# Generation Function For One-loop Tensor Reduction

Bo Fengab111Emails<EMAIL_ADDRESS>
aBeijing Computational Science Research Center, Beijing 100084, China
bPeng Huanwu Center for Fundamental Theory, Hefei, Anhui, 230026, China

###### Abstract:

For loop integrals, the reduction is the standard method. Having an efficient
way to find reduction coefficients is an important topic in scattering
amplitudes. In this paper, we present the generation functions of reduction
coefficients for general one-loop integrals with arbitrary tensor rank.

Generation function, Tensor reduction, One-loop integral

##  1 Motivation

As the bridge connecting the theoretical frame and the experiment data,
scattering amplitude is always one of central concepts in the quantum field
theory. Its efficient computation (including higher order contributions in
perturbation) becomes the necessary, especially with the coming of the LHC
experiment [1]. The on-shell program222There are many works in this topic. For
introduction, please see following two books [2, 3]. Some early works with on-
shell concept are [4, 5, 6, 7, 8, 9]. in the scattering amplitudes, as the
outcome of the challenge, has made the calculations of one-loop amplitude
straightforward [10].

In general the loop computation can be divided into two parts: the
construction of integrands and then doing the integration. Although the
construction of integrands using Feynman diagrams are well established,
sometimes it is not the economic way to do so. Looking for a better way to
construct integrands is one of current research directions333For example, the
unitarity cut method proposed in [4, 5, 7] uses on-shell tree level amplitudes
as input. . For the second part, i.e., doing the integration, a very useful
method is the reduction. It has been shown that any integral can be written as
the linear combination of some basis integrals (called the master integrals)
with coefficients as the rational function of external momenta, masses and
spacetime dimension. Using the idea of reduction, loop integration can be
separated into two parallel tasks: the computation of master integrals and the
algorithm to efficiently extract the reduction coefficients. Progresses in any
one task will enable us to do more and more complicated integrations (for a
nice introduction of recent developments, see [11]).

For the reduction, it can be classified into two categories: the reduction at
the integrand level and the reduction at the integral level. The reduction at
the integrand level can be systematically solved using the computational
algebraic geometry [12, 13, 14, 15]. For the reduction at the integral level,
the first proposal is the famous Passarino-Veltman reduction (PV-reduction)
method [16]. There are other proposals, such as the Integration-by-Part (IBP)
method [17, 18, 19, 20, 21, 22, 23], the unitarity cut method [4, 5, 7, 24,
25, 26, 27, 28, 29, 30], Intersection number [31, 32, 33, 34]. Although there
are a lot of developments for the reduction at the integral level, it is still
desirable to improve them by current complexity of computations.

In recent papers [35, 36, 37, 39, 38] we have introduced the auxiliary vector
$R$ to improve the traditional PV-reduction method. Using $R$ we can construct
the differential operators and then establish algebraic recurrence relation to
determine reduction coefficients analytically. This method has also been
generalized to two-loop sunset diagram (see [38]) where the original PV-
reduction method is hard to be used. When using the auxiliary vector $R$ in
the IBP method, the efficiency of reduction has also been improved as shown in
[40, 41].

Although the advantage of using auxiliary vector $R$ has been demonstrated
from various aspects, the algebraic recursive structure makes it still hard to
have a general understanding of reduction coefficients for higher and higher
tensor ranks. Could we get more by this method? As we will show in this paper,
indeed we can get more if we probe the reduction problem from a new angle. The
key idea is the concept of generation function. In fact, generation function
is well known in physics and mathematics. Sometimes the coefficients of a
series is hard to imagine, but the series itself is easy to write down. For
example, the Hermite Polynomial $H_{n}(x)$ can be read out from the generation
function

$\displaystyle e^{2tx-t^{2}}=\sum_{n=0}^{\infty}H_{n}(x){t^{n}\over
n!}\,.~{}~{}~{}~{}$ (1)

Thus we can ask that if we sum the reduction coefficients of different tensor
ranks together, could we get a simpler answer? For reduction problem, the
numerator of tensor rank $k$ is given by $(2\ell\cdot R)^{k}$ in our method
and we need to see how to sum them together. There are many ways to sum them.
Two typical ones are

$\displaystyle\psi_{1}(t)=\sum_{n=0}^{\infty}t^{n}(2\ell\cdot R)^{n}={1\over
1-t(2\ell\cdot
R)},~{}~{}~{}~{}~{}~{}\psi_{2}(t)=\sum_{n=0}^{\infty}{(2\ell\cdot
R)^{n}t^{n}\over n!}=e^{t(2\ell\cdot R)}\,.~{}~{}~{}~{}$ (2)

In this paper, we will focus on the generation function of the type
$\psi_{2}(t)$ because it is invariant under the differential action, i.e,
${de^{x}\over dx}=e^{x}$. We will see that the generation functions satisfy
simple differential equations, which can be solved analytically.

The plan of the paper is following. In the section two, we present the
generation function of reduction coefficients of tadpole integral. The tadpole
example is a little bit trivial, thus in the section three, we discuss
carefully how to find generation functions for the bubble integral, which is
the simplest nontrivial example. With the experience obtained for bubble, we
present the solution for general one-loop integrals in the section four. To
demonstrate the frame established in section four, we discuss briefly the
triangle example in the section five. Finally, a brief summary and discussion
is given in the section six. Some technical details have been collected in the
Appendix, where in the Appendix A, the solution of two typical differential
equations is presented, while the solution of recursive relation for the
bubble has been explained in the Appendix B.

##  2 Tadpole

With above brief discussion, in this section we start from the simplest case,
i.e., the tadpole topology to discuss the generation function. Summing all
tensor ranks properly we have444The mass dimension of parameter $t$ is $-2$.

$\displaystyle I_{tad}(t,R)\equiv\int d\ell{e^{t(2\ell\cdot
R)}\over\ell^{2}-M^{2}}=c_{1\to 1}(t,R,M)\int
d\ell{1\over\ell^{2}-M^{2}}~{}~{}~{}~{}$ (3)

where $c_{1\to 1}(t,R,M)$ is the generation function of reduction coefficients
and for simplicity we have defined $\int
d\ell_{i}(\bullet)\equiv\int{d^{D}\ell_{i}\over i\pi^{D/2}}(\bullet)$. To find
closed analytic expression for $c_{1\to 1}(t,R,M)$ we establish the
corresponding differential equation. Acting with $\partial_{R}$ we have

$\displaystyle{\partial\over\partial R}\cdot{\partial\over\partial
R}I_{tad}(t,R)=\int d\ell{4t^{2}\ell^{2}e^{t(2\ell\cdot
R)}\over\ell^{2}-M^{2}}=4t^{2}M^{2}I_{tad}(t,R)~{}~{}~{}~{}$ (4)

at one side, and $\left({\partial\over\partial R}\cdot{\partial\over\partial
R}c_{1\to 1}(t,R,M)\right)\int d\ell{1\over\ell^{2}-M^{2}}$ at another side,
thus we get

$\displaystyle{{\partial\over\partial R}\cdot{\partial\over\partial R}c_{1\to
1}(t,R,M)=4t^{2}M^{2}c_{1\to 1}(t,R,M)}~{}\,.~{}~{}~{}$ (5)

By the Lorentz invariance, $c_{1\to 1}(t,R,M)$ is the function of $r=R\cdot R$
only, i.e., $c_{1\to 1}(t,R,M)=f(r)$. It is easy to see that differential
equation (5) becomes

$\displaystyle\boxed{4rf^{\prime\prime}+{2D}f^{\prime}-{4t^{2}M^{2}}f=0}~{}\,,~{}~{}~{}$
(6)

which is the form (161) studied in the Appendix A. This second order
differential equation has two singular points $r=0$ and $r=\infty$, where the
singular point $r=0$ is canonical. The solution has been given in (180).
Putting $A=4,B=2D,C=-4t^{2}M^{2}$ in (176) and the boundary condition
$c_{0}=1$, we get immediately

$\displaystyle c_{1\to 1}(t,R,M)=\sum_{n=0}^{\infty}{(t^{2}M^{2}r)^{n}\over
n!\left({D\over 2}\right)_{n}}=~{}_{0}F_{1}(\emptyset;{D\over
2};t^{2}M^{2}r)~{}\,.~{}~{}~{}$ (7)

Before ending this section, let us mention that when we do the reduction for
other topologies, we will meet the reduction of $\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell-K)^{2}-M^{2}}$. Using the momentum shifting, it is easy to see
that

$\displaystyle\int d\ell{e^{t(2\ell\cdot R)}\over(\ell-K)^{2}-M^{2}}=\int
d\ell{e^{t(2(\ell+K)\cdot R)}\over\ell^{2}-M^{2}}=e^{2t(K\cdot R)}c_{1\to
1}(t,R,M)\int d\ell{1\over\ell^{2}-M^{2}}~{}\,.~{}~{}~{}$ (8)

The results shows the advantage of using generation function with exponential
form.

##  3 Bubble

Having found the generation function of tadpole reduction, we move to the
first nontrivial example, i.e., the generation function of bubble reduction,
which is defined through

$\displaystyle I_{bub}(t,R)\equiv\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}$ (9) $\displaystyle=$
$\displaystyle c_{2\to 2}\int
d\ell{1\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}+c_{2\to
1;\widehat{1}}\int d\ell{1\over(\ell^{2}-M_{0}^{2})}+c_{2\to
1;\widehat{0}}\int d\ell{1\over((\ell-K)^{2}-M_{1}^{2})}~{}\,.~{}~{}~{}$

For simplicity, we have not written down variables of reduction coefficients
explicitly. If written explicitly, it will be $c(t,R,K;M_{0},M_{1})$ or
$c(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$ if using the Lorentz contraction form.

The generation form (9) can produce some nontrivial relations among these
generation functions of reduction coefficients easily. Noticing
that555Comparing to the shifting symmetry discussed in (10), one can also
consider the symmetry with $\ell\to-\ell$. For this one, we have $R\to-R$ and
$K\to-K$. If using the variables $R^{2},K\cdot R$, it is invariant. In other
words, the symmetry $\ell\to-\ell$ is trivial.

$\displaystyle I_{bub}(t,R)\equiv\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}=e^{t(2K\cdot R)}\int
d\widetilde{\ell}{e^{t(2\widetilde{\ell}\cdot
R)}\over((\widetilde{\ell}+K)^{2}-M_{0}^{2})(\widetilde{\ell}^{2}-M_{1}^{2})}$
(10) $\displaystyle=$ $\displaystyle e^{t(2K\cdot R)}\left\\{c_{2\to
2}(t,R^{2},-K\cdot R,K^{2};M_{1},M_{0})\int
d\ell{1\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}\right.$
$\displaystyle+c_{2\to 1;\widehat{0}}(t,R^{2},-K\cdot R,K^{2};M_{1},M_{0})\int
d\ell{1\over(\ell^{2}-M_{0}^{2})}$ $\displaystyle\left.+c_{2\to
1;\widehat{1}}(t,R^{2},-K\cdot R,K^{2};M_{1},M_{0})\int
d\ell{1\over((\ell-K)^{2}-M_{1}^{2})}\right\\}~{}\,,~{}~{}~{}$

we have

$\displaystyle c_{2\to 2}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$
$\displaystyle=$ $\displaystyle e^{t(2K\cdot R)}c_{2\to 2}(t,R^{2},-K\cdot
R,K^{2};M_{1},M_{0})~{}\,,~{}~{}~{}$ (11) $\displaystyle c_{2\to
1;\widehat{1}}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$ $\displaystyle=$
$\displaystyle e^{t(2K\cdot R)}c_{2\to 1;\widehat{0}}(t,R^{2},-K\cdot
R,K^{2};M_{1},M_{0})~{}\,,~{}~{}~{}$ (12) $\displaystyle c_{2\to
1;\widehat{0}}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$ $\displaystyle=$
$\displaystyle e^{t(2K\cdot R)}c_{2\to 1;\widehat{1}}(t,R^{2},-K\cdot
R,K^{2};M_{1},M_{0})~{}~{}~{}~{}$ (13)

by comparing (10) with (9). The first relation (11) can be a consistent check
for $c_{2\to 2}$ while the second relation (12) and the third relation (13)
tell us that we need to compute only one of $c_{2\to 1;\widehat{i}}$
functions. Another useful check is the mass dimension. Since the mass
dimension of $t$ is $(-2)$, we have

$\displaystyle[c_{2\to 2}]=0,~{}~{}~{}~{}[c_{2\to 1}]=-2~{}~{}$ (14)

###  3.1 Differential equations

Now we will write down differential equations for these generation functions.
Acting $\partial_{R}\cdot\partial_{R}$ on both sides of (9) we have

$\displaystyle\partial_{R}\cdot\partial_{R}I_{bub}(t,R)=\int
d\ell{4t^{2}\ell^{2}e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}$ (15) $\displaystyle=$
$\displaystyle 4t^{2}M_{0}^{2}I_{bub}(t,R)+4t^{2}e^{t(2K\cdot R)}\int
d\ell{e^{t(2\ell\cdot R)}\over(\ell^{2}-M_{1}^{2})}~{}\,,~{}~{}~{}$

thus we derive

$\displaystyle\partial_{R}\cdot\partial_{R}c_{2\to 2}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})$ $\displaystyle=$ $\displaystyle 4t^{2}M_{0}^{2}c_{2\to
2}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})~{}\,,~{}~{}~{}$ (16)
$\displaystyle\partial_{R}\cdot\partial_{R}c_{2\to
1;\widehat{1}}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$ $\displaystyle=$
$\displaystyle 4t^{2}M_{0}^{2}c_{2\to 1;\widehat{1}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})~{}\,,~{}~{}~{}$ (17)
$\displaystyle\partial_{R}\cdot\partial_{R}c_{2\to
1;\widehat{0}}(t,R^{2},K\cdot R,K^{2};M_{0},M_{1})$ $\displaystyle=$
$\displaystyle 4t^{2}M_{0}^{2}c_{2\to 1;\widehat{0}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})$ (18) $\displaystyle~{}~{}~{}+4t^{2}e^{t(2K\cdot
R)}c_{1\to 1}(t,R^{2},M_{1})~{}\,.~{}~{}~{}$

Acting with $K\cdot\partial_{R}$ we have

$\displaystyle K\cdot\partial_{R}I_{bub}(t,R)=\int
d\ell{t(2K\cdot\ell)e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}=\int
d\ell{t(D_{0}-D_{1}+f)e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})}$ (19) $\displaystyle=$
$\displaystyle tfI_{bub}(t,R)-t\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})}+te^{t(2K\cdot R)}\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{1}^{2})}~{}\,,~{}~{}~{}$

thus we derive

$\displaystyle K\cdot\partial_{R}c_{2\to 2}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})=tfc_{2\to 2}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})~{}\,,~{}~{}~{}$ (20) $\displaystyle
K\cdot\partial_{R}c_{2\to 1;\widehat{1}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})=tfc_{2\to 1;\widehat{1}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})-tc_{1\to 1}(t,R^{2},M_{0})~{}\,,~{}~{}~{}$ (21)
$\displaystyle K\cdot\partial_{R}c_{2\to 1;\widehat{0}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})=tfc_{2\to 1;\widehat{0}}(t,R^{2},K\cdot
R,K^{2};M_{0},M_{1})+te^{t(2K\cdot R)}c_{1\to
1}(t,R^{2},M_{1})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (22)

where $f=K^{2}-M_{1}^{2}+M_{0}^{2}$. Above two groups of differential
equations can be uniformly written as

