\section{Introduction
\noindent
Future experimental programs at
the High-Luminosity Large Hadron
Collider (HL-LHC)~\cite{ATLAS:2013hta,
CMS:2013xfa} and the International
Linear Collider (ILC)~\cite{Baer:2013cma}
aim to measure precisely the properties
of Higgs boson, top quark and vector
bosons for discovering the nature of
the Higgs sector as well as for finding
the effects of physics beyond the standard
model. In order to match the high-precision
of experimental data in the near future,
theoretical predictions including
high-order corrections are required.
In this framework, detailed evaluations
of one-loop multi-leg and higher-loop at
general scale and mass assignments
are necessary.

One-loop Feynman integrals in general
space-time dimension play a crucial
role for several reasons. Within the
general framework for computing
two-loop or higher-loop corrections,
higher-terms in
the $\varepsilon$-expansion
(with $\varepsilon =2-d/2$)
from one-loop integrals are necessary for
building blocks. Moreover, one-loop
integrals in $d>4$ may be taken into
account in a reduction for tensor one-loop
\cite{Davydychev:1991va, Fleischer:2010sq},
two-loop and higher-loop integrals~\cite{IBP}.
There have been available many calculations
for scalar one-loop functions in general
dimension $d$~\cite{Boos:1990rg,
Davydychev:1990cq,Davydychev:1997wa,
Anastasiou:1999ui,Suzuki:2003jn,Abreu:2015zaa,
Phan:2017xsj}.
However, not all of the calculations
cover general $\varepsilon$-expansion at general
scale and internal mass assignments.
Furthermore, a recurrence relation
in $d$ for Feynman loop integrals has been proposed
\cite{Tarasov:1996br} and solved for scalar one-loop
integrals which have been expressed in terms of
generalized hypergeometric series
\cite{Fleischer:2003rm}.
However, the general solutions for arbitrary
kinematics have not been found,
as pointed out in \cite{Bluemlein:2015sia}.
More recently, scalar 1-loop Feynman
integrals as meromorphic functions in general
space-time dimension, for arbitrary kinematics
has been presented
in~\cite{Bluemlein:2017rbi,Phan:2018cnz}.
In the present paper, new
analytic formulas for one-loop three-point
Feynman integrals in general space-time
dimension are reported by following
an alternative approach.
The analytic results are expressed
in terms of $_2F_1$, $_3F_2$ and
Appell $F_1$ hypergeometric
functions. The evaluations are performed
in general configuration of internal masses
and external momenta. Last but not least,
our results are cross-checked to other papers
which have been available in several
special cases.

The layout of the paper is as follows: In section $2$,
we present in detail the method for evaluating scalar
one-loop three-point functions. In this section,
we first introduce notations used for this work.
We next consider the case of one-loop triangle diagram
with two light-like external momenta and
generalize the calculations for general case.
Finally, tensor one-loop three-point integrals are
discussed. Conclusions and outlooks are devoted in
section $3$. Several useful formulas applied
in this calculation can be found in the appendixes.
 \section{Analytic formulas
Detailed evaluations for one-loop
three-point integrals
are presented in this section.
\subsection{Definitions
We arrive at notations for
the calculations in this subsection.
Feynman integrals of scalar
one-loop three-point functions
are defined:
\begin{eqnarray}
\label{feynmanintegral}
&&\hspace{-1.5cm}
J_3(d; \{\nu_1,\nu_2,\nu_3\})
\equiv J_3(d; \{\nu_1,\nu_2,\nu_3\}
;p_1^2,p_2^2,p_3^2;
m_1^2,m_2^2,m_3^2) =\\
&=& \int \frac{d^d k}{i\pi^{d/2}}
\dfrac{1}{[(k+q_1)^2 -m_2^2 + i\rho]^{\nu_1}
[(k+q_2)^2 -m_3^2 + i\rho]^{\nu_2}
[(k+q_3)^2 -m_1^2 + i\rho]^{\nu_3}}. \nonumber
\end{eqnarray}
We refer hereafter
$J_3 \equiv J_3(d; \{1,1,1\})$.
In this definition,
the term $i\rho$ is Feynman's
prescription and $d$ is space-time dimension.
The internal (loop) momentum is $k$ and the
external momenta are $p_1$, $p_2$, $p_3$.
They are inward as described in
Fig.~\ref{j3diagram}. We use the momenta
$q_1 = p_1$, $q_2 =p_1+p_2$ and
$q_3 = p_1+p_2+p_3 =0$ following
momentum conservation.
The internal masses are $m_1$, $m_2$ and
$m_3$. $J_3$ is a function of
$p_1^2$, $p_2^2$, $p_3^2$ and $m_1^2$, $m_2^2$,
$m_3^2$.

It has known that an algebraically compact
expression and numerically stable
representation for Feynman diagrams
can be obtained by using kinematic variables such as
the determinants of Caylay and Gram
matrices~\cite{kajantie}. The expression
also reflects the symmetry of the corresponding
topologies.
\begin{figure}[ht]
\begin{center}
\begin{pspicture}(-3, -3)(3.5, 3)
\psset{linewidth=0.1pt}
\psset{unit = 0.6}
\psline(-2.5,-2)(0,2)
\psline{->}(-2.5, -2)(-1.25, 0)
\psline(0,2)(2.5,-2)
\psline{->}(0, 2)(1.25, 0)
\psline(-2.5,-2)(2.5,-2)
\psline{->}(2.5, -2)(0, -2)
\psline(-2.5,-2)(-5,-4)
\psline{->}(-5, -4)(-3.8, -3)
\psline(2.5,-2)(5,-4)
\psline{->}(5,-4)(3.8,-3)
\psline(0,2)(0,5)
\psline{->}(0,5)(0,3)
\rput(0.8, 3.5){$p_2^2$}
\rput(-4.2,-2.5){$ p_1^2$}
\rput(4.2, -2.5){$ p_3^2$}
\rput(0, -2.8){$m_3^2$}
\rput(-2.5,0){$m_1^2$}
\rput(2.5, 0){$ m_2^2$}
\psset{dotsize=3pt}
\psdots(-2.5, -2)(2.5, -2)(0, 2)
\end{pspicture}
\end{center}
\caption{\label{j3diagram}One-loop
triangle diagrams.}
\end{figure}
We hence review the
mentioned kinematic variables
in the following paragraphs.
The determinant of Cayley
matrix of one-loop
triangle diagrams is given
\begin{eqnarray}
\label{deltan}
S_3 &=&
\left|
\begin{array}{ccc}
2m_1^2& -p_1^2 +m_1^2 +m_2^2 & -p_3^2 +m_1^2 +m_3^2   \\
-p_1^2 +m_1^2 +m_2^2 & 2m_2^2 & -p_2^2 +m_2^2 +m_3^2  \\
-p_3^2 +m_1^2 +m_3^2 & -p_2^2 +m_2^2 +m_3^2 & 2 m_3^2 \\
\end{array}
\right|.
