regarding (4). First, according to our iterative
strategy, since $I_{n+1;\widehat{S}}^{R}$ has fewer number of propagators,
when solving $\vec{{\alpha}}^{gen}$ defined in (1), we treat
$\vec{{\alpha}}^{gen}_{n+1;\widehat{S}}$ as known functions. Secondly, the
right hand side of (4) has been written in the expansion of larger basis
$\bm{J}_{n+1}$ instead of $\bm{J}_{n+1;\widehat{S}}$. Thus the column vector
$\vec{{\alpha}}^{gen}_{n+1;\widehat{S}}$ should be extended also, i.e., the
component of vector $\vec{{\alpha}}^{gen}_{n+1;\widehat{S}}$ should be written
as $(\vec{{\alpha}}^{gen}_{n+1;\widehat{S}})_{\widehat{S}_{1}}$ by our
convention given in (3). One obvious result from the extension is that
$(\vec{{\alpha}}^{gen}_{n+1;\widehat{S}})_{\widehat{S}_{1}}=0$ as long as
$S\not\subseteq S_{1}$.

Next we consider a little bit more general generating functions defined by

$I_{n+1;\\{\bm{a}\\}}^{R}\equiv\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot
R}\over\prod_{i=0}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})^{a_{i}}}=\bm{J}_{n+1}\cdot\vec{{\alpha}}^{gen}_{\bm{a}}~{}~{}~{}$
(5)

where $\bm{a}$ is the vector of powers of propagators with $a_{i}\geq
1,\forall i$. There are two different approaches to the problem (5). The first
approach bases on the observation

$I^{R}_{n+1;\\{\bm{a}\\}}=\left(\prod_{j=0}^{n}{1\over(a_{j}-1)!}{\partial^{a_{j}-1}\over\partial(M_{j}^{2})^{a_{j}-1}}\right)I^{R}_{n+1}~{}.~{}~{}$
(6)

Using the result

${\partial\over\partial(M_{j}^{2})}I^{R}_{n+1}=\left({\partial\over\partial(M_{j}^{2})}\bm{J}_{n+1}\right)\cdot\vec{{\alpha}}^{gen}+\bm{J}_{n+1}\cdot\left({\partial\over\partial(M_{j}^{2})}\vec{{\alpha}}^{gen}\right)\equiv\bm{J}_{n+1}\cdot{\cal
D}_{n+1;j}\vec{{\alpha}}^{gen}~{},~{}~{}$ (7)

where we have rewritten the differential action over basis as

${\partial\over\partial(M_{j}^{2})}\bm{J}_{n+1}\equiv\bm{J}_{n+1}{\cal
H}_{n+1;j}~{}~{}~{}$ (8)

and defined the ”covariant derivative” as

${\cal D}_{n+1;j}\equiv{\partial\over\partial(M_{j}^{2})}+{\cal
H}_{n+1;j}~{},~{}~{}$ (9)

we can obtain the reduction of general integrals in (5) as

$I^{R}_{n+1;\\{\bm{a}\\}}=\bm{J}_{n+1}\cdot\left\\{\left(\prod_{j=0}^{n}{1\over(a_{j}-1)!}{\cal
D}^{a_{j}-1}_{n+1;j}\right)\vec{{\alpha}}^{gen}\right\\}~{}.~{}~{}$ (10)

Thus to complete solve the problem by the first approach, we just need to find
the matrix ${\cal H}_{n+1;j}$. Before presenting the systematical algorithm to
write down it in later section, we give some general discussions:

  * •

(a) Since the differential operators ${\partial\over\partial(M_{j}^{2})}$ are
commutative with each other on the left hand side of (7), we have following
”integrability conditions” as

$[{\cal D}_{n+1;i},{\cal D}_{n+1;j}]=0,~{}~{}~{}\forall i,j~{},~{}~{}$ (11)

which is equivalent to

$\left({\partial\over\partial(M_{i}^{2})}{\cal H}_{n+1;j}\right)+{\cal
H}_{n+1;i}{\cal H}_{n+1;j}-\left({\partial\over\partial(M_{j}^{2})}{\cal
H}^{n+1}_{i}\right)-{\cal H}_{n+1;j}{\cal H}_{n+1;i}=0~{}.~{}~{}$ (12)

Relations (12) will give a strong consistent check for matrices ${\cal
H}_{n+1;j}$.

  * •

(b) Elements of the matrix can be represented as $({\cal
H}_{n+1;j})_{\widehat{S}_{1},\widehat{S}_{2}}$ where the list $S_{i}$
indicates the removal of propagators from the $(n+1)$-gon (notably, for the
first column $S_{2}=\emptyset$). Thus according to the definition (8), The
$\widehat{S}_{2}$-th column of the matrix $({\cal H}_{n+1;j})$ represents the
reduction of scalar integral obtained from $J_{n+1;\widehat{S}_{2}}$ by
increasing the power of $j$-th propagator from one to two. With this picture,
we immediately obtain following results:

    * –

When $j\in S_{2}$, $({\cal
H}_{n+1;j})_{\widehat{S}_{1},\widehat{S}_{2}}=0,\forall S_{1}$, since
${\partial\over\partial M_{j}^{2}}J_{n+1;\widehat{S}_{2}}=0$.

    * –

When $S_{2}\not\subseteq S_{1}$ ($\emptyset\subseteq S$ for any list $S$),
$({\cal H}_{n+1;j})_{\widehat{S}_{1},\widehat{S}_{2}}=0$ since the reduction
of ${\partial\over\partial M_{j}^{2}}J_{n+1;\widehat{S}_{2}}$ can not reach
the scalar basis $J_{n+1;\widehat{S}_{1}}$.

  * •

(c) By our iterative construction, only the first column of the matrix ${\cal
H}_{n+1;j}$ is unknown, since all other columns are the reduction of lower
gons, after naturally embedding to the larger basis $\bm{J}_{n+1}$ by the
exactly same remark after the equation (4).

The second approach to solve (5) is to sum over all $\bm{a}$ as

$\sum_{a_{0},\cdots,a_{n}=1}^{\infty}t_{0}^{a_{0}-1}\cdot t_{1}^{a_{1}-1}\cdot
t_{2}^{a_{2}-1}\cdot\cdots\cdot t_{n}^{a_{n}-1}I_{n+1;\\{\bm{a}\\}}^{R}$ (13)

to reach

$I_{n+1}^{R}(\bm{t})\equiv\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot
R}\over\prod_{i=0}^{n}((\ell-K_{i})^{2}-M_{i}^{2}-t_{i})}~{}.~{}~{}$ (14)

In fact, $I_{n+1}^{R}(\bm{t})$ is the most general generating function for one
loop reduction we are looking for. It gives the reduction coefficients of one
loop integrals with arbitrary numerator and denominator after the expansion of
$R,\bm{t}$.

Expression (14) is very similar to (1) except the mass shifting $M_{i}^{2}\to
M_{i}^{2}+t_{i}$. Thus using (1) we have immediately

$\displaystyle
I_{n+1}^{R}(\bm{t})=\bm{J}_{n+1}(\bm{t})\cdot\vec{{\alpha}}^{gen}(R,\bm{t})~{}~{}~{}$
(15)

where the scalar basis $\bm{J}_{n+1}(\bm{t})$ is the one in (2) with the mass
shifting $M_{i}^{2}\to M_{i}^{2}+t_{i}$. To complete the reduction (15), we
need to find the reduction

$\bm{J}_{n+1}(\bm{t})=\bm{J}_{n+1}~{}{\cal G}(\bm{t})~{}~{}~{}$ (16)

where the matrix ${\cal G}$ depends on $\bm{t}$ only, but not $R$. In other
words, to compute ${\cal G}(\bm{t})$, we can start from (14) by setting $R=0$.
Assembling all together we finally have

$I_{n+1}^{R}(\bm{t})=\bm{J}_{n+1}\cdot\vec{\gamma}^{gg}(R,\bm{t}),~{}~{}~{}~{}~{}~{}~{}~{}~{}\vec{\gamma}^{gg}(R,\bm{t})\equiv{\cal
G}(\bm{t})\vec{{\alpha}}^{gen}(R,\bm{t})~{}.~{}~{}$ (17)

Although $\vec{\gamma}^{gg}(R,\bm{t})$ is the most general generating function
for the reduction, it is more complicated comparing to $\vec{{\alpha}}^{gen}$
in (1) with the appearance of multiple variables $t_{i}$.

Above two approaches are, in fact, related to each other as following. Using
(10) one finds

$\displaystyle\sum_{a_{0},\cdots,a_{n}=1}^{\infty}{t_{0}^{a_{0}-1}\cdot
t_{1}^{a_{1}-1}\cdot\cdots\cdot
t_{n}^{a_{n}-1}}I^{R}_{n+1;\\{\bm{a}\\}}=\prod_{j=0}^{n}\left(\sum_{a_{j}=1}^{\infty}{t_{j}^{a_{j}-1}\over(a_{j}-1)!}\left({\partial\over\partial
M_{j}^{2}}\right)^{a_{j}-1}\right)I^{R}_{n+1}$ (18) $\displaystyle=$
$\displaystyle\bm{J}_{n+1}\left(\prod_{j=0}^{n}\left(\sum_{a_{j}=1}^{\infty}{t_{j}^{a_{j}-1}\over(a_{j}-1)!}\left({\cal
D}_{n+1;j}\right)^{a_{j}-1}\right)\vec{{\alpha}}^{gen}\right)=\bm{J}_{n+1}\cdot\left(e^{\sum_{j=0}^{n}t_{j}{\cal
D}_{n+1;j}}\vec{{\alpha}}^{gen}\right)~{}~{}~{}$

thus one gets

$\vec{\gamma}^{gg}(R,\bm{t})=e^{\sum_{j=0}^{n}t_{j}{\cal
D}_{n+1;j}}\vec{{\alpha}}^{gen}(R)~{}~{}~{}$ (19)

Although knowing the matrices ${\cal H}_{n+1;j}$ is sufficient to compute
$\vec{\gamma}^{gg}(R,\bm{t})$ via equation (19), the evaluation of
differential action $e^{\sum_{j=0}^{n}t_{j}{\cal D}_{n+1;j}}$ is still a non-
trivial task. For the completion of the paper, we will systematically and
analytically solve $\vec{{\alpha}}^{gen}(R)$, ${\cal H}_{n+1;j}$ and ${\cal
G}(\bm{t})$ one by one in subsequent sections.

Before ending this section, it is important to emphasize that explicit
expressions of generating functions in (1), (5) and (14) rely on the choice of
loop momentum $\ell$. For the convenient in subsequent discussions, unless
explicitly stated, we adopt the convention that $K_{0}=0$.

##  3 The IBP relations

In this section, we will derive differential equations for generating function
(1) coming from IBP relations. The results presented in this section form the
foundation for all discussions in later sections.

###  3.1 Fundamental IBP relations

For fundamental IBP relations, the first one is

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int{d^{D}\ell\over
i\pi^{D/2}}{d\over d\ell^{\mu}}\left\\{\ell^{\mu}{e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}\right\\}$ (20) $\displaystyle=$
$\displaystyle\int{d^{D}\ell\over i\pi^{D/2}}{De^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}+\int{d^{D}\ell\over i\pi^{D/2}}{2\ell\cdot
Re^{2\ell\cdot R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}$ $\displaystyle+\int{d^{D}\ell\over
i\pi^{D/2}}{-2\ell^{2}e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})^{2}\prod_{i=1}^{n}((\ell-K_{i})^{2}-M_{i}^{2})}$
$\displaystyle+\sum_{j=1}^{n}\int{d^{D}\ell\over i\pi^{D/2}}{-2\ell\cdot(\ell-
K_{j})e^{2\ell\cdot R}\over((\ell-
K_{j})^{2}-M_{j}^{2})^{2}(\ell^{2}-M_{0}^{2})\prod_{i=1,i\neq j}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}~{}.~{}~{}$

Among these four terms at the right hand side of (20), the first term is just
$D\bm{J}_{n+1}\cdot\vec{{\alpha}}^{gen}$. The second term is
$\left(R\cdot{\partial\over\partial
R}\right)I_{n+1}^{R}=\bm{J}_{n+1}\cdot\left(\left(R\cdot{\partial\over\partial
R}\right)\vec{{\alpha}}^{gen}\right)$. For the third term, after writing
$2\ell^{2}=2((\ell^{2}-M_{0}^{2})+M_{0}^{2})$ we get

$\int{d^{D}\ell\over i\pi^{D/2}}{-2M_{0}^{2}e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})^{2}\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}+\int{d^{D}\ell\over i\pi^{D/2}}{-2e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}~{}.~{}~{}$ (21)

By discussions from (5) to (10), the first term is just
$\bm{J}_{n+1}\cdot(-2M_{0}^{2}{\cal D}_{n+1;0}\vec{{\alpha}}^{gen})$. To
simplify the notation, we will define

${\cal D}_{n+1;i}\vec{{\alpha}}^{gen}\equiv\vec{X}_{i}~{}~{}~{}$ (22)

and the third term of (20) can be written as
$\bm{J}_{n+1}\cdot\left(-2M_{0}^{2}\vec{X}_{0}-2\vec{{\alpha}}^{gen}\right)$.
By similar manipulations, the fourth term of (20) is

$\bm{J}_{n+1}\cdot\left(\sum_{j=1}^{n}f_{j}\vec{X}_{j}-\vec{{\alpha}}^{gen}-{\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)~{}~{}~{}$ (23)

where $\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}$ is known as defined in (4) and
$f_{j}\equiv K_{j}^{2}-M_{j}^{2}-M_{0}^{2}$. It is worth to emphasize that
although naively ${\cal D}_{n+1;j}$ contains the unknown matrix ${\cal
H}_{n+1;j}$, the entire combination ${\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}$ is, in fact, known due to
the following observation

${\partial\over\partial(M_{j}^{2})}I^{R}_{n+1;\widehat{0}}=\bm{J}_{n+1}\cdot\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)~{}~{}~{}$ (24)

with our iterative assumption. Putting all together, the IBP relation (20)
becomes

$0=\bm{J}_{n+1}\cdot\left\\{\left(D-2-n+R\cdot{d\over
dR}\right)\vec{{\alpha}}^{gen}-2M_{0}^{2}\vec{X}_{0}+\sum_{j=1}^{n}\left(f_{j}\vec{X}_{j}-{\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)\right\\}~{}~{}~{}~{}~{}~{}$
(25)

From it we read out the differential equation for generating function

$0=\left\\{\left(D-2-n+R\cdot{d\over
dR}\right)\vec{{\alpha}}^{gen}-2M_{0}^{2}\vec{X}_{0}+\sum_{j=1}^{n}\left(f_{j}\vec{X}_{j}-{\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)\right\\}~{}~{}~{}~{}~{}~{}$
(26)

Having worked through the IBP relation (20), we merely write down other
fundamental IBP relations and corresponding differential equations without
details. From IBP relations444In fact, instead of using ${d\over
d\ell^{\mu}}K_{r}^{\mu}$, it is better to use ${d\over d\ell^{\mu}}(\ell-
K_{r})^{\mu}$ as shown in [54] (see also [53]), which is the origin of the
simplification used in (36).

$0=\int{d^{D}\ell\over i\pi^{D/2}}{d\over
d\ell^{\mu}}\left\\{K_{r}^{\mu}{e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}\right\\}~{}~{}~{}$ (27)

with $r=1,...,n$, we have

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\left\\{(2K_{r}\cdot
R)\vec{{\alpha}}^{gen}+\left({\cal
D}_{n+1;0}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)-g_{r}\vec{X}_{0}-\left({\cal
D}_{n+1;r}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)+(-g_{r}+2K^{2}_{r})\vec{X}_{r}\right.$
(28) $\displaystyle\left.+\sum_{j=1,j\neq r}^{n}\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)-\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)+(-g_{r}+2K_{r}\cdot
K_{j})\vec{X}_{j}\right\\}~{}~{}~{}$

with $g_{r}=K_{r}^{2}-M_{r}^{2}+M_{0}^{2}$. From IBP relation

$0=\int{d^{D}\ell\over i\pi^{D/2}}{d\over
d\ell^{\mu}}\left\\{R^{\mu}{e^{2\ell\cdot
R}\over(\ell^{2}-M_{0}^{2})\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2})}\right\\}~{}~{}~{}$ (29)

we have

$0=2R^{2}\vec{{\alpha}}^{gen}+\sum_{j=0}^{n}\left(\left(2R\cdot
K_{j}-R\cdot{\partial\over\partial R}\right)\vec{X}_{j}\right)~{}~{}~{}$ (30)

where the $K_{0}=0$ convention in (30) should be remembered. Compared to the
familiar IBP relations (20) and (27), the (29) is a new one by the
introduction of the auxiliary vector $R$. It will play crucial role when
solving the generating function $\vec{{\alpha}}^{gen}$.