$\displaystyle\partial_{R}\cdot\partial_{R}c_{T}$ $\displaystyle=$
$\displaystyle 4t^{2}M_{0}^{2}c_{T}+4t^{2}\xi_{R}h_{T}~{}\,,~{}~{}$ (23)
$\displaystyle K\cdot\partial_{R}c_{T}$ $\displaystyle=$ $\displaystyle
tfc_{T}+t\xi_{K}h_{T}~{}~{}~{}$ (24)

where $h_{T}$ is the possible non-homogenous contribution coming from lower
topology (tadpole). For different $T$ we have

$\displaystyle T=\\{2\to 2\\}:$ $\displaystyle h_{T}=0~{}~{}~{}~{}~{}{\rm
or}~{}~{}~{}~{}~{}\xi_{R}=\xi_{K}=0~{}\,,~{}~{}$ (25) $\displaystyle T=\\{2\to
1;\widehat{1}\\}:$ $\displaystyle h_{T}=c_{1\to
1}(t,R^{2},M_{0}),~{}~{}~{}\xi_{R}=0,~{}~{}~{}\xi_{K}=-1~{}\,,~{}~{}$ (26)
$\displaystyle T=\\{2\to 1;\widehat{0}\\}:$ $\displaystyle h_{T}=e^{t(2K\cdot
R)}c_{1\to
1}(t,R^{2},M_{1}),~{}~{}~{}\xi_{R}=1,~{}~{}~{}\xi_{K}=+1~{}\,.~{}~{}$ (27)

Generation functions are functions of $f(r,p)$ with $r=R^{2}$ and $p=K\cdot
R$. It is easy to work out that

$\displaystyle{\partial\over\partial R^{\mu}}$ $\displaystyle=$ $\displaystyle
2\eta_{\rho\mu}R^{\rho}\partial_{r}+K_{\mu}\partial_{p},~{}~{}~{}~{}K\cdot{\partial\over\partial
R^{\mu}}=2p\partial_{r}+K^{2}\partial_{p}~{}~{}~{}~{}$ (28)
$\displaystyle\eta^{\mu\nu}{\partial\over\partial
R^{\nu}}{\partial\over\partial R^{\mu}}$ $\displaystyle=$
$\displaystyle(4r\partial^{2}_{r}+4p\partial_{p}\partial_{r}+K^{2}\partial^{2}_{p}+2D\partial_{r})~{}~{}~{}~{}$
(29)

thus (23) and (24) can be written as

$\displaystyle(4r\partial^{2}_{r}+4p\partial_{p}\partial_{r}+K^{2}\partial^{2}_{p}+2D\partial_{r}-4t^{2}M_{0}^{2})c_{T}=4t^{2}\xi_{R}h_{T}~{}\,,~{}~{}$
(30)
$\displaystyle(2p\partial_{r}+K^{2}\partial_{p}-tf)c_{T}=t\xi_{K}h_{T}~{}\,.~{}~{}$
(31)

Equations (30) and (31) are the differential equations we need to solve. We
will present two ways to solve them. One is by the series expansion of the
naive variables $r,p$. This is the method used in [35, 36]. However, as we
will show, using the idea of generation function, the powers of $r,p$ are
independent to each other, thus the recursion relations become simpler and can
be solved explicitly. Another method is to solve the differential equation
directly and get more compact and analytical expression. An important lesson
from the second method is that the right variables to do the series expansion
are not $r,p$ but their proper combination.

###  3.2 The series expansion

In this subsection, we will present the series expansion of $r,p$.
Writing666Since we are considering the generation functions, the $n,m$ are
free indices, while in previous works [35, 36] with fixed tensor rank $K$,
$n,m$ are constrained by $2n+m=K$. One can see that many manipulations are
simplified using the idea of generation functions.

$\displaystyle
c=\sum_{n,m=0}^{\infty}c_{n,m}r^{n}p^{m},~{}~{}~{}~{}~{}h_{c}=\sum_{n,m=0}^{\infty}h_{n,m}r^{n}p^{m}~{}~{}~{}$
(32)

and putting them to (30) and (31) we get following equations

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2(n+1)(2n+2m+D)c_{n+1,m}+K^{2}(m+2)(m+1)c_{n,m+2}-4t^{2}M_{0}^{2}c_{n,m}-4t^{2}\xi_{R}h_{n,m}~{}~{}n,m\geq
0~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (33) $\displaystyle 0$ $\displaystyle=$
$\displaystyle
2(n+1)c_{n+1,m}+K^{2}(m+2)c_{n,m+2}-tfc_{n,m+1}-t\xi_{K}h_{n,m+1}~{}~{}~{}~{}m,n\geq
0~{}~{}~{}$ (34) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
K^{2}c_{n,1}-tfc_{n,0}-t\xi_{K}h_{n,0}~{}~{}~{}~{}~{}n\geq 0~{}~{}~{}$ (35)

Using (33) and (34) we can solve

$\displaystyle c_{n+1,m}$ $\displaystyle=$ $\displaystyle{4M_{0}^{2}t^{2}\over
2(n+1)(D+2n+m-1)}c_{n,m}-{tf(m+1)\over 2(n+1)(D+2n+m-1)}c_{n,m+1}$ (36)
$\displaystyle+{4t^{2}\xi_{R}h_{n,m}-t\xi_{K}(m+1)h_{n,m+1}\over
2(n+1)(D+2n+m-1)}~{}~{}~{}~{}~{}~{}~{}~{}~{}$ $\displaystyle c_{n,m+2}$
$\displaystyle=$
$\displaystyle{-4M_{0}^{2}t^{2}\over(m+2)(D+2n+m-1)K^{2}}c_{n,m}+{tf(D+2n+2m)\over(m+2)(D+2n+m-1)K^{2}}c_{n,m+1}$
(37) $\displaystyle+{-4t^{2}\xi_{R}h_{n,m}+t\xi_{K}(D+2n+2m)h_{n,m+1}\over
2(n+1)(D+2n+m-1)}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$

Now setting $m=0$ in (36) and combining (35) we can solve

$\displaystyle c_{n+1,0}$ $\displaystyle=$
$\displaystyle{(-t^{2}f^{2}+4M_{0}^{2}t^{2}K^{2})\over
2(n+1)(D+2n-1)K^{2}}c_{n,0}-{t^{2}(\xi_{K}f-4\xi_{R}K^{2})h_{n,0}+t\xi_{K}K^{2}h_{n,1}\over
2(n+1)(D+2n-1)K^{2}}~{}~{}~{}~{}~{}~{}$ (38) $\displaystyle c_{n,1}$
$\displaystyle=$ $\displaystyle{tf\over K^{2}}c_{n,0}+{t\xi_{K}h_{n,0}\over
K^{2}}~{}~{}~{}~{}~{}~{}$ (39)

Using equation (38) we can recursively solve all $c_{n,0}$ starting from the
boundary condition $c_{0,0}=1$ for $c_{2\to 2}$ or $c_{0,0}=0$ for $c_{2\to
1}$. Knowing all $c_{n,0}$ we can use (39) to get all $c_{n,1}$. To solve all
$c_{n,m}$, we use (33) and (34) again, but now solve

$\displaystyle c_{n,m+1}$ $\displaystyle=$ $\displaystyle{4M_{0}^{2}t\over
f(m+1)}c_{n,m}-{2(n+1)(D+2n+m-1)\over
tf(m+1)}c_{n+1,m}+{4t\xi_{R}h_{n,m}-\xi_{K}(m+1)h_{n,m+1}}~{}~{}~{}~{}~{}~{}$
(40) $\displaystyle c_{n,m+2}$ $\displaystyle=$
$\displaystyle{-2(n+1)(D+2n+2m)\over(m+1)(m+2)K^{2}}c_{n+1,m}+{4t^{2}M_{0}^{2}\over(m+1)(m+2)K^{2}}c_{n,m}+{4t^{2}\xi_{R}h_{n,m}\over(m+1)(m+2)K^{2}}~{}~{}~{}~{}~{}~{}~{}$
(41)

Both equations can be used recursively to solve $c_{n,m}$. After using one of
them to get all $c_{n,m}$, another one becomes nontrivial consistent check.
Among them (40) is better, since it solves $(m+1)$ from $m$.

Using above algorithm, we present the first few terms of generation functions
for comparison. For $c_{2\to 2}$ we have

$\displaystyle c_{2,2}$ $\displaystyle=$ $\displaystyle 1+{ft\over
K^{2}}p+{(Df^{2}-4K^{2}M_{0}^{2})t^{2}\over
2(D-1)(K^{2})^{2}}p^{2}+{(4K^{2}M_{0}^{2}-f^{2})t^{2}\over 2(D-1)K^{2}}r$ (42)
$\displaystyle+{f((2+D)f^{2}-12K^{2}M_{0}^{2})t^{3}\over
6(D-1)(K^{2})^{3}}p^{3}-{f(f^{2}-4K^{2}M_{0}^{2})t^{3}\over
2(D-1)(K^{2})^{2}}rp+...~{}~{}~{}$

and

$\displaystyle c_{2\to 1;\widehat{1}}$ $\displaystyle=$ $\displaystyle
0-{t\over K^{2}}p-{Dft^{2}\over 2(D-1)(K^{2})^{2}}p^{2}+{ft^{2}\over
2(D-1)K^{2}}r$ (43)
$\displaystyle-{(D(2+D)f^{2}-8(D-1)K^{2}M_{0}^{2})t^{3}\over
6(D-1)D(K^{2})^{3}}p^{3}+{(Df^{2}-4(D-1)K^{2}M_{0}^{2})t^{3}\over
2(D-1)D(K^{2})^{2}}rp+...~{}~{}~{}$

and

$\displaystyle c_{2\to 1;\widehat{0}}$ $\displaystyle=$ $\displaystyle
0+{t\over K^{2}}p+{(-D\widetilde{f}+4(D-1)K^{2})t^{2}\over
2(D-1)(K^{2})^{2}}p^{2}+{\widetilde{f}t^{2}\over
2(D-1)K^{2}}r+{(Df\widetilde{f}+4(D-1)K^{2}M_{1}^{2})t^{3}\over
2(D-1)D(K^{2})^{2}}rp$ (44)
$\displaystyle+{(6DK^{2}(K^{2}(D-2)+DM_{0}^{2})-2(D+2)(3D-2)K^{2}M_{1}^{2}+D(D+2)\widetilde{f}^{2})t^{3}\over
6(D-1)(K^{2})^{2}}p^{3}+..~{}~{}~{}$

where we have defined $\widetilde{f}=K^{2}-M_{0}^{2}+M_{1}^{2}=-f+2K^{2}$.

Here we have presented the general recursive algorithm. In the Appendix B, we
will show that these recursion relations can be solved explicitly, i.e., we
find explicit expressions for all coefficients $c_{n,m}$.

###  3.3 The analytic solution

In previous subsection, we have present the solution using the series
expansion. In this subsection, we will solve the two differential equations
(30) and (31) directly.

Let us start from (31) first. To solve it, we define following new variables

$\displaystyle x$ $\displaystyle=$ $\displaystyle
K^{2}r-p^{2},~{}~{}~{}~{}~{}y=p~{}\,,~{}~{}~{}$ (45)

then (31) becomes

$\displaystyle(2p\partial_{r}+K^{2}\partial_{p}-tf)c(r,p)=(K^{2}\partial_{y}-tf)c(x,y)=t\xi_{K}h_{T}~{}\,,~{}~{}$
(46)

where we have used

$\displaystyle p$ $\displaystyle=$ $\displaystyle
y,~{}~{}~{}~{}~{}~{}r={x+y^{2}\over
K^{2}},~{}~{}~{}~{}\partial_{r}=K^{2}\partial_{x},~{}~{}~{}\partial_{p}=-2y\partial_{x}+\partial_{y}~{}\,.~{}~{}$
(47)

The differential equation (46) can be solved as (see the discussion in the
Appendix, for example, (152))

$\displaystyle c(x,y)={1\over K^{2}}e^{{tf\over
K^{2}}y}\left(G(x)+\int_{0}^{y}dwe^{-{tf\over
K^{2}}w}t\xi_{K}h_{T}(x,w)\right)~{}\,,~{}~{}$ (48)

where the function $G$ depends only on $x$, while $h_{T}\equiv h_{T}(x,y)$ is
the function of both $x,y$.

Now we consider the equation (30). The first step is to simplify it by writing

$\displaystyle(4r\partial^{2}_{r}+4p\partial_{p}\partial_{r}+K^{2}\partial^{2}_{p}+2D\partial_{r})={1\over
K^{2}}(2p\partial_{r}+K^{2}\partial_{p})(2p\partial_{r}+K^{2}\partial_{p})+(4r-{4p^{2}\over
K^{2}})\partial_{r}^{2}+2(D-1)\partial_{r}\,.~{}~{}~{}~{}~{}~{}~{}$ (49)

Thus using (31), (30) becomes

$\displaystyle\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)c=\left(-t\xi_{K}\partial_{y}+{4t^{2}\xi_{R}K^{2}-t^{2}\xi_{K}f\over
K^{2}}\right)h_{T}\,.~{}~{}~{}$ (50)

Putting (48) to (50) and simplifying we get

$\displaystyle\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)G(x)$ (51) $\displaystyle=$ $\displaystyle K^{2}e^{-{tf\over
K^{2}}y}\left(-t\xi_{K}\partial_{y}+{4t^{2}\xi_{R}K^{2}-t^{2}\xi_{K}f\over
K^{2}}\right)h_{T}(x,y)$
$\displaystyle-\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)\int_{0}^{y}dwe^{-{tf\over
K^{2}}w}t\xi_{K}h_{T}(x,w)\,.~{}~{}~{}$

Equation (51) is the form of (161) which has been discussed in the Appendix.
One interesting point is that since the left hand side is independent of $y$,
the right hand side should be zero under the action of $\partial_{y}$. One can
check that it is indeed true.

Having laid out the frame, we can use it to solve various generation
functions.

####  3.3.1 The generation function $c_{2\to 2}$

For this case, we have $h_{T}=0$, thus using the result (176) we can
immediately written down

$\displaystyle c_{2\to 2}(t,r,p,K^{2};M_{0},M_{1})$ $\displaystyle=$
$\displaystyle~{}_{0}F_{1}(\emptyset;{D-1\over
2};\left({t^{2}(4K^{2}M_{0}^{2}-f^{2})x\over 4(K^{2})^{2}}\right)e^{{tf\over
K^{2}}y}|_{x\to K^{2}r-p^{2},y\to p}~{}~{}~{}$ (52)

One can check it with the series expansion (188) given in the Appendix B.
Comparing to it, the result (52) is very simple and compact. This shows the
power of using the generation function. Also, the differential equations (30)
and (31) tell us the right variables for the series expansion should be $x,y$
instead of the naive variables $r,p$.