\end{eqnarray}
In the same manner, the Cayley determinants
of one-loop two-point Feynman
diagrams which are obtained by shrinking an
propagator in the three-point integrals.
The determinants are written explicitly as:
\begin{eqnarray}
S_{12}&=&
\left|
\begin{array}{cc}
2m_1^2& -p_1^2 +m_1^2 +m_2^2 \\
-p_1^2 +m_1^2 +m_2^2 & 2m_2^2 \\
\end{array}
         \right|  =
-\lambda(p_1^2, m_1^2,m_2^2),  \\
S_{13}&=& \left|
\begin{array}{cc}
2m_1^2 & -p_3^2 +m_1^2 +m_3^2 \\
-p_3^2 +m_1^2 +m_3^2  & 2 m_3^2 \\
\end{array}
\right|
=-\lambda(p_3^2, m_1^2,m_3^2), \\
S_{23}
&=&\left|
\begin{array}{cc}
 2m_2^2 & -p_2^2 +m_2^2 +m_3^2 \\
 -p_2^2 +m_2^2 +m_3^2 & 2 m_3^2 \\
\end{array}
\right|
=-\lambda(p_2^2, m_2^2,m_3^2).
\end{eqnarray}
Where $ \lambda(x, y,z)
= x^2 +y^2 +z^2 -2 xy -2 xz -2 yz$
is so-called K\"allen function.
Next, the determinant of Gram matrix
of one-loop three-point functions is
given
\begin{eqnarray}
 G_3 = -8
 \left|
\begin{array}{cc}
 p_1^2    & p_1p_2 \\
 p_1p_2   & p_2^2
\end{array}
  \right|
 = 2 \lambda(p_1^2,p_3^2,p_2^2).
\end{eqnarray}
In the same above definitions, we also get
the Gram determinants of two-point functions
as follows
\begin{eqnarray}
 G_{12} = - 4p_1^2, \quad G_{13}
 = - 4p_3^2, \quad G_{23} = -4p_2^2.
\end{eqnarray}
In this work, the analytic formulas
for scalar one-loop three-point integrals
are expressed as functions with
arguments of the ratio of the above kinematic
determinants. Therefore, it is worth to
introduce the following index
variables
\begin{eqnarray}
M_{3} &=&
-\dfrac{S_3}{G_3}, \quad \text{for}
\quad G_3 \neq 0,                                 \\
M_{ij} &=&
-\dfrac{S_{ij}}{G_{ij}}, \quad \text{for}
\quad G_{ij} \neq 0, \quad \text{with}
\quad i,j =1,2,3.
\end{eqnarray}

By introducing Feynman parameters, we then integrate
over the loop-momentum. The resulting integral after
taking over one of Feynman parameters reads
\begin{eqnarray}
\label{feynj3}
\dfrac{J_3}{\Gamma\left (\frac{6-d}{2} \right)}
&=& -
\int \limits_0^1 dx\int \limits_0^{1-x} dy
\dfrac{1}{(Ax^2 + By^2+ 2C xy+ Dx + Ey + F
- i\rho)^{\frac{6-d}{2}} }.
\end{eqnarray}
The corresponding coefficients
$A,B,C,\;\cdots,F$ are shown:
\begin{eqnarray}
A &=& p_1^2,\quad\quad\quad \quad
\quad \quad \quad \quad   D \;=\;-(p_1^2+m_1^2-m_2^2), \\
B &=& p_3^2, \quad   \quad \quad
\quad \quad \quad   \quad
\quad E \;=\; -(p_3^2+m_1^2-m_3^2),\\
C &=& -p_1p_3,\quad   \quad
\quad \quad   \quad    \quad\; F \;=\; m_1^2.
\end{eqnarray}
Detailed calculation for $J_3$ in (\ref{feynj3})
is presented in next subsections.
\subsection{Two light-like momenta
We first consider the simple case which is
two light-like external momenta. Without loss
of the generality, we can take
$p_1^2 =0,~p_3^2 =0$.
The Feynman parameter integral
in Eq.~(\ref{feynj3})
is now casted into the simpler form
\begin{eqnarray}
\label{twolight-like}
J_3
&=&
-\Gamma\left (\frac{6-d}{2} \right)
\int \limits_0^1 dx
\int \limits_0^{1-x}dy
\dfrac{1}
{(2Cxy + Dx + Ey+ F
- i\rho)^{\frac{6-d}{2}} }.
\end{eqnarray}
It finds that the
denominator function of
the integrand in (\ref{twolight-like})
depends linearly on both $x$ and $y$. Hence,
the integral can be taken
easily. The resulting integral
reads after performing the $y$-integration
\begin{eqnarray}
\label{twolight-like1}
\dfrac{J_3 }
{\Gamma\left (\frac{4-d}{2} \right)}
&=&
\int \limits_0^1 dx \;
\dfrac{\left[ (m_2^2-m_1^2)x
+ m_1^2 -i\rho
\right]^{\frac{d-4}{2}}
-
\left[\; p_2^2x^2
-(p_2^2 +m_3^2 -m_2^2)x
+m_3^2 -i\rho
\right]^{\frac{d-4}{2} }
}
{ p_2^2 x - m_3^2 +m_1^2 }.
\nonumber\\
\end{eqnarray}
Both the integrands are singularity at
$x_1 = (m_3^2-m_1^2)/p_2^2$.
However, it is verified that
\begin{eqnarray}
p_2^2x_1^2 -(p_2^2 +m_3^2 -m_2^2)x_1 + m_3^2
= (m_2^2-m_1^2) x_1  + m_1^2 =M_{3}.
\end{eqnarray}
It means that
the residue contributions at this pole
of two integrations in Eq.~(\ref{twolight-like1})
will be cancelled.
As a result, $J_3$ stays finite at this point.
The first integral in Eq.~(\ref{twolight-like1})
can be formulated by mean of Appell $F_1$ functions.