###  3.2 Simplifying differential equations

Now we try to simplify the differential equations (26), (28) and (30). With
definitions

$\displaystyle\vec{Y}_{0}$ $\displaystyle\equiv$
$\displaystyle\left(D-2-n+R\cdot{\partial\over\partial
R}\right)\vec{{\alpha}}^{gen}-\sum_{j=1}^{n}{\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}~{}~{}~{}~{}~{}~{}$ (31)
$\displaystyle\vec{Y}_{r}$ $\displaystyle\equiv$ $\displaystyle(2K_{r}\cdot
R)\vec{{\alpha}}^{gen}+\left({\cal
D}_{n+1;0}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)-\left({\cal
D}_{n+1;r}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)$ (32)
$\displaystyle+\sum_{j=1,j\neq r}^{n}\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)-\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{0}}^{gen}\right)~{}~{}~{}$

differential equations (26) and (28) can be cast to

$\displaystyle\vec{Y}_{0}$ $\displaystyle=$ $\displaystyle
2M_{0}^{2}\vec{X}_{0}-\sum_{j=1}^{n}(K_{j}^{2}-M_{j}^{2}-M_{0}^{2})\vec{X}_{j}~{}~{}~{}~{}~{}$
(33) $\displaystyle\vec{Y}_{r}$ $\displaystyle=$
$\displaystyle(K_{r}^{2}-M_{r}^{2}+M_{0}^{2})\vec{X}_{0}-(-(K_{r}^{2}-M_{r}^{2}+M_{0}^{2})+2K^{2}_{r})\vec{X}_{r}$
(34) $\displaystyle-\sum_{j=1,j\neq
r}^{n}(-(K_{r}^{2}-M_{r}^{2}+M_{0}^{2})+2K_{r}\cdot
K_{j})\vec{X}_{j}~{}~{}~{}$

With the definition of the modified Caylay matrix $F$ with elements

$f_{ij}=(K_{i}-K_{j})^{2}-M_{i}^{2}-M_{j}^{2},~{}~{}~{}~{}i,j=0,...,n;~{}~{}~{}K_{0}=0~{}~{}~{}~{}~{}$
(35)

and

$\vec{\cal Y}_{r}\equiv\vec{Y}_{r}-\vec{Y}_{0},~{}~{}~{}~{}\vec{\cal
Y}_{0}\equiv-\vec{Y}_{0}~{},~{}~{}~{}~{}$ (36)

(33) and (34) can be written beautifully as555Above simplification is
motivated by Eq(16) of [53] (see also [54])

$\vec{\cal
Y}_{r}=\sum_{j=0}^{n}f_{rj}\vec{X}_{j}~{},~{}~{}~{}r=0,1...,n~{}~{}~{}~{}$
(37)

One immediately solves

$\boxed{\vec{X}_{i}=(F^{-1})_{ij}\vec{\cal Y}_{j}}~{}~{}~{}~{}~{}$ (38)

where $\vec{\cal Y}_{r}$, $r=0,1,...,n$ can be simplified as

$\vec{\cal Y}_{r}=-\left(D-2-n-2K_{r}\cdot R+R\cdot{\partial\over\partial
R}\right)\vec{{\alpha}}^{gen}+\sum_{j=0,j\neq r}^{n}\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)~{}~{}~{}~{}~{}$ (39)

by using (31) and (32).

Now we put (38) back to (30) to get

$\displaystyle\boxed{0=2R^{2}\vec{{\alpha}}^{gen}+\sum_{j=0}^{n}\left(\left(2R\cdot
K_{j}-R\cdot{\partial\over\partial R}\right)(F^{-1})_{jr}\vec{\cal
Y}_{r}\right)}~{}~{}~{}$ (40)

Equations (40) and (38) are two main results of the section. Using (40) we can
solve $\vec{{\alpha}}^{gen}$ and using (38) we can solve the unknown matrix
${\cal H}_{n+1;j}$. Before ending this section, let us remark that so far we
have only used fundamental IBP relations for differential equations. In fact,
one can consider other IBP relations as done in [44, 52]. We will give more
discussions of the issue in the conclusion.

##  4 Solving $\vec{{\alpha}}^{gen}$

In this section, we will use differential equations presented in previous
section to solve the generating function $\vec{{\alpha}}^{gen}$. The same
generating function has been studied in [51]. However, differential equations
in [51] are established using the PV-reduction method, thus they are $(n+1)$
partial differential equations. Opposite to it, as we will show in this
section, using the IBP method a single second order ordinary differential
equation for generating function can be established, thus solving it is much
simpler.

###  4.1 Solving by series expansion

First we solve it using series expansion. To simplify notations, let us define

$B_{r}=\sum_{j=0,j\neq r}^{n}\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)~{}~{}~{}~{}$ (41)

which is known by our iterative construction (see the explanation (24)). In
fact, using the definition (22) each term at the right hand side of (41) can
be solved by (38) with reduced modified Caylay matrix and the generating
function of $n$-gon. Then Eq (40) can be rearranged to

$\displaystyle\left\\{2R^{2}-\sum_{j=0}^{n}\left(2R\cdot
K_{j}-R\cdot{\partial\over\partial
R}\right)(F^{-1})_{jr}\left(D-2-n-2K_{r}\cdot R+R\cdot{\partial\over\partial
R}\right)\right\\}\vec{{\alpha}}^{gen}$ (42) $\displaystyle=$
$\displaystyle-\sum_{j=0}^{n}\left(2R\cdot K_{j}-R\cdot{\partial\over\partial
R}\right)(F^{-1})_{jr}B_{r}~{}.~{}~{}$

Now we expand

$\displaystyle\vec{{\alpha}}^{gen}$ $\displaystyle=$
$\displaystyle\sum_{i_{0},...,i_{n}=0}^{\infty}\vec{{\alpha}}_{i_{0}i_{1}...i_{n}}(R^{2})^{i_{0}}(2K_{1}\cdot
R)^{i_{1}}...(2K_{n}\cdot R)^{i_{n}}$ $\displaystyle B_{r}$ $\displaystyle=$
$\displaystyle\sum_{i_{0},...,i_{n}=0}^{\infty}\vec{b}_{r;i_{0}i_{1}...i_{n}}(R^{2})^{i_{0}}(2K_{1}\cdot
R)^{i_{1}}...(2K_{n}\cdot R)^{i_{n}}~{}~{}~{}$ (43)

Putting (43) to (42) and matching coefficients of independent tensor
structures at both sides, we can solve

$\displaystyle\vec{{\alpha}}_{i_{0}i_{1}...i_{n}}(2i_{0}+\sum_{t=1}^{n}i_{t})(D-2-n+(2i_{0}+\sum_{t=1}^{n}i_{t}))\sum_{j,r=0}^{n}(F^{-1})_{jr}$
(44) $\displaystyle=$
$\displaystyle-\left[\sum_{j=1}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}\vec{b}_{r;i_{0}i_{1}...(i_{j}-1)...i_{n}}-(2i_{0}+\sum_{t=1}^{n}i_{t})\sum_{j,r=0}^{n}(F^{-1})_{jr}\vec{b}_{r;i_{0}i_{1}...i_{n}}\right]$
$\displaystyle+\sum_{j=0}^{n}\sum_{r=1}^{n}(F^{-1})_{jr}\vec{{\alpha}}_{i_{0}i_{1}...(i_{r}-1)...i_{n}}(2i_{0}+\sum_{t=1}^{n}i_{t})$
$\displaystyle+\sum_{j=1}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...i_{n}}(D-2-n+2i_{0}+\sum_{t=1}^{n}i_{t}-1)$
$\displaystyle-\sum_{j,r=1}^{n}(F^{-1})_{jr}\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...(i_{r}-1)...i_{n}}-2\vec{{\alpha}}_{(i_{0}-1)i_{1}...i_{n}}~{}~{}~{}$

If we consider the symmetry of matrix $F$, it can be simplified as

$\displaystyle\vec{{\alpha}}_{i_{0}i_{1}...i_{n}}(2i_{0}+\sum_{t=1}^{n}i_{t})(D-2-n+(2i_{0}+\sum_{t=1}^{n}i_{t}))\sum_{j,r=0}^{n}(F^{-1})_{jr}$
(45) $\displaystyle=$
$\displaystyle-\left[\sum_{j=1}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}\vec{b}_{r;i_{0}i_{1}...(i_{j}-1)...i_{n}}-(2i_{0}+\sum_{t=1}^{n}i_{t})\sum_{j,r=0}^{n}(F^{-1})_{jr}\vec{b}_{r;i_{0}i_{1}...i_{n}}\right]$
$\displaystyle+\sum_{j=1}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...i_{n}}(D-3-n+4i_{0}+2\sum_{t=1}^{n}i_{t})$
$\displaystyle-\sum_{j,r=1}^{n}(F^{-1})_{jr}\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...(i_{r}-1)...i_{n}}-2\vec{{\alpha}}_{(i_{0}-1)i_{1}...i_{n}}~{}~{}~{}$

There are a few remarks for the results (44) and (45). First we make the
convention that coefficients $({\alpha}/b_{r})_{i_{0}...i_{n}}=0$ if any
subscript $i_{k}<0$. Secondly, for the first term in the last line of (44) and
(45), if $j=r$, we should interpret
$\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...(i_{r}-1)...i_{n}}$ as
$\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-2)...i_{n}}$. Thirdly since for
$\vec{{\alpha}}$, any subscript at the right hand side of (44) and (45) is
always equal or at least one less than those at the left hand side, we can
solve $\vec{{\alpha}}_{i_{0}i_{1}...i_{n}}$ recursively using (45) with the
obviously initial condition $\vec{{\alpha}}_{00...0}=\\{1,0,0,...,0\\}$.
Fourthly for $n\geq 1$, the summation is666see Eq(B.6) in [55]

$\sum_{i,j=0}^{n}(F^{-1})_{i,j}=(-)^{n}2^{n}{{\rm Gram}\over{\rm
det}(F)},~{}~{}~{}$ (46)

where the ${\rm Gram}$ is defined as

${\rm Gram}={\rm det}((K_{a}-K_{*})\cdot(K_{b}-K_{*})),~{}~{}~{}a,b\in{\cal
K}\setminus\\{K_{*}\\}~{}~{}~{}$ (47)

with $K_{*}$ any fixed moment in the set ${\cal
K}=\\{K_{0},K_{1},...,K_{n}\\}$. For our convention, $K_{0}=0$ so it is
convenient to take $K_{*}=K_{0}$. Using (46) one can simplify the calculation
of (44) and (45). Fifthly when we consider the first component of the vector
$\vec{{\alpha}}$, i.e., $(\vec{{\alpha}})_{\emptyset}$ , since all
contributions from $\vec{b}$ is zero, (45) can be further simplified to

$\displaystyle(\vec{{\alpha}}_{i_{0}i_{1}...i_{n}})_{\emptyset}(2i_{0}+\sum_{t=1}^{n}i_{t})(D-2-n+(2i_{0}+\sum_{t=1}^{n}i_{t}))\sum_{j,r=0}^{n}(F^{-1})_{jr}$
(48) $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}(\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...i_{n}})_{\emptyset}(D-3-n+4i_{0}+2\sum_{t=1}^{n}i_{t})$
$\displaystyle-\sum_{j,r=1}^{n}(F^{-1})_{jr}(\vec{{\alpha}}_{i_{0}i_{1}...(i_{j}-1)...(i_{r}-1)...i_{n}})_{\emptyset}-2(\vec{{\alpha}}_{(i_{0}-1)i_{1}...i_{n}})_{\emptyset}~{}.~{}~{}$

It gives the reduction coefficients from $(n+1)$-gon to $(n+1)$-gon.

Now we use several examples to demonstrate the use of (45).

####  4.1.1 The tadpole

For this case, since all $B_{r}=0$, (45) is simplified to

${{\alpha}}_{i_{0}}(2i_{0})(D-2+2i_{0})F_{00}^{-1}=-2{{\alpha}}_{(i_{0}-1)}~{}~{}~{}$
(49)

with

$F_{00}=-2M_{0}^{2}~{}.$ (50)

Thus it is easy to solve

$\displaystyle{{\alpha}}_{i_{0}}={M_{0}^{2}\over i_{0}({D\over
2}+i_{0}-1)}{{\alpha}}_{i_{0}-1},~{}~{}~{}{{\alpha}}_{0}=1$ (51)

which reproduces the result given in [51], i.e.,

${\alpha}^{gen}(M_{0},R)=\sum_{n=0}^{\infty}{(M_{0}^{2}R^{2})^{n}\over
n!\left({D\over 2}\right)_{n}}=~{}_{0}F_{1}(\emptyset;{D\over
2};M_{0}^{2}R^{2})~{}~{}~{}~{}$ (52)

where the Pochhammer symbol is defined by777From the definition one can see
that $(x)_{n=0}=1,\forall x$.

$(x)_{n}={\Gamma(x+n)\over\Gamma(x)}=\prod_{i=1}^{n}(x+(i-1))~{}.~{}~{}~{}$
(53)

Before ending this part, let us remark that using the momentum shifting, it is
easy to see that

$\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot R}\over((\ell-
K_{1})^{2}-M_{1}^{2})}=e^{2K_{1}\cdot
R}{\alpha}^{gen}(M_{1},R)\int{d^{D}\ell\over i\pi^{D/2}}{1\over((\ell-
K_{1})^{2}-M_{1}^{2})}~{}~{}~{}~{}$ (54)

which will be useful for later iterative computation.

####  4.1.2 The bubble

Now let us consider the first nontrivial case, i.e., the bubble. According to
our convention, the basis is

$\bm{J}_{2}=\\{I_{2},I_{2;\widehat{0}},I_{2;\widehat{1}}\\}~{}~{}~{}~{}~{}$
(55)

The modified Caylay matrix $F$ is

$\displaystyle
F=\left(\begin{array}[]{cc}-2M_{0}^{2}&K_{1}^{2}-M_{0}^{2}-M_{1}^{2}\\\
K_{1}^{2}-M_{0}^{2}-M_{1}^{2}&-2M_{1}^{2}\end{array}\right)~{}.~{}~{}~{}$ (58)

Using (24) one can find888To carry out the computation, one need to know the
matrix ${\cal H}_{n+1;j}$, which will be discussed in the section 5. for the
expansion
$B_{i}=\sum_{i_{0}i_{1}}\vec{b}_{i;i_{0}i_{1}}(R^{2})^{i_{0}}(2K_{1}\cdot
R)^{i_{1}}$ with

$\displaystyle\vec{b}_{0;i_{0}i_{1}}={(D-2+2i_{0})\over 2M_{1}^{2}}{1\over
i_{1}!}\left(\begin{array}[]{c}0\\\ {(M_{1}^{2})^{i_{0}}\over
i_{0}!\left({D\over 2}\right)_{i_{0}}}\\\
0\end{array}\right)~{},~{}~{}~{}\vec{b}_{1;i_{0}i_{1}}={(D-2+2i_{0})\over
2M_{0}^{2}}\delta_{i_{1},0}\left(\begin{array}[]{c}0\\\ 0\\\
{(M_{0}^{2})^{i_{0}}\over i_{0}!\left({D\over
2}\right)_{i_{0}}}\end{array}\right)~{}~{}~{}~{}$ (65)

Although it is not difficult to write down the analytic expressions using (44)
or (45), since these results have been presented in various literatures many
times (see, for example, [46] ), here we will just give some numerical checks.
We choose numeric values of various kinetic variable as follows

$M_{0}^{2}\to\frac{123}{100},\ M_{1}^{2}\to\frac{64}{25},\
K_{1}^{2}\to\frac{69}{50},~{}~{}R^{2}\to\frac{133}{10},\ K_{1}\cdot
R\to\frac{367}{100},\ D\to\frac{969}{50}.$ (66)

Then we have

$\displaystyle{\rm Rank~{}One}:$ $\displaystyle(2R\cdot
K_{1})\vec{\alpha}_{0,1}=(0.132971,\ 2.65942,\ -2.65942)^{T}$
$\displaystyle{\rm Rank~{}Two}:$ $\displaystyle 2!((2R\cdot
K_{1})^{2}\vec{\alpha}_{0,2}+R^{2}\vec{\alpha}_{1,0})=(0.964908,\ 20.252,\
-0.346648)^{T}$ $\displaystyle{\rm Rank~{}Three}:$ $\displaystyle 3!((2R\cdot
K_{1})^{3}\vec{\alpha}_{0,3}+R^{2}(2R\cdot
K_{1})\vec{\alpha}_{1,1})=(0.380212,\ 174.709,\ -13.8036)^{T}.$ (67)

They agree with results by FIRE6 [20].

####  4.1.3 Triangle

Again, we give only the numerical check. We choose numeric values of various
kinetic variable as follows:

$\displaystyle M_{0}^{2}\to\frac{123}{100},\ M_{1}^{2}\to\frac{64}{25},\
M_{2}^{2}\to\frac{357}{100},\ K_{1}^{2}\to\frac{69}{50},\
K_{2}^{2}\to\frac{61}{25}$ $\displaystyle R^{2}\to\frac{133}{10},\ K_{1}\cdot
R\to\frac{367}{100},\ K_{2}\cdot R\to\frac{523}{100},\ K_{1}\cdot
K_{2}\to\frac{67}{100},\ D\to\frac{969}{50}.$ (68)

Then we have

$\displaystyle{\rm Rank~{}One}:$ $\displaystyle(2R\cdot
K_{1})\vec{\alpha}_{0,1,0}+(2R\cdot K_{2})\vec{\alpha}_{0,0,1}=(0.256446,\
3.49834,\ -1.86777,\ -1.63054,\ 0,\ 0,\ 0)^{T}.$ $\displaystyle{\rm
Rank~{}Two}:$ $\displaystyle 2!((2R\cdot
K_{1})^{2}\vec{\alpha}_{0,2,0}+(2R\cdot
K_{2})^{2}\vec{\alpha}_{0,0,2}+(2R\cdot K_{1})(2R\cdot
K_{1})\vec{\alpha}_{0,1,1}+R^{2}\vec{\alpha}_{1,0,0})$
$\displaystyle=(-0.523243,\ 29.5767,\ -0.881583,\ -0.639263,\ 8.33983,\
-6.53695,\ -1.80288)^{T}.$ $\displaystyle{\rm Rank~{}Three}:$ $\displaystyle
3!((2R\cdot K_{1})^{3}\vec{\alpha}_{0,3,0}+(2R\cdot
K_{2})^{3}\vec{\alpha}_{0,0,3}+(2R\cdot K_{1})^{2}(2R\cdot
K_{2})\vec{\alpha}_{0,2,1}$ (69) $\displaystyle+(2R\cdot K_{1})(2R\cdot
K_{2})^{2}\vec{\alpha}_{0,1,2}+R^{2}(2R\cdot
K1)\vec{\alpha}_{1,1,0}+R^{2}(2R\cdot K_{2})\vec{\alpha}_{1,0,1})$
$\displaystyle=(-0.43628,\ 260.958,\ 0.723717,\ 0.0778428,\ 3.58409,\
-67.3923,\ 0.605462)^{T}.$

They agree with results by FIRE6 [20].