####  3.3.2 The generation function $c_{2\to 1;\widehat{1}}$

For this case we have $\xi_{R}=0,\xi_{K}=-1$ and

$\displaystyle h_{T}(r)=c_{1\to 1}(t,r={x+y^{2}\over
K^{2}},M_{0})=\sum_{n=0}^{\infty}{(t^{2}M_{0}^{2})^{n}\left({x+y^{2}\over
K^{2}}\right)^{n}\over n!\left({D\over 2}\right)_{n}}~{}~{}~{}$ (53)

which satisfies the differential equation (6). The (51) becomes

$\displaystyle\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)G(x)$ (54) $\displaystyle=$ $\displaystyle K^{2}e^{-{tf\over
K^{2}}y}t\left(\partial_{y}+{tf\over K^{2}}\right)h_{T}(x,y)$
$\displaystyle+t\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)\int_{0}^{y}dwe^{-{tf\over K^{2}}w}h_{T}(x,w)~{}~{}~{}$

A first important check is that the right hand side of (54) is
$y$-independent. Acting ${\partial\over\partial y}$ at the right hand side we
will get

$\displaystyle te^{-{tf\over K^{2}}y}\left\\{-tf\left(\partial_{y}+{tf\over
K^{2}}\right)h_{T}(x,y)+K^{2}\left(\partial^{2}_{y}+{tf\over
K^{2}}\partial_{y}\right)h_{T}(x,y)\right.$
$\displaystyle\left.+\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)h_{T}(x,y)\right\\}~{}~{}~{}$ (55)

Using

$\displaystyle\partial_{x}h_{T}(r)={\partial{x+y^{2}\over K^{2}}\over\partial
x}\partial_{r}h_{T}={1\over
K^{2}}\partial_{r}h_{T},~{}~{}~{}~{}~{}\partial_{y}h_{T}(r)={\partial{x+y^{2}\over
K^{2}}\over\partial y}\partial_{r}h_{T}={2y\over
K^{2}}\partial_{r}h_{T}~{}~{}~{}$ (56)

one can check that (55) is reduced to the differential equation (6), thus we
have proved the $y$-independent of (54).

Setting $y=0$ in (54) we get

$\displaystyle\left({4xK^{2}}\partial_{x}^{2}+2(D-1)K^{2}\partial_{x}+{t^{2}(f^{2}-4K^{2}M_{0}^{2})\over
K^{2}}\right)G(x)=t^{2}fh_{T}(x,y=0)~{}~{}~{}$ (57)

where we have used $\partial_{y}h_{T}(x,y=0)={2y\over
K^{2}}\partial_{r}h_{T}=0$. The differential equation (57) is the form of
(161) and we get the solution

$\displaystyle G(x)$ $\displaystyle=$ $\displaystyle{t^{2}f\over
4K^{2}}G_{0}(x)\int_{0}^{x}dww^{-{(D-1)\over 2}}e^{-2G_{0}(w)}\int_{0}^{w}d\xi
h_{T}(\xi,y=0)G_{0}^{-1}(\xi)e^{2G_{0}(\xi)}\xi^{{(D-1)\over 2}-1}~{}~{}~{}$
(58)

where

$\displaystyle G_{0}(x)$ $\displaystyle=$
$\displaystyle~{}_{0}F_{1}(\emptyset;{(D-1)\over
2};{t^{2}(4K^{2}M_{0}^{2}-f^{2})\over 4(K^{2})^{2}}x)~{}~{}~{}$ (59)

Putting all together we finally have

$\displaystyle c_{2\to 1;\widehat{1}}(x,y)={1\over K^{2}}e^{{tf\over
K^{2}}y}\left(G(x)-t\int_{0}^{y}dwe^{-{tf\over
K^{2}}w}h_{T}(x,w)\right)~{}~{}~{}$ (60)

Although we have a very compact expression (60) for generation function, in
practice it is more desirable to have the series expansion form. In the
Appendix A we have introduced three ways. Here we work out the expansion by
direct integration. Using (53) we have

$\displaystyle\int_{0}^{y}dwe^{-{tf\over
K^{2}}w}h_{T}(x,w)=\sum_{n=0}^{\infty}{(t^{2}M_{0}^{2})^{n}\over
n!\left({D\over 2}\right)_{n}}\int_{0}^{y}dwe^{-{tf\over
K^{2}}w}\left({x+w^{2}\over K^{2}}\right)^{n}~{}~{}~{}$ (61)

To work out the integration, we see that

$\displaystyle
R({\alpha})\equiv\int_{0}^{T}due^{{\alpha}u}={1\over{\alpha}}e^{{\alpha}u}|_{0}^{T}={e^{{\alpha}T}-1\over{\alpha}}~{}~{}~{}~{}$
(62)

thus

$\displaystyle{d^{n}\over d{\alpha}^{n}}R({\alpha})$ $\displaystyle=$
$\displaystyle\int_{0}^{T}due^{{\alpha}u}u^{n}={(-)^{n}n!\over{\alpha}^{n+1}}\left(e^{{\alpha}T}\lfloor
e^{-{\alpha}T}\rfloor_{{\alpha}^{n}}-1\right)$ (63) $\displaystyle=$
$\displaystyle{(-)^{n}n!\over{\alpha}^{n+1}}\left(e^{{\alpha}T}\left(e^{-{\alpha}T}-\sum_{i=n+1}^{\infty}{(-{\alpha}T)^{i}\over
i!}\right)-1\right)={-(-)^{n}n!\over{\alpha}^{n+1}}e^{{\alpha}T}\sum_{i=n+1}^{\infty}{(-{\alpha}T)^{i}\over
i!}~{}~{}~{}~{}$

where the symbol $\lfloor Y(x)\rfloor_{x^{n-1}}$ means to keep the Taylor
expansion up to the order of $x^{(n-1)}$. Using (63) we have

$\displaystyle\int_{0}^{T}due^{{\alpha}u}({\beta}+\gamma
u^{2})^{N}=\sum_{i=0}^{N}{N!\over
i!(N-i)!}{\beta}^{N-i}\gamma^{i}\int_{0}^{T}due^{{\alpha}u}u^{2i}$ (64)
$\displaystyle=$ $\displaystyle\sum_{i=0}^{N}{N!\over
i!(N-i)!}{\beta}^{N-i}\gamma^{i}{-(2i)!\over{\alpha}^{2i+1}}e^{{\alpha}T}\sum_{j=2i+1}^{\infty}{(-{\alpha}T)^{j}\over
j!}~{}~{}~{}~{}$

Using (64) we can evaluate (61) as

$\displaystyle\int_{0}^{y}dwe^{-{tf\over K^{2}}w}h_{T}(x,w)=ye^{-{tf\over
K^{2}}y}\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{n}{\left({t^{2}M_{0}^{2}\over
K^{2}}\right)^{n}\over\left({D\over 2}\right)_{n}}{(2i)!x^{n-i}y^{2i}\over
i!(n-i)!}{\left({tfy\over K^{2}}\right)^{j}\over(j+2i+1)!}~{}~{}~{}$ (65)

The evaluation of $G(x)$ can be found in (170) as

$\displaystyle G(x)$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\sum_{i=0}^{n-1}{\left({D-1\over
2}\right)_{i}\over n!\left({D-1\over 2}\right)_{n}\left({D\over
2}\right)_{i}}{f(M_{0}^{2})^{i}(t^{2}x)^{n}(4K^{2}M_{0}^{2}-f^{2})^{n-i-1}\over
4^{n-i}(K^{2})^{2n-i-1}}~{}~{}~{}$ (66)

Collecting all pieces together, we finally have

$\displaystyle c_{2\to 1;\widehat{1}}(t,r,p,K^{2};M_{0},M_{1})$
$\displaystyle=$ $\displaystyle{1\over K^{2}}e^{{tf\over
K^{2}}y}\sum_{n=0}^{\infty}\sum_{i=0}^{n-1}{\left({D-1\over 2}\right)_{i}\over
n!\left({D-1\over 2}\right)_{n}\left({D\over
2}\right)_{i}}{f(M_{0}^{2})^{i}(t^{2}x)^{n}(4K^{2}M_{0}^{2}-f^{2})^{n-i-1}\over
4^{n-i}(K^{2})^{2n-i-1}}$ (67) $\displaystyle-{ty\over
K^{2}}\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{n}{\left({t^{2}M_{0}^{2}\over
K^{2}}\right)^{n}\over\left({D\over 2}\right)_{n}}{(2i)!x^{n-i}y^{2i}\over
i!(n-i)!}{\left({tfy\over K^{2}}\right)^{j}\over(j+2i+1)!}~{}~{}~{}$

This result can be checked with the one given in (199). One can see that the
formula (67) is much more compact and manifest with various analytic
structures.

##  4 The general frame

Having the detail computations in the bubble, in this section we will set up
the general frame to find generation functions for general one-loop integrals
with $(n+1)$ propagators. The system has $n$ external momenta
$K_{i},i=1,...,n$ and $(n+1)$ masses $M_{j}^{2},j=0,1,...,n$. Using the
auxiliary vector $R$ we have $(n+1)$ auxiliary scalar products (ASP) $r=R\cdot
R$, $p_{i}=K_{i}\cdot R,i=1,...,n$. From the experience in the bubble, we know
that these ASP’s are not good variables to solve differential equations
produced by $\partial_{R}\cdot\partial_{R}$ and
$K_{i}\cdot\partial_{R},i=1,...,n$. Thus we will discuss how to find these
good variables in the first subsection. Then we discuss the differential
equations in these new variables in the second subsection and finally their
solutions in the third subsection.

###  4.1 Finding good variables

We will denote the good variables by $x$ and $y_{i},i=1,...,n$. To simplify
the differential equations, we need to impose following conditions

$\displaystyle(K_{i}\cdot\partial_{R})x=0,~{}~{}~{}~{}(K_{i}\cdot\partial_{R})y_{j}\sim\delta_{ij},~{}~{}~{}~{}~{}\forall
i,j=1,...,n~{}\,.~{}~{}$ (68)

To see there is indeed a solution for (68), let us define the Gram matrix $G$
and the row vector $P^{T}$ as

$\displaystyle G_{ij}=K_{i}\cdot K_{j},~{}~{}~{}~{}(P^{T})_{i}=K_{i}\cdot
R\,.~{}~{}~{}$ (69)

Putting $y_{j}=\sum_{t}{\beta}_{jt}p_{t}$ to (68), it is easy to see that the
condition becomes

$\displaystyle K_{i}\cdot\partial_{R}y_{j}=\sum_{t}{\beta}_{jt}(K_{t}\cdot
K_{i})\sim\delta_{ij}\,,~{}~{}~{}$ (70)

thus the matrix ${\beta}$ can be solved as

$\displaystyle{\beta}=|G|G^{-1}~{}~{}~{}$ (71)

where $|G|$ is the Gram determinant. For $x$, let us assume

$\displaystyle x=|G|r+P^{T}AP,~{}~{}~{}~{}A^{T}=A~{}\,.~{}~{}$ (72)

Since

$\displaystyle K_{i}\cdot\partial_{R}x$ $\displaystyle=$
$\displaystyle|G|2p_{i}+2(P^{T})_{R\to K_{i}}AP,~{}~{}~{}$ (73)

where $P_{R\to K_{i}}$ means to replace vector $R$ by the vector $K_{i}$.
Collecting all $i$ together, the right hand side of (73) is just
$(2|G|I+2GA)P$, thus we have the solution

$\displaystyle A=-|G|G^{-1}\,.~{}~{}~{}$ (74)

Putting everything together, we finally have

$\displaystyle x$ $\displaystyle=$
$\displaystyle|G|(r-P^{T}G^{-1}P),~{}~{}~{}~{}~{}~{}~{}Y=|G|G^{-1}P~{}~{}~{}$
(75)

where the mass dimensions of various quantities are

$\displaystyle[|G|]=2n,~{}~{}~{}[(G^{-1})_{ij}]=-2,~{}~{}~{}[A_{ij}]=2(n-1),~{}~{}~{}[x]=2(n+1),~{}~{}~{}[y_{i}]=2n~{}\,.~{}~{}$
(76)

From (75) we can solve

$\displaystyle
P={1\over|G|}GY,~{}~{}~{}~{}r={|G|x+Y^{T}GY\over|G|^{2}}\,.~{}~{}~{}$ (77)

###  4.2 The differential equations

Having found the good variables, we express differential operators
$\partial_{R}\cdot\partial_{R}$ and $K_{i}\cdot\partial_{R},i=1,...,n$ using
them. The first step is to use (75) to write

$\displaystyle{\partial\over\partial R^{\mu}}$ $\displaystyle=$
$\displaystyle\left(2|G|R_{\mu}-2|G|{\cal
K}^{T}_{\mu}G^{-1}P\right)\partial_{x}+|G|{\cal
K}^{T}_{\mu}G^{-1}\partial_{Y}~{}~{}~{}$ (78)

where we have defined ${\cal K}^{T}=(K_{1},...,K_{n})$ and
$\partial_{Y}^{T}=(\partial_{y_{1}},...,\partial_{y_{n}})$. Thus we find

$\displaystyle{\cal K}\cdot{\partial\over\partial
R^{\mu}}=|G|\partial_{Y},~{}~{}~{}~{}~{}~{}\partial_{R}\cdot\partial_{R}=2|G|(D-n)\partial_{x}+4|G|x\partial^{2}_{x}+|G|^{2}\partial_{Y}^{T}G^{-1}\partial_{Y}~{}\,.~{}~{}$
(79)

The differential equations for $c_{T}$, where $T$ denotes different types of
generation functions, have following pattern

$\displaystyle K_{i}\cdot\partial_{R}c_{T}$ $\displaystyle=$
$\displaystyle{\alpha}_{i}c_{T}+H_{T;i},~{}~{}i=1,2,...,n~{}$ (80)
$\displaystyle\partial_{R}\cdot\partial_{R}c_{T}$ $\displaystyle=$
$\displaystyle{\alpha}_{R}c_{T}+H_{T;R}~{}~{}~{}$ (81)

where ${\alpha}_{R},{\alpha}_{i}$ are constant (which are independent of $T$)
and $H_{T;R},H_{T;i}$ are known functions coming from lower topologies. Using
the result (79) and (80), we find

$\displaystyle|G|^{2}\partial_{Y}^{T}G^{-1}\partial_{Y}c_{T}={\alpha}^{T}_{\cal
K}G^{-1}{\alpha}_{\cal K}c_{T}+H^{T}_{T;\cal K}G^{-1}{\alpha}_{\cal
K}+|G|\partial_{Y}^{T}G^{-1}H_{T;\cal K}~{}~{}~{}$ (82)

where ${\alpha}^{T}_{\cal K}=({\alpha}_{1},...,{\alpha}_{n})$ and
$H^{T}_{T;\cal K}=(H_{T;1},...,H_{T;n})$. Thus the differential equations (80)
can be written as

$\displaystyle\left(\partial_{y_{i}}-{{\alpha}_{i}\over|G|}\right)c_{T}$
$\displaystyle=$ $\displaystyle{1\over|G|}H_{T;i}~{}~{}i=1,2,...,n~{}\,,$ (83)

while (81) becomes

$\displaystyle\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right)c_{T}$
$\displaystyle=$ $\displaystyle{\cal H}_{T;R}~{}~{}~{}$ (84)

with

$\displaystyle\widetilde{{\alpha}}_{R}={\alpha}^{T}_{\cal
K}G^{-1}{\alpha}_{\cal K}-{\alpha}_{R},~{}~{}~{}~{}~{}~{}{\cal
H}_{T;R}=-H^{T}_{T;\cal K}G^{-1}{\alpha}_{\cal
K}-|G|\partial_{Y}^{T}G^{-1}H_{T;\cal K}+H_{T;R}~{}\,.~{}~{}$ (85)

Having given the differential equations (83) and (84), there is an important
point to be mentioned. For (83) and (84) to have solution, functions $H$ are
not arbitrary, but must satisfy the integrability conditions, which are

$\displaystyle\left(\partial_{y_{j}}-{{\alpha}_{j}\over|G|}\right)H_{T;i}=\left(\partial_{y_{i}}-{{\alpha}_{i}\over|G|}\right)H_{T;j},~{}~{}~{}\forall
i,j=1,2,...,n~{}~{}~{}$ (86)

and

$\displaystyle\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right){1\over|G|}H_{T;i}=\left(\partial_{y_{i}}-{{\alpha}_{i}\over|G|}\right){\cal
H}_{T;R},~{}~{}~{}\forall i=1,2,...,n~{}\,.~{}~{}$ (87)

Differential equations (83) and (84) are the type of (148) and (161)
respectively in the Appendix A, for which the solution has been presented. In
next subsection, we will solve them analytically.