For the second integral in
Eq.~(\ref{twolight-like1}),
we can apply the formula for master integral
as Eq.~(\ref{K2}). The result for $J_3$ then
reads
\begin{eqnarray}
\label{twolightlike1}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2}\right)}
&=& -\dfrac{ (m_1^2)^{\frac{d-4}{2}} }{m_3^2-m_1^2}
\; F_1 \left( 1; 1,\frac{4-d}{2}; 2;
\frac{p_2^2}{m_3^2-m_1^2 },
\frac{m_1^2-m_2^2 +i\rho}{m_1^2} \right)         \\
&&\hspace{-1.8cm} +
\left(\dfrac{\partial_1 S_3}{p_2^2\;G_{23}} \right)
\Bigg[
\left(\dfrac{\partial_2 S_{23}}{G_{23}} \right)
\dfrac{(m_3^2)^{\frac{d-4}{2} } }{M_{3} -m_3^2}
\; F_1\left(1; 1,\frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_3^2}{ M_{3} -m_3^2},
1-\frac{M_{23}}{m_3^2}  \right)\nonumber\\
&&\hspace{0cm} +
\left( \dfrac{\partial_3 S_{23}}{G_{23} } \right)
\dfrac{(m_2^2)^{\frac{d-4}{2} } }{M_{3} -m_2^2}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
 1-\frac{M_{23}}{m_2^2}  \right)   \Bigg] \nonumber \\
&&\hspace{-1.8cm} +
\Bigg[
\dfrac{M_{23} - m_3^2}{2p_2^2(M_{3} -m_3^2)}
\left(m^2_{3} \right)^{\frac{d-4}{2} }
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{23} -m_3^2}{ M_{3} -m_3^2},
1-\frac{M_{23}}{m_3^2}  \right)    \nonumber\\
&&\hspace{-1.3cm}
-  \dfrac{M_{23} - m_2^2}{2p_2^2(M_{3} -m_2^2)}
\left(m^2_{2} \right)^{\frac{d-4}{2}}
\; F_1\left(1; 1,\frac{4-d}{2}; 2;
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{23}}{m_2^2}  \right)   \Bigg].
\nonumber
\end{eqnarray}
Another representation for $J_3$ can be
obtained by using Eq.~(\ref{K1})
\begin{eqnarray}
\label{twolightlike2}
 \dfrac{J_3}{\Gamma\left(\frac{4-d}{2}\right)} &=&
 -\dfrac{ (m_1^2)^{\frac{d-4}{2}} }{m_3^2-m_1^2}
 \; F_1 \left( 1; 1, \frac{4-d}{2}; 2;
 \frac{p_2^2}{m_3^2-m_1^2 },
 \frac{m_1^2-m_2^2 +i\rho}{m_1^2} \right)
 \\
&&\hspace{-1.8cm} +
\left( \dfrac{\partial_1 S_3}{p_2^2\;G_{23}} \right)
\Bigg[
\left( \dfrac{\partial_2 S_{23}}{G_{23}} \right)
\dfrac{(M_{23} -i\rho)^{\frac{d-4}{2}} }{M_{3} -M_{23}}
\; F_1\left(\frac{1}{2}; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_3^2}{ M_{23} -M_{3} },
1-\frac{m_3^2}{M_{23}}  \right)\nonumber\\
&&\hspace{0.1cm}
+ \left( \dfrac{\partial_3 S_{23}}{G_{23} } \right)
\dfrac{(M_{23}-i\rho)^{\frac{d-4}{2}} }{M_{3} -M_{23}}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_2^2}{ M_{23} -M_{3}},
1-\frac{m_2^2}{M_{23}}  \right)  \Bigg] \nonumber \\
&&\hspace{-1.8cm}
+ \Bigg[
\dfrac{ \left(M_{23} - i\rho\right)^{\frac{d-4}{2}} }
{2(M_{3} -M_{23} )}
\; F_1\left(1; 1, \frac{4-d}{2}; 2; \dfrac{M_{23} -m_3^2}
{ M_{3} -M_{23} },
1-\frac{m_3^2}{M_{23}}  \right)
\nonumber\\
&&\hspace{0cm}-
\dfrac{\left(M_{23}-i\rho \right)^{\frac{d}{2} -2} }
{2p_2^2(M_{3} -M_{23} )}
\; F_1\left(1; 1, \frac{4-d}{2}; 2; \dfrac{M_{23} -m_2^2}
{M_{3} -M_{23} },
1-\frac{m_2^2}{M_{23}}  \right)   \Bigg]. \nonumber
\end{eqnarray}
The results are shown in
Eqs.~(\ref{twolightlike1}, \ref{twolightlike2})
are new hypergeometric representations for
scalar one-loop three-point functions in general
space-time dimension. In the next paragraphs,
we consider $J_3$ in several special cases.
\begin{enumerate}
\item $\underline{m_1^2 = m_2^2 = m_3^2=0:}$ \\
One-loop triangle
diagrams with all massless
internal lines are considered.
In this case, the integral $J_3$
in (\ref{twolight-like}) becomes
\begin{eqnarray}
J_3 &=& -\Gamma\left(\frac{6-d}{2} \right)
\int \limits_0^1 dx
\int \limits_0^{1-x} dy
\dfrac{1}{\left(-p_2^2\; x\; y
-i\rho \right)^{\frac{6-d}{2} }}
\\
&=&
\dfrac{\Gamma\left(\frac{4-d}{2}  \right)
\Gamma\left(\frac{d-4}{2}\right)
\Gamma\left(\frac{d-2}{2}\right) }
{\Gamma(d-3)} \left(-p_2^2 -i\rho
\right)^{\frac{d-6}{2}}. \label{case11}
\end{eqnarray}
It is important to note that we use
$-p_2^2 x y -i\rho \equiv (-p_2^2-i\rho ) x y $.
Therefore, for analytic continuation the result
when $p_2^2>0$, we should keep $i\rho$-term
together with external momentum
like $p_2^2 \rightarrow p_2^2 +i\rho$.
The result in (\ref{case11}) shows
full agreement with Eq.~$(4.5)$ in
\cite{Ellis:2007qk}.
\\
\item $\underline{m_2^2 = m_3^2 =0, m_1^2=m^2:}$ \\
If internal mass configuration
takes $m_2^2 = m_3^2 =0, m_1^2=m^2$, $J_3$ becomes
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=&
\int \limits_0^1 dx \;
\dfrac{[m^2(1-x)]^{ \frac{d-4}{2}}}{p_2^2 x + m^2}
-\int \limits_0^1 dx \;
\dfrac{[p^2_2x^2 - p_2^2 x -i\rho]
^{ \frac{d-4}{2} }}{p_2^2 x + m^2} \\
&&\hspace{-2.2cm}
=
\dfrac{\Gamma\left(\frac{d-2}{2}\right)}
{\Gamma\left( \frac{d}{2} \right)}
(m^2)^{ \frac{d-6}{2}}
\Fh21\Fz{1, 1}{\frac{d}{2} }{\dfrac{-p_2^2}{m^2}}
-\dfrac{\Gamma\left(\frac{d-2}{2}\right)^2 }
{\Gamma(d-2)}
\dfrac{(-p_2^2 -i\rho)^{\frac{d-4}{2}} }{m^2}
\Fh21\Fz{1, \frac{d-2}{2}}{d-2}{\dfrac{-p_2^2}{m^2}}.
\nonumber \\
\end{eqnarray}
This result is in agreement with
Eq.~($B2$) in Ref.~\cite{Abreu:2015zaa}.
In the limit of $p_2^2 \rightarrow  -m^2<0$,
we arrive at
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=& -\left\{
\dfrac{\Gamma\left( \frac{d-2}{2}\right)
\Gamma\left(\frac{d-4}{2} \right) }
{\Gamma\left(d-3\right) }
-\dfrac{\Gamma\left(\frac{d-4}{2}\right) }
 {\Gamma\left( \frac{d-2}{2}\right) }
\right\}(m^2)^{ \frac{d-6}{2}}.