###  4.2 Solving by differential equation

The equation (42) is the differential equation for generating function
$\vec{{\alpha}}^{gen}$ already. However, it is a partial differential
equation, which is harder to solve than the ordinary differential equation.
Fortunately, the differential operator on the left hand side of (42) is
special and we can transfer it into an ordinary differential equation by
defining a new variable

$R=t\widetilde{R}~{}.~{}~{}~{}$ (70)

With this definition, it is easy to see that for any function $f(R)=\sum
f_{i_{0}...i_{n}}(R^{2})^{i_{0}}(2K_{1}\cdot R)^{i_{1}}...(2K_{n}\cdot
R)^{i_{n}}\equiv\widetilde{f}(t,\widetilde{R})$

$R\cdot{d\over dR}f(R)=t{d\over
dt}\widetilde{f}(t,\widetilde{R})~{}~{},~{}~{}~{}(2K_{r}\cdot
R)f(R)=t(2K_{r}\cdot\widetilde{R})\widetilde{f}(t,\widetilde{R})~{}.~{}~{}~{}$
(71)

Using above results and with some algebraic computations (42) becomes999 A
similar first order ordinary differential equation has also been given in
[52]. However, since the definition of generating function is different
between [51] and [52], the equation and solution are also different.

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\left[\sum_{j=0}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}\right]t{\partial^{2}\over\partial
t^{2}}\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R})$ (72)
$\displaystyle+\left[(D-2-n+1)\sum_{j=0}^{n}\sum_{r=0}^{n}(F^{-1})_{jr}-2\sum_{j=0}^{n}\sum_{r=1}^{n}(F^{-1})_{jr}(2K_{r}\cdot\widetilde{R})t\right]{\partial\over\partial
t}\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R})$
$\displaystyle+\left[2\widetilde{R}^{2}t-(D-2-n+1)\sum_{j=1}^{n}\sum_{r=0}^{n}(2\widetilde{R}\cdot
K_{j})(F^{-1})_{jr}+\sum_{j=1}^{n}\sum_{r=1}^{n}(2\widetilde{R}\cdot
K_{j})(F^{-1})_{jr}(2K_{r}\cdot\widetilde{R})t\right]\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R})$
$\displaystyle+\sum_{j=0}^{n}\left((2\widetilde{R}\cdot
K_{j})-{\partial\over\partial
t}\right)(F^{-1})_{jr}\widetilde{B}_{r}(t,\widetilde{R})~{}~{}~{}~{}$

which belongs to the following type of second order ordinary differential
equation

$At{d^{2}\over dt^{2}}W+(B_{0}+B_{1}t){d\over dt}W+(C_{0}+C_{1}t)W+{\cal
B}(t)=0~{}.~{}~{}~{}$ (73)

To express coefficients more compactly, let us defined following two row
vectors with $(n+1)$ components

${\bm{I}}=(1,1,...,1),~{}~{}~{}{\cal
P}=(2K_{0}\cdot\widetilde{R},2K_{1}\cdot\widetilde{R},...,2K_{n}\cdot\widetilde{R})~{}.~{}~{}~{}$
(74)

Using them we have

$\displaystyle A$ $\displaystyle=$
$\displaystyle{\bm{I}}\cdot(F^{-1})\cdot{\bm{I}}^{T},~{}~{}~{}~{}~{}B_{0}=(D-(n+1))A$
$\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle-2{\bm{I}}\cdot(F^{-1})\cdot{\cal P}^{T}=-2{\cal
P}\cdot(F^{-1})\cdot{\bm{I}}^{T},~{}~{}~{}~{}C_{0}={(D-(n+1))\over 2}B_{1}$
$\displaystyle C_{1}$ $\displaystyle=$ $\displaystyle 2\widetilde{R}^{2}+{\cal
P}\cdot(F^{-1})\cdot{\cal P}^{T}~{}.~{}~{}~{}$ (75)

We want to point out that although previous derivation has used the convention
$K_{0}=0$, results (73), (74) and (75) can be freely applied to the case
$K_{0}\neq 0$.

The detail discussion of equation (73) has been given in the Appendix, where
both analytic expression and series expansion have been given. Using the
results (180), (187) and (188) we write down the solution of (73) as

$\displaystyle\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R})$
$\displaystyle=$ $\displaystyle e^{{-B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over
2}t}(~{}_{1}F_{1}(a_{1};b_{1};x))\left\\{\vec{c}_{1}+\int_{0}^{x}ds{s^{-b_{1}}e^{s}\over(~{}_{1}F_{1}(a_{1};b_{1};s))^{2}}\right.$
(76)
$\displaystyle~{}~{}~{}~{}\left.\left(\vec{c}_{2}+\int_{0}^{s}dyg(y)y^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};y)e^{-y}\right)\right\\}~{}~{}~{}~{}$

with

$\displaystyle x$ $\displaystyle=$
$\displaystyle{-t\sqrt{B_{1}^{2}-4C_{1}A}\over A},~{}~{}~{}b_{1}={B_{0}\over
A},~{}~{}a_{1}={1\over\sqrt{B_{1}^{2}-4C_{1}A}}\left({-B_{1}+\sqrt{B_{1}^{2}-4C_{1}A}\over
2}{B_{0}\over A}+{C_{0}}\right)$ $\displaystyle\vec{g}(x)$ $\displaystyle=$
$\displaystyle{1\over\sqrt{B_{1}^{2}-4AC_{1}}}e^{{-(B_{1}-\sqrt{B_{1}^{2}-4AC_{1}})x\over
2\sqrt{B_{1}^{2}-4AC_{1}}}}\vec{\cal
B}({-xA\over\sqrt{B_{1}^{2}-4C_{1}A}})~{}~{}~{}~{}$ (77)

Since when $t=0$, $~{}_{1}F_{1}(a_{1};b_{1};0)=1$, we find
$\vec{c}_{1}=\\{1,0,0,...,0\\}^{T}$. The finiteness of
${d\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R})\over dt}|_{t=0}$ forces
$\vec{c}_{2}=\vec{0}$. The result of (76) is given in the vector form. When we
consider the first component, i.e., the reduction coefficient of $(n+1)$-gon,
the generating function is just

$\displaystyle(\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R}))_{\emptyset}=e^{{-B_{1}+\sqrt{B_{1}^{2}-4C_{1}A}\over
2A}t}~{}_{1}F_{1}(a_{1};b_{1};x)~{}~{}~{}~{}$ (78)

Using the analytic result (78) (or the simplified version (80) or (85) ), one
can do the similar computations like the one done in [52] to find other
components of generating function using the result (78). Here we will not
repeat the computation.

For our case with special values of $A,B_{i},C_{i}$ given in (75), one can
find that

$b_{1}=(D-(n+1)),~{}~{}~{}~{}~{}a_{1}={b_{1}\over 2}={(D-(n+1))\over
2}~{}~{}~{}~{}$ (79)

for (76), thus (78) can be simplified to

$(\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R}))_{\emptyset}=e^{{-B_{1}+\sqrt{B_{1}^{2}-4C_{1}A}\over
2A}t}~{}_{1}F_{1}({(D-(n+1))\over
2};(D-(n+1));{-t\sqrt{B_{1}^{2}-4C_{1}A}\over A})~{}~{}~{}~{}$ (80)

Next we present some examples.

####  4.2.1 The tadpole

For the tadpole without the choice $K_{0}=0$, we have

$\displaystyle F$ $\displaystyle=$ $\displaystyle-2M_{0}^{2},$ $\displaystyle
A$ $\displaystyle=$ $\displaystyle{-1\over
2M_{0}^{2}},~{}~{}~{}~{}~{}~{}B_{0}=(D-1){-1\over
2M_{0}^{2}},~{}~{}~{}~{}~{}B_{1}={2K_{0}\cdot\widetilde{R}\over M_{0}^{2}},$
$\displaystyle C_{0}$ $\displaystyle=$
$\displaystyle{(D-1)2K_{0}\cdot\widetilde{R}\over
2M_{0}^{2}},~{}~{}~{}~{}~{}~{}C_{1}=2\widetilde{R}^{2}-{(2K_{0}\cdot\widetilde{R})^{2}\over
2M_{0}^{2}}~{}~{}~{}$ (81)

and

$\vec{\widetilde{{\alpha}}}_{tad}^{gen}(t,\widetilde{R})=e^{{(2K_{0}\cdot\widetilde{R}-2\sqrt{M_{0}^{2}\widetilde{R}^{2}})t}}~{}~{}_{1}F_{1}({(D-1)\over
2};D-1;4t\sqrt{M_{0}^{2}\widetilde{R}^{2}})~{}~{}~{}$ (82)

Comparing with (52) and (54), we get following relation

$e^{{-2t\sqrt{M_{0}^{2}\widetilde{R}^{2}}}}~{}~{}_{1}F_{1}({(D-1)\over
2};D-1;4t\sqrt{M_{0}^{2}\widetilde{R}^{2}})=~{}~{}_{0}F_{1}(0;{D\over
2};M_{0}^{2}\widetilde{R}^{2}t^{2})~{}~{}~{}$ (83)

In fact, the identity (83) is the special case of the identity

$e^{-x}~{}~{}_{1}F_{1}(a;2a;2x)=~{}~{}_{0}F_{1}(\emptyset;a+{1\over
2};{x^{2}\over 4})~{}~{}~{}$ (84)

Using (84), (80) can be further simplified as

$\displaystyle(\vec{\widetilde{{\alpha}}}^{gen}(t,\widetilde{R}))_{\emptyset}=e^{{-B_{1}t\over
2A}}~{}_{0}F_{1}(\emptyset;{(D+1-(n+1))\over 2};{(B_{1}^{2}-4C_{1}A)t^{2}\over
16A^{2}})~{}~{}~{}~{}$ (85)

where the square root in (80) disappear naturally using the form (85).

####  4.2.2 The massless bubble

For this one, we just check the result (78). Using

$\displaystyle F=\left(\begin{array}[]{cc}0&K_{1}^{2}\\\
K_{1}^{2}&0\end{array}\right)~{},~{}~{}~{}F^{-1}=\left(\begin{array}[]{cc}0&{1\over
K_{1}^{2}}\\\ {1\over K_{1}^{2}}&0\end{array}\right)~{}~{}~{}$ (90)

we have

$A={2\over K^{2}},~{}~{}~{}B_{0}={2(D-2)\over
K^{2}},~{}~{}B_{1}={-4K\cdot\widetilde{R}\over
K^{2}},~{}~{}C_{0}={-2(D-2)K\cdot\widetilde{R}\over
K^{2}},~{}~{}C_{1}=2\widetilde{R}^{2}~{}~{}~{}$ (91)

and

$x=-2t\sqrt{(K\cdot\widetilde{R})^{2}-K^{2}\widetilde{R}^{2}},~{}~{}~{}b_{1}=(D-2),~{}~{}~{}a_{1}={(D-2)\over
2}~{}~{}~{}$ (92)

The generating function of bubble to bubble is

$\displaystyle
e^{(K\cdot\widetilde{R}+\sqrt{(K\cdot\widetilde{R})^{2}-K^{2}\widetilde{R}^{2}})t}~{}~{}_{1}F_{1}\left({(D-2)\over
2};(D-2);-2t\sqrt{(K\cdot\widetilde{R})^{2}-K^{2}\widetilde{R}^{2}}\right)$
(93) $\displaystyle=$ $\displaystyle
e^{(K\cdot\widetilde{R})t}~{}~{}_{0}F_{1}\left(\emptyset;{(D-1)\over
2};{1\over
4}t^{2}((K\cdot\widetilde{R})^{2}-K^{2}\widetilde{R}^{2})\right)~{}~{}~{}$

One can easily expand (93) and check with known results:

$\displaystyle
1+(K\cdot\widetilde{R})t+{(D(K\cdot\widetilde{R})^{2}-K^{2}\widetilde{R}^{2})t^{2}\over
2(D-1)}+{((D+2)(K\cdot\widetilde{R})^{3}-3K^{2}\widetilde{R}^{2}(K\cdot\widetilde{R}))t^{3}\over
6(D-1)}$
$\displaystyle+{((D+2)(D+4)(K\cdot\widetilde{R})^{3}-6(D+2)K^{2}\widetilde{R}^{2}(K\cdot\widetilde{R})^{2}+4(K^{2}\widetilde{R}^{2})^{2})t^{4}\over
24(D-1)(D+1)}+...$ (94)

##  5 Solving ${\cal H}_{n+1;j}$

In this section, we will solve the matrix ${\cal H}_{n+1;j}$ defined in (8),
which is the reduction of scalar integrals with the $j$-th propagator having
power two while others, power one. The element of the matrix can be
represented as $({\cal H}_{n+1;j})_{\widehat{S}_{1},\widehat{S}_{2}}$ where
the list $S$ indicates the removing propagators from the $(n+1)$-gon.
Furthermore, by our iterative construction, only the first column of the
matrix ${\cal H}_{n+1;j}$ is unknown, since all other columns are naturally
embedding of the matrix ${\cal H}_{(n+1;\widehat{S});j}$ where
$(n+1;\widehat{S})$ is the $m$-gon obtained from $(n+1)$-gon by removing
propagators from the non-empty list $S$.

Having recalled above facts, now we solve ${\cal H}_{n+1;j}$ using (38), which
expands to

$\left({\partial\over\partial(M_{i}^{2})}+{\cal
H}_{n+1;i}\right)\vec{{\alpha}}^{gen}=\sum_{r=0}^{n}(F^{-1}_{ir})\left(-\left(D-2-n-2K_{r}\cdot
R+R\cdot{\partial\over\partial
R}\right)\vec{{\alpha}}^{gen}+B_{r}\right)~{}.~{}~{}~{}$ (95)

If we put the tensor expansion (43) into (95) and focus on terms independent
of $R$, we will find

$\left({\partial\over\partial(M_{i}^{2})}+{\cal
H}_{n+1;i}\right)\vec{{\alpha}}_{00...0}=\sum_{r=0}^{n}(F^{-1}_{ir})\left(-\left(D-2-n\right)\vec{{\alpha}}_{00...0}+\vec{b}_{r;00...0}\right)~{}.~{}~{}~{}$
(96)

Using the boundary condition, i.e.,
$\vec{{\alpha}}_{00...0}=\\{1,0,0,...,0\\}^{T}$, we arrive

${\cal
H}_{n+1;i}\cdot\\{1,0,0,...,0\\}^{T}=\sum_{r=0}^{n}(F^{-1}_{ir})\left(-\left(D-2-n\right)\\{1,0,0,...,0\\}^{T}+\vec{b}_{r;00...0}\right)~{}.~{}~{}~{}$
(97)

The left hand side of (97) picks the unknown first column of the $N\times N$
matrix ${\cal H}_{n+1;j}$, thus $N$ equations101010Since we assume the general
masses, $N=2^{n+1}-1$. (97) will determine ${\cal H}_{n+1;j}$ completely.

To have a better picture of (97), we need to understand $B_{r}$ better.
Recalling the definition of (41), each term $\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)$ is, in fact, the
reduction of (see (7))

$\int{d^{D}\ell\over i\pi^{D/2}}{e^{2\ell\cdot R}\over((\ell-
K_{j})^{2}-M_{j}^{2})^{2}\prod_{t\neq j,r}((\ell-
K_{t})^{2}-M_{t}^{2})}=\bm{J}_{n+1}\cdot\left({\cal
D}_{n+1;j}\vec{{\alpha}}_{n+1;\widehat{r}}^{gen}\right)~{}.~{}~{}~{}~{}$ (98)

To find the $R$-independent term in (98), we set $R=0$ and the left hand side
becomes

$\displaystyle\int{d^{D}\ell\over i\pi^{D/2}}{1\over((\ell-
K_{j})^{2}-M_{j}^{2})^{2}\prod_{t\neq j,r}((\ell-K_{t})^{2}-M_{t}^{2})}$ (99)
$\displaystyle=$ $\displaystyle{\partial\over\partial
M_{j}^{2}}\left\\{\int{d^{D}\ell\over i\pi^{D/2}}{1\over((\ell-
K_{j})^{2}-M_{j}^{2})\prod_{t\neq j,r}((\ell-K_{t})^{2}-M_{t}^{2})}\right\\}$
$\displaystyle=$ $\displaystyle{\partial\over\partial
M_{j}^{2}}(\bm{J}_{(n+1;\widehat{r})}\cdot\vec{{\alpha}}_{\widehat{r}})=\bm{J}_{(n+1;\widehat{r})}\cdot{\cal
H}_{(n+1;\widehat{r});j}\cdot\vec{{\alpha}}_{\widehat{r}}~{}.~{}~{}~{}~{}$

In (99), $\bm{J}_{(n+1;\widehat{r})}$ is the standard scalar basis according
to our convention (3) after removing the $r$-th propagator, thus we have the
vector $\vec{{\alpha}}_{\widehat{r}}=\\{1,0,0,...,0\\}^{T}$. From this
argument, the $R$-independent term is nothing, but the first row of the matrix
${\cal H}_{(n+1;\widehat{r});j}$ after the natural embedding into the basis
$\bm{J}_{n+1}$. Now the meaning of $\vec{b}_{r;00...0}$ is clear: it is the
sum of the reduction of scalar integrals with $r$-th propagator removed and
$j$-th propagator ($j=0,1,...,n;j\neq r$) having power two. Using the
component form, we find

$(\vec{b}_{r;00...0})_{\widehat{S}}=\sum_{j=0,j\neq r}^{n}({\cal
H}_{n+1;j})_{\widehat{S},\widehat{r}}~{}.~{}~{}~{}$ (100)

Having this understanding, (97) gives the solution of each component of the
first column as

$({\cal
H}_{n+1;i})_{\widehat{S},\widehat{\emptyset}}=-\left\\{\left(D-2-n\right)\sum_{r=0}^{n}(F^{-1}_{ir})\right\\}\delta_{S,\emptyset}+\sum_{r=0}^{n}(F^{-1}_{ir})\sum_{j=0,j\neq
r}^{n}({\cal H}_{n+1;j})_{\widehat{S},\widehat{r}}~{},~{}~{}~{}$ (101)

or in vector form

${\cal
H}_{n+1;i}|_{1C}=-\left\\{\left(D-2-n\right)\sum_{r=0}^{n}(F^{-1}_{ir})\right\\}\\{1,0,0,..,0\\}^{T}+\sum_{r=0}^{n}(F^{-1}_{ir})\sum_{j=0,j\neq
r}^{n}({\cal H}_{n+1;j})|_{(r+2)C}~{}~{}~{}~{}$ (102)

where $iC$ means the $i$-th column. Result (101) (or (102)) gives a very clear
recursive construction of the matrix ${\cal H}_{n+1;i}$. Now based on (101),
we can write down the recursive algorithm for all components of the matrix
${\cal H}_{n+1;i}$:

  * •

(1) Let us consider the arbitrary element $({\cal H}_{n+1;i})_{\widehat{\cal
R},\widehat{\cal C}}$ where the removed lists ${\cal R},{\cal C}$ indicate the
corresponding row and column. The first obvious result is when $i\in{\cal C}$,
$({\cal H}_{n+1;i})_{\widehat{\cal R},\widehat{\cal C}}=0,~{}~{}\forall{\cal
R}$. Now we will assume that $i\not\in{\cal C}$. The second obvious result is
that when ${\cal C}\not\subseteq{\cal R}$, the $({\cal
H}_{n+1;i})_{\widehat{\cal R},\widehat{\cal C}}=0$.