###  4.3 Analytic solution

In this part, we will present the necessary steps of solving above
differential equations (83) and (84). Let us solve them one by one. For
differential equation (83) with $i=1$, using the result (152) in the Appendix
A, we have

$\displaystyle c_{T}(x,y)$ $\displaystyle=$ $\displaystyle
e^{{{\alpha}_{1}\over|G|}y_{1}}\left(F_{T}(x,y_{2},...,y_{n})+{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}H_{T;1}(x,w_{1},y_{2},...,y_{n})\right)~{}~{}~{}$
(88)

where $F_{T}(x,y_{2},...,y_{n})$ does not depend on the variable $y_{1}$. Now
we act with $\left(\partial_{y_{2}}-{{\alpha}_{2}\over|G|}\right)$ at the both
sides of (88) to get the differential equation

$\displaystyle\left(\partial_{y_{2}}-{{\alpha}_{2}\over|G|}\right)F_{T}(x,y_{2},...,y_{n})$
$\displaystyle=$
$\displaystyle-{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}\left(\partial_{y_{2}}-{{\alpha}_{2}\over|G|}\right)H_{T;1}(x,w_{1},y_{2},...,y_{n})$
(89)
$\displaystyle+e^{-{{\alpha}_{1}\over|G|}y_{1}}{1\over|G|}H_{T;2}(x,y_{1},y_{2},...,y_{n})\,.~{}~{}~{}$

Using (152) we find

$\displaystyle
F_{T}(x,y_{2},...,y_{n})=e^{{{\alpha}_{2}\over|G|}y_{2}}F_{T}(x,y_{3},...,y_{n})+e^{{{\alpha}_{2}\over|G|}y_{2}}e^{-{{\alpha}_{1}\over|G|}y_{1}}{1\over|G|}\int_{0}^{y_{2}}dw_{2}e^{-{{\alpha}_{2}\over|G|}w_{2}}H_{T;2}(x,y_{1},w_{2},...,y_{n})$
$\displaystyle-e^{{{\alpha}_{2}\over|G|}y_{2}}{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}\left(e^{-{{\alpha}_{2}\over|G|}y_{2}}H_{T;1}(x,w_{1},y_{2},...,y_{n})-H_{T;1}(x,w_{1},y_{2}=0,...,y_{n})\right)\,.~{}~{}~{}$
(90)

Putting (90) back to (88) and doing some algebraic manipulations, we get

$\displaystyle c_{T}(x,y)$ $\displaystyle=$ $\displaystyle
e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}F_{T}(x,y_{3},...,y_{n})$
(91)
$\displaystyle+e^{{{\alpha}_{2}\over|G|}y_{2}}{1\over|G|}\int_{0}^{y_{2}}dw_{2}e^{-{{\alpha}_{2}\over|G|}w_{2}}H_{T;2}(x,y_{1},w_{2},...,y_{n})$
$\displaystyle+e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}H_{T;1}(x,w_{1},y_{2}=0,...,y_{n})~{}~{}~{}$

Repeating above procedure with the action
$\left(\partial_{y_{2}}-{{\alpha}_{2}\over|G|}\right)$ we can solve
$F_{T}(x,y_{3},...,y_{n})$ and then find

$\displaystyle
c_{T}(x,y)=e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}e^{{{\alpha}_{3}\over|G|}y_{3}}F_{T}(x,y_{4},...,y_{n})$
$\displaystyle+e^{{{\alpha}_{3}\over|G|}y_{3}}{1\over|G|}\int_{0}^{y_{3}}dw_{3}e^{-{{\alpha}_{3}\over|G|}w_{3}}H_{T;3}(x,y_{1},y_{2},w_{3},...,y_{n})$
$\displaystyle+e^{{{\alpha}_{2}\over|G|}y_{2}}e^{{{\alpha}_{3}\over|G|}y_{3}}{1\over|G|}\int_{0}^{y_{2}}dw_{2}e^{-{{\alpha}_{2}\over|G|}w_{2}}H_{T;2}(x,y_{1},w_{2},y_{3}=0,...,y_{n})$
$\displaystyle+e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}e^{{{\alpha}_{3}\over|G|}y_{3}}{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}H_{T;1}(x,w_{1},y_{2}=0,y_{3}=0,...,y_{n})~{}~{}~{}$
(92)

By checking (91) and (92) we can see that after solving $n$ first order
differential equations (83) we get

$\displaystyle
c_{T}(x,y_{1},...,y_{n})=e^{\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}F(x)+{\cal
H}_{T;K}~{}~{}~{}~{}~{}~{}$ (93)

with

$\displaystyle{\cal
H}_{T;K}={1\over|G|}\widetilde{\sum}_{i=n}^{1}e^{\widetilde{\sum}_{j=n}^{i}{\alpha}_{j}y_{j}\over|G|}\int_{0}^{y_{i}}dw_{i}e^{-{\alpha}_{i}w_{i}\over|G|}H_{T;i}(x,y_{1},...,y_{i-1},w_{i},0,...,0)~{}~{}~{}~{}~{}~{}$
(94)

where for simplicity we have defined the sum $\widetilde{\sum}_{i=a}^{b}$ to
mean to take the sum over $(a,a-1,a-2,...,b)$ with $a\geq b$.

Before going to solve the only unknown function $F(x)$, let us check that the
form (93) does satisfy the differential equations (83). When acting with
$\left(\partial_{y_{k}}-{{\alpha}_{k}\over|G|}\right)$ on the both sides, it
is easy to see that the first term at the right hand side of (93) and the
terms in ${\cal H}_{T;K}$ with $i<k$ give zero contributions since they
contain only the factor $e^{-{\alpha}_{k}y_{k}\over|G|}$ depending on $y_{k}$.
For the term $i=k$ in ${\cal H}_{T;K}$, the action gives

$\displaystyle{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+1}{\alpha}_{j}y_{j}\over|G|}H_{T;k}(x,y_{1},...,y_{k-1},y_{k},0,...,0)~{}\,.~{}~{}~{}~{}~{}$
(95)

For the term $i=k+1$ in ${\cal H}_{T;K}$, the action gives

$\displaystyle{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+1}{\alpha}_{j}y_{j}\over|G|}\int_{0}^{y_{k+1}}dw_{k+1}e^{-{\alpha}_{k+1}w_{k+1}\over|G|}\left(\partial_{y_{k}}-{{\alpha}_{k}\over|G|}\right)H_{T;k+1}(x,y_{1},...,y_{k},w_{k+1},0,...,0)$
(96) $\displaystyle=$
$\displaystyle{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+1}{\alpha}_{j}y_{j}\over|G|}\int_{0}^{y_{k+1}}dw_{k+1}e^{-{\alpha}_{k+1}w_{k+1}\over|G|}\left(\partial_{w_{k+1}}-{{\alpha}_{k+1}\over|G|}\right)H_{T;k}(x,y_{1},...,y_{k},w_{k+1},0,...,0)~{}~{}~{}~{}~{}~{}$

where in the second line we have used the integrability condition (86). After
partial integration we get

$\displaystyle-{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+1}{\alpha}_{j}y_{j}\over|G|}H_{T;k}(x,y_{1},...,y_{k-1},y_{k},0,...,0)+{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+2}{\alpha}_{j}y_{j}\over|G|}H_{T;k}(x,y_{1},...,y_{k},y_{k+1},0,...,0)\,.~{}~{}~{}~{}~{}~{}$
(97)

The first term in (97) cancels the term in (95) and we are left with the
second term in (97). Now the pattern is clear. The $i=k+2$ term in ${\cal
H}_{T;K}$ will produce two terms after using the integrability condition and
partial integration, the first term will cancel the second term in (97), while
the second term will be the form

$\displaystyle{1\over|G|}e^{\widetilde{\sum}_{j=n}^{k+3}{\alpha}_{j}y_{j}\over|G|}H_{T;k}(x,y_{1},...,y_{k},y_{k+1},y_{k+2},0,...,0)~{}\,.~{}~{}~{}~{}~{}$
(98)

Continuing to the term $i=n$ in ${\cal H}_{T;K}$ we will be left with
${1\over|G|}H_{T;k}(x,y_{1},...,y_{n})$, thus we have proved that (93) does
satisfy the differential equations (83).

Now we consider the differential equation (84). Using the form (93), we derive

$\displaystyle\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right)F(x)$
$\displaystyle=$ $\displaystyle
e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}\left({\cal
H}_{T;R}-\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right){\cal
H}_{T;K}\right)~{}~{}~{}~{}~{}~{}~{}~{}$ (99)

One important point of (99) is that the right hand side must be
$y_{i}$-independent. To check this point, we act $\partial_{y_{k}}$ at the
right hand side to give

$\displaystyle-{{\alpha}_{k}\over|G|}e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}\left({\cal
H}_{T;R}-\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right){\cal
H}_{T;K}\right)$
$\displaystyle+e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}\left(\partial_{y_{k}}{\cal
H}_{T;R}-\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right)\partial_{y_{k}}{\cal
H}_{T;K}\right)\,.~{}~{}~{}~{}~{}~{}~{}~{}$ (100)

Since we have proved

$\displaystyle\left(\partial_{y_{k}}-{{\alpha}_{k}\over|G|}\right){\cal
H}_{T;K}={1\over|G|}H_{T;k}~{}\,,~{}~{}~{}~{}~{}~{}~{}$ (101)

(100) is simplified to

$\displaystyle-{{\alpha}_{k}\over|G|}e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}{\cal
H}_{T;R}+e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}\left(\partial_{y_{k}}{\cal
H}_{T;R}-\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right){1\over|G|}H_{T;k}\right)~{}\,.~{}~{}~{}~{}~{}~{}~{}$
(102)

Using the integrability condition (87) we get

$\displaystyle-{{\alpha}_{k}\over|G|}e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}{\cal
H}_{T;R}+e^{-\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}\left(\partial_{y_{k}}{\cal
H}_{T;R}-\left(\partial_{y_{k}}-{{\alpha}_{k}\over|G|}\right){\cal
H}_{T;R}\right)=0~{}\,.~{}~{}~{}~{}~{}~{}~{}$ (103)

Having checked the $y$-independent, we can take $y_{i}$ to be any values at
the right hand side of (99). From the expression (94) one can see that if we
take $y_{1}=y_{2}=...=y_{n}=0$, we have ${\cal H}_{T;K}=0$, thus (99) is
simplified to

$\displaystyle\left(4|G|x\partial^{2}_{x}+2|G|(D-n)\partial_{x}+\widetilde{{\alpha}}_{R}\right)F(x)$
$\displaystyle=$ $\displaystyle{\cal
H}_{T;R}(x,0,0,...,0)\,.~{}~{}~{}~{}~{}~{}~{}~{}$ (104)

Above differential equation is the form of (161) in the Appendix A and we can
write down the solution immediately (see (180))

$\displaystyle F(x)=F_{0}(x)\left(f_{0}+\int_{0}^{x}dww^{-{(D-n)\over
2}}e^{-2F_{0}(w)}\int_{0}^{w}d\xi{1\over A}{\cal
H}_{T;R}(x,0,0,...,0)F_{0}^{-1}(\xi)e^{2F_{0}(\xi)}\xi^{{(D-n)\over
2}-1}\right)~{}~{}~{}~{}~{}~{}~{}~{}$ (105)

where

$\displaystyle F_{0}(x)=~{}_{0}F_{1}(\emptyset;{(D-n)\over
2};{-\widetilde{{\alpha}}_{R}x\over 4|G|})~{}\,.~{}~{}~{}~{}~{}~{}~{}$ (106)

Putting (106) back to (93) we get the final analytic expression of generation
functions

$\displaystyle c_{T}(x,y_{1},...,y_{n})={\cal
H}_{T;K}+e^{\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}F_{0}(x)$
$\displaystyle\left(f_{0}+\int_{0}^{x}dww^{-{(D-n)\over
2}}e^{-2F_{0}(w)}\int_{0}^{w}d\xi{1\over A}{\cal
H}_{T;R}(x,0,0,...,0)F_{0}(\xi)^{-1}e^{2F_{0}(\xi)}\xi^{{(D-n)\over
2}-1}\right)~{}\,.~{}~{}~{}~{}~{}$ (107)

When we consider the generation functions of reduction coefficients of one-
loop integrals with $(n+1)$ propagators, there is a special case, where all
$H_{T;i},H_{T;R}$ are zero. For this case, we can write down immediately the
generation function

$\displaystyle c_{n+1\to
n+1}(R,K_{1},...,K_{n})=~{}_{0}F_{1}(\emptyset;{(D-n)\over
2};{-\widetilde{{\alpha}}_{R}x\over
4|G|})e^{\sum_{i=1}^{n}{\alpha}_{i}y_{i}\over|G|}~{}\,.~{}~{}~{}~{}$ (108)

##  5 Triangle

In this part we present another example, i.e., the triangle, to demonstrate
the general frame laid down in previous section. The seven generation
functions are defined by

$\displaystyle I_{tri}(t,R)\equiv\int d\ell{e^{t(2\ell\cdot
R)}\over(\ell^{2}-M_{0}^{2})((\ell-K_{1})^{2}-M_{1}^{2})((\ell-
K_{2})^{2}-M_{2}^{2})}\equiv\int d\ell{e^{t(2\ell\cdot R)}\over
D_{0}D_{1}D_{2}}$ (109) $\displaystyle=$ $\displaystyle c_{3\to 3}\int
d\ell{1\over D_{0}D_{1}D_{2}}+c_{3\to 2;\widehat{i}}\int
d\ell{1\over\prod_{j\neq i,0}^{2}D_{j}}+c_{3\to 1;i}\int d\ell{1\over
D_{i}}\,.~{}~{}~{}~{}$