\end{eqnarray}

When $\mathcal{R}$e$\left(d-6\right)>0$
and $m^2 \rightarrow 0$,
$J_3 \rightarrow 0$.\\
\item $\underline{m_1^2 = m_3^2:}$\\%
We are also interested in the case of
$m_1^2 = m_3^2 $.
In such case, one should exchange the
order of integration
in  Eq.~(\ref{twolight-like}).
The $x$-integration can be taken first.
Subsequently, we arrive at
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=&
-\int \limits_0^1 dy \;
\dfrac{[p^2_2y^2 - (p_2^2+m_2^2-m_1^2 )y
+m_2^2 -i\rho]^{ \frac{d-4}{2}}
-(m_1^2)^{ \frac{d-4}{2}}
}
 {p_2^2 y -m_2^2 + m_1^2}.
\end{eqnarray}
It is easy to find out that
\begin{eqnarray}
\left(p^2_2y^2
- (p_2^2+m_2^2-m_1^2 )y
+m_2^2 \right)|_{y=(m_2^2 -m_1^2)/p_2^2}
= m_1^2.
\end{eqnarray}
As previous explanation,
the integral $J_3$ in this case also stays
finite at $y=(m_2^2 -m_1^2)/p_2^2$.
Using the analytical solution for master
integral as Eq.~(\ref{K2})
in appendix $A$, the result reads
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)  }
&=& -\dfrac{(m_1^2)^{ \frac{d-4}{2} }}{m_2^2-m_1^2}
  \Fh21\Fz{1, 1}{2}{-\dfrac{p_2^2}{m_2^2-m_1^2}}     \\
&&\hspace{-1.8cm} +
\left( \dfrac{\partial_3 S_3}{p_2^2\; G_{23}} \right)
\Bigg[
\left( \dfrac{\partial_3 S_{23}}{G_{23}} \right)
\dfrac{(m_2^2)^{\frac{d-4}{2}} }{M_{3} -m_2^2}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{23}}{m_2^2}  \right)\nonumber\\
&&\hspace{0cm} +
\left(\dfrac{\partial_2 S_{23}}{G_{23} } \right)
\dfrac{(m_1^2)^{\frac{d-4}{2}} }{M_{3} -m_1^2}
\; F_1\left(1; 1,\frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_1^2}{ M_{3} -m_1^2},
1-\frac{M_{23}}{m_2^2}  \right) \Bigg]
\nonumber \\
&&\hspace{-1.8cm} +
\Bigg[
\dfrac{M_{23} - m_2^2}{2p_2^2(M_{3} -m_2^2)}
\left(m^2_{2} \right)^{\frac{d-4}{2}}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{23}}{m_2^2}  \right) \nonumber\\
&&\hspace{-0.5cm}
-\dfrac{M_{23} - m_1^2}{2p_2^2(M_{3} -m_1^2)}
\left(m^2_{1} \right)^{\frac{d-4}{2} }
\; F_1\left(1; 1,\frac{4-d}{2}; 2;
\dfrac{M_{23} -m_1^2}{ M_{3} -m_1^2},
1-\frac{M_{23}}{m_1^2}  \right)   \Bigg],
\nonumber
\end{eqnarray}
provided that
$\left|\frac{M_{23} -m_{1,2}^2}
{ M_{3} -m_{1,2}^2} \right|<1$,
$\left| 1-\frac{M_{23}}{m_{1,2}^2} \right|<1$.\\
\item $\underline{m_1^2 = m_2^2=0, m_3^2 =m^2,
p_2^2 =q^2:}$\\
This case has been calculated
in~\cite{Abreu:2015zaa}. In this
particular configuration, $J_3$ becomes
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&
- \int \limits_0^1 dx
\; \dfrac{\left[ q^2x^2 -(q^2 +m^2)x +m^2
-i\rho  \right]^{\frac{d-4}{2}}}
{q^2 x - m^2 }.
\end{eqnarray}
It is confirmed that
\begin{eqnarray}
q^2x^2
- (q^2 +m^2)x
+ m^2|_{x= m^2/q^2}=0.
\end{eqnarray}
As a matter of this fact, $J_3$ stays
finite at $x= m^2/q^2$.
By shifting $x \rightarrow 1-x$,
the resulting integral reads
\begin{eqnarray}
\dfrac{J_3}
{\Gamma\left (\frac{4-d}{2} \right)}
&=&
\left(m^2-q^2-i\rho\right)^{\frac{d-6}{2} }
\int \limits_0^1 dx \; x^{\frac{d-4}{2} }
\left(1-\dfrac{q^2}{q^2 -m^2}x\right)^{\frac{d-4}{2}}.
\end{eqnarray}
Following Eq.~(\ref{gauss-int}) in appendix $B$,
we derive this integral in terms of Gauss
hypergeometric functions
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&\dfrac{\Gamma\left(\frac{d-2}{2}\right)  }
{\Gamma\left(\frac{d}{2}\right)}
\left(m^2-q^2-i\rho \right)^{\frac{d-6}{2}}\;
\Fh21\Fz{\frac{6-d}{2},\frac{d-2}{2} }
{\frac{d}{2}} {\dfrac{q^2}{q^2 - m^2}}
\nonumber\\
&=&\dfrac{\Gamma\left(\frac{d-2}{2}\right) }
{\Gamma\left(\frac{d}{2}\right)}
\left(m^2\right)^{\frac{d-6}{2}}\;
\Fh21\Fz{1,\frac{6-d}{2}}{ \frac{d}{2}}
{\dfrac{q^2}{ m^2}},
\end{eqnarray}
provided that $\left|\frac{q^2}{m^2} \right|<1$
and $\mathcal{R}$e$\left(d-2\right) >0$.
This gives full agreement result with Eq.~($B7$) in
Ref~\cite{Abreu:2015zaa}. In the limit of
$q^2 \rightarrow m^2$, one arrives at
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right) }
& = &\dfrac{\Gamma\left(d-4\right)  }
{ \Gamma\left(d-3\right)}\;
\left( m^2 \right)^{\frac{d-6}{2}},
\end{eqnarray}
provided that $\mathcal{R}$e$\left(d-4\right) >0$.
In additional,
if $\mathcal{R}$e$\left(d-6\right)>0$
and $m^2 \rightarrow 0$,
$J_3 \rightarrow 0$.\\
\item $\underline{m_1^2 = m_2^2 = m_3^2 =m^2:}
\\
This case has been performed in
Ref~\cite{Davydychev:2003mv}.
One confirms that $M_{3} = m^2$
in this kinematic configuration.