  * •

(2) Now we define some notations for later convenience. First we use ${\cal
P}=\\{0,1,...,n\\}$ to represent the complete list of propagators. Given
${\cal C}$ we use $n_{{\cal C}}$ to denote the number of elements in the list
and define the reduced modified Caylay matrix $F_{{\cal P}\setminus{\cal C}}$
by removing the corresponding rows and columns indicated by the list (see
(35)).

  * •

(3) When ${\cal R}={\cal C}$, we have

$({\cal H}_{n+1;i})_{\widehat{\cal C},\widehat{\cal C}}=-(D-1-(n+1-n_{{\cal
C}}))\sum_{r\in{\cal P}\setminus{\cal C}}(F_{{\cal P}\setminus{\cal
C}})^{-1}_{ir}~{}.~{}~{}~{}$ (103)

  * •

(4) When ${\cal C}\subset{\cal R}$, we have

$({\cal H}_{n+1;i})_{\widehat{\cal R},\widehat{\cal C}}=\sum_{r\in{\cal
P}\setminus{\cal C}}(F_{{\cal P}\setminus{\cal C}})^{-1}_{ir}\sum_{j\in{\cal
P}\setminus\left({\cal C}\bigcup\\{r\\}\right)}({\cal
H}_{n+1;j})_{\widehat{\cal R},{\cal C}\bigcup\\{r\\}}~{}.~{}~{}~{}$ (104)

  * •

(5) Equation (104) shows explicitly the recursive structure. With fixed row
list ${\cal R}$, the column list at the right hand side is larger than the
left hand side by one element. Iteratively using (104) will reach two possible
terminations: either ${\cal R}={\cal C}$ with the expression (103) and either
${\cal C}\not\subseteq{\cal R}$ with zero contribution.

Now we demonstrate the use of (101) (or (103) and (104)) with several
examples.

###  5.1 Tadpole

For the tadpole, the second term in (101) does not exist, thus we have
immediately

${\cal H}_{1;0}=-(D-2)F_{00}^{-1}={(D-2)\over 2M_{0}^{2}}~{}~{}~{}$ (105)

If we use (103) and (104), the only case is ${\cal R}={\cal C}=\emptyset$,
which reduces to (105) again.

###  5.2 Bubble

For the bubble, the second term in (101) does contribute. Using (105), we have

$\vec{b}_{0;00...0}=\left(\begin{array}[]{c}0\\\ {(D-2)\over 2M_{1}^{2}}\\\
0\end{array}\right)~{},~{}~{}~{}\vec{b}_{1;00...0}=\left(\begin{array}[]{c}0\\\
0\\\ {(D-2)\over 2M_{0}^{2}}\end{array}\right)~{},~{}~{}~{}$ (106)

according to the basis (55). Using (102) and the modified Caylay matrix $F$
given in (58) we get the first column as

$\displaystyle({\cal
H}_{2;0})|_{1C}={1\over\Delta_{bub}}\left(\begin{array}[]{c}{(D-3)(K_{1}^{2}+M_{1}^{2}-M_{0}^{2})}\\\
{-(D-2)}\\\ -(K_{1}^{2}-M_{0}^{2}-M_{1}^{2}){(D-2)\over
2M_{0}^{2}}\end{array}\right)~{},~{}~{}({\cal
H}_{2;1})|_{1C}={1\over\Delta_{bub}}\left(\begin{array}[]{c}{(D-3)(K_{1}^{2}+M_{0}^{2}-M_{1}^{2})}\\\
-(K_{1}^{2}-M_{0}^{2}-M_{1}^{2}){(D-2)\over 2M_{1}^{2}}\\\
-(D-2)\end{array}\right)~{}.~{}~{}~{}$ (113)

where $\Delta_{bub}={\rm det}(F)$. The other columns are natural embedding of
(105), i.e.,

$\displaystyle{\cal H}_{2;0}=\left(\begin{array}[]{ccc}*&0&0\\\ &0&0\\\
&0&{(D-2)\over 2M_{0}^{2}}\end{array}\right),~{}~{}~{}{\cal
H}_{2;1}=\left(\begin{array}[]{ccc}*&0&0\\\ &{(D-2)\over 2M_{1}^{2}}&0\\\
&0&0\end{array}\right)~{}~{}$ (120)

###  5.3 Triangle

To simplify the notation, let us use $\Delta_{I}$ to represent the determinant
of the sub-matrix, which is obtained from the modified Caylay matrix $F$
defined in (35) by extracting the rows and columns corresponding to the index
set $I$. Thus the first column of $\left(\mathcal{H}_{3;0}\right)$ can be
expressed as:

$\left.\left(\mathcal{H}_{3;0}\right)\right|_{1C}=\left(\begin{array}[]{l}-\frac{(D-4)\left(f_{01}\left(f_{12}+2M_{2}^{2}\right)+f_{02}\left(f_{12}+2M_{1}^{2}\right)-f_{12}^{2}+4M_{1}^{2}M_{2}^{2}\right)}{\Delta_{tri}}\\\
-\frac{2(D-3)\left(f_{12}+M_{1}^{2}+M_{2}^{2}\right)\left(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}\right)}{\Delta_{23}\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{02}+M_{0}^{2}+M_{2}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)}{\Delta_{13}\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{01}+M_{0}^{2}+M_{1}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)}{\Delta_{12}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{13}\left(f_{12}+2M_{2}^{2}\right)\left(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}\right)-\Delta_{23}\left(f_{02}+2M_{2}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)\right)}{2M_{2}^{2}\Delta_{13}\Delta_{23}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{12}\left(f_{12}+2M_{1}^{2}\right)\left(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}\right)-\Delta_{23}\left(f_{01}+2M_{1}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)\right)}{2M_{1}^{2}\Delta_{12}\Delta_{23}\Delta_{tri}}\\\
\frac{(2-D)\left(\Delta_{13}\left(f_{01}+2M_{0}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)+\Delta_{12}\left(f_{02}+2M_{0}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)\right)}{2M_{0}^{2}\Delta_{12}\Delta_{13}\Delta_{tri}}\\\
\end{array}\right)$ (121)

Similarly, the first columns of other matrices are

$\left.\left(\mathcal{H}_{3;1}\right)\right|_{1C}\left(\begin{array}[]{l}-\frac{(D-4)\left(f_{01}\left(f_{02}+2M_{2}^{2}\right)+f_{02}f_{12}-f_{02}^{2}+2M_{0}^{2}\left(f_{12}+2M_{2}^{2}\right)\right)}{\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{12}+M_{1}^{2}+M_{2}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)}{\Delta_{23}\Delta_{tri}}\\\
-\frac{2(D-3)\left(f_{02}+M_{0}^{2}+M_{2}^{2}\right)\left(f_{02}^{2}-4M_{0}^{2}M_{2}^{2}\right)}{\Delta_{13}\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{01}+M_{0}^{2}+M_{1}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)}{\Delta_{12}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{23}\left(f_{02}+2M_{2}^{2}\right)\left(f_{02}^{2}-4M_{0}^{2}M_{2}^{2}\right)-\Delta_{13}\left(f_{12}+2M_{2}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)\right)}{2M_{2}^{2}\Delta_{13}\Delta_{23}\Delta_{tri}}\\\
\frac{(2-D)\left(\Delta_{23}\left(f_{01}+2M_{1}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)+\Delta_{12}\left(f_{12}+2M_{1}^{2}\right)\left(2M_{2}^{2}f_{01}+f_{02}f_{12}\right)\right)}{2M_{1}^{2}\Delta_{12}\Delta_{23}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{12}\left(f_{02}+2M_{0}^{2}\right)\left(f_{02}^{2}-4M_{0}^{2}M_{2}^{2}\right)-\Delta_{13}\left(f_{01}+2M_{0}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)\right)}{2M_{0}^{2}\Delta_{12}\Delta_{13}\Delta_{tri}}\\\
\end{array}\right)$ (122)

and

$\left.\left(\mathcal{H}_{3;2}\right)\right|_{1C}\left(\begin{array}[]{l}\frac{(D-4)\left(-f_{01}\left(f_{02}+f_{12}\right)+f_{01}^{2}-2M_{1}^{2}\left(f_{02}+2M_{0}^{2}\right)-2M_{0}^{2}f_{12}\right)}{\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{12}+M_{1}^{2}+M_{2}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)}{\Delta_{23}\Delta_{tri}}\\\
\frac{2(D-3)\left(f_{02}+M_{0}^{2}+M_{2}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)}{\Delta_{13}\Delta_{tri}}\\\
-\frac{2(D-3)\left(f_{01}+M_{0}^{2}+M_{1}^{2}\right)\left(f_{01}^{2}-4M_{0}^{2}M_{1}^{2}\right)}{\Delta_{12}\Delta_{tri}}\\\
\frac{(2-D)\left(\Delta_{13}\left(f_{12}+2M_{2}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)+\Delta_{23}\left(f_{02}+2M_{2}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)\right)}{2M_{2}^{2}\Delta_{13}\Delta_{23}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{23}\left(f_{01}+2M_{1}^{2}\right)\left(f_{01}^{2}-4M_{0}^{2}M_{1}^{2}\right)-\Delta_{12}\left(f_{12}+2M_{1}^{2}\right)\left(f_{01}f_{12}+2M_{1}^{2}f_{02}\right)\right)}{2M_{1}^{2}\Delta_{12}\Delta_{23}\Delta_{tri}}\\\
\frac{(D-2)\left(\Delta_{13}\left(f_{01}+2M_{0}^{2}\right)\left(f_{01}^{2}-4M_{0}^{2}M_{1}^{2}\right)-\Delta_{12}\left(f_{02}+2M_{0}^{2}\right)\left(f_{01}f_{02}+2M_{0}^{2}f_{12}\right)\right)}{2M_{0}^{2}\Delta_{12}\Delta_{13}\Delta_{tri}}\\\
\end{array}\right)$ (123)

Starting from this subsection, we will not write down the remaining columns of
$\mathcal{H}_{n;i}$, which are natural extension of known results (for
example, the (120) for $n=3$) according to the basis ordering defined in (3).

###  5.4 The box

We have used our formula to compute the matrices $\mathcal{H}_{n;i}$ up to
pentagon $n=5$ and check numerically with FIRE [20] up to some $i$ of box. In
this part, we will present only the result $\mathcal{H}_{4;0}$ to demonstrate
the iterative structure of denominators and the increasing complexity of
numerators when going to lower and lower gon:

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{0,1}=\frac{1}{\Delta_{\text{box}}}$
$\displaystyle\Big{(}(D-5)(f_{01}f_{12}f_{23}+2M_{3}^{2}(f_{12}(f_{01}-f_{12})+2M_{2}^{2}(f_{01}+2M_{1}^{2}))+f_{01}f_{13}f_{23}+2M_{2}^{2}f_{01}f_{13}-$
$\displaystyle
f_{01}f_{23}^{2}+f_{02}(f_{12}(f_{13}+2M_{3}^{2})+f_{13}f_{23}-f_{13}^{2}+2M_{1}^{2}(f_{23}+2M_{3}^{2}))+f_{03}(f_{12}(f_{13}+$
$\displaystyle
f_{23})-f_{12}^{2}+2M_{2}^{2}(f_{13}+2M_{1}^{2})+2M_{1}^{2}f_{23})-2f_{12}f_{13}f_{23}-2M_{2}^{2}f_{13}^{2}-2M_{1}^{2}f_{23}^{2})\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{1,1}=\frac{1}{\Delta_{234}\Delta_{\text{box}}}$
$\displaystyle\Big{(}2(D-4)(-2f_{12}(f_{13}+f_{23}+2M_{3}^{2})+f_{12}^{2}-2f_{13}(f_{23}+2M_{2}^{2})+f_{13}^{2}-4M_{1}^{2}f_{23}+f_{23}^{2}-$
$\displaystyle
4M_{1}^{2}M_{2}^{2}-4(M_{1}^{2}+M_{2}^{2})M_{3}^{2})(f_{12}f_{13}f_{23}+M_{3}^{2}f_{12}^{2}+M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2})\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{2,1}=\frac{1}{\Delta_{134}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-((D-4)(-2f_{02}(f_{03}+f_{23}+2M_{3}^{2})+f_{02}^{2}-2f_{03}(f_{23}+2M_{2}^{2})+f_{03}^{2}-4M_{0}^{2}f_{23}+f_{23}^{2}-$
$\displaystyle
4M_{0}^{2}M_{2}^{2}-4(M_{0}^{2}+M_{2}^{2})M_{3}^{2})(2M_{3}^{2}(2M_{2}^{2}f_{01}+f_{02}f_{12})+f_{23}(f_{02}f_{13}-f_{01}f_{23})+$
$\displaystyle f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13})))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{3,1}=$
$\displaystyle\frac{1}{\Delta_{124}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-((D-4)(-2f_{01}(f_{03}+f_{13}+2M_{3}^{2})+f_{01}^{2}-2f_{03}(f_{13}+2M_{1}^{2})+f_{03}^{2}-4M_{0}^{2}f_{13}+f_{13}^{2}-$
$\displaystyle
4M_{0}^{2}M_{1}^{2}-4(M_{0}^{2}+M_{1}^{2})M_{3}^{2})(2M_{3}^{2}(f_{01}f_{12}+2M_{1}^{2}f_{02})+f_{13}(f_{01}f_{23}-f_{02}f_{13})+$
$\displaystyle f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23})))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{4,1}=$
$\displaystyle\frac{1}{\Delta_{123}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-((D-4)(-2f_{01}(f_{02}+f_{12}+2M_{2}^{2})+f_{01}^{2}-2f_{02}(f_{12}+2M_{1}^{2})+f_{02}^{2}-4M_{0}^{2}f_{12}+f_{12}^{2}-$
$\displaystyle
4M_{0}^{2}M_{1}^{2}-4(M_{0}^{2}+M_{1}^{2})M_{2}^{2})(f_{01}(f_{12}f_{23}+2M_{2}^{2}f_{13})+f_{02}(f_{12}f_{13}+2M_{1}^{2}f_{23})-$
$\displaystyle f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2})))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{5,1}=$
$\displaystyle\frac{1}{\Delta_{134}\Delta_{234}\Delta_{34}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-2(D-3)(f_{23}+M_{2}^{2}+M_{3}^{2})(\Delta_{234}(f_{02}f_{23}+2M_{3}^{2}(f_{02}+2M_{2}^{2})+f_{03}f_{23}+2M_{2}^{2}f_{03}-$
$\displaystyle
f_{23}^{2})(-f_{01}(f_{23}^{2}-4M_{2}^{2}M_{3}^{2})+f_{02}f_{13}f_{23}+2M_{2}^{2}f_{03}f_{13})+f_{12}(\Delta_{234}(2M_{3}^{2}f_{02}+$
$\displaystyle
f_{03}f_{23})(f_{02}f_{23}+2M_{3}^{2}(f_{02}+2M_{2}^{2})+f_{03}f_{23}+2M_{2}^{2}f_{03}-f_{23}^{2})-2\Delta_{134}(f_{13}^{2}(3M_{2}^{2}f_{23}+$
$\displaystyle
f_{23}^{2}+2M_{2}^{2}M_{3}^{2})-f_{13}(f_{23}^{3}-4M_{2}^{2}M_{3}^{2}f_{23})+M_{1}^{2}(f_{23}+2M_{3}^{2})(f_{23}^{2}-4M_{2}^{2}M_{3}^{2})))-$
$\displaystyle
2f_{12}^{2}\Delta_{134}(M_{3}^{2}(f_{13}(3f_{23}+2M_{2}^{2})-f_{23}^{2})+f_{13}f_{23}^{2}+4M_{2}^{2}M_{3}^{4})-2M_{3}^{2}f_{12}^{3}\Delta_{134}(f_{23}+$
$\displaystyle
2M_{3}^{2})-2\Delta_{134}(f_{13}(f_{23}+2M_{2}^{2})-f_{23}^{2}+4M_{2}^{2}M_{3}^{2})(M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2}))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{6,1}=$
$\displaystyle\frac{1}{\Delta_{124}\Delta_{234}\Delta_{24}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-2(D-3)(f_{13}+M_{1}^{2}+M_{3}^{2})(\Delta_{234}(f_{01}f_{13}+2M_{3}^{2}(f_{01}+2M_{1}^{2})+f_{03}f_{13}+2M_{1}^{2}f_{03}-$
$\displaystyle
f_{13}^{2})(f_{23}(f_{01}f_{13}+2M_{1}^{2}f_{03})-f_{02}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2}))+f_{12}(\Delta_{234}(2M_{3}^{2}f_{01}+$
$\displaystyle
f_{03}f_{13})(f_{01}f_{13}+2M_{3}^{2}(f_{01}+2M_{1}^{2})+f_{03}f_{13}+2M_{1}^{2}f_{03}-f_{13}^{2})+2\Delta_{124}(-$
$\displaystyle
2M_{3}^{2}(M_{1}^{2}f_{23}(2f_{13}+f_{23})+M_{2}^{2}f_{13}(f_{13}-2M_{1}^{2}))+f_{13}f_{23}(-f_{13}f_{23}+f_{13}^{2}-$
$\displaystyle
3M_{1}^{2}f_{23})-M_{2}^{2}f_{13}^{3}+8M_{1}^{2}M_{2}^{2}M_{3}^{4}))-2f_{12}^{2}\Delta_{124}(f_{13}^{2}(f_{23}-M_{3}^{2})+3M_{3}^{2}f_{13}f_{23}+$
$\displaystyle
2M_{1}^{2}M_{3}^{2}(f_{23}+2M_{3}^{2}))-2M_{3}^{2}f_{12}^{3}\Delta_{124}(f_{13}+2M_{3}^{2})+2\Delta_{124}(-f_{13}f_{23}+$
$\displaystyle
f_{13}^{2}-2M_{1}^{2}(f_{23}+2M_{3}^{2}))(M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2}))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{7,1}=$
$\displaystyle\frac{1}{\Delta_{123}\Delta_{23}\Delta_{234}\Delta_{\text{box}}}$
$\displaystyle\Big{(}-2(D-3)(f_{12}+M_{1}^{2}+M_{2}^{2})(-\Delta_{234}(f_{01}(f_{12}+2M_{2}^{2})+f_{02}(f_{12}+2M_{1}^{2})-f_{12}^{2}+4M_{1}^{2}M_{2}^{2})(-$
$\displaystyle
f_{01}(f_{12}f_{23}+2M_{2}^{2}f_{13})-f_{02}(f_{12}f_{13}+2M_{1}^{2}f_{23})+f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}))-2\Delta_{123}(f_{12}(f_{13}+$
$\displaystyle
f_{23})-f_{12}^{2}+2M_{2}^{2}(f_{13}+2M_{1}^{2})+2M_{1}^{2}f_{23})(f_{12}f_{13}f_{23}+M_{3}^{2}f_{12}^{2}+M_{2}^{2}(f_{13}^{2}-$
$\displaystyle 4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2}))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{8,1}=$
$\displaystyle\frac{1}{\Delta_{124}\Delta_{134}\Delta_{14}\Delta_{\text{box}}}$
$\displaystyle\Big{(}2(D-3)(f_{03}+M_{0}^{2}+M_{3}^{2})(-(\Delta_{124}(f_{02}(f_{03}+2M_{3}^{2})+f_{03}f_{23}-f_{03}^{2}+2M_{0}^{2}(f_{23}+$
$\displaystyle
2M_{3}^{2}))(2M_{3}^{2}(2M_{2}^{2}f_{01}+f_{02}f_{12})+f_{23}(f_{02}f_{13}-f_{01}f_{23})+f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13})))-$
$\displaystyle\Delta_{134}(f_{01}(f_{03}+2M_{3}^{2})+f_{03}f_{13}-f_{03}^{2}+2M_{0}^{2}(f_{13}+2M_{3}^{2}))(2M_{3}^{2}(f_{01}f_{12}+$
$\displaystyle
2M_{1}^{2}f_{02})+f_{13}(f_{01}f_{23}-f_{02}f_{13})+f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23})))\Big{)}$
$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{9,1}=$
$\displaystyle\frac{1}{\Delta_{123}\Delta_{13}\Delta_{134}\Delta_{\text{box}}}$
$\displaystyle\Big{(}2(D-3)(f_{02}+M_{0}^{2}+M_{2}^{2})(f_{02}^{2}(\Delta_{123}(-2M_{3}^{2}(f_{12}f_{23}-2M_{2}^{2}f_{01})-(f_{23}^{2}(f_{01}+$
$\displaystyle
f_{13}))+f_{03}(f_{12}(f_{23}-2M_{3}^{2})-f_{13}(f_{23}-2M_{2}^{2})))-\Delta_{134}(f_{01}(f_{12}(f_{13}-f_{23})-$
$\displaystyle
2M_{2}^{2}f_{13}+2M_{1}^{2}f_{23})+f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2})+f_{12}(f_{12}f_{13}+2M_{1}^{2}f_{23})))+f_{02}(f_{03}(\Delta_{123}(f_{23}(f_{23}(f_{01}-$
$\displaystyle
f_{12})-4M_{2}^{2}f_{13})-4M_{2}^{2}M_{3}^{2}(f_{01}+f_{12}))+\Delta_{134}(f_{01}+f_{12})(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}))+$
$\displaystyle\Delta_{123}(f_{01}(f_{23}^{3}-4M_{2}^{2}M_{3}^{2}f_{23})-2M_{0}^{2}(f_{23}+2M_{2}^{2})(2M_{3}^{2}f_{12}+f_{13}f_{23}))-$
$\displaystyle\Delta_{134}(f_{01}^{2}(f_{12}f_{23}+2M_{2}^{2}f_{13})+f_{01}(4M_{2}^{2}(f_{12}f_{13}+M_{1}^{2}f_{23})+f_{12}^{2}f_{23})+$
$\displaystyle
2M_{0}^{2}(f_{12}+2M_{2}^{2})(f_{12}f_{13}+2M_{1}^{2}f_{23}))-(f_{03}^{2}\Delta_{123}(f_{12}f_{23}+2M_{2}^{2}f_{13})))-$
$\displaystyle
2\Delta_{123}(M_{2}^{2}(f_{03}+2M_{0}^{2})+M_{0}^{2}f_{23})(f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13})-f_{01}(f_{23}^{2}-$
$\displaystyle
4M_{2}^{2}M_{3}^{2}))+2\Delta_{134}(M_{2}^{2}(f_{01}+2M_{0}^{2})+M_{0}^{2}f_{12})(f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2})-f_{01}(f_{12}f_{23}+$
$\displaystyle
2M_{2}^{2}f_{13}))+f_{02}^{3}(\Delta_{123}(2M_{3}^{2}f_{12}+f_{13}f_{23})+\Delta_{134}(f_{12}f_{13}+2M_{1}^{2}f_{23})))\Big{)}$