Using the permutation symmetry and the shifting of loop momentum we can find
nontrivial relations among these seven generation functions. The first group
of relations is

$\displaystyle c_{3\to 3}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle c_{3\to 3}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$
$\displaystyle c_{3\to 2;\widehat{0}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$
$\displaystyle=$ $\displaystyle c_{3\to
2;\widehat{0}}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$ $\displaystyle c_{3\to
2;\widehat{1}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle c_{3\to 2;\widehat{2}}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$
$\displaystyle c_{3\to 2;\widehat{2}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$
$\displaystyle=$ $\displaystyle c_{3\to
2;\widehat{1}}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$ $\displaystyle c_{3\to
1;0}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$ $\displaystyle
c_{3\to 1;0}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$ $\displaystyle c_{3\to
1;1}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$ $\displaystyle
c_{3\to 1;2}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,,$ $\displaystyle c_{3\to
1;2}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$ $\displaystyle
c_{3\to 1;1}(t,R;M_{0};K_{2},M_{2};K_{1},M_{1})\,.~{}~{}~{}~{}$ (110)

The second group of relations is

$\displaystyle c_{3\to 3}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{1}\cdot R}c_{3\to
3}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$ $\displaystyle c_{3\to
2;\widehat{0}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{1}\cdot R}c_{3\to
2;\widehat{1}}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$ $\displaystyle
c_{3\to 2;\widehat{1}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{1}\cdot R}c_{3\to
2;\widehat{0}}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$ $\displaystyle
c_{3\to 2;\widehat{2}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{1}\cdot R}c_{3\to
2;\widehat{1}}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$ $\displaystyle
c_{3\to 1;0}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{1}\cdot R}c_{3\to
1;1}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$ $\displaystyle c_{3\to
1;1}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$ $\displaystyle
e^{2K_{1}\cdot R}c_{3\to 1;0}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,,$
$\displaystyle c_{3\to 1;2}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$
$\displaystyle=$ $\displaystyle e^{2K_{1}\cdot R}c_{3\to
1;1}(t,R;M_{1};-K_{1},M_{0};K_{2}-K_{1},M_{2})\,.~{}~{}~{}~{}$ (111)

The third group of relations is

$\displaystyle c_{3\to 3}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{2}\cdot R}c_{3\to
3}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$ $\displaystyle c_{3\to
2;\widehat{0}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{2}\cdot R}c_{3\to
2;\widehat{2}}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$ $\displaystyle
c_{3\to 2;\widehat{1}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{2}\cdot R}c_{3\to
2;\widehat{1}}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$ $\displaystyle
c_{3\to 2;\widehat{2}}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{2}\cdot R}c_{3\to
2;\widehat{0}}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$ $\displaystyle
c_{3\to 1;0}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$
$\displaystyle e^{2K_{2}\cdot R}c_{3\to
1;2}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$ $\displaystyle c_{3\to
1;1}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$ $\displaystyle=$ $\displaystyle
e^{2K_{2}\cdot R}c_{3\to 1;1}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,,$
$\displaystyle c_{3\to 1;2}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})$
$\displaystyle=$ $\displaystyle e^{2K_{2}\cdot R}c_{3\to
1;0}(t,R;M_{2};K_{1}-K_{2},M_{1};-K_{2},M_{0})\,.~{}~{}~{}~{}$ (112)

Using these three groups of relations, we just need to compute three
generation functions, for example, $c_{3\to 3}$, $c_{3\to 2;\widehat{1}}$ and
$c_{3\to 1;0}$. The mass dimensions of them are

$\displaystyle[t]=-2,~{}~{}~{}~{}~{}~{}[c_{3\to 3}]=0,~{}~{}~{}~{}[c_{3\to
2;\widehat{i}}]=-2,~{}~{}~{}~{}~{}[c_{3\to 1;i}]=-4~{}\,.~{}~{}~{}$ (113)

###  5.1 The differential equations

Since we have given enough details in the section of bubble, here we will be
more briefly. Using
$\partial_{R}\cdot\partial_{R},K_{1}\cdot\partial_{R},K_{2}\cdot\partial_{R}$
operators, we can find

$\displaystyle\partial_{R}\cdot\partial_{R}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})=4t^{2}M_{0}^{2}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})+4t^{2}\xi_{T}h_{T}~{}\,,~{}~{}~{}$
(114) $\displaystyle
K_{1}\cdot\partial_{R}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})=tf_{1}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})-t\widetilde{\xi}_{T}\widetilde{h}_{T}+t\xi_{T}h_{T}~{}\,,~{}~{}~{}$
(115) $\displaystyle
K_{2}\cdot\partial_{R}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})=tf_{2}c_{T}(t,R;M_{0};K_{1},M_{1};K_{2},M_{2})-t\widehat{\xi}_{T}\widehat{h}_{T}+t\xi_{T}h_{T}~{}\,,~{}~{}~{}$
(116)

where $T$ is the index for different generation functions and
$f_{1}=K_{1}^{2}-M_{1}^{2}+M_{0}^{2}$, $f_{2}=K_{2}^{2}-M_{2}^{2}+M_{0}^{2}$.
The various constant $\xi,\widetilde{\xi},\widehat{\xi}$ are given in the
table

(125)

while $h_{T}$ are given by

$\displaystyle h_{3\to 2;\widehat{0}}$ $\displaystyle=$ $\displaystyle
e^{2K_{1}\cdot R}c_{2\to 2}(t,R^{2},(K_{2}-K_{1})\cdot
R,(K_{2}-K_{1})^{2};M_{1},M_{2})\,,$ $\displaystyle h_{3\to 1;1}$
$\displaystyle=$ $\displaystyle e^{2K_{1}\cdot R}c_{2\to
2;\widehat{1}}(t,R^{2},(K_{2}-K_{1})\cdot R,(K_{2}-K_{1})^{2};M_{1},M_{2})\,,$
$\displaystyle h_{3\to 1;2}$ $\displaystyle=$ $\displaystyle e^{2K_{1}\cdot
R}c_{2\to 2;\widehat{0}}(t,R^{2},(K_{2}-K_{1})\cdot
R,(K_{2}-K_{1})^{2};M_{1},M_{2})\,,~{}~{}~{}~{}$ (126)

and $\widetilde{h}_{T}$ are given by

$\displaystyle\widetilde{h}_{3\to 2;\widehat{1}}$ $\displaystyle=$
$\displaystyle c_{2\to 2}(t,R^{2},K_{2}\cdot R,K_{2}^{2};M_{0},M_{2})\,,$
$\displaystyle\widetilde{h}_{3\to 1;0}$ $\displaystyle=$ $\displaystyle
c_{2\to 1;\widehat{1}}(t,R^{2},K_{2}\cdot R,K_{2}^{2};M_{0},M_{2})\,,$
$\displaystyle\widetilde{h}_{3\to 1;2}$ $\displaystyle=$ $\displaystyle
c_{2\to 1;\widehat{0}}(t,R^{2},K_{2}\cdot
R,K_{2}^{2};M_{0},M_{2})\,,~{}~{}~{}~{}$ (127)

and $\widehat{h}_{T}$ are given by

$\displaystyle\widehat{h}_{3\to 2;\widehat{2}}$ $\displaystyle=$
$\displaystyle c_{2\to 2}(t,R^{2},K_{1}\cdot R,K_{1}^{2};M_{0},M_{1})\,,$
$\displaystyle\widehat{h}_{3\to 1;0}$ $\displaystyle=$ $\displaystyle c_{2\to
1;\widehat{1}}(t,R^{2},K_{1}\cdot R,K_{1}^{2};M_{0},M_{1})\,,$
$\displaystyle\widehat{h}_{3\to 1;1}$ $\displaystyle=$ $\displaystyle c_{2\to
1;\widehat{0}}(t,R^{2},K_{1}\cdot R,K_{1}^{2};M_{0},M_{1})\,.~{}~{}~{}~{}$
(128)

The differential equations (114), (115) and (116) are indeed the form (80) and
(81) in previous section.

For triangle, the natural variables for generation functions are

$\displaystyle r=R\cdot R,~{}~{}~{}p_{1}=K_{1}\cdot
R,~{}~{}~{}~{}p_{2}=K_{2}\cdot R~{}\,.~{}~{}~{}$ (129)

However, the good variables for differential equations (114), (115) and (116)
are $x,y_{1},y_{2}$ as

$\displaystyle x$ $\displaystyle=$
$\displaystyle[K_{1}^{2}K_{2}^{2}-(K_{1}\cdot
K_{2})^{2}]r-K_{2}^{2}p_{1}^{2}-K_{1}^{2}p_{2}^{2}+(2K_{1}\cdot
K_{2})p_{1}p_{2}$ $\displaystyle=$
$\displaystyle|G|r-p_{1}y_{1}-p_{2}y_{2}\,,$ $\displaystyle y_{2}$
$\displaystyle=$ $\displaystyle-(K_{1}\cdot
K_{2})p_{1}+K_{1}^{2}p_{2},~{}~{}~{}~{}~{}y_{1}=-(K_{1}\cdot
K_{2})p_{2}+K_{2}^{2}p_{1}~{}\,,~{}~{}~{}$ (130)

which are defined in (75) with Gram matrix

$\displaystyle
G(K_{1},K_{2})\equiv\left(\begin{array}[]{cc}K_{1}^{2}&K_{1}\cdot K_{2}\\\
K_{1}\cdot
K_{2}&K_{2}^{2}\end{array}\right),~{}~{}~{}~{}G^{-1}={1\over|G|}\left(\begin{array}[]{cc}K_{2}^{2}&-K_{1}\cdot
K_{2}\\\ -K_{1}\cdot K_{2}&K_{1}^{2}\end{array}\right)~{}\,.~{}~{}~{}$ (135)

Using the new variables, the differential equations are

$\displaystyle\left(\partial_{y_{1}}-{{\alpha}_{1}\over|G|}\right)c_{T}={1\over|G|}H_{T;1},~{}~{}~{}~{}~{}~{}~{}\left(\partial_{y_{2}}-{{\alpha}_{2}\over|G|}\right)c_{T}={1\over|G|}H_{T;2}~{}~{}~{}~{}$
(136)
$\displaystyle\left(4|G|x\partial^{2}_{x}+2(D-2)|G|\partial_{x}+\widetilde{{\alpha}}_{R}\right)c_{T}={\cal
H}_{T;R}~{}~{}~{}~{}$ (137)

where

$\displaystyle{\alpha}_{1}=tf_{1},~{}~{}~{}{\alpha}_{2}=tf_{2},~{}~{}~{}~{}\widetilde{{\alpha}}_{R}={t^{2}f_{1}^{2}K_{2}^{2}+t^{2}f_{2}^{2}K_{1}^{2}-2t^{2}f_{1}f_{2}(K_{1}\cdot
K_{2})\over|G|}-4t^{2}M_{0}^{2}~{}~{}~{}~{}$ (138)

and

$\displaystyle{\cal H}_{T;R}=-\left(K_{2}^{2}\partial_{y_{1}}-(K_{1}\cdot
K_{2})\partial_{y_{2}}+{{\alpha}_{1}K_{2}^{2}-{\alpha}_{2}(K_{1}\cdot
K_{2})\over|G|}\right)H_{T;1}$
$\displaystyle-\left(K_{1}^{2}\partial_{y_{2}}-(K_{1}\cdot
K_{2})\partial_{y_{1}}+{{\alpha}_{2}K_{1}^{2}-{\alpha}_{1}(K_{1}\cdot
K_{2})\over|G|}\right)H_{T;2}+H_{T;R}~{}\,.~{}~{}~{}~{}~{}~{}$ (139)

The solution is given by

$\displaystyle c_{T}(x,y_{1},y_{2})$ $\displaystyle=$ $\displaystyle
e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}F^{tri}_{T}(x)+e^{{{\alpha}_{2}\over|G|}y_{2}}{1\over|G|}\int_{0}^{y_{2}}dw_{2}e^{-{{\alpha}_{2}\over|G|}w_{2}}H_{T;2}(x,y_{1},w_{2})$
(140)
$\displaystyle+e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}{1\over|G|}\int_{0}^{y_{1}}dw_{1}e^{-{{\alpha}_{1}\over|G|}w_{1}}H_{T;1}(x,w_{1},0)~{}~{}~{}~{}~{}~{}~{}$

where

$\displaystyle F^{tri}_{0}(x)=~{}_{0}F_{1}(\emptyset;{(D-2)\over
2};{-\widetilde{{\alpha}}_{R}x\over 4|G|})~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (141)

and

$\displaystyle
F^{tri}(x)=F^{tri}_{0}(x)\left(f_{0}+\int_{0}^{x}dww^{-{(D-2)\over
2}}e^{-2F^{tri}_{0}(w)}\int_{0}^{w}d\xi{1\over A}{\cal
H}_{T;R}(x,0,0,...,0)F^{tri}_{0}(\xi)^{-1}e^{2F^{tri}_{0}(\xi)}\xi^{{(D-2)\over
2}-1}\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (142)

The generation function $c_{3\to 3}$: For this one, from the Table (125), we
see that $H_{T;1}=H_{T;2}=H_{T;R}=0$, thus we write down immediately

$\displaystyle c_{3\to
3}=e^{{{\alpha}_{1}\over|G|}y_{1}}e^{{{\alpha}_{2}\over|G|}y_{2}}~{}_{0}F_{1}(\emptyset;{(D-2)\over
2};{-\widetilde{{\alpha}}_{R}x\over 4|G|})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(143)

The generation function $c_{3\to 2}$: There are three generation functions
$c_{3\to 2;\widehat{i}}$. We want to choose one of them with simplest
$H_{1},H_{2},H_{R}$. Checking with Table (125), we see that if we consider
$c_{3\to 2;\widehat{1}}$, we will have $H_{R}=0,H_{2}=0$ and

$\displaystyle H_{3\to 2;\widehat{1};1}(x,y_{1},y_{2})$ $\displaystyle=$
$\displaystyle-c_{2\to 2}(t,R^{2},K_{2}\cdot R,K_{2}^{2};M_{0},M_{2})$ (144)
$\displaystyle=$ $\displaystyle-~{}_{0}F_{1}\left(\emptyset;{D-1\over
2};\left({t^{2}(4K_{2}^{2}M_{0}^{2}-f_{2}^{2})(K_{2}^{2}r-p_{2}^{2})\over
4(K_{2}^{2})^{2}}\right)\right)e^{{tf_{2}\over
K_{2}^{2}}p_{2}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$

where we have used the result (52) and expressed $r,p_{1},p_{2}$ using (77)

$\displaystyle p_{1}={K_{1}^{2}y_{1}+(K_{1}\cdot
K_{2})y_{2}\over|G|},~{}~{}~{}~{}p_{2}={K_{2}^{2}y_{2}+(K_{1}\cdot
K_{2})y_{1}\over|G|},~{}~{}~{}r={|G|x+K_{1}^{2}y_{1}^{2}+K_{2}^{2}y_{2}^{2}+2(K_{1}\cdot
K_{2})y_{1}y_{2}\over|G|^{2}}\,.~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (145)

Thus by (139) we can find

$\displaystyle{\cal H}_{3\to
2;\widehat{1};R}=\left({{\alpha}_{1}K_{2}^{2}-{\alpha}_{2}(K_{1}\cdot
K_{2})\over|G|}\right)e^{{tf_{2}\over
K_{2}^{2}}p_{2}}~{}_{0}F_{1}\left(\emptyset;{D-1\over
2};\left({t^{2}(4K_{2}^{2}M_{0}^{2}-f_{2}^{2})(K_{2}^{2}r-p_{2}^{2})\over
4(K_{2}^{2})^{2}}\right)\right)$
$\displaystyle+{t^{2}(4K_{2}^{2}M_{0}^{2}-f_{2}^{2})\over
4(K_{2}^{2})^{2}}{2K_{2}^{2}y_{1}\over|G|}e^{{tf_{2}\over
K_{2}^{2}}p_{2}}~{}_{0}F_{1}^{(1)}\left(\emptyset;{D-1\over
2};\left({t^{2}(4K_{2}^{2}M_{0}^{2}-f_{2}^{2})(K_{2}^{2}r-p_{2}^{2})\over
4(K_{2}^{2})^{2}}\right)\right)$ (146)

where we have defined

$\displaystyle~{}_{A}F_{B}^{(n)}(a_{1},...,a_{A};b_{1},...,b_{B};x)={d^{n}\over
dx^{n}}~{}_{A}F_{B}(a_{1},...,a_{A};b_{1},...,b_{B};x)$ (147)

Putting them back to (140) we can find the analytic expression. Using it we
can get the explicit series expansion as discussed in the Appendix A.