We derive analytic formula
for $J_3$ as follows
\begin{eqnarray}
\label{rijkM}
\dfrac{J_3}{\Gamma\left (\frac{6-d}{2} \right)}
= -\int \limits_0^1 dx\; \int \limits_0^{1-x} dy\;
\dfrac{1}{( p_2^2 \;x\;y+m^2 - i\rho)^{\frac{6-d}{2}} }.
\end{eqnarray}
Using Mellin-Barnes relation one has
\begin{eqnarray}
J_3  = -\frac{1}{2\pi i}
\int\limits_{-i \infty}^{i\infty} ds \;
\Gamma\left(-s\right) \Gamma\left(\frac{6-d+2s}{2}\right)
\int\limits_0^1 dx\; \int \limits_0^{1-x} dy\;
\dfrac{[ (p_2^2 -i\rho) \;x\;y]^s }{ (m^2)^{3-\frac{d}{2} +s} }.
\end{eqnarray}
After taking over $x,y$-integrations,
the contour integral is taken the form of
\begin{eqnarray}
J_3 &=& -
\frac{\sqrt{\pi}}{2\pi i}
~(m^2)^{\frac{d-6}{2}}
\int\limits_{-i \infty}^{i\infty} ds \;
\dfrac{ \Gamma\left(-s\right)
\Gamma\left(\frac{6-d+2s}{2}\right)
\Gamma(s+1)^2}{4\;\Gamma(s+2)
\Gamma\left(s+\frac{3}{2} \right)}
\left(- \dfrac{p_2^2 -i\rho}{4m^2}\right)^s.
\nonumber\\
\end{eqnarray}
By closing the integration contour to
the right side of imaginary axis
in the $s$-complex plane, we then take
into account the residua of sequence poles
from $\Gamma(-s)$. The result
is presented in terms of series of
generalized hypergeometric function
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{6-d}{2} \right)}
&=&-\dfrac{(m^2)^{\frac{d-6}{2} }}{2}\;
\Fh32\Fz{1,1,\frac{6-d}{2} }{ 2,
\frac{3}{2}}{ \dfrac{p_2^2 -i\rho}{4m^2}},
\end{eqnarray}
provided that
$\left|\frac{p_2^2 -i\rho}{4m^2} \right|<1$.
This result is in
agreement with Ref~\cite{Davydychev:2003mv}.
\end{enumerate}
\subsection{One light-like momentum
We are going to proceed the method for one
light-like momentum case. Without any loss of
the generality, we can choose $p_3^2=0$.
Let us use $q^2 = -2p_1p_3$.
Applying the same previous procedure,
one obtains Feynman parameter
integral
\begin{eqnarray}
\label{onelightlike}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right) }
&=&
\int \limits_0^1 dx\;
\dfrac{\left[p_2^2x^2 -(p_2^2 +m_3^2 -m_2^2)x
+m_3^2 -i\rho \right]^{\frac{d-4}{2} }}
{ q^2 x + m_3^2 -m_1^2 } \\
&&-
\int \limits_0^1 dx\;
\dfrac{\left[p_1^2x^2 -(p_1^2 +m_1^2 -m_2^2)x
+ m_1^2-i\rho  \right]^{\frac{d-4}{2}} }
{ q^2 x + m_3^2 -m_1^2 }.  \nonumber
\end{eqnarray}
We find that two integrands have same
singularity pole at $x= (m_1^2-m_3^2)/q^2$.
It is easy to verify that
the residue contributions from this pole will
be cancelled out. As a result, $J_3$ stays finite
at this point. The analytic result for $J_3$
can be presented as a compact form:
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
=  J_{123} +J_{231}.
\end{eqnarray}
Where the terms $J_{123}$, $J_{231}$ are
obtained by using Eq.~(\ref{K2}).
These terms are written in terms of
$F_1$ as follows
\begin{eqnarray}
\label{onelightlike1}
J_{123} &=&
-\dfrac{\partial_3 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_2S_{12}}{G_{12}} \right)
\dfrac{(m_1^2)^{\frac{d-4}{2} }}{M_{3} -m_1^2}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{12} -m_1^2}{ M_{3} -m_1^2},
1-\frac{M_{12}}{m_1^2}  \right)  \nonumber\\
&&-\dfrac{\partial_3 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_1S_{12}}{G_{12}} \right)
\dfrac{(m_2^2)^{\frac{d-4}{2} } }{M_{3} -m_2^2}
\; F_1\left(1; 1,\frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{12} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{12}}{m_2^2}  \right)
\nonumber\\
&& +\left(\dfrac{M_{12} -m_1^2 }{2(M_{3} -m_1^2)} \right)
\dfrac{(m_1^2)^{\frac{d-4}{2}}  }{p_1^2-p_2^2}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_1^2}{ M_{3} -m_1^2},
1-\frac{M_{12}}{m_1^2}  \right)  \nonumber\\
&&-\left(\dfrac{M_{12} -m_2^2 }{2(M_{3} -m_2^2) }\right)
\dfrac{(m_2^2)^{\frac{d-4}{2} }}{p_1^2-p_2^2}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{12}}{m_2^2}  \right),
\end{eqnarray}
and
\begin{eqnarray}
\label{onelightlike12}
J_{231} &=&
-\dfrac{\partial_1 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_2S_{23}}{G_{23}} \right)
\dfrac{(m_3^2)^{\frac{d-4}{2} } }{M_{3} -m_3^2}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_3^2}{ M_{3} -m_3^2},
1-\frac{M_{23}}{m_3^2}  \right)  \nonumber\\
&&-\dfrac{\partial_1 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_3S_{23}}{G_{23}} \right)
\dfrac{(m_2^2)^{\frac{d-4}{2} } }{M_{3} -m_2^2}
\; F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{23}}{m_2^2}  \right)
\nonumber\\
&& +\left(\dfrac{M_{23} -m_3^2 }{2(M_{3} -m_3^2)} \right)
\dfrac{(m_3^2)^{\frac{d-4}{2} }}{p_2^2-p_1^2}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{23} -m_3^2}{ M_{3} -m_3^2},
1-\frac{M_{23}}{m_3^2}  \right)  \nonumber\\
&&-\left(\dfrac{M_{23} -m_2^2 }{2(M_{3} -m_2^2)}\right)
\dfrac{(m_2^2)^{\frac{d-4}{2} }}{ p_2^2-p_1^2}
\; F_1\left(1; 1,\frac{4-d}{2}; 2;
\dfrac{M_{23} -m_2^2}{ M_{3} -m_2^2},
1-\frac{M_{23}}{m_2^2}  \right).
\end{eqnarray}
It is important to note that the result for $J_3$
in (\ref{onelightlike1}, \ref{onelightlike12})
are only valid if  the
absolute value of  the arguments of
{the Appell $F_1$ functions} in this presentation
are less  than $1$. If these kinematic
variables do not meet this condition.
One has to perform analytic {continuations}
for Appell $F_1$
functions~\cite{olsson}.