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{10,1}=$
$\displaystyle\frac{1}{\Delta_{12}\Delta_{123}\Delta_{124}\Delta_{\text{box}}}$
$\displaystyle\Big{(}2(D-3)(f_{01}+M_{0}^{2}+M_{1}^{2})(f_{01}^{2}(\Delta_{123}(-2M_{3}^{2}(f_{12}f_{13}-2M_{1}^{2}f_{02})-(f_{13}^{2}(f_{02}+$
$\displaystyle
f_{23}))+f_{03}(f_{12}(f_{13}-2M_{3}^{2})-f_{23}(f_{13}-2M_{1}^{2})))-\Delta_{124}(2M_{2}^{2}(f_{13}(f_{02}+f_{12})-$
$\displaystyle
2M_{1}^{2}f_{03})-f_{02}f_{12}f_{13}+f_{02}f_{12}f_{23}-2M_{1}^{2}f_{02}f_{23}+f_{03}f_{12}^{2}+f_{12}^{2}f_{23}))+$
$\displaystyle
f_{01}(f_{03}(\Delta_{123}(f_{13}(f_{13}(f_{02}-f_{12})-4M_{1}^{2}f_{23})-4M_{1}^{2}M_{3}^{2}(f_{02}+f_{12}))+\Delta_{124}(f_{02}+$
$\displaystyle
f_{12})(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}))+\Delta_{123}(f_{02}(f_{13}^{3}-4M_{1}^{2}M_{3}^{2}f_{13})-2M_{0}^{2}(f_{13}+2M_{1}^{2})(2M_{3}^{2}f_{12}+$
$\displaystyle
f_{13}f_{23}))-\Delta_{124}(f_{02}^{2}(f_{12}f_{13}+2M_{1}^{2}f_{23})+f_{02}(f_{12}^{2}f_{13}+4M_{1}^{2}f_{12}f_{23}+$
$\displaystyle
4M_{1}^{2}M_{2}^{2}f_{13})+2M_{0}^{2}(f_{12}+2M_{1}^{2})(f_{12}f_{23}+2M_{2}^{2}f_{13}))-(f_{03}^{2}\Delta_{123}(f_{12}f_{13}+$
$\displaystyle
2M_{1}^{2}f_{23})))+f_{01}^{3}(f_{12}(2M_{3}^{2}\Delta_{123}+\Delta_{124}f_{23})+f_{13}(\Delta_{123}f_{23}+2M_{2}^{2}\Delta_{124}))-$
$\displaystyle
2\Delta_{123}(M_{1}^{2}(f_{03}+2M_{0}^{2})+M_{0}^{2}f_{13})(f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23})-f_{02}(f_{13}^{2}-$
$\displaystyle
4M_{1}^{2}M_{3}^{2}))+2\Delta_{124}(M_{1}^{2}(f_{02}+2M_{0}^{2})+M_{0}^{2}f_{12})(f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2})-f_{02}(f_{12}f_{13}+$
$\displaystyle 2M_{1}^{2}f_{23})))\Big{)}$

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{11,1}=$
$\displaystyle\frac{1}{2M_{3}^{2}\Delta_{124}\Delta_{134}\Delta_{14}\Delta_{234}\Delta_{24}\Delta_{34}\Delta_{\text{box}}}$
$\displaystyle\Big{(}(2-D)(-\Delta_{124}\Delta_{234}\Delta_{24}(2M_{3}^{2}(2M_{2}^{2}f_{01}+f_{02}f_{12})+f_{23}(f_{02}f_{13}-$
$\displaystyle
f_{01}f_{23})+f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13}))(\Delta_{14}(f_{23}+2M_{3}^{2})(f_{02}(f_{23}+2M_{3}^{2})+$
$\displaystyle
f_{03}(f_{23}+2M_{2}^{2})-f_{23}^{2}+4M_{2}^{2}M_{3}^{2})+\Delta_{34}(f_{03}+2M_{3}^{2})(f_{02}(f_{03}+2M_{3}^{2})+f_{03}f_{23}-$
$\displaystyle
f_{03}^{2}+2M_{0}^{2}(f_{23}+2M_{3}^{2})))-(\Delta_{134}\Delta_{234}\Delta_{34}(2M_{3}^{2}(f_{01}f_{12}+2M_{1}^{2}f_{02})+$
$\displaystyle
f_{13}(f_{01}f_{23}-f_{02}f_{13})+f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23}))(\Delta_{14}(f_{13}+2M_{3}^{2})(f_{01}(f_{13}+$
$\displaystyle
2M_{3}^{2})+f_{03}(f_{13}+2M_{1}^{2})-f_{13}^{2}+4M_{1}^{2}M_{3}^{2})+\Delta_{24}(f_{03}+2M_{3}^{2})(f_{01}(f_{03}+2M_{3}^{2})+$
$\displaystyle
f_{03}f_{13}-f_{03}^{2}+2M_{0}^{2}(f_{13}+2M_{3}^{2}))))+2\Delta_{124}\Delta_{134}\Delta_{14}(f_{12}f_{13}f_{23}+$
$\displaystyle
M_{3}^{2}f_{12}^{2}+M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2})(\Delta_{24}(f_{23}+2M_{3}^{2})(f_{12}(f_{23}+2M_{3}^{2})+$
$\displaystyle
f_{13}(f_{23}+2M_{2}^{2})-f_{23}^{2}+4M_{2}^{2}M_{3}^{2})+\Delta_{34}(f_{13}+2M_{3}^{2})(f_{12}(f_{13}+2M_{3}^{2})+f_{13}f_{23}-$
$\displaystyle f_{13}^{2}+2M_{1}^{2}(f_{23}+2M_{3}^{2}))))\Big{)}$

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{12,1}=$
$\displaystyle\frac{1}{2M_{2}^{2}\Delta_{123}\Delta_{13}\Delta_{134}\Delta_{23}\Delta_{234}\Delta_{34}\Delta_{\text{box}}}$
$\displaystyle\Big{(}(2-D)(-\Delta_{123}\Delta_{23}\Delta_{234}(2M_{3}^{2}(2M_{2}^{2}f_{01}+f_{02}f_{12})+f_{23}(f_{02}f_{13}-$
$\displaystyle
f_{01}f_{23})+f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13}))(\Delta_{13}(f_{23}+2M_{2}^{2})(f_{02}(f_{23}+2M_{3}^{2})+$
$\displaystyle
f_{03}(f_{23}+2M_{2}^{2})-f_{23}^{2}+4M_{2}^{2}M_{3}^{2})+\Delta_{34}(f_{02}+2M_{2}^{2})(f_{02}(f_{03}+f_{23})-f_{02}^{2}+$
$\displaystyle
2M_{2}^{2}(f_{03}+2M_{0}^{2})+2M_{0}^{2}f_{23}))+\Delta_{134}\Delta_{234}\Delta_{34}(-f_{01}(f_{12}f_{23}+2M_{2}^{2}f_{13})-$
$\displaystyle
f_{02}(f_{12}f_{13}+2M_{1}^{2}f_{23})+f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}))(\Delta_{13}(f_{12}+2M_{2}^{2})(f_{01}(f_{12}+$
$\displaystyle
2M_{2}^{2})+f_{02}(f_{12}+2M_{1}^{2})-f_{12}^{2}+4M_{1}^{2}M_{2}^{2})+\Delta_{23}(f_{02}+2M_{2}^{2})(f_{01}(f_{02}+2M_{2}^{2})+$
$\displaystyle
f_{02}f_{12}-f_{02}^{2}+2M_{0}^{2}(f_{12}+2M_{2}^{2})))+2\Delta_{123}\Delta_{13}\Delta_{134}(f_{12}f_{13}f_{23}+$
$\displaystyle
M_{3}^{2}f_{12}^{2}+M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2})(\Delta_{34}(f_{12}+2M_{2}^{2})(f_{12}(f_{13}+f_{23})-$
$\displaystyle
f_{12}^{2}+2M_{2}^{2}(f_{13}+2M_{1}^{2})+2M_{1}^{2}f_{23})+\Delta_{23}(f_{23}+2M_{2}^{2})(f_{12}(f_{23}+2M_{3}^{2})+f_{13}(f_{23}+$
$\displaystyle 2M_{2}^{2})-f_{23}^{2}+4M_{2}^{2}M_{3}^{2})))\Big{)}$

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{13,1}=$
$\displaystyle\frac{1}{2M_{1}^{2}\Delta_{12}\Delta_{123}\Delta_{124}\Delta_{23}\Delta_{234}\Delta_{24}\Delta_{\text{box}}}$
$\displaystyle\Big{(}(2-D)(-\Delta_{123}\Delta_{23}\Delta_{234}(2M_{3}^{2}(f_{01}f_{12}+2M_{1}^{2}f_{02})+f_{13}(f_{01}f_{23}-$
$\displaystyle
f_{02}f_{13})+f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23}))(\Delta_{12}(f_{13}+2M_{1}^{2})(f_{01}(f_{13}+2M_{3}^{2})+$
$\displaystyle
f_{03}(f_{13}+2M_{1}^{2})-f_{13}^{2}+4M_{1}^{2}M_{3}^{2})+\Delta_{24}(f_{01}+2M_{1}^{2})(f_{01}(f_{03}+f_{13})-f_{01}^{2}+$
$\displaystyle
2M_{1}^{2}(f_{03}+2M_{0}^{2})+2M_{0}^{2}f_{13}))+\Delta_{124}\Delta_{234}\Delta_{24}(\Delta_{23}(f_{01}+2M_{1}^{2})(f_{01}(f_{02}+$
$\displaystyle
f_{12})-f_{01}^{2}+2M_{1}^{2}(f_{02}+2M_{0}^{2})+2M_{0}^{2}f_{12})+\Delta_{12}(f_{12}+2M_{1}^{2})(f_{01}(f_{12}+2M_{2}^{2})+$
$\displaystyle
f_{02}(f_{12}+2M_{1}^{2})-f_{12}^{2}+4M_{1}^{2}M_{2}^{2}))(-f_{01}(f_{12}f_{23}+2M_{2}^{2}f_{13})-f_{02}(f_{12}f_{13}+$
$\displaystyle
2M_{1}^{2}f_{23})+f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2}))+2\Delta_{12}\Delta_{123}\Delta_{124}(f_{12}f_{13}f_{23}+$
$\displaystyle
M_{3}^{2}f_{12}^{2}+M_{2}^{2}(f_{13}^{2}-4M_{1}^{2}M_{3}^{2})+M_{1}^{2}f_{23}^{2})(\Delta_{24}(f_{12}+2M_{1}^{2})(f_{12}(f_{13}+f_{23})-$
$\displaystyle
f_{12}^{2}+2M_{2}^{2}(f_{13}+2M_{1}^{2})+2M_{1}^{2}f_{23})+\Delta_{23}(f_{13}+2M_{1}^{2})(f_{12}(f_{13}+2M_{3}^{2})+f_{13}f_{23}-$
$\displaystyle f_{13}^{2}+2M_{1}^{2}(f_{23}+2M_{3}^{2}))))\Big{)}$

$\displaystyle\left.\left(\mathcal{H}_{4;0}\right)\right|_{14,1}=$
$\displaystyle\frac{1}{2M_{0}^{2}\Delta_{12}\Delta_{123}\Delta_{124}\Delta_{13}\Delta_{134}\Delta_{14}\Delta_{\text{box}}}$
$\displaystyle\Big{(}(2-D)(-\Delta_{12}\Delta_{123}\Delta_{124}(2M_{3}^{2}(2M_{2}^{2}f_{01}+f_{02}f_{12})+f_{23}(f_{02}f_{13}-$
$\displaystyle
f_{01}f_{23})+f_{03}(f_{12}f_{23}+2M_{2}^{2}f_{13}))(\Delta_{13}(f_{03}+2M_{0}^{2})(f_{02}(f_{03}+2M_{3}^{2})+$
$\displaystyle
f_{03}f_{23}-f_{03}^{2}+2M_{0}^{2}(f_{23}+2M_{3}^{2}))+\Delta_{14}(f_{02}+2M_{0}^{2})(f_{02}(f_{03}+f_{23})-f_{02}^{2}+$
$\displaystyle
2M_{2}^{2}(f_{03}+2M_{0}^{2})+2M_{0}^{2}f_{23}))-\Delta_{123}\Delta_{13}\Delta_{134}(2M_{3}^{2}(f_{01}f_{12}+2M_{1}^{2}f_{02})+$
$\displaystyle
f_{13}(f_{01}f_{23}-f_{02}f_{13})+f_{03}(f_{12}f_{13}+2M_{1}^{2}f_{23}))(\Delta_{12}(f_{03}+2M_{0}^{2})(f_{01}(f_{03}+$
$\displaystyle
2M_{3}^{2})+f_{03}f_{13}-f_{03}^{2}+2M_{0}^{2}(f_{13}+2M_{3}^{2}))+\Delta_{14}(f_{01}+2M_{0}^{2})(f_{01}(f_{03}+f_{13})-$
$\displaystyle
f_{01}^{2}+2M_{1}^{2}(f_{03}+2M_{0}^{2})+2M_{0}^{2}f_{13}))+\Delta_{124}\Delta_{134}\Delta_{14}(\Delta_{13}(f_{01}+$
$\displaystyle
2M_{0}^{2})(f_{01}(f_{02}+f_{12})-f_{01}^{2}+2M_{1}^{2}(f_{02}+2M_{0}^{2})+2M_{0}^{2}f_{12})+\Delta_{12}(f_{02}+2M_{0}^{2})(f_{01}(f_{02}+$
$\displaystyle
2M_{2}^{2})+f_{02}f_{12}-f_{02}^{2}+2M_{0}^{2}(f_{12}+2M_{2}^{2})))(-f_{01}(f_{12}f_{23}+2M_{2}^{2}f_{13})-f_{02}(f_{12}f_{13}+$
$\displaystyle
2M_{1}^{2}f_{23})+f_{03}(f_{12}^{2}-4M_{1}^{2}M_{2}^{2})))\Big{)}$

##  6 The most general generating function

Now we consider the most general generating function (14) for both arbitrary
tensor in numerator and arbitrary power distribution of propagators in
denominator. As discussed in section 2, knowing the generating function (1) it
can be reduced to the study of the reduction of scalar integral
${{J}}_{n}(t)$111111The notation in (124) is slightly different from the one
in (14) with the region of index: in previous sections it is from $0$ to $n$,
while in this section it is from $1$ to $n$. 121212The reduction of (124) has
also been studied independently in [69].