For generation functions $c_{3\to 1;i}$, we can do the similar calculations.
The key is to find $H_{T;i}$ and $H_{T;R}$, which is the generation functions
of one order lower topology. Thus we see the recursive structures of
generation functions from lower topologies to higher topologies. The logic is
clear although working out details takes some effects.

##  6 Conclusion

In this paper, we have introduced the concept of generation function for
reduction coefficients of loop integrals. For one-loop integrals, using the
recent proposal of auxiliary vector $R$, we can construct two types of
differential operators ${\partial\over\partial R}\cdot{\partial\over\partial
R}$ and $K_{i}\cdot{\partial\over\partial R}$. Using these operators, we can
establish corresponding differential equations for generation functions. By
proper changing of variables, these differential equations can be written into
the decoupled form, thus one can solve them one by one analytically.
Obviously, one could try to apply the same idea to discuss the reduction
problem for two and higher loop integrals. But with the appearance of
irreducible scalar products, the problem become harder. One can try to use the
IBP relation in the Baikov representation [43] as did in [40, 41].

## Acknowledgments

We would like to thank Jiaqi Chen, Tingfei Li and Rui Yu for useful discussion
and early participant. This work is supported by Chinese NSF funding under
Grant No.11935013, No.11947301, No.12047502 (Peng Huanwu Center).

##  Appendix A Solving differential equations

As shown in the paper, differential equations for generation functions can be
reduced to two typical ones. In this appendix, we present details of solving
these typical differential equations.

The first order differential equation: The first typical differential equation
is following first order differential equation

$\displaystyle\boxed{\left(A{d\over dx}+B\right)F(x)=H(x)}~{}~{}~{}~{}$ (148)

where $A,B$ are independent of $x$ and $H(x)$ is the known function of $x$. To
solve it, first we solve the homogenous part

$\displaystyle\left(A{d\over
dx}+B\right)F_{0}(x)=0,~{}~{}~{}~{}~{}\Longrightarrow~{}~{}~{}F_{0}(x)=e^{-{B\over
A}x}\,.~{}~{}~{}~{}$ (149)

Then we write $F(x)=F_{0}(x)F_{1}(x)$ in (148) to get the differential
equation for $F_{1}(x)$ as

$\displaystyle A{d\over dx}F_{1}(x)=F_{0}^{-1}H(x)~{}\,,~{}~{}~{}$ (150)

thus we have

$\displaystyle F_{1}(x)=F_{1}(x=0)+\int_{0}^{x}dt{1\over A}e^{{B\over
A}t}H(t)~{}\,.~{}~{}~{}$ (151)

Knowing the special solution of $F(x)$ in (151), the general solution of (148)
is given by

$\displaystyle\boxed{F(x)=F_{0}(x)F_{1}(x)+\widetilde{{\alpha}}F_{0}(x)=e^{-{B\over
A}x}\left({\alpha}+\int_{0}^{x}dw{1\over A}e^{{B\over
A}w}H(w)\right)}~{}~{}~{}~{}$ (152)

where the constant ${\alpha}$ is determined by the boundary condition.

If we want to have the series expansion of $x$, there are several ways to do.
The first one is to carry out the integration

$\displaystyle\int_{0}^{x}dt{1\over A}e^{{B\over A}t}H(t)={x\over
A}\sum_{n=0}^{\infty}\sum_{a=0}^{\infty}h_{n}x^{n}{1\over(n+a+1)}{\left({Bx\over
A}\right)^{a}\over a!}~{}~{}~{}~{}$ (153)

where we have used the expansion of $H(x)=\sum_{n=0}^{\infty}h_{n}x^{n}$. The
second way is to put $F(x)=\sum_{n=0}^{\infty}f_{n}x^{n}$ directly to (148) to
arrive the recursion relation

$\displaystyle
f_{n+1}={-B\over(n+1)A}f_{n}+{h_{n}\over(n+1)A}\equiv\gamma(n)f_{n}+\rho(n),~{}~{}~{}~{}n\geq
0~{}\,.~{}~{}~{}$ (154)

The recursive relation (154) can be solved as

$\displaystyle f_{n+1}$ $\displaystyle=$ $\displaystyle
f_{0}\prod_{i=0}^{n}\gamma(i)+\sum_{i=0}^{n}\rho(i)\prod_{j=i+1}^{n}\gamma(j)~{}\,.~{}~{}~{}$
(155)

It is easy to compute

$\displaystyle\Xi[n+1]$ $\displaystyle=$
$\displaystyle\prod_{i=0}^{n}\gamma(i)={(-B)^{n+1}\over(n+1)!A^{n+1}}~{}~{}~{}~{}~{}$
(156)

and

$\displaystyle\rho(i)\prod_{j=i+1}^{n-1}\gamma(j)={h_{i}\over(i+1)A}\prod_{j=i+1}^{n-1}\gamma(j)={h_{i}\over-B}\prod_{j=i}^{n-1}\gamma(j)={h_{i}\over-B}{\Xi[n]\over\Xi[i]}={h_{i}\over-B}{i!(-B)^{n-i}\over
n!A^{n-i}}~{}\,,~{}~{}~{}~{}$ (157)

thus we have

$\displaystyle F(x)$ $\displaystyle=$
$\displaystyle\sum_{n=0}\left(f_{0}{(-B)^{n}\over
n!A^{n}}+\sum_{i=0}^{n-1}{h_{i}\over A}{i!(-B)^{n-1-i}\over
n!A^{n-1-i}}\right)x^{n}~{}\,.~{}~{}~{}~{}$ (158)

The third way is to use analytic expression (152). We need to compute
${d^{n}F(x)\over dx^{n}}$ and then set $x=0$. One can see that, for example,

$\displaystyle{dF(x)\over dx}$ $\displaystyle=$ $\displaystyle-{B\over
A}e^{-{B\over A}x}\left({\alpha}+\int_{0}^{x}dw{1\over A}e^{{B\over
A}w}H(w)\right)+{1\over A}H(x)~{}\,,~{}~{}~{}$ (159)

thus

$\displaystyle{dF(x)\over dx}|_{x=0}$ $\displaystyle=$ $\displaystyle-{B\over
A}{\alpha}+{1\over A}H(x=0)~{}\,.~{}~{}~{}$ (160)

The important point is that when $x=0$ the integration $\int_{0}^{x}dw...=0$,
thus we have got rid of integration and all we need to do are the
differentiations over $e^{{B\over A}x}$ and $H(x)$.

The second order differential equation: The second typical differential
equation met in this paper is the following second order differential equation

$\displaystyle\boxed{\left(Ax{d^{2}\over dx^{2}}+B{d\over
dx}+C\right)F(x)=H(x)}~{}~{}~{}~{}~{}$ (161)

where $A,B,C$ are independent of $x$ and $H(x)$ is a known function of $x$.
Let us to solve it using the series expansion. Writing

$\displaystyle
F(x)=\sum_{n=0}^{\infty}f_{n}x^{n},~{}~{}~{}~{}~{}~{}~{}H(x)=\sum_{n=0}^{\infty}h_{n}x^{n}~{}~{}~{}~{}~{}$
(162)

and putting back to (161) we have

$\displaystyle\sum_{n=0}^{\infty}h_{n}x^{n}=\sum_{n=0}^{\infty}(An(n+1)f_{n+1}x^{n}+B(n+1)f_{n+1}x^{n}+Cf_{n}x^{n})~{}\,,~{}~{}~{}~{}$
(163)

thus we have the relation

$\displaystyle h_{n}=(n+1)(B+An)f_{n+1}+Cf_{n},~{}~{}~{}n\geq
0~{}\,.~{}~{}~{}~{}$ (164)

Using it, we can solve777An important point is that although (161) is second
order differential equations, but because the $x$ in front of ${d^{2}\over
dx^{2}}$, around $x=0$ it is essentially the first order differential
equation. This explains why using only $f_{0}$ and known $H(x)$ we can
determine $F(x)$ using (165).

$\displaystyle f_{n+1}$ $\displaystyle=$
$\displaystyle{h_{n}\over(n+1)(B+An)}-C{1\over(n+1)(B+An)}f_{n}\equiv\gamma(n)f_{n}+\rho(n),~{}~{}~{}~{}n\geq
0~{}\,.~{}~{}~{}~{}$ (165)

The recursive relation (165) can be solved as

$\displaystyle f_{n+1}$ $\displaystyle=$ $\displaystyle
f_{0}\prod_{i=0}^{n}\gamma(i)+\sum_{i=0}^{n}\rho(i)\prod_{j=i+1}^{n}\gamma(j)~{}\,.~{}~{}~{}~{}$
(166)

It is easy to compute

$\displaystyle\Xi[n+1]$ $\displaystyle=$
$\displaystyle\prod_{i=0}^{n}\gamma(i)={(-C)^{n+1}\over(n+1)!A^{n+1}\prod_{i=0}^{n}(B/A+i)}={(-C)^{n+1}\over(n+1)!A^{n+1}\left({B\over
A}\right)_{n+1}}~{}~{}~{}~{}~{}$ (167)

where we have used the Pochhammer symbol to simplify the expression888From the
definition one can see that $(x)_{n=0}=1,\forall x$.

$\displaystyle(x)_{n}={\Gamma(x+n)\over\Gamma(x)}=\prod_{i=1}^{n}(x+(i-1))~{}\,.~{}~{}~{}$
(168)

Using

$\displaystyle\rho(i)\prod_{j=i+1}^{n-1}\gamma(j)={h_{i}\over(i+1)(B+Ai)}\prod_{j=i+1}^{n-1}\gamma(j)={-h_{i}\over
C}\prod_{j=i}^{n-1}\gamma(j)={-h_{i}\over
C}{\Xi[n]\over\Xi[i]}~{}~{}~{}~{}~{}$ (169)

we have

$\displaystyle F(x)$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}f_{n}x^{n}=\sum_{n=0}^{\infty}x^{n}\Xi[n]\left\\{f_{0}+\sum_{i=0}^{n-1}{-h_{i}\over
C\Xi[i]}\right\\}~{}\,.~{}~{}~{}~{}~{}$ (170)

Let us define the following function

$\displaystyle H_{A,B,C}(x)=\sum_{i=0}^{\infty}{-h_{i}x^{i}\over
C\Xi[i]}=\sum_{i=0}^{\infty}{i!A^{i}\over(-C)^{i+1}}\left({B\over
A}\right)_{i}h_{i}x^{i}~{}~{}~{}~{}~{}~{}$ (171)

which can be considered as the ”dual function” of $H(x)$ corresponding to the
differential equation (161), then we can write

$\displaystyle\sum_{i=0}^{n-1}{-h_{i}\over C\Xi[i]}=\lfloor
H_{A,B,C}(x=1)\rfloor_{x^{n-1}}~{}~{}~{}~{}~{}~{}$ (172)

where the symbol $\lfloor Y(x)\rfloor_{x^{n-1}}$ means to keep the Taylor
series up to the order of $x^{(n-1)}$. Thus $F(x)$ can be written compactly as

$\displaystyle F(x)$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}f_{n}x^{n}=\sum_{n=0}^{\infty}x^{n}\Xi[n]\left\\{f_{0}+\lfloor
H_{A,B,C}(x=1)\rfloor_{x^{n-1}}\right\\}~{}\,.~{}~{}~{}~{}~{}$ (173)

For the special case $H(x)=0$, it is easy to see that

$\displaystyle
F(x)=f_{0}F_{0},~{}~{}~{}~{}~{}~{}F_{0}(x)\equiv\sum_{n=0}^{\infty}{(-C)^{n}x^{n}\over
n!A^{n}\left({B\over A}\right)_{n}}~{}\,.~{}~{}~{}~{}~{}$ (174)

The expression (174) is nothing, but the special case of generalized
hypergeometric function (see (C.2) of [11]), which is defined as

$\displaystyle~{}_{A}F_{B}(a_{1},...,a_{A};b_{1},...,b_{B};x)=\sum_{n=0}^{\infty}{(a_{1})_{n}...(a_{A})_{n}\over(b_{1})_{n}...(b_{B})_{n}}{x^{n}\over
n!}~{}\,,~{}~{}~{}$ (175)

thus we have

$\displaystyle
F(x)=f_{0}F_{0}(x),~{}~{}~{}~{}~{}~{}F_{0}=~{}_{0}F_{1}(\emptyset;{B\over
A};{-Cx\over A})~{}\,.~{}~{}~{}~{}~{}$ (176)

Solution in (170) is given in the series expansion. We can also write it in
the analytic expression. Writing $F(x)=F_{0}(x)F_{1}(x)$ we can find the
differential equation of $F_{1}(x)$ as

$\displaystyle H(x)=F_{0}(x)\left(Ax{d^{2}\over dx^{2}}+(B+2Ax{dF_{0}(x)\over
dx}){d\over dx}\right)F_{1}(x)~{}\,,~{}~{}~{}~{}~{}$ (177)

which is the first order differential equation of ${dF_{1}(x)\over dx}=U(x)$.
Using the similar method as for the differential equation (148) we can solve