We can get another representation for
$J_{123}, J_{231}$ by using a transformation
for $F_1$ (seen appendix $B$). Taking $J_{123}$
as a example, one has
\begin{eqnarray}
\label{onelightlike2}
J_{123}&=&
\dfrac{\partial_3 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_2S_{12}}{G_{12}} \right)
\dfrac{(M_{12}-i\rho)^{\frac{d-4}{2} } }{M_{12}-M_{3}}
\; F_1\left(\frac{1}{2}; 1,\frac{4-d}{2};
\frac{3}{2}; \dfrac{M_{12} -m_1^2}{ M_{12} -M_{3}},
1-\frac{m_1^2}{M_{12}}  \right)  \nonumber\\
&&\hspace{-0.5cm}
+\dfrac{\partial_3 S_3}{4(p_2^2-p_1^2)^2}
\left(\dfrac{\partial_1 S_{12}}{G_{12}} \right)
\dfrac{(M_{12} -i\rho )^{\frac{d-4}{2} } }{M_{12} -M_{3}}
\; F_1\left(\frac{1}{2}; 1,\frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{12} -m_2^2}{ M_{12} -M_{3} },
1-\frac{m_2^2}{M_{12}}  \right)
\nonumber\\
&& \hspace{-0.5cm}
+ \left(\dfrac{M_{12} -m_1^2 }{2(M_{3} -M_{12} )}\right)
\dfrac{(M_{12} -i\rho)^{\frac{d-4}{2} }}{p_2^2-p_1^2}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_1^2}{ M_{12} -M_{3} },
1-\frac{m_1^2}{M_{12}}  \right)  \nonumber\\
&&\hspace{-0.5cm}
-\left(\dfrac{M_{12} -m_2^2 }{2(M_{3} -M_{12}) }\right)
\dfrac{(M_{12} -i\rho)^{\frac{d-4}{2} }}{p_2^2-p_1^2}
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_2^2}{ M_{12} -M_{12}},
1-\frac{m_2^2}{M_{12}}  \right).
\end{eqnarray}
From Eqs.~(\ref{onelightlike1}, \ref{onelightlike2}),
we can perform analytic
continuation the result in the limits of
$M_{ij}= 0, m_i^2, m_j^2$ (for $i,j=1,2,3$)
and $M_{3} =0$, etc. This can be worked out
by applying the transformations for Appell $F_1$
functions, seen appendix $B$.

The results in Eqs.~(\ref{onelightlike1},
\ref{onelightlike2}) are new hypergeometric
representations for scalar one-loop three-point
functions for this case in general $d$.
Several special cases for $J_3$ are considered
in the next paragraphs.
\begin{enumerate}
\item $\underline{m_1^2 = m_2^2 =m_3^2=0:}$\\
One first arrives at the case of all
massless internal lines. In this case,
Eq.~(\ref{onelightlike}) becomes
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=& -\int \limits_0^1 dx\;
\dfrac{\left[p_2^2x^2
- p_2^2x -i\rho \right]^{\frac{d-4}{2} } }
{ (p_2^2-p_1^2)x  }
+
\int \limits_0^1 dx\;
\dfrac{\left[p_1^2x^2
-p_1^2x -i\rho  \right]^{\frac{d-4}{2} } }
{ (p_2^2-p_1^2) x}
\\
&=&\sum\limits_{k=1}^2
\dfrac{(-1)^k}
{p_1^2-p_2^2}  \int \limits_0^1 dx\;
(-p_k^2 +i\rho)^{\frac{d-4}{2} }
x^{\frac{d-6}{2}} (1-x)^{\frac{d-4}{2}}
\\
&=&\dfrac{\Gamma\left (\frac{d-4}{2}\right)
\Gamma\left (\frac{d-2}{2}\right)}
{ \Gamma(d-3)}\sum\limits_{k=1}^2
(-1)^k\;\dfrac{ (-p_k^2
+i\rho)^{\frac{d-4}{2}}}{ p_1^2-p_2^2}.
\end{eqnarray}
We have already used
$p_2^2x^2 -p_2^2x -i\rho = (-p_2^2 -i\rho) x(1-x) $.
This result coincides with Eq.~($4.6$) in
Ref.~\cite{Ellis:2007qk}.
In the limit of $p_2^2 \rightarrow p_1^2$,
one arrives at
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=&  \dfrac{
\Gamma^2\left (\frac{d-4}{2}\right)}{\Gamma(d-3)}
\; (-p_2^2 -i\rho)^{ \frac{d-6}{2}}.
\end{eqnarray}
If $\mathcal{R}$e$\left(d-6\right)>0$
and $p_2^2 \rightarrow 0$,
the integral $J_3 \rightarrow 0 $.
\\
\item $\underline{m_1^2 =m_2^2 =0, m_3^2 = m^2:}$\\
We are concerning the case in which the internal masses
have $m_1^2 =m_2^2 =0, m_3^2 = m^2$. Let us note that
$q^2 = 2p_1p_3$, Eq.~(\ref{onelightlike}) now gets
the form of
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=& -
\int \limits_0^1 dx\;
\dfrac{\left[p_2^2x^2 -(p_2^2 +m^2 )x +
m^2 -i\rho  \right]^{\frac{d-4}{2} }
-\left[p_1^2x^2 -p_1^2 x
 -i\rho  \right]^{\frac{d-4}{2}}
} {q^2x - m^2 }.
\nonumber\\
\end{eqnarray}
We make a change variable like $x \rightarrow 1-x$
for the first integral. Subsequently, it is
presented in terms of Appell $F_1$ functions.
While the second integral is formulated
by mean of Gauss hypergeometric functions.
The result is shown in concrete as follows
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&
-\dfrac{\Gamma\left (\frac{d-2}{2} \right)^2  }
{\Gamma\left (d-2\right)}
\dfrac{ (-p_1^2 -i\rho)^{ \frac{d-4}{2}}}{m^2}
\Fh21\Fz{1, \frac{d-2}{2}}{d-2}{ \dfrac{q^2}{m^2} }
\\
&&
\hspace{-1.8cm} + \dfrac{
\Gamma\left (\frac{d-2}{2} \right)}
{\Gamma\left (\frac{d}{2} \right)}
\dfrac{(m^2-p_2^2 -i\rho)^{\frac{d-4}{2} } }
{q^2-m^2}
F_1\left(\frac{d-2}{2}; 1,\frac{4-d}{2}; \frac{d}{2};
\dfrac{q^2}{q^2-m^2}; \dfrac{p_2^2}{p_2^2 -m^2+i\rho}
\right) \nonumber
\end{eqnarray}
provided that $\left|q^2/m^2 \right|<1$ and
$\mathcal{R}$e$\left(d-2\right)>0$.