$J_{n}(\bm{t})\equiv\int{d^{D}\ell\over
i\pi^{D/2}}{1\over\prod_{i=1}^{n}((\ell-
K_{i})^{2}-M_{i}^{2}-t_{i})}~{}.~{}~{}$ (124)

The IBP relations are given by (see (38) with $R=0$)

${\partial\over\partial
t_{i}}J_{n}(\bm{t})=\sum_{j}(F^{-1})_{ij}(\bm{t})\left\\{-(D-1-n)J_{n}(\bm{t})+\sum_{r\neq
j}{\partial\over\partial
t_{r}}J_{n;\widehat{j}}(\bm{t})\right\\}~{},~{}~{}~{}~{}i=1,...,n~{}$ (125)

where the elements of the matrix $F(\bm{t})$ is given by

$F_{ij}(\bm{t})=(K_{i}-K_{j})^{2}-M_{i}^{2}-t_{i}-M_{j}^{2}-t_{j}~{}~{}~{}~{}$
(126)

and

$J_{n;\widehat{S}}(\bm{t})\equiv\int{d^{D}\ell\over
i\pi^{D/2}}{1\over\prod_{i=1,i\not\in S}^{n}((\ell-
K_{i})^{2}-M_{i}^{2}-t_{i})}~{}.~{}~{}$ (127)

Using the expansion (16), i.e.,
$J_{n;\widehat{S}_{2}}(\bm{t})=\sum_{S_{1}}J_{n;\widehat{S}_{1}}{\cal
G}_{\widehat{S}_{1},\widehat{S}_{2}}(\bm{t})$, (125) gives following partial
differential equations

${\partial\over\partial t_{i}}{\cal
G}_{\widehat{S}_{1},\emptyset}(\bm{t})=\sum_{j}(F^{-1})_{ij}(\bm{t})\left\\{-(D-1-n){\cal
G}_{\widehat{S}_{1},\emptyset}(\bm{t})+\sum_{r\neq j}{\partial\over\partial
t_{r}}{\cal G}_{\widehat{S}_{1},\widehat{j}}(\bm{t})\right\\}~{}.~{}$ (128)

There are two different logics regarding to differential equations (128). The
first logic is to use the iterative construction. Assuming the general
generating functions for $m$-gon with $m<n$ are known, i.e., ${\cal
G}_{\widehat{S}_{1},\widehat{j}}(\bm{t})$ is known, we can utilize (128) to
solve the unknown reduction coefficient ${\cal
G}_{\widehat{S}_{1},\emptyset}(\bm{t})$. We call this logic as ”bottom up”
method. The second logic is called the ”top down” method, which has been
demonstrated in [52]. In the first step we solve the case $S_{1}=\emptyset$ of
(128). The result gives also the expression ${\cal
G}_{\widehat{i},\widehat{j}}$ with proper redefinition of modified Caylay
matrix. With the knowledge we can solve the case $S_{1}=i$ of (128). By
repeating the process we can get the general result of ${\cal
G}_{\widehat{S}_{1},\emptyset}$. In fact, in the second subsection, we utilize
this logic to solve ${\cal G}_{\widehat{n},\emptyset}$.

Another remark is that compared to the ordinary differential equation (72)
with only one variable, now we have $n$ variables, so solving them will be
difficult in general. However, as we will show, the one-loop case is special
and solving them is relatively easier.

###  6.1 The reduction from $n$-gon to $n$-gon

To demonstrate the key idea of solving (128), let us consider the simplest
case $S_{1}=\emptyset$, i.e., the reduction from $n$-gon to $n$-gon. The
equation (128) is simplified to the homogenous equation

${\partial\over\partial t_{i}}{\cal
G}_{\emptyset,\emptyset}(\bm{t})=-\sum_{j}(F^{-1})_{ij}(\bm{t})(D-1-n){\cal
G}_{\emptyset,\emptyset}(\bm{t})~{}.~{}$ (129)

We can combine (129) to give

$\displaystyle\sum_{i}\left({\partial\over\partial t_{i}}{\cal
G}_{\emptyset,\emptyset}(\bm{t})\right)dt_{i}$ $\displaystyle=$
$\displaystyle-(D-1-n){\cal
G}_{\emptyset,\emptyset}(\bm{t})\sum_{i,j}(F^{-1})_{ij}(\bm{t})dt_{i}$ (130)
$\displaystyle=$ $\displaystyle-{(D-1-n)\over 2}{\cal
G}_{\emptyset,\emptyset}(\bm{t})\sum_{i,j}(F^{-1})_{ij}(\bm{t})(dt_{i}+dt_{j})$
$\displaystyle=$ $\displaystyle{(D-1-n)\over 2}{\cal
G}_{\emptyset,\emptyset}(\bm{t})\sum_{i,j}(F^{-1})_{ij}(\bm{t})dF_{ij}(\bm{t})~{},~{}$

where we have used the symmetry of $F$-matrix and the expression (126). To
continue, let us recall following result: for matrix $g_{\mu\nu}$ and
determinant $g=|g_{\mu\nu}|$ we have

$dg=g\left(g^{-1}\right)^{\mu\nu}dg_{\mu\nu}~{}.~{}~{}~{}$ (131)

Thus (130) becomes

$d{\cal G}_{\emptyset,\emptyset}(\bm{t})={(D-1-n)\over 2}{\cal
G}_{\emptyset,\emptyset}(\bm{t}){d|F|\over|F|}~{},~{}$ (132)

which can be rewritten as

$d\ln{\cal G}_{\emptyset,\emptyset}(\bm{t})={(D-1-n)\over 2}d\ln|F|~{}.~{}$
(133)

Now we can integrate (133) easily to reach

$\ln{\cal G}_{\emptyset,\emptyset}(\bm{t})-\ln{\cal
G}_{\emptyset,\emptyset}(\bm{t}=0)={(D-1-n)\over
2}d\ln|F|(\bm{t})-{(D-1-n)\over 2}d\ln|F|(\bm{t}=0)~{}.~{}$ (134)

Using the boundary condition ${\cal G}_{\emptyset,\emptyset}(\bm{t}=0)=1$ we
finally obtain

${\cal
G}_{\emptyset,\emptyset}(\bm{t})=\left({|F|(\bm{t})\over|F|(\bm{t}=0)}\right)^{{(D-1-n)\over
2}}~{}.~{}$ (135)

One can also use Lemma 02 given in the Appendix (B) to check directly that
(135) does satisfy the equation (129). It is very nice to see that under this
construction, the generating function from $n$-gon to $n$-gon is surprisingly
concise.

####  6.1.1 Examples

Here we give some examples for the result (135). For tadpole

$I_{1}(a)=\int\frac{d^{D}l}{i\pi^{D/2}}\frac{1}{(l^{2}-M^{2})^{a}}.$ (136)

it is easy to find the IBP relation

$I_{1}(a)=\frac{D-2a+2}{2(a-1)M^{2}}I_{1}(a-1),$ (137)

From it we can solve the reduction coefficient as

$C_{1}(a)=\frac{(-1)^{a-1}(1-\frac{D}{2})_{a-1}}{(a-1)!(M^{2})^{a-1}}.~{}~{}$
(138)

The generating function of this example is

${\cal
G}_{\emptyset,\emptyset}(t_{1})=\left(1+\frac{t_{1}}{M^{2}}\right)^{{(D-2)\over
2}}~{}~{}$ (139)

It is easy to check the coefficients of its Taylor series expansion in terms
of $t_{1}$ exactly match results in (138). It can be seen that, even though
the expression for $C_{1}(a)$ is not complicated, the form of the generating
function is still simpler.

The second example is massless bubble

$I_{2}(a_{1},a_{2})=\int\frac{d^{D}l}{i\pi^{D/2}}\frac{1}{\left(l^{2}\right)^{a_{1}}\left((l-K)^{2}\right)^{a_{2}}}.$
(140)

The IBP relations give

$\begin{aligned}
&I_{2}(a_{1},a_{2})=-\frac{1}{(a_{2}-1)K^{2}}[(D-2a_{1}-a_{2}+1)I_{2}(a_{1},a_{2}-1)-(a_{2}-1)I_{2}(a_{1}-1,a_{2})],\\\
&I_{2}(a_{1},1)=-\frac{D-a_{1}-1}{(a_{1}-1)K^{2}}I_{2}(a_{1}-1,1).\end{aligned}~{}~{}~{}$
(141)

The generating function of this example is

$\mathcal{G}_{\emptyset,\emptyset}(t_{1},t_{2})=\left(\frac{(K^{2})^{2}+2(t_{1}+t_{2})K^{2}+(t_{1}-t_{2})^{2}}{(K^{2})^{2}}\right)^{\frac{D-3}{2}}~{}~{}~{}$
(142)

From it one can read out the reduction coefficient as

$\begin{aligned}
&C_{2}(a_{1},a_{2})=\frac{2^{a_{1}+a_{1}-2}\left(\frac{D-3}{2}\right)_{\lceil\frac{a_{1}+a_{2}-2}{2}\rceil}\left(a_{1}+1-\frac{D}{2}\right)_{a_{2}-1}\left(a_{2}+1-\frac{D}{2}\right)_{\lfloor\frac{a_{1}-a_{2}}{2}\rfloor}}{(a_{1}-1)!(a_{2}-1)!(K^{2})^{a_{1}+a_{2}-2}},\
\text{for}\ a_{1}\geq a_{2},\\\ &C_{2}(a_{1},a_{2})=C_{2}(a_{2},a_{1}),\
\text{for}\ a_{1}<a_{2}.\end{aligned}~{}~{}~{}$ (143)

where $\lfloor x\rfloor$ represents the floor function of $x$, and $\lceil
x\rceil$ represents the ceiling function of $x$. It can be checked that
results (143) do solve (141).

The third example is the triangle

$I_{3}(a_{1},a_{2},a_{3})=\int{d^{D}\ell\over
i\pi^{D/2}}{\frac{1}{(l^{2}-M_{1}^{2})^{a_{1}}((l-K_{2})^{2}-M_{2}^{2})^{a_{2}}((l-K_{3})^{2}-M_{3}^{3})^{a_{3}}}}~{}.$
(144)

For this example, we give the numerical check in the Table 1, where the
comparison between results calculated using generating functions and those
obtained with the Kira program [19]. The numerical values we have chosen are

$\displaystyle D=6.32,\ M^{2}_{1}=1.25,\ M^{2}_{2}=2.56,\ M^{2}_{3}=3.57,$
(145) $\displaystyle K_{1}=\vec{0},\ K^{2}_{2}=1.38,\ K_{2}\cdot K_{3}=2.05,\
K^{2}_{3}=5.16.$

$(a_{1},a_{2},a_{3})$ | The numerical result of Kira | The numerical result of generating function
---|---|---
(3,2,2) | -11.619 | -11.619
(5,2,1) | 19.9606 | 19.9606
(3,3,1) | 9.81226 | 9.81226
(4,3,1) | -32.4261 | -32.4261
(2,2,4) | 11.1596 | 11.1596
(1,6,1) | 2.03967 | 2.03967
(2,1,4) | 1.36335 | 1.36335
(3,5,6) | -193588 | -193588
(6,4,2) | 27906.2 | 27906.2
Table 1: Reduction Result of Triangle to Triangle by Kira and generating
function

###  6.2  The reduction from $n$-gon to $(n-1)$-gon

Now we use (128) to find the generating function from $n$-gon to $(n-1)$-gon.
For simplicity, let us focus on the case $S_{1}=n$. For this case, (128) is
not homogenous differential equation anymore. Due to
$\mathcal{G}_{\widehat{n},\widehat{j}}(\bm{t})=0$ for $j\neq n$, there is only
one non-zero term in the non-homogeneous part. Expressing the inverse matrix
of $F(\bm{t})$ in terms of its adjoint matrix $F^{*}(\bm{t})$, (128) becomes

$\frac{\partial}{\partial
t_{i}}\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})=\left\\{-\frac{\sum_{j}F^{*}_{ij}(\bm{t})}{|F|(\bm{t})}(D-1-n)\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})\right\\}+\frac{F^{*}_{in}(\bm{t})}{|F|(\bm{t})}\sum_{r\neq
n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t}),$ (146)

where $\left(F^{*}\right)_{ij}(\bm{t})$ is the algebraic cofactor of the
element $F_{ij}(\bm{t})$ in the matrix $F(\bm{t})$. Since
$\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})$ gives the reduction of
$(n-1)$-gon to $(n-1)$ gon, using the result (135) we learn

$\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})=\left({F^{*}_{nn}(\bm{t})\over
F^{*}_{nn}(\bm{t}=0)}\right)^{{(D-n)\over 2}}.~{}~{}~{}$ (147)

Now we will solve $N$ partial differential equations (146) one by one.

Let us start from the equation $i=n$. Using the method of constant variation,
the general solution of following first-order partial differential equation

$\frac{\partial}{\partial
x}f(x,\vec{y})=a(x,\vec{y})f(x,\vec{y})+h(x,\vec{y}),$ (148)

is given by

$f(x,\vec{y})=e^{\int a(x,\vec{y})dx}\left(c(\vec{y})+\int e^{-\int
a(x,\vec{y})dx}\cdot h(x,\vec{y})dx\right),$ (149)

where $\vec{y}$ represents all other variables except $x$. Back to our goal
(146), first we notice that the $e^{\pm\int a(x,\vec{y})dx}$ part is nothing,
but $|F|(\bm{t})^{\pm\frac{D-1-n}{2}}$. This claim can be derived by the exact
same computations done in previous subsection. Secondly, the second term
within braces in equation (149) is

$\displaystyle g_{\widehat{n},\emptyset}(\bm{t})$ $\displaystyle\equiv$
$\displaystyle\int dt_{n}\left(|F|(\bm{t})^{-\frac{D-1-n}{2}-1}\cdot
F^{*}_{nn}(\bm{t})\cdot\sum_{r\neq n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})\right)$ $\displaystyle=$
$\displaystyle F^{*}_{nn}(\bm{t})\cdot\sum_{r\neq n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})\mathcal{T}(\bm{t}),$
$\displaystyle\mathcal{T}(\bm{t})$ $\displaystyle\equiv$ $\displaystyle\int
dt_{n}\left(|F|(\bm{t})^{-\frac{D-1-n}{2}-1}\right)$ (150)

where we have used the fact that $F^{*}_{nn}(\bm{t})\cdot\sum_{r\neq
n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})$ is independent of
$t_{n}$. The part that needs to be integrated is only
$|F|(\bm{t})^{-\frac{D-1-n}{2}-1}$. Using results given in Lemma 01 and Lemma
02 in the Appendix B, we have

$\displaystyle|F|(\bm{t})=a_{0}+a_{1}t_{n}+a_{2}t_{n}^{2}$ (151)

with coefficients $a_{0},a_{1},a_{2}$

$a_{0}=|F|(\bm{t})\big{|}_{t_{n}=0},~{}~{}~{}~{}a_{1}=-2\sum_{i=1}^{n}F^{*}_{ni}(\bm{t})\big{|}_{t_{n}=0},~{}~{}~{}~{}a_{2}=-\sum_{i\neq
n,j\neq n}F^{*}_{in,jn}(\bm{t}).$ (152)

While $a_{0}$ and $a_{1}$ depend on $t_{1},t_{2},\cdots,t_{n-1}$, $a_{2}$ is a
pure constant. Then according to the integral formula,

$\int dx\ \left((x-x_{+})(x-x_{-})\right)^{c}={1\over
1+c}\left(x-x_{+}\right)^{c+1}\left(x-x_{-}\right)^{c}\cdot\
_{2}F_{1}\left(1,-c;2+c;\frac{x-x_{+}}{x-x_{-}}\right),$ (153)

integral in (150) is

$\mathcal{T}(\bm{t})={-2\over
D-n-1}a_{2}^{-\frac{D-n+1}{2}}(t_{n}-x_{+})^{-\frac{D-n-1}{2}}(t_{n}-x_{-})^{-\frac{D-n+1}{2}}\cdot\
_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right)$
(154)

with $x_{\pm}=\frac{-a_{1}\pm\sqrt{a^{2}_{1}-4a_{0}a_{2}}}{2a_{2}}$. Putting
it back we have

$\displaystyle g_{\widehat{n},\emptyset}(\textbf{t})=$ $\displaystyle{-2\over
D-n-1}\left(\left(F^{*}_{nn}(\bm{t})\right)\cdot\sum_{r\neq
n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})\right)a_{2}^{-\frac{D-n+1}{2}}(t_{n}-x_{+})^{-\frac{D-n-1}{2}}(t_{n}-x_{-})^{-\frac{D-n+1}{2}}$
(155) $\displaystyle\times$
$\displaystyle{}_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right).$

Up to now the generating function
$\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})$ from $n$-gon to $(n-1)$-gon is