$\displaystyle U(x)=x^{-{B\over
A}}e^{-2F_{0}(x)}\left({\alpha}_{1}+\int_{0}^{x}dw{1\over
A}H(w)F^{-1}_{0}(w)e^{2F_{0}(w)}w^{{B\over A}-1}\right)~{}\,,~{}~{}~{}~{}~{}$
(178)

thus we obtain

$\displaystyle F(x)$ $\displaystyle=$ $\displaystyle
F_{0}(x)\left({\alpha}_{2}+\int_{0}^{x}dww^{-{B\over
A}}e^{-2F_{0}(w)}\left({\alpha}_{1}+\int_{0}^{w}d\xi{1\over
A}H(\xi)F^{-1}_{0}(\xi)e^{2F_{0}(\xi)}\xi^{{B\over
A}-1}\right)\right)~{}~{}~{}~{}~{}~{}$ (179)

where the ${\alpha}_{1},{\alpha}_{2}$ can be determined using the initial
condition of $F(x=0)$ and ${dF(x=0)\over dx}$. Using the expansion of $F(x)$
we see that ${\alpha}_{2}=f_{0}$ and ${\alpha}_{1}=0$ thus we have

$\displaystyle\boxed{F(x)=F_{0}(x)\left(f_{0}+\int_{0}^{x}dww^{-{B\over
A}}e^{-2F_{0}(w)}\int_{0}^{w}d\xi{1\over
A}H(\xi)F^{-1}_{0}(\xi)e^{2F_{0}(\xi)}\xi^{{B\over
A}-1}\right)}~{}\,.~{}~{}~{}~{}~{}$ (180)

The analytic form (180) is very compact, but it is hard to carry out the
integration in general. However, we can use it to get series expansion just
like before. One can see that

$\displaystyle{dF(x)\over dx}$ $\displaystyle=$ $\displaystyle{dF_{0}(x)\over
dx}\left(f_{0}+\int_{0}^{x}dww^{-{B\over
A}}e^{-2F_{0}(w)}\int_{0}^{w}dt{1\over
A}H(t)F_{0}(t)^{-1}e^{2F_{0}(t)}t^{{B\over A}-1}\right)$ (181)
$\displaystyle+F_{0}(x)x^{-{B\over A}}e^{-2F_{0}(x)}\int_{0}^{x}dt{1\over
A}H(t)F_{0}(t)^{-1}e^{2F_{0}(t)}t^{{B\over A}-1}~{}~{}~{}~{}~{}$

When taking the value at the $x=0$, we need to be careful for the second line
and evaluate it as

$\displaystyle\int_{0}^{x}dt{1\over A}H(t)F_{0}(t)^{-1}e^{2F_{0}(t)}t^{{B\over
A}-1}\to{1\over A}H(x=0)F_{0}^{-1}(x=0)e^{2F_{0}(x=0)}\int_{0}^{x}dtt^{{B\over
A}-1}$ (182) $\displaystyle=$ $\displaystyle{1\over
A}H(x=0)F_{0}^{-1}(x=0)e^{2F_{0}(x=0)}{A\over B}x^{B\over
A}|_{0}^{x}=H(x=0)F_{0}^{-1}(x=0)e^{2F_{0}(x=0)}{1\over B}x^{B\over
A}~{}~{}~{}~{}~{}$

With the result (182), (181) gives

$\displaystyle{dF(x)\over dx}$ $\displaystyle=$ $\displaystyle
f_{0}{dF_{0}(x)\over dx}|_{x=0}+{1\over B}H(x=0)$ (183)

From (161) we have

$\displaystyle{dF\over dx}|_{x=0}={1\over B}H(x=0)-{C\over B}F(x=0)$ (184)

thus we see that

$\displaystyle f_{0}=-{C\over B}F(x=0)\left({dF_{0}(x)\over
dx}|_{x=0}\right)^{-1}$ (185)

For general ${d^{n}F(x)\over dx^{n}}$ we do same thing to get the series
expansion of $x$. At each step, we use (182) and there is no integration to be
done for $x=0$.

##  Appendix B The explicit solutions of $c_{n,m}$ for bubble reduction

In this part, we will show how to get explicit solutions for the recursion
relations (38), (39), (40) and (41).

###  B.1 The generation function $c_{2\to 2}$

For this case, we have $h_{T}=0$, i.e., all $h_{n,m}=0$ in (38), (39), (40)
and (41). Using (38) it is easy to find

$\displaystyle c_{N,0}$ $\displaystyle=$
$\displaystyle\prod_{n=1}^{N}{(-f^{2}+4K^{2}M_{0}^{2})t^{2}\over
2K^{2}n(D+2n-3)}=\prod_{n=0}^{N-1}{K^{2}({\beta}-{\alpha}^{2})\over
2(n+1)(D+2n-1)}~{}~{}~{}~{}$ (186)

where we have used the initial condition $c_{0,0}=1$ and defined

$\displaystyle{\alpha}={tf\over
K^{2}},~{}~{}~{}~{}{\beta}={4t^{2}M_{0}^{2}\over K^{2}}\,.~{}~{}~{}~{}$ (187)

Using (40) to compute first few $c_{n,m}$, one can see the pattern

$\displaystyle c_{N,m}={1\over m!}d_{N,m}c_{N,0}~{}~{}~{}~{}$ (188)

where $d_{N,m}$ depends on both $N,m$ and first few $d_{N,m}$ are

$\displaystyle
d_{N,0}=1,~{}d_{N,1}={\alpha},~{}~{}~{}d_{N,2}={(D+2N){\alpha}^{2}-{\beta}\over(D+2N-1)}\,.~{}~{}~{}$
(189)

Putting the form (188) to (40) we get the recursion relation

$\displaystyle d_{N,m+1}$ $\displaystyle=$
$\displaystyle{{\beta}\over{\alpha}}d_{N,m}-{({\beta}-{\alpha}^{2})(D+2N+m-1)\over{\alpha}(D+2N-1)}d_{N+1,m}~{}~{}~{}~{}$
(190)

Using the formally defined operator $\widehat{P}$ such that

$\displaystyle\widehat{P}f(N,m)=f(N+1,m)$ (191)

the solution of (190) can be formally given by

$\displaystyle d_{N,M}$ $\displaystyle=$
$\displaystyle\prod_{m=1}^{M}\left({{\beta}\over{\alpha}}-{({\beta}-{\alpha}^{2})(D+2N+m-2)\over{\alpha}(D+2N-1)}\widehat{P}\right),~{}~{}~{}M\geq
1,$ (192) $\displaystyle:=$
$\displaystyle\left({{\beta}\over{\alpha}}-{({\beta}-{\alpha}^{2})(D+2N+M-2)\over{\alpha}(D+2N-1)}\widehat{P}\right)\left({{\beta}\over{\alpha}}-{({\beta}-{\alpha}^{2})(D+2N+(M-1)-2)\over{\alpha}(D+2N-1)}\widehat{P}\right)$
$\displaystyle\times......\left({{\beta}\over{\alpha}}-{({\beta}-{\alpha}^{2})(D+2N+2-2)\over{\alpha}(D+2N-1)}\widehat{P}\right)\left({{\beta}\over{\alpha}}-{({\beta}-{\alpha}^{2})(D+2N+1-2)\over{\alpha}(D+2N-1)}\widehat{P}\right)d_{N,0}~{}~{}~{}~{}$

where since the appearance of the operator $\widehat{P}$, the ordering of the
multiple factors is given explicitly. With a little computation, one can see
that

$\displaystyle d_{N,M}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{M}{(-)^{i}{\beta}^{M-i}({\beta}-{\alpha}^{2})^{i}\over{\alpha}^{M}}d_{N,M;i}~{}~{}~{}~{}$
(193)

where

$\displaystyle d_{N,M;i}$ $\displaystyle=$
$\displaystyle\sum_{1;[t_{1},...,t_{i}]}^{M}~{}\prod_{s=1}^{i}{D+2(N+s-1)+(M+1-t_{s})-2\over
D+2(N+s-1)-1},~{}~{}~{}i\geq 1;~{}~{}d_{N,M;0}=1~{}~{}~{}~{}$ (194)

and the summation sign is defined as

$\displaystyle\sum_{1\leq t_{1}<t_{2}...<t_{i}\leq
M}:=\sum_{1;[t_{1},...,t_{i}]}^{M}~{}~{}~{}~{}$ (195)

Combining (188) with (186) and (193), we have the explicit solution for the
generation function of $c_{2\to 2}(t,r,p,K^{2};M_{0},M_{1})$.

####  B.1.1 The generation function $c_{2\to 1;\widehat{1}}$

For this case, we have $h_{T}=c_{1\to 1}(t,r,M_{0})$. Using (7) to (38) one
can write down

$\displaystyle c_{n+1,0}=\gamma(n)c_{n,0}+\rho(n)~{}~{}~{}$ (196)

with

$\displaystyle\gamma(n)={K^{2}({\beta}-{\alpha}^{2})\over
2(n+1)(D+2n-1)},~{}~{}~{}~{}\rho(n)={t^{2n+1}(M_{0}^{2})^{n}{\alpha}\over
2(D+2n-1)(n+1)!\left({D\over 2}\right)_{n}}~{}~{}~{}$ (197)

thus we can solve

$\displaystyle c_{n+1,0}$ $\displaystyle=$ $\displaystyle
c_{0,0}\prod_{i=0}^{n}\gamma(i)+\sum_{i=0}^{n}\rho(i)\prod_{j=i+1}^{n}\gamma(j)~{}~{}~{}$
(198)

where for current case, the initial condition is $c_{0,0}=0$. To find
$c_{n,m}$ we write

$\displaystyle c_{N,m}$ $\displaystyle=$ $\displaystyle{1\over
m!}d_{N,m}c_{N,0}-{1\over m!}{t{\beta}^{N}(K^{2})^{N-1}\over
N!4^{N}\left({D\over 2}\right)_{N}}b_{N,m}~{}~{}~{}$ (199)

with the first few $d_{N,m}$ and $b_{N,m}$

$\displaystyle d_{N,0}$ $\displaystyle=$ $\displaystyle
1,~{}~{}~{}~{}~{}~{}~{}b_{N,0}=0$ $\displaystyle d_{N,1}$ $\displaystyle=$
$\displaystyle{\alpha},~{}~{}~{}~{}~{}~{}~{}b_{N,1}=1$ $\displaystyle d_{N,2}$
$\displaystyle=$
$\displaystyle{-{\beta}+{\alpha}^{2}(D+2N)\over(D+2N-1)},~{}~{}~{}~{}~{}~{}~{}~{}~{}b_{N,2}={{\alpha}(D+2N)\over(D+2N-1)}~{}~{}~{}$
(200)

Using the form (199) to (40) we get the recursion relations

$\displaystyle d_{N,m+1}$ $\displaystyle=$
$\displaystyle{{\beta}\over{\alpha}}d_{N,m}-{(D+2N+m-1)({\beta}-{\alpha}^{2})\over{\alpha}(D+2N-1)}d_{N+1,m}~{}~{}~{}$
(201) $\displaystyle b_{N,m+1}$ $\displaystyle=$
$\displaystyle{{\beta}\over{\alpha}}b_{N,m}-{{\beta}(D+2N+m-1)\over{\alpha}(D+2N)}b_{N+1,m}+{(D+2N+m-1)\over(D+2N-1)}d_{N+1,m}~{}~{}~{}$
(202)

The solution of $d_{N,m}$ can be similarly solved using the operator language
as in previous subsubsection and we find

$\displaystyle d_{N,M}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{M}{(-)^{i}{\beta}^{M-i}({\beta}-{\alpha}^{2})^{i}\over{\alpha}^{M}}d_{N,M;i}~{}~{}~{}$
(203)

with

$\displaystyle d_{N,M;i}$ $\displaystyle=$
$\displaystyle\sum_{1;[t_{1},...,t_{i}]}^{M}~{}\prod_{s=1}^{i}{D+2(N+s-1)+(M+1-t_{s})-2\over
D+2(N+s-1)-1},~{}~{}~{}i\geq 1;~{}~{}d_{N,M;0}=1~{}~{}~{}$ (204)

The solution for $b_{N,m}$ is a little bit complicated because the third term
at the right hand side of (202). To solve it, we write

$\displaystyle b_{N,m+1}$ $\displaystyle=$
$\displaystyle\widetilde{\gamma}_{N}(m)b_{N,m}+\widetilde{\rho}_{N}(m)~{}~{}~{}$
(205)

where

$\displaystyle\widetilde{\gamma}_{N}(m)=\left({{\beta}\over{\alpha}}-{{\beta}(D+2N+m-1)\over{\alpha}(D+2N)}\widehat{P}\right),~{}~{}~{}~{}\widetilde{\rho}_{N}(m)={(D+2N+m-1)\over(D+2N-1)}d_{N+1,m}~{}~{}~{}$
(206)

Iterating (205) with proper ordering we have

$\displaystyle b_{N,m+1}$ $\displaystyle=$
$\displaystyle\widetilde{\gamma}_{N}(m)\widetilde{\gamma}_{N}(m-1)...\widetilde{\gamma}_{N}(0)b_{N,0}+\widetilde{\gamma}_{N}(m)\widetilde{\gamma}_{N}(m-1)...\widetilde{\gamma}_{N}(1)\widetilde{\rho}_{N}(0)$
(207)
$\displaystyle+\widetilde{\gamma}_{N}(m)\widetilde{\gamma}_{N}(m-1)...\widetilde{\gamma}_{N}(2)\widetilde{\rho}_{N}(1)+...+\widetilde{\gamma}_{N}(m)\widetilde{\rho}_{N}(m-1)+\widetilde{\rho}_{N}(m)~{}~{}~{}$

Using

$\displaystyle(\widehat{P})^{a}\widetilde{\rho}_{N}(m)={(D+2(N+a)+m-1)\over(D+2(N+a)-1)}d_{N+1+a,m}~{}~{}~{}$
(208)

and

$\displaystyle\widetilde{\gamma}_{N}(m)\widetilde{\gamma}_{N}(m-1)...\widetilde{\gamma}_{N}(m-k)\widetilde{\rho}_{N}(m-k-1)$
(209) $\displaystyle=$
$\displaystyle\sum_{i=0}^{k+1}(-)^{i}\left({{\beta}\over{\alpha}}\right)^{k+1}\left(\sum_{1;[t_{1},...,t_{i}]}^{k+1}\prod_{s=1}^{i}{D+2(N+s-1)+(m+1-t_{s})-1\over
D+2(N+s-1)}\right)$
$\displaystyle\times{(D+2(N+i)+(m-k-1)-1)\over(D+2(N+i)-1)}d_{N+1+i,m-k-1}~{}~{}~{}$

we finally reach

$\displaystyle b_{N,m+1}$ $\displaystyle=$
$\displaystyle{(D+2N+m-1)\over(D+2N-1)}d_{N+1,m}+\sum_{k=0}^{m-1}\left({{\beta}\over{\alpha}}\right)^{k+1}\sum_{i=0}^{k+1}(-)^{i}$
(210)
$\displaystyle\times{(D+2(N+i)+(m-k-1)-1)\over(D+2(N+i)-1)}d_{N+1+i,m-k-1}$
$\displaystyle\times\left(\sum_{1;[t_{1},...,t_{i}]}^{k+1}\prod_{s=1}^{i}{D+2(N+s-1)+(m+1-t_{s})-1\over
D+2(N+s-1)}\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}$

where the condition $b_{N,0}=0$ has been used. The formula (199) plus (198),
(203) and (210) gives the explicit solution for generation function $c_{2\to
1;\widehat{1}}$.