One finds another representation for $J_3$
by applying Eq.~(\ref{f1relation1}) in appendix $B$
\begin{eqnarray}
\label{j300m}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&
-\dfrac{\Gamma\left(\frac{d-2}{2} \right)^2  }
{\Gamma\left (d-2\right)}
\dfrac{ (-p_1^2 -i\rho)^{ \frac{d-4}{2}}}{m^2}
\Fh21\Fz{1, \frac{d-2}{2}}{d-2}{ \dfrac{q^2}{m^2} }
\\
&&
 - \dfrac{
\Gamma\left (\frac{d-2}{2} \right)}
{\Gamma\left (\frac{d}{2} \right)}
(m^2)^{\frac{d-6}{2} } \;
F_1\left(1; 1,\frac{4-d}{2}; \frac{d}{2};
\dfrac{q^2}{m^2}; \dfrac{p_2^2}{m^2-i\rho}
\right). \nonumber
\end{eqnarray}
This representation gives agreement
result with Eq.~($C6$) in \cite{Abreu:2015zaa}.
In the limit of $q^2 \rightarrow m^2$, one gets
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&-\dfrac{\Gamma\left(\frac{d-4}{2} \right)}
{\Gamma\left (\frac{d-2}{2} \right)}
(m^2)^{\frac{d-6}{2} } \;
\Fh21\Fz{1,\frac{4-d}{2} }{ \frac{d-2}{2}}
{ \dfrac{p_2^2}{m^2-i\rho} }
\nonumber\\
&&
-\dfrac{\Gamma\left (\frac{4-d}{2} \right)
\Gamma\left (\frac{d-2}{2} \right)^2
 \Gamma\left (\frac{d}{2} \right)^2 }
 {\Gamma\left (d-3\right)}
 \dfrac{ (-p_1^2 -i\rho)^{\frac{d-4}{2}}}{m^2}.
 \end{eqnarray}
When $p_2^2 \rightarrow m^2$, the result in
(\ref{j300m}) reads
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&-\dfrac{\Gamma\left (d-3\right)}
{\Gamma\left (d-2 \right)}
(m^2)^{\frac{d-6}{2} }
\;\Fh21\Fz{1, 1}{d-2}{ \dfrac{q^2}{m^2} } \\
&&-\dfrac{
\Gamma\left (\frac{d-2}{2} \right)^2  }
{\Gamma\left (d-2\right)}
\dfrac{ (-p_1^2 -i\rho)^{ \frac{d-4}{2}}   }{m^2}
\Fh21\Fz{1, \frac{d-2}{2}}{d-2}
{ \dfrac{q^2}{m^2} }. \nonumber
 \end{eqnarray}
 \item $\underline{m_1^2 = m_3^2 =0, m_2^2 =m^2:}$\\
This case has been performed
in Ref~\cite{Fleischer:2003rm}.
The Feynman parameter
integral for $J_3$ in this case reads
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)  }
&=&\int \limits_0^1 dx\;
\dfrac{\left[p_2^2x^2 -(p_2^2 -m^2)x -i\rho
\right]^{\frac{d-4}{2}}
- \left[p_1^2x^2 -(p_1^2  -m^2)x -i\rho  \right]
^{\frac{d-4}{2}}  }{ -2p_1p_3 \; x }\nonumber\\
&& \\
&=&\sum\limits_{k=1}^2
(-1)^k \dfrac{(m^2 - p_k^2 -i\rho)
^{\frac{d-4}{2} }}{p_1^2 -p_2^2}\;
\int \limits_0^1 dx\; x^{\frac{d-6}{2}}
\left[1- \dfrac{p_k^2}{p_k^2 -m^2 +i\rho}
\;x \right]^{ \frac{d-4}{2}}
\nonumber\\
&& \\
&=& \dfrac{ \Gamma\left(\frac{d-4}{2}\right) }
{ \Gamma\left( \frac{d-2}{2}\right)} \;
\sum\limits_{k=1}^2 (-1)^k
\frac{(m^2 - p_k^2 -i\rho)^{\frac{d-4}{2}}}{p_1^2 -p_2^2}\;
\Fh21\Fz{\frac{4-d}{2}; \frac{d-4}{2} }{\frac{d-2}{2}}
{\dfrac{p_k^2}{p_k^2 -m^2 +i\rho}   }
\nonumber\\
&& \\
&=&\dfrac{\Gamma\left(\frac{d-4}{2}\right) }
{ \Gamma\left( \frac{d-2}{2}\right)} \;
\sum\limits_{k=1}^2 (-1)^k
\frac{(m^2)^{\frac{d-4}{2}}}{p_1^2 -p_2^2}\;
\Fh21\Fz{1;\frac{4-d}{2}}
{\frac{d-2}{2}}{\dfrac{p_k^2}{m^2-i\rho}   }.
 \end{eqnarray}
This gives a perfect agreement with
Ref~\cite{Fleischer:2003rm}. In the limit
of $p_2^2\rightarrow p_1^2$, one first uses
\begin{eqnarray}
\dfrac{d}{dp_2^2}
\left\{\Fh21\Fz{1;\frac{4-d}{2} }
{\frac{d-2}{2}}{\dfrac{p_2^2}{m^2-i\rho} } \right\}
= -\dfrac{1}{m^2}\dfrac{d-4}{d-2}
\;\Fh21\Fz{2;\frac{6-d}{2} }{\frac{d}{2} }
{ \dfrac{p_2^2}{m^2-i\rho} }.
\end{eqnarray}
The result reads
\begin{eqnarray}
J_3 &=& 2
\dfrac{\Gamma\left(\frac{4-d}{2} \right) }
{2-d} \; (m^2)^{\frac{d-6}{2} }
\Fh21\Fz{2;\frac{6-d}{2} }{\frac{d}{2}}
{\dfrac{p_2^2}{m^2-i\rho}   }.
\end{eqnarray}
\item  $\underline{m_2^2 = m_3^3 =0,
m_1^2=m^2:}$
\\
Let us note that $q^2 = 2p_1p_3$,
the integral $J_3$ in (\ref{onelightlike})
is written by
 \begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=& \int \limits_0^1 dx\; \Bigg\{
\dfrac{\left[p_2^2x^2
- p_2^2 x  -i\rho  \right]^{\frac{d-4}{2}}}
{-q^2 x -m^2 } - \dfrac{\left[p_1^2x^2 -(p_1^2 -m^2)x
-i\rho  \right]^{\frac{d-4}{2} }}
{ q^2\; x -q^2  -m^2 } \Bigg\}.  \nonumber\\
\end{eqnarray}
Here we have already performed
a shift $x \rightarrow 1-x $ for
the second integral.
It is then expressed in terms of
Appell $F_1$ functions. While
the first integral is presented in
terms of Gauss hypergeometric functions.