$\displaystyle\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})=$
$\displaystyle{-2\over(D-n-1)a_{2}}\cdot\frac{1}{t_{n}-x_{-}}\
_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right)\left(\left(F^{*}_{nn}(\bm{t})\right)\cdot\sum_{r\neq
n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})\right)$ (156)
$\displaystyle+$ $\displaystyle
c_{1}(t_{1},\cdots,t_{n-1})\cdot\left(|F|(\bm{t})\right)^{\frac{D-1-n}{2}}.$

To solve the undetermined function $c(t_{1},\cdots,t_{n-1})$, we can
substitute the above expression into the differential equation (146) with
$i=n-1$. Doing similar computations, we will find expression with undetermined
function $c_{2}(t_{1},\cdots,t_{n-2})$ in principle. Iterating the procedure
until $i=1$ and using the boundary condition we should be able to fix
$\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})$ completely. However, in the
Appendix C we will prove that  the $c_{1}(t_{1},\cdots,t_{n-1})$ in expression
(156) is actually a constant independent of $t_{1},\cdots,t_{n-1}$. Therefore,
using the boundary conditions, the expression for the very concise generating
function $\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})$ from $n$-gon to
$(n-1)$-gon is

$\displaystyle\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})=$
$\displaystyle\frac{-2}{(D-n-1)a_{2}}\frac{1}{t_{n}-x_{-}}\
_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right)\left(F^{*}_{nn}(\bm{t})\cdot\sum_{r\neq
n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})\right)$ (157)
$\displaystyle-$ $\displaystyle
g_{\widehat{n},\emptyset}(\bm{t}=0)\cdot\left(|F|(\bm{t})\right)^{\frac{D-1-n}{2}}.$

where $g_{\widehat{n},\emptyset}(\bm{t}=0)$ is the numerical value obtained by
setting all $t_{i}$ to zero in equation (155). Substituting the expression of
$\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})$, the above formula can be
further simplified as:

$\displaystyle\mathcal{G}_{\widehat{n},\emptyset}(\bm{t})=$
$\displaystyle-\frac{2(D-n)}{D-n-1}\Bigg{(}\frac{1}{t_{n}-x_{-}}\
_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right)\left({F^{*}_{nn}(\bm{t})\over
F^{*}_{nn}(\bm{t}=0)}\right)^{{(D-n)\over 2}}$ (158) $\displaystyle-$
$\displaystyle\left\\{\frac{1}{t_{n}-x_{-}}\
_{2}F_{1}\left(1,\frac{D-n+1}{2};\frac{n+3-D}{2};\frac{t_{n}-x_{+}}{t_{n}-x_{-}}\right)\right\\}\Bigg{|}_{\bm{t}\to
0}\left({|F|(\bm{t})\over|F|(\bm{t}=0)}\right)^{{(D-1-n)\over 2}}\Bigg{)}.$

####  6.2.1 Numerical check

Here we present the numerical results of the triangle reduction to verify our
calculations. We choose the same numerical values as (145). The Table 2 and
Table 3 lists the comparison between the results calculated using generating
functions and those obtained with the Kira program.

$(a_{1},a_{2},a_{3})$ | The numerical result of Kira | The numerical result of generating function
---|---|---
(3,2,2) | 10.2631 | 10.2631
(5,2,1) | -17.6323 | -17.6323
(3,3,1) | -8.65731 | -8.65731
(4,3,1) | 28.6463 | 28.6463
(2,2,4) | -9.85984 | -9.85984
(1,6,1) | -1.80302 | -1.80302
(2,1,4) | -1.20463 | -1.20463
Table 2: Reduction Result of Triangle to $D_{1},D_{2}$ Bubble by Kira and generating function $(a_{1},a_{2},a_{3})$ | The numerical result of Kira | The numerical result of generating function
---|---|---
(3,2,2) | 7.67615 | 7.67615
(5,2,1) | -13.1893 | -13.1893
(3,3,1) | -6.48029 | -6.48029
(4,3,1) | 21.4263 | 21.4263
(2,2,4) | -7.37398 | -7.37398
(1,6,1) | -1.34816 | -1.34816
(2,1,4) | -0.900785 | -0.900785
Table 3: Reduction Result of Triangle to $D_{2},D_{3}$ Bubble by Kira and
generating function

###  6.3 The reduction from $n$-gon to general $m$-gon

Now we solve the differential equations of (128) for general cases. Mimic the
derivation of (130) we get

$\displaystyle d{\cal G}_{\widehat{S}_{1},\emptyset}(\bm{t})$ $\displaystyle=$
$\displaystyle{(D-1-n)\over 2}{\cal
G}_{\widehat{S}_{1},\emptyset}(\bm{t}){d\ln|F|}+\sum_{i}{\cal
B}_{i;\widehat{S}_{1}}(\bm{t})dt_{i}$ $\displaystyle{\cal
B}_{i;\widehat{S}_{1}}(\bm{t})$ $\displaystyle=$
$\displaystyle\sum_{j}(F^{-1})_{ij}(\bm{t})\sum_{r\neq
j}{\partial\over\partial t_{r}}{\cal
G}_{\widehat{S}_{1},\widehat{j}}(\bm{t})~{}.~{}$ (159)

By the constant variation method we write

${\cal
G}_{\widehat{S}_{1},\emptyset}(\bm{t})=\left({|F|(\bm{t})\over|F|(\bm{t}=0)}\right)^{{(D-1-n)\over
2}}\widetilde{g}_{\widehat{S}_{1},\emptyset}(\bm{t})~{}.~{}$ (160)

Putting it back to (159) and doing some manipulations we get

$d{\widetilde{g}}_{\widehat{S}_{1},\emptyset}(\bm{t})=\left({|F|(\bm{t})\over|F|(\bm{t}=0)}\right)^{-{(D-1-n)\over
2}}\sum_{i}{\cal B}_{i;\widehat{S}_{1}}(\bm{t})dt_{i}~{}.~{}$ (161)

To solve ${\widetilde{g}}_{\widehat{S}_{1},\emptyset}(\bm{t})$, we just need
to do multi-variable integrations. Noticing that the existence of solution of
(161) means the integrability condition

$\displaystyle
d\left\\{\left({|F|(\bm{t})\over|F|(\bm{t}=0)}\right)^{-{(D-1-n)\over
2}}\sum_{i}{\cal B}_{i;\widehat{S}_{1}}(\bm{t})dt_{i}\right\\}=0~{},~{}$ (162)

thus we have the freedom to choose integration pathes. The procedure done in
previous subsection is, in fact, equivalent to following choice of integration
path, i.e., beginning at point $(0,0,\cdots,0)$, proceeding to point
$(0,0,\cdots,t_{n})$, then to point $(0,\cdots,t_{n-1},t_{n})$, and so forth,
until finally reaching the integration endpoint $(t_{1},t_{2},\cdots,t_{n})$.
Another choice is the path that connecting point $(0,0,\cdots,0)$ to
$(t_{1},t_{2},\cdots,t_{n})$ by a straight line. This line is parameterized as
$(\lambda t_{1},\lambda t_{2},\cdots,\lambda t_{n})$, thus the solution of
(161) is

$\displaystyle{\widetilde{g}}_{\widehat{S}_{1},\emptyset}(\bm{t})-{\widetilde{g}}_{\widehat{S}_{1},\emptyset}(\bm{t}=0)=\int_{0}^{1}d\lambda\left({|F|(\lambda\bm{t})\over|F|(\bm{t}=0)}\right)^{-{(D-1-n)\over
2}}\sum_{i}{\cal B}_{i;\widehat{S}_{1}}(\lambda\bm{t})t_{i}~{}~{}$ (163)

where the boundary condition is that

$\displaystyle{\widetilde{g}}_{\widehat{S}_{1},\emptyset}(\bm{t}=0)=\left\\{\begin{array}[]{ll}1,{}{}&S_{1}=\emptyset\\\
0,{}{}&S_{1}\neq\emptyset\end{array}\right.~{}~{}$ (166)

In principle, knowing the general solution (135) from $n$-gon to $n$-gon, one
can use (163) to get the general solution from $n$-gon to $(n-1)$-gon.
Iterating the procedure, one can find the general solution from $n$-gon to
arbitrary $m$-gon, although carrying out the integration explicitly will need
some work.

##  7 Conclusion

In this paper we have studied the most general generating function for
reduction of one-loop integrals with arbitrary tensor structure in numerator
and arbitrary power distribution of propagators in denominator. To achieve the
goal, we have separated the task to two parts: the reduction with only
arbitrary numerator and the reduction of scalar integrals with arbitrary
denominator. Using the IBP method, we have derived and solved corresponding
differential equations for these two generating functions. Combining these two
results, we finally obtain the most general generating function.

Knowing the analytic expression of generating function is only the first step
of the study. Understanding the analytic structure and fully using it in
various situations are more interesting. Here we give some brief discussions.
Although in this paper we have only focus on the generating function of
reduction, First our results should be useful for the evaluation of master
integrals. For example, since the parameters ${\bm{t}}$ have played the role
of the mass, if we set $M_{i}=0$ in (124), the generating function ${\cal G}$
studied in the section 6 will give the expansion of massive scalar integrals
by the massless integrals. Another example is following. Currently the
standard way to compute master integrals is to use the differential equation
[56, 57, 58, 59], i.e., finding

$\displaystyle{\partial\over\partial s}{\cal I}=A_{s}{\cal I}~{}~{}~{}$ (167)

When acting ${\partial\over\partial s}$ on the master integrals, we get
integrals with nontrivial numerators and higher power of propagators at the
left hand side of (167). After reducing them to the basis, we get the right
hand side of (167). Since the generating function contains all information of
reduction, one could read out the matrix $A_{s}$ directly from it.
Furthermore, one could try to find canonical basis [60] as well as the symbol
[61, 62, 63] directly using generating function. Working out these details
will be our future project.

Secondly, in this paper we have used only fundamental IBP relations to
establish differential equations. As demonstrated in [64, 65, 66] (i.e., the
module intersection method), we could use non-fundamental IBP relations to
avoid the appearance of propagators with higher powers. In fact, Kosower has
used exact the same idea to discuss the generating function for some
particular tensor structures in [44]. While the module intersection method
will simplify resulted IBP relations, finding the syzygy does take some
considerable computations. It will be interesting to investigate when applying
same idea to generating functions, which kinds of simplification we will get.
Furthermore, there are other reduction methods, such as unitarity cut method,
intersection theory method or recent proposed linear relation method in [67].
Thus it is natural to ask which insight we will obtain if using them to study
the generating function131313In [53] the generating function has been studied
using linear relation method. .

Thirdly, as we have mentioned in the introduction, one of our main motivations
is to generalize the generating function to higher loop integrals. Although
the experience of the one-loop in this paper will be very helpful, we expect
more complicated technical problems will appear for higher loop. Solving them
will be an exciting challenge.

## Acknowledgements

We would like to thank Yang Zhang, Tingfei Li, Xiang Li and Yongqun Xu for
useful discussion. Chang Hu and Jiyuan Shen thank Hangzhou Institute for
Advanced Study for financial support. Yaobo Zhang thanks Ningxia University
for providing the startup fund for young faculty research. This work is
supported by Chinese NSF funding under Grant No.11935013, No.11947301,
No.12047502 (Peng Huanwu Center) and NSAF grant in National Natural Science
Foundation of China (NSFC) with grant No. U2230402.

##  Appendix A Solving the differential equation

In this appendix, for self-contained, we give some details of solving the
differential equation.

###  A.1 The generalized hypergeometric functions

First we collect some mathematical results which are used in the paper. The
generalized hypergeometric function (see (C.2) of [8] or the section 2.9 of
[68]) is defined as

$~{}_{A}F_{B}(a_{1},...,a_{A};b_{1},...,b_{B};x)=\sum_{n=0}^{\infty}{(a_{1})_{n}...(a_{A})_{n}\over(b_{1})_{n}...(b_{B})_{n}}{x^{n}\over
n!}~{}\,,~{}~{}~{}$ (168)

where the Pochhammer symbol is given in (53). Let $\theta\equiv x{d\over dx}$,
then $w=~{}_{A}F_{B}$ satisfies the differential equation

$\left(\theta(\theta+b_{1}-1)...(\theta+b_{B}-1)-x(\theta+a_{1})...(\theta+a_{A})\right)w=0~{}.~{}~{}$
(169)

Let us consider the special case. When $A=B=1$, the solution is called the
Kummer’s function. It is easy to work out that the differential equation (169)
is

$x\left(x{d^{2}\over dx^{2}}+(b_{1}-x){d\over dx}-a_{1}\right)w=0~{}~{}~{}$
(170)

and the solution is $~{}_{1}F_{1}(a_{1};b_{1};x)$. If we rescale
$x={\alpha}z$, (170) will becomes

$\left(z{d^{2}\over dz^{2}}+{(b_{1}-{\alpha}z)}{d\over
dz}-a_{1}{\alpha}\right)w=0~{}~{}~{}$ (171)

and the solution will be $~{}_{1}F_{1}(a_{1};b_{1};{\alpha}z)$.

###  A.2 The inhomogeneous differential equation

Now we consider the inhomogeneous differential equation of following type

$\left(x{d^{2}\over dx^{2}}+(b_{1}-x){d\over dx}-a_{1}\right)w=g(x)~{}.~{}~{}$
(172)

Using the result (170), we write

$w=~{}_{1}F_{1}(a_{1};b_{1};x)\widetilde{H}(x)~{},~{}~{}$ (173)

thus we have the differential equation for $\widetilde{H}(x)$

$x~{}_{1}F_{1}\widetilde{H}^{\prime\prime}+\widetilde{H}^{\prime}(2x~{}_{1}F^{\prime}_{1}+(b_{1}-x)~{}_{1}F_{1})=g(x)~{}.~{}~{}$
(174)

We can further simplify by defining

$\displaystyle\widetilde{H}^{\prime}$ $\displaystyle=$ $\displaystyle
h(x)e^{-\int_{0}^{x}dt{(2t~{}_{1}F^{\prime}_{1}(a_{1};b_{1};t)+(b_{1}-t)~{}_{1}F_{1}(a_{1};b_{1};t))\over
t~{}_{1}F_{1}(a_{1};b_{1};t)}}$ (175) $\displaystyle=$ $\displaystyle
h(x)e^{-\left(2\ln(~{}_{1}F_{1}(a_{1};b_{1};x))+b_{1}\ln
x-x\right)}={h(x)x^{-b_{1}}e^{x}\over(~{}_{1}F_{1}(a_{1};b_{1};x))^{2}}~{}.~{}~{}$

Putting (175) to (174) we find

$h^{\prime}=g(x)x^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};x)e^{-x}~{},~{}~{}$ (176)

thus we can solve

$h(x)=C_{2}+\int_{0}^{x}dyg(y)y^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};y)e^{-y}~{}~{}~{}$
(177)

and

$\widetilde{H}(x)=C_{1}+\int_{0}^{x}dt{t^{-b_{1}}e^{t}\over(~{}_{1}F_{1}(a_{1};b_{1};t))^{2}}\left(C_{2}+\int_{0}^{t}dyg(y)y^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};y)e^{-y}\right)~{}.~{}~{}$
(178)

Finally we have the solution141414In fact, to require the finiteness at $x=0$,
one must take $C_{2}=0$. Furthermore, (179) can have a nice formal expression
as discussed in (193).

$\displaystyle
w=~{}_{1}F_{1}(a_{1};b_{1};x)\left\\{C_{1}+\int_{0}^{x}dt{t^{-b_{1}}e^{t}\over(~{}_{1}F_{1}(a_{1};b_{1};t))^{2}}\left(C_{2}+\int_{0}^{t}dyg(y)y^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};y)e^{-y}\right)\right\\}.~{}~{}~{}~{}~{}$
(179)

In (179) there are two unknown constants $C_{1},C_{2}$ which is the typical
character for second order differential equation. We should use the boundary
condition to determine them.

###  A.3 Solving equation

Now we use the result in previous subsections to solve equation (73), which,
after proper scaling, can be simplified as

$t{d^{2}\over dt^{2}}\vec{W}+(B_{0}+B_{1}t){d\over
dt}\vec{W}+(C_{0}+C_{1}t)\vec{W}+\vec{\cal B}(t)=0~{}~{}~{}~{}$ (180)

To solve it, we need to do following two steps. Since (180) is not the
standard form (171) yet, we do the rescaling

$\vec{W}=f(t)\vec{U}~{}~{}~{}$ (181)

with

$f^{\prime\prime}+B_{1}f^{\prime}+C_{1}f=0,~{}~{}~{}\Longrightarrow
f=e^{{-B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over 2}t}~{}~{}~{}$ (182)

to reach the wanted form of differential equation

$\displaystyle 0$ $\displaystyle=$ $\displaystyle t{d^{2}\over
dt^{2}}\vec{U}+\left(B_{0}+t\sqrt{B_{1}^{2}-4C_{1}}\right){d\over dt}\vec{U}$
(183) $\displaystyle+\left({-B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over
2}B_{0}+C_{0}\right)\vec{U}+e^{{B_{1}-\sqrt{B_{1}^{2}-4C_{1}}\over
2}t}\vec{\cal B}(t)~{}~{}~{}$

Let us define

$\displaystyle x=-t\sqrt{B_{1}^{2}-4C_{1}}~{}~{}~{}$ (184)

then above equation becomes

$\displaystyle 0$ $\displaystyle=$ $\displaystyle x{d^{2}\over
dx^{2}}\vec{U}+\left(B_{0}-x\right){d\over
dx}\vec{U}+{-1\over\sqrt{B_{1}^{2}-4C_{1}}}\left({-B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over
2}B_{0}+C_{0}\right)\vec{U}$ (185)
$\displaystyle+{-1\over\sqrt{B_{1}^{2}-4C_{1}}}e^{{B_{1}-\sqrt{B_{1}^{2}-4C_{1}}\over
2}{-x\over\sqrt{B_{1}^{2}-4C_{1}}}}\vec{\cal
B}({-x\over\sqrt{B_{1}^{2}-4C_{1}}})~{}~{}~{}$

Comparing it with (172), we see that

$\displaystyle a_{1}$ $\displaystyle=$
$\displaystyle{1\over\sqrt{B_{1}^{2}-4C_{1}}}\left({-B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over
2}B_{0}+C_{0}\right),~{}~{}~{}~{}~{}~{}b_{1}=B_{0},$ $\displaystyle g(x)$
$\displaystyle=$
$\displaystyle{1\over\sqrt{B_{1}^{2}-4C_{1}}}e^{{B_{1}-\sqrt{B_{1}^{2}-4C_{1}}\over
2}{-x\over\sqrt{B_{1}^{2}-4C_{1}}}}\vec{\cal
B}({-x\over\sqrt{B_{1}^{2}-4C_{1}}})~{}~{}~{}$ (186)

Thus using (179), we have

$\displaystyle\vec{U}(x)$ $\displaystyle=$
$\displaystyle(~{}_{1}F_{1}(a_{1};b_{1};x))\left\\{c_{1}+\int_{0}^{x}ds{s^{-b_{1}}e^{s}\over(~{}_{1}F_{1}(a_{1};b_{1};s))^{2}}\left(c_{2}+\int_{0}^{s}dyg(y)y^{b_{1}-1}~{}_{1}F_{1}(a_{1};b_{1};y)e^{-y}\right)\right\\}~{}~{}~{}$
(187)

and finally

$\displaystyle\vec{W}(t)$ $\displaystyle=$ $\displaystyle
e^{-{B_{1}+\sqrt{B_{1}^{2}-4C_{1}}\over 2}t}\vec{U}(x)~{}~{}~{}$ (188)

###  A.4 Series expansion

Although we have obtained the closed analytic expression, it is still useful
to give the series expansion to demonstrate the vanishing of square root for
each order of $R$.