Knowing $c_{2\to 1;\widehat{1}}$, we can use (13) to get the generation
function $c_{2\to 1;\widehat{0}}$ or directly compute it using (38), (39),
(40) and (41).

####  B.1.2 The proof of one useful relation

When we use the improved PV-reduction method with auxiliary vector $R$ to
discuss the reduction of sunset topology, an important reduction relation
between different tensor ranks has been observed in [38]. Later this relation
has been studied in [39, 40, 42]. For bubble it is given explicitly by

$\displaystyle I_{2}^{(r)}$ $\displaystyle=$
$\displaystyle{(D+2r-4)fp\over(D+r-3)K^{2}}I_{2}^{(r-1)}-{(r-1)(4M_{0}^{2}p^{2}+(f^{2}-4M_{0}^{2}K^{2})r)\over(D+r-3)K^{2}}I_{2}^{(r-2)}$
(211) $\displaystyle+{p\over
K^{2}}I_{2;\widehat{0}}^{(r-1)}+{(r-1)((K^{2}-M_{0}^{2}+M_{1}^{2})r-2p^{2})\over(D+r-3)K^{2}}I_{2;\widehat{0}}^{(r-2)}+{-p\over
K^{2}}I_{2;\widehat{1}}^{(r-1)}+{(r-1)fr\over(D+r-3)K^{2}}I_{2;\widehat{1}}^{(r-2)}~{}~{}~{}~{}$

where

$\displaystyle I_{2}^{(r)}=\int d\ell{(2\ell\cdot
R)^{r}\over(\ell^{2}-M_{0}^{2})((\ell-K)^{2}-M_{1}^{2})},~{}~{}~{}I_{2;\widehat{1}}^{(r)}=\int
d\ell{(2\ell\cdot
R)^{r}\over(\ell^{2}-M_{0}^{2})},~{}~{}~{}I_{2;\widehat{0}}^{(r)}=\int
d\ell{(2\ell\cdot R)^{r}\over(\ell-K)^{2}-M_{1}^{2}}~{}~{}~{}~{}$ (212)

We can use the series form to check the relation (211). Let us define

$\displaystyle F[N+2]$ $\displaystyle=$ $\displaystyle
c^{(N+2)}-\left({(D+2N)fpc^{(N+1)}-(N+1)(4M_{0}^{2}p^{2}+(f^{2}-4M_{0}^{2}K^{2})r)c^{(N)}\over(D+N-1)K^{2}}\right)~{}~{}~{}$
(213)

where

$\displaystyle c^{(N)}={N!\over
t^{N}}\sum_{n=0}^{[N/2]}c_{n,N-2n}r^{n}p^{N-2n}~{}~{}~{}$ (214)

and $c$ can be $c_{2\to 2}$ or $c_{2\to 1}$. Depending on if $N$ is even or
odd, the computation details have some differences, thus we consider $N$ even
only, while $N$ odd will be similar. Expanding (213) we have

$\displaystyle F[2N+2]$ $\displaystyle=$ $\displaystyle{(2N+2)!\over
t^{2N+2}}r^{0}p^{2N+2}\left(c_{0,2N+2}-\left({(D+4N)tfc_{0,2N+1}\over(2N+2)(D+2N-1)K^{2}}-{4M_{0}^{2}t^{2}c_{0,2N}\over
2(N+1)(D+2N-1)K^{2}}\right)\right)$ (215) $\displaystyle+{(2N+2)!\over
t^{2N+2}}r^{N+1}p^{0}\left(c_{N+1,0}-\left(-{t^{2}(f^{2}-4M_{0}^{2}K^{2})\over(2N+2)(D+2N-1)K^{2}}c_{N,0}\right)\right)$
$\displaystyle+{(2N+2)!\over
t^{2N+2}}\sum_{n=1}^{N}r^{n}p^{2N+2-2n}\left(c_{n,2N+2-2n}-\left({(D+4N)tf\over(2N+2)(D+2N-1)K^{2}}c_{n,2N+1-2n}\right.\right.$
$\displaystyle\left.\left.-{4M_{0}^{2}t^{2}\over(2N+2)(D+2N-1)K^{2}}c_{n,2N-2n}-{t^{2}(f^{2}-4M_{0}^{2}K^{2})\over(2N+2)(D+2N-1)K^{2}}c_{n-1,2N+2-2n}\right)\right)~{}~{}~{}$

For the term with $r^{0}p^{2N+2}$, using the relation (37) with $n=0,m=2N$, we
find the part inside the bracket is simplified to

$\displaystyle{-4\xi_{R}t^{2}h_{0,2N}+\xi_{K}t(D+4N)h_{0,2N+1}\over
2(N+1)(D+2N-1)K^{2}}~{}~{}~{}$ (216)

For the term with $r^{N+1}p^{0}$, using the relation (38) with $n=N$, we find
the part inside the bracket is simplified to

$\displaystyle-{t^{2}(\xi_{K}f-\xi_{R}4K^{2})h_{N,0}+\xi_{K}tK^{2}h_{N,1}\over
2(N+1)(D+2N-1)K^{2}}~{}~{}~{}$ (217)

For the term with $r^{n}p^{2N+2-2n}$, the computation is a little bit
complicated. First we use (36) with $n\to n-1$ and $m\to
2N-2n+2,2N-2n+1,2N-2n$ to write all $c_{i,j}$ with $i=n-1$. Then we use (37)
with $n\to n-1$ and $m\to 2N-2n+1,2N-2n$. After doing above two steps and
making algebraic simplification, we get

$\displaystyle{t^{3}(\xi_{K}M_{0}^{2}-\xi_{R}f)\over(N+1)(D+2N-1)nK^{2}}h_{n-1,2N-2n+1}+{t^{2}(-\xi_{K}nf+\xi_{R}4K^{2}(N+1))\over
2(N+1)(D+2N-1)nK^{2}}h_{n-1,2N-2n+2}$ $\displaystyle+{-\xi_{K}(2N-2n+3)t\over
2n(D+2N-1)}h_{n-1,2N-2n+3}~{}~{}~{}$ (218)

For different $c$ we use the different known $h_{T}$ as given in (25), (26)
and (27), thus one can see the relation (211) is satisfied.

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kern[1]#1@kerna

Temporal Properties of Enumerative Shaping: Autocorrelation and Energy
Dispersion Index

Yunus Can Gültekin, Kaiquan Wu, and Alex Alvarado

Department of Electrical Engineering, Eindhoven University of Technology, The
Netherlands.

<EMAIL_ADDRESS>
###### Abstract

We study the effective SNR behavior of various enumerative amplitude shaping
algorithms. We show that their relative behavior can be explained via the
temporal autocorrelation function or via the energy dispersion index.

##  1 Introduction

Probabilistic amplitude shaping (PAS) has become an indispensable technique
for optical communication systems in recent years [2, 3]. PAS provides signal-
to-noise ratio (SNR) gains more than 1 dB over uniform quadrature amplitude
modulation (QAM) constellations and enables rate adaptivity. Thanks to these
properties, PAS has been numerically and experimentally demonstrated to yield
more than 30% reach increase for a wide range of transmission rates [4, 5, 6].
To secure these gains with low complexity, recently-developed enumerative
coding algorithms such as enumerative sphere shaping (ESS), kurtosis-limited
ESS (K-ESS), and band-trellis ESS (B-ESS) have been widely investigated as
efficient amplitude shaping techniques [7, 8, 9, 10, 11]. It has been observed
that they offer different effective SNRs. Moreover, ESS and K-ESS result in
SNR penalties against uniform signaling, while B-ESS provides SNR gains. In
this work, we seek to understand why this is the case. We study enumerative
shaping techniques via two metrics that characterize temporal properties of
channel inputs: auto-correlation function (ACF) of [12, Ch. 6] and recently-
introduced energy dispersion index (EDI) of [13, Defn. 5]. We show that B-ESS
is the most efficient in both ACF- and EDI-sense, and this is the reason why
it provides the largest SNR.

##  2 System Model: Probabilistic Amplitude Shaping

PAS combines an amplitude shaper with a channel code. Here, we focus on the
shaping blocks implemented via enumerative coding techniques: ESS [6], K-ESS
[9], and B-ESS [10]. With ESS, amplitude $N$-sequences
$\underline{a}=(a_{1},a_{2},\dotsc,a_{N})$ that satisfy a maximum-energy
constraint $\sum_{k=1}^{N}|a_{k}|^{2}<E_{\max}$ are generated. With K-ESS,
amplitude sequences also satisfy a maximum-kurtosis constraint
$\sum_{k=1}^{N}|a_{k}|^{4}<K_{\max}$. Finally, with B-ESS, sequences have
limited windowed energy variations. Then the amplitude sequences are converted
into a channel input sequence $\underline{x}$ of QAM symbols. At the receiver,
the channel output $\underline{y}$ is observed. We want to find insights into
the memory-related effects of the channel on $\underline{x}$. In what follows,
we discuss two metrics to study these effects.

In [12], the temporal correlations are described via the ACF
$R_{\theta}[\tau]$ of the moving-window average angle $\theta_{k}$ of past $w$
symbols, see (1) and (2), resp. Here, $w$ is the moving average window length,
$\angle z$ is the phase of $z$, and $R_{\theta}[\tau]$ is normalized to have
$R_{\theta}[0]=1$. The correlations $R_{\theta}[\tau]$ for $\tau\leq w$ are
introduced by the moving average in (2). After $w$ symbols, the correlations
$R_{\theta}[\tau]$ give insight about the effective memory observed in the
channel.

In [13], the time-varying characteristics of the energy in the transmitted
signal are quantified via the EDI $\Psi$ in (3). EDI is the normalized average
of the variations in windowed energy $G^{W}$ in (4). Here, $W$ is the window
length. It has been shown in the literature that there are strong correlations
between EDI and effective SNR [13, 14]. The optimum window length $W$—in the
sense that the absolute value of the correlation between EDI and SNR is
maximized—depends on the length of the link and gives an insight into the
memory of the channel [13, Fig. 9].

$R_{\theta}[\tau]=\frac{E[\theta_{k}\theta_{k-\tau}]}{n-\tau}.\hskip
4.26773pt(1)\hskip
11.38092pt\theta_{k}=\angle\sum_{m=-(w-1)/2}^{(w-1)/2}\boldsymbol{x}^{*}_{k-m}\boldsymbol{y}_{k-m}.\hskip
4.26773pt(2)\hskip
11.38092pt\Psi=\frac{\macc@depth\@nebgroup\@emptyegroup\macc@set@skewchar\z@\frozen@everymath{\macc@group}\macc@set@skewchar@nested@a\macc@nested@a111{\text{Var}}(G^{W})}{\macc@depth\@nebgroup\@emptyegroup\macc@set@skewchar\z@\frozen@everymath{\macc@group}\macc@set@skewchar@nested@a\macc@nested@a111{E}(G^{W})}.\hskip
4.26773pt(3)\hskip
11.38092ptG^{W}_{k}=\sum_{i=k-W/2}^{k+W/2}|x_{i}|^{2}.\hskip 5.69046pt(4)$

In the next section, we will provide numerical results showing that these two
metrics, ACF in (1) and EDI in (3), can be used to predict the effective SNRs
of shaped signaling schemes.

##  3 Results

We simulate the transmission of a single-polarized single-channel waveform
over 1 and 4 spans of 80 km fiber with an attenuation of 0.19 dB/km, a
dispersion of 17 ps/nm/km, and a nonlinear (NL) parameter of 1.3 1/W/km using
the split-step Fourier method (SSFM). A root-raised-cosine pulse with 10%
roll-off is used. The symbol rate is 56 GBd. The fiber is followed by an
erbium-doped fiber amplifier with a noise figure of 5.5 dB. These parameters
are compatible with the ones used in [13]. At the receiver, first, the
chromatic dispersion, then the phase rotation induced by the NL are ideally
compensated. PAS is realized based on ESS, K-ESS, and B-ESS with the
parameters used to generate [10, Fig. 3] with 64-QAM. Shaping blocklength is
$N=108$ amplitudes. The shaping rate is 1.5 bit/1D, which leads to a net
information rate of 8 bit/4D when combined with a rate-5/6 channel code.
However here, we do not realize channel coding and select the signs for the
amplitudes uniformly at random.

Fig. 1: Left: Effective SNR after 1 span, inset: after 4 spans. Middle: ACF
with $w=5$. Right: EDI.

Fig. 1 (left) shows the effective SNR vs. the launch power $P$ after 1 and 4
spans of transmission. In Fig. 1 (middle), we show the ACF $R_{\theta}[\tau]$
as a function of $\tau$ for $w=5$. This ACF is computed for input-output pairs
belonging to the 1-span scenario at the corresponding optimum launch powers.
We see that ESS and K-ESS result in ACFs very similar to uniform signaling
(except at the tail ends). On the other hand, B-ESS has an ACF function which
vanishes more quickly. This implies that when input sequences generated via
B-ESS are transmitted, the observed effective memory is shorter than it is for
ESS, K-ESS, and uniform signaling. We believe this explains (at least
partially) the significantly-improved end-to-end decoding performance provided
by B-ESS [10, Figs. 3-4] based on typical mismatched bit-metric decoding in
which the memory in the channel is neglected.

In Fig. 1 (right), we show the EDI of considered schemes as a function of the
moving average length $W$. We also indicate the optimum window lengths
$W^{*}=30$ and $150$ for the considered setup after 1 and 4 spans of
transmission [13, Figs. 9a-b], resp. The relative behaviors of EDIs at $W=30$
and $150$ are the same as that of SNRs in Fig. 1 (left): a smaller EDI results
in a larger SNR. Since B-ESS was designed specifically to limit the variations
in channel input signal energy, it provides the smallest EDI for most of the
$W$ region (especially around $W=N$). This can also be used to explain the
significant SNR improvement provided by B-ESS in Fig. 1 (left). Moreover, this
correlation between EDI and SNR can be exploited to design B-ESS for different
link setups without resorting to computationally-expensive SSFM-based
simulations.

Finally, we have checked numerically and discovered that K-ESS has smaller
average energy, smaller average energy variance, and smaller kurtosis than
B-ESS, i.e., it is better in terms of average statistics which neglect
temporal characteristics. This implies that these statistics are insufficient
to capture the properties of shaping schemes that influence the resulting NL
interference and hence, the effective SNR. EDI, on the other hand, focuses on
temporal characteristics, evaluates windowed energy variations, and
qualitatively predicts the SNR accurately.

##  4 Conclusions

We have shown via fiber simulations that the relative effective SNR behaviors
of uniform vs. enumeratively shaped signaling (via ESS, K-ESS, and B-ESS) can
be explained via arguments based on autocorrelation function (ACF) and energy
dispersion index (EDI). B-ESS—limiting windowed energy variations in input
sequences—has the most quickly-vanishing ACF and the smallest EDI and hence,
provides the largest SNR.

_
The work of Y.C. Gültekin, K. Wu, and A. Alvarado has received funding from
the NWO via the VIDI grant ICONIC (grant ID: 15685), and from the ERC under
the EU’s Horizon 2020 research and innovation programme via the Starting grant
FUN-NOTCH (grant ID: 757791) and via the Proof of Concept grant SHY-FEC (grant
ID: 963945)._

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