Combining all these terms, analytic
result for $J_3$ reads
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left(\frac{4-d}{2} \right)}
&=& \dfrac{\Gamma\left (\frac{d-2}{2} \right)  }
{\Gamma\left (\frac{d}{2} \right)}
\dfrac{(m^2-p_1^2 -i\rho)^{\frac{d-4}{2} }}
{q^2+m^2} \times\\
&& \hspace{1cm}\times
F_1\left(\dfrac{d-2}{2}; 1,\frac{4-d}{2}; \frac{d}{2};
\dfrac{q^2}{q^2 + m^2}; \dfrac{p_1^2}{p_1^2 -m^2+i\rho}
 \right) \nonumber\\
&&-
\dfrac{\Gamma\left(\frac{d-2}{2} \right)^2  }
{\Gamma\left(d-2\right)}
\dfrac{(-p_2^2 -i\rho)^{ \frac{d-4}{2} } }{m^2}
\Fh21\Fz{1, \frac{d-2}{2} }{d-2}{ -\dfrac{q^2}{m^2} }.
\nonumber
\end{eqnarray}
Another representation for $J_3$ is derived
by using Eq.~(\ref{f1relation1}) in appendix $B$.
It is
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&\dfrac{\Gamma\left (\frac{d-2}{2} \right)  }
{\Gamma\left(\frac{d}{2} \right)}\; (m^2)^{\frac{d-6}{2} }
F_1\left(1; 1, \frac{4-d}{2}; \frac{d}{2};
-\dfrac{q^2}{m^2}; \dfrac{p_1^2}{m^2-i\rho}
 \right) \nonumber\\
&&-
 \dfrac{\Gamma\left (\frac{d-2}{2} \right)^2  }
 {\Gamma\left (d-2\right)}
 \dfrac{ (-p_1^2 -i\rho)^{ \frac{d-4}{2} } }{m^2}
 \Fh21\Fz{1, \frac{d-2}{2} }{d-2}{ -\dfrac{q^2}{m^2} }.
 \end{eqnarray}
From this representation, one can perform
analytic continuation this result
in the limits of $q^2 \rightarrow -m^2$
and $p_1^2 \rightarrow m^2$.
\\
\item $\underline{p_1^2=p_2^2 \neq 0:}$
\\
We are going to consider an interesting case in which is
$p_1^2=p_2^2 \neq 0$. In such the case, one recognizes
that $G_3=0$. We can present $J_3$ in terms of two scalar
one-loop two-point functions as follows
\begin{eqnarray}
\label{j3gram}
J_3 &=& -\Gamma\left (\frac{4-d}{2} \right)
\int \limits_0^1 dx\; \Bigg\{
\dfrac{\left[p_2^2x^2 -(p_2^2 +m_3^2 -m_2^2)x +m_3^2 -i\rho
\right]^{\frac{d-4}{2} }}{m_3^2 -m_1^2 }   \\
&&\hspace{3cm}
+ \dfrac{\left[p_1^2x^2 -(p_1^2 +m_1^2 -m_2^2)x +m_1^2
-i\rho \right]^{\frac{d-4}{2} }}{m_1^2 -m_3^2 }
\Bigg\} \nonumber\\
&=&-\Gamma\left (\frac{4-d}{2}\right)
(J_{231} +J_{123}).
\end{eqnarray}
Both terms in right
hand side of Eq.~(\ref{j3gram}) are
determined as Feynman parameter integrals
of scalar one-loop two-point functions. They are
calculated in detail as follows.
Let us consider $J_{231}$
as a example. We can rewrite
$J_{231}$ in  the following form:
\begin{eqnarray}
 J_{231} &=&\dfrac{1}{m_3^2 -m_1^2} \int \limits_0^1 dx\;
 \left[p_2^2(x -x_{23})^2 +M_{23} -i\rho \right]^{\frac{d-4}{2} }.
\end{eqnarray}
The integral will be worked out by
applying Mellin-Barnes relation which is
\begin{eqnarray}
 J_{231} &=&
 \dfrac{2}{m_3^2 -m_1^2}\; \frac{1}{2\pi i} \int\limits_{-i\infty}^{i\infty}
 \dfrac{\Gamma(-s) \Gamma\left(\frac{4-d+2s}{2} \right) \Gamma(s+\frac{1}{2})}
 {\Gamma\left(\frac{4-d}{2} \right)\Gamma(s+\frac{3}{2})}
\\
 && \hspace{0cm}\times
 \left( M_{23} -i\rho \right)^{\frac{d-4}{2} }
 \left[
 x_{23} \left( \dfrac{p_2^2x_{23}^2}{M_{23} -i\rho}\right)^s
 + x_{32} \left( \dfrac{p_2^2x_{32}^2}{M_{23} -i\rho}\right)^s
\;\; \right]           \nonumber\\
 &=& \dfrac{\left( M_{23}-i\rho\right)^{\frac{d-4}{2}} }{ m_3^2 -m_1^2}
 \left\{
 x_{23}\; \Fh21\Fz{\frac{4-d}{2}; \frac{1}{2}}{\frac{3}{2}}{
 \dfrac{p_2^2x_{23}^2}{M_{23} -i\rho} }
 +(2 \leftrightarrow 3)
 \right\}.
\end{eqnarray}
The $J_3$ now is casted into the form of
\begin{eqnarray}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=&\Bigg\{
\left(\dfrac{\partial_2 S_{23} }{G_{23}}\right)
\dfrac{(M_{23} -i\rho)^{ \frac{d-4}{2}}}{m_3^2-m_1^2}
\Fh21\Fz{\frac{4-d}{2}; \frac{1}{2} }
{\frac{3}{2}}{\dfrac{M_{23}-m_3^2}{M_{23} -i\rho} }
+(2\leftrightarrow3) \Bigg\} \nonumber
\\
&&
\hspace{-0.2cm}
+\Bigg\{
\left(\dfrac{\partial_2 S_{12} }{G_{23}}\right)
\dfrac{(M_{12} -i\rho)^{ \frac{d-4}{2} }} {m_1^2-m_3^2}
\Fh21\Fz{\frac{4-d}{2}; \frac{1}{2} }{\frac{3}{2}}
{\dfrac{M_{12}-m_1^2}{M_{12} -i\rho} }
+(2\leftrightarrow 1) \Bigg\}
\nonumber\\
\end{eqnarray}
\end{enumerate}
provided that
$\left|\frac{M_{12}-m_{1,3}^2}{M_{12} -i\rho}\right|<1$.
\subsection{General case
We are going to generalize the method
for the general case in which
$p_i^2 \neq 0$ for $i=1,2,3$. Following
an idea in~\cite{'tHooft:1978xw},
we first apply the Euler transformation
like $y \rightarrow y - \beta x$,
the polynomial written in terms of $x,y$ in
$J_3$'s integrand will becomes
\begin{eqnarray}
 &&\hspace{-0.5cm}
 A x^2+ By^2+2Cxy+ Dx+Ey+F- i\rho =\\
 &&= (B\beta^2 - 2C\beta + A) x^2
 + B y^2 +2(C-B\beta) xy + (D - E\beta)x + E y + F -i\rho.
 \nonumber