Putting the series expansion

$W=\sum_{n=0}^{\infty}\vec{w}_{n}t^{n},~{}~{}~{}~{}\vec{\cal
B}(t)=\sum_{n=0}^{\infty}\vec{b}_{n}t^{n}~{}~{}~{}$ (189)

to (180), we get the recursion relation

$0=\vec{w}_{n+1}(n+1)(n+B_{0})+\vec{w}_{n}(B_{1}n+C_{0})+\vec{w}_{n-1}C_{1}+\vec{b}_{n}~{}~{}~{}$
(190)

where $w_{n<0}=0$ has been assumed. When $n=0$, it is reduced to

$0=\vec{w}_{1}B_{0}+\vec{w}_{0}C_{0}+\vec{b}_{0}~{}~{}~{}$ (191)

Using the boundary condition
$\vec{w}_{0}=\vec{{\alpha}}_{0}\equiv\\{1,0,0,...,0\\}^{T}$ and
$\vec{b}_{0}=0$, we can solve

$\vec{w}_{1}={-C_{0}\over B_{0}}\vec{\alpha}_{0}~{}~{}~{}$ (192)

For $n\geq 1$ we have

$\vec{w}_{n+1}={-(B_{1}n+C_{0})\over(n+1)(n+B_{0})}\vec{w}_{n}+{-C_{1}\over(n+1)(n+B_{0})}\vec{w}_{n-1}+{-1\over(n+1)(n+B_{0})}\vec{b}_{n}~{}~{}~{}$
(193)

Thus all $\vec{w}_{n}$ have been determined.

##  Appendix B Some properties about the matrix F(t)

In this section, we will present some properties of the matrix $F(\bm{t})$
defined in (126). Utilizing these properties, we can derive several very
useful results for the reduction of $n$-gon to the $(n-1)$-gon. We will not
provide proofs for these properties, since readers can easily verify their
correctness with basic linear algebra techniques.

  * •

Lemma 01: With the explicit form (126), the determinant of $F(\bm{t})$ is
following form

$|F|(\bm{t})=\sum_{i\leq
j}c_{ij}t_{i}t_{j}+\sum_{i=1}^{n}c_{i}t_{i}+c_{0}.~{}~{}~{}$ (194)

where $c_{ij},c_{i},c_{0}$ are independent of $t_{1},\cdots,t_{n}$.

  * •

Lemma 02: For $i,j=1,2,\cdots,n$ we have,

$\displaystyle\frac{\partial}{\partial
t_{i}}|F|\left(\bm{t}\right)=-2\left(F_{i1}^{*}(\bm{t})+F_{i2}(\bm{t})^{*}+\cdots+F_{in}^{*}(\bm{t})\right),$
(195) $\displaystyle\frac{\partial^{2}|F|\left(\bm{t}\right)}{\partial
t^{2}_{i}}=-\sum_{k_{1},k_{2}\neq i}F^{*}_{ik_{1},ik_{1}}(\bm{t}),$
$\displaystyle\frac{\partial^{2}|F|\left(\bm{t}\right)}{\partial t_{i}\partial
t_{j}}=+\sum_{k_{1}\neq i,k_{2}\neq j}F^{*}_{ik_{1},jk_{1}}(\bm{t})\ \
\text{for}\ i\neq j.~{}~{}~{}$

where $F_{i1}^{*}$ is the cofactor and $F^{*}_{ik_{1},ik_{1}}$ the second
order cofactor. More explicitly, if we define
$Q^{\\#}_{i_{1}i_{2},j_{1}j_{2}}$ as the $(n-2)\times(n-2)$ matrix obtained by
removing the $i_{1}$-th row and $i_{2}$-th row, as well as the $j_{1}$-th
column and $j_{2}$-th column from the matrix $Q$, then the second order
cofactor
$Q^{*}_{i_{1}i_{2},j_{1}j_{2}}=(-1)^{i_{1}+i_{2}+j_{1}+j_{2}}|Q^{\\#}_{i_{1}i_{2},j_{1}j_{2}}|$.
Similarly, we can define the third-order algebraic cofactor
$Q_{i_{1}i_{2}i_{3},j_{1}j_{2}j_{3}}$ as well as the higher order cofactor.

  * •

Lemma 03: For the algebraic cofactors of matrix $F(\bm{t})$ and $i\neq j\neq
k$, we have

$\displaystyle\frac{\partial F_{ij}^{*}(\bm{t})}{\partial
t_{i}}+\frac{1}{2}\frac{\partial F_{ii}^{*}(\bm{t})}{\partial t_{j}}=0,$ (196)
$\displaystyle\frac{\partial F^{*}_{ij}(\bm{t})}{\partial t_{k}}=\sum_{l\neq
i}F^{*}_{il,jk}(\bm{t})+\sum_{l\neq j}F^{*}_{ik,jl}(\bm{t})$
$\displaystyle\frac{\partial}{\partial
t_{k}}\sum_{i,j=1}^{n}F^{*}_{ij}(\bm{t})=0,\ \text{for}\ k=1,2,\cdots,n.$

  * •

Lemma 04: For any matrix $A$, and $i_{1}<i_{2}$, $j_{1}<j_{2}$, The Desnanot-
Jacobi identity is given by

$\displaystyle
A_{i_{1}j_{1}}^{*}A_{i_{2}j_{2}}^{*}-A_{i_{1}j_{2}}^{*}A_{i_{2}j_{1}}^{*}=A_{i_{1}i_{2},j_{1}j_{2}}^{*}|A|.$
(197)

##  Appendix C Proof of the correctness of the generating function from
$n$-gon to $(n-1)$-gon

In this section, we will proof that the $c_{1}(t_{1},\cdots,t_{n-1})$ in
expression (156) is actually a constant independent of
$t_{1},t_{1},\cdots,t_{n-1}$. In fact, this claim is equivalent to the claim
that $g_{\widehat{n},\emptyset}(\bm{t})$ given by (150) and (154) satisfies

$\frac{\partial}{\partial
t_{i}}g_{\widehat{n},\emptyset}(\bm{t})=-|F|(\bm{t})^{-\frac{D-1-n}{2}-1}\cdot
F^{*}_{in}(\bm{t})\sum_{r\neq n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t}),$ (198)

for $i=1,2,\cdots,n$. Without loss of generality, we provide the proof for the
case $i=1$. We rewrite the determinant of matrix $F(\bm{t})$ in the following
form

$\displaystyle|F|(\bm{t})=A_{1}t_{1}^{2}+A_{2}t_{n}^{2}+A_{3}t_{1}t_{n}+A_{4}t_{1}+A_{5}t_{n}+A_{6}.$
(199)

Then, utilizing the Euler transformation formula of hypergeometric functions

${}_{2}F_{1}(a,b;c;z)=(1-z)^{-b}\
_{2}F_{1}\left(c-a,b;c;\frac{z}{z-1}\right),$ (200)

we can rewrite (154) as

$\displaystyle\mathcal{T}(\bm{t})=$
$\displaystyle\frac{2}{1-D+n}\frac{-\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}+2A_{2}t_{n}+A_{3}t_{1}+A_{5}}{2A_{2}}$
(201) $\displaystyle\times$
$\displaystyle\left(\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}+2A_{2}t_{n}+A_{3}t_{1}+A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\right)^{\frac{1}{2}(D-n-1)+1}$
$\displaystyle\times$
$\displaystyle\left(A_{1}t_{1}^{2}+A_{2}t_{n}^{2}+A_{3}t_{1}t_{n}+A_{4}t_{1}+A_{5}t_{n}+A_{6}\right)^{\frac{1}{2}(-D+n+1)-1}$
$\displaystyle\times$
$\displaystyle{}_{2}F_{1}\Bigg{(}\frac{1}{2}(D-n-1)+1,\frac{1}{2}(-D+n+1);\frac{1}{2}(-D+n+1)+1;$
$\displaystyle\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}-2A_{2}t_{n}-A_{3}t_{1}-A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\Bigg{)}.$

The advantage of this transformation is that we can utilize the following
derivative formula for hypergeometric functions

$\frac{d}{dx}\
{}_{2}F_{1}\left(-c,c+1;c+2;x\right)=\frac{(c+1)\left((1-x)^{c}-{}_{2}F_{1}(-c,c+1;c+2;x)\right)}{x},$
(202)

thus it ensures that the parameters of the hypergeometric function remain
unchanged during the differentiation process of
$g_{\widehat{n},\emptyset}(\bm{t})$ with respect to $t_{1}$.

$\displaystyle\frac{\partial\mathcal{T}(\bm{t})}{\partial t_{1}}=$
$\displaystyle\left(-\frac{2(n-D)}{1-D+n}\right)\frac{4A_{1}A_{2}t_{1}+2A_{2}A_{4}-A^{2}_{3}t_{1}-A_{3}A_{5}}{\left((A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})\right)^{3/2}}$
(203) $\displaystyle\times$
$\displaystyle\left(\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}+2A_{2}t_{n}+A_{3}t_{1}+A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\right)^{\frac{1}{2}(D-n-1)}$
$\displaystyle\times$
$\displaystyle\left(A_{1}t_{1}^{2}+A_{2}t_{n}^{2}+A_{3}t_{1}t_{n}+A_{4}t_{1}+A_{5}t_{n}+A_{6}\right)^{\frac{1}{2}(-D+n+1)}$
$\displaystyle\times$
$\displaystyle{}_{2}F_{1}\Bigg{(}\frac{1}{2}(D-n-1)+1,\frac{1}{2}(-D+n+1);\frac{1}{2}(-D+n+1)+1;$
$\displaystyle\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}-2A_{2}t_{n}-A_{3}t_{1}-A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\Bigg{)}$
$\displaystyle+$
$\displaystyle\frac{(2A_{1}x(2A_{2}t_{n}+A_{5})+A_{4}(2A_{2}t_{n}-A_{3}x+A_{5})-A_{3}t_{n}(A_{3}t_{1}+A_{5})-2A_{3}A_{6})}{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}$
$\displaystyle\left(A_{1}t_{1}^{2}+A_{2}t_{n}^{2}+A_{3}t_{1}t_{n}+A_{4}t_{1}+A_{5}t_{n}+A_{6}\right)^{\frac{1}{2}(-D+n+1)-1}.$

Substituting (147), we have

$\displaystyle\sum_{r\neq n}\frac{\partial}{\partial
t_{r}}\mathcal{G}_{\widehat{n},\widehat{n}}(\bm{t})=\sum_{r\neq
n}\frac{\partial}{\partial t_{r}}\left({F^{*}_{nn}(\bm{t})\over
F^{*}_{nn}(\bm{t}=0)}\right)^{{(D-n)\over 2}}$ (204) $\displaystyle=$
$\displaystyle\frac{D-n}{2}F^{*}_{nn}(\bm{t})^{\frac{D-n}{2}-1}\left((F^{*}_{nn}(\bm{t}=0))^{\frac{n-D}{2}}\sum_{r\neq
n}\frac{\partial}{\partial t_{r}}F^{*}_{nn}(\bm{t})\right).$

From Lemma 01 and Lemma 03 of Appendix B, one can see
$\left((F^{*}_{nn}(\bm{t}=0))^{\frac{n-D}{2}}\sum_{r\neq
n}\frac{\partial}{\partial t_{r}}F^{*}_{nn}(\bm{t})\right)$ is a constant
independent of $\bm{t}$. Thus, substituting (201) and (203), with some
computations (198) becomes

$\displaystyle 0=$ $\displaystyle
H_{1}\cdot{}_{2}F_{1}\Bigg{(}\frac{1}{2}(D-n-1)+1,\frac{1}{2}(-D+n+1);\frac{1}{2}(-D+n+1)+1;$
(205)
$\displaystyle\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}-2A_{2}t_{n}-A_{3}t_{1}-A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\Bigg{)}+H_{0},$

where

$\displaystyle H_{0}$ $\displaystyle=$
$\displaystyle\Bigg{(}\frac{(2A_{1}x(2A_{2}t_{n}+A_{5})+A_{4}(2A_{2}t_{n}-A_{3}x+A_{5})-A_{3}t_{n}(A_{3}t_{1}+A_{5})-2A_{3}A_{6})}{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}F^{*}_{nn}(\bm{t})-F^{*}_{1n}(\bm{t})\Bigg{)}$
(206) $\displaystyle\times$ $\displaystyle|F|(\bm{t})^{-\frac{D-1-n}{2}-1},$

and

$\displaystyle H_{1}$ $\displaystyle=$
$\displaystyle\frac{D-n}{1-D+n}|F|(\bm{t})^{-\frac{D-1-n}{2}}\frac{1}{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}$
(207)
$\displaystyle\times\left(\frac{\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}+2A_{2}t_{n}+A_{3}t_{1}+A_{5}}{2\sqrt{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}}\right)^{\frac{1}{2}(D-n-1)}$
$\displaystyle\times\left(\frac{\partial F^{*}_{nn}(\bm{t})}{\partial
t_{1}}+F^{*}_{nn}(\bm{t})\frac{8A_{1}A_{2}t_{1}+4A_{2}A_{4}-2A_{3}^{2}t_{1}-2A_{3}A_{5}}{(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})}\right)$

Next we only need to prove that $H_{1}=H_{0}=0$. One can check

$\displaystyle A_{2}=\frac{1}{2}\frac{\partial|F|(\bm{t})}{\partial
t_{n}^{2}},~{}~{}~{}~{}A_{3}t_{1}+A_{5}=\frac{\partial^{2}|F|(\bm{t})}{\partial
t_{n}}-2A_{2}\cdot t_{n},$ (208) $\displaystyle
A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6}=|Q|-\frac{\partial|F|(\bm{t})}{\partial
t_{n}}\cdot t_{n}+A_{2}t_{n}^{2}$

Thus we have

$\displaystyle(A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})=\left(\frac{\partial|F|(\bm{t})}{\partial
t_{n}}-2At_{n}\right)^{2}-4A_{2}\left(|F|(\bm{t})-\frac{\partial|F|(\bm{t})}{\partial
t_{n}}t_{n}+A_{2}t_{n}^{2}\right)$ (209) $\displaystyle=$
$\displaystyle\frac{\partial|F|(\bm{t})}{\partial
t_{n}}\frac{\partial|F|(\bm{t})}{\partial
t_{n}}-4A_{2}|F|(\bm{t})=4\left(\sum_{i,j=1}^{n}F_{in}^{*}\cdot
F_{jn}^{*}+\sum_{i,j=1}^{n-1}F_{in,jn}^{*}|Q|\right)=4\left(\sum_{i,j=1}^{n}F_{ij}^{*}\right)F_{nn}^{*},$

where the last line of the above equation is obtained from Lemma 04 of
Appendix B. Furthermore, we have

$\displaystyle(2A_{1}x(2A_{2}t_{n}+A_{5})+A_{4}(2A_{2}t_{n}-A_{3}x+A_{5})-A_{3}t_{n}(A_{3}t_{1}+A_{5})-2A_{3}A_{6})$
(210) $\displaystyle=$
$\displaystyle\left(A_{5}+A_{3}t_{1}+2A_{2}t_{n}\right)\left(A_{4}+2A_{1}t_{1}+A_{3}t_{n}\right)-2A_{3}|F|(\bm{t})$
$\displaystyle=$ $\displaystyle\frac{\partial|F|(\bm{t})}{\partial
t_{n}}\frac{\partial|F|(\bm{t})}{\partial t_{1}}-2\frac{\partial}{\partial
t_{1}}\left(\frac{\partial}{\partial
t_{n}}|F|(\bm{t})\right)\cdot|Q|=4\left(\sum_{i,j=1}^{n}F_{1i}^{*}F_{jn}^{*}-\sum_{j\neq
1,i\neq n}F^{*}_{1j,in}|F|(\bm{t})\right)$ $\displaystyle=$ $\displaystyle
4F_{1n}^{*}\cdot\sum_{i,j=1}^{n}F_{ij}^{*}$

and

$\displaystyle
8A_{1}A_{2}t_{1}+4A_{2}A_{4}-2A_{3}^{2}t_{1}-2A_{3}A_{5}=-\frac{\partial}{\partial
t_{1}}\left((A_{3}t_{1}+A_{5})^{2}-4A_{2}(A_{1}t_{1}^{2}+A_{4}t_{1}+A_{6})\right)$
(211) $\displaystyle=$
$\displaystyle-4\left(\sum_{i,j=1}^{n}F_{ij}^{*}\right)\frac{\partial
F_{nn}^{*}}{\partial t_{1}}.$

The last line of (211) is obtained by that $\sum_{i,j=1}^{n}F_{ij}^{*}$ is a
constant due to Lemma 03 of Appendix B. By (209) (210) and (211), one can
easily check that $H_{1}=H_{0}=0$. Thus we have completed the proof.

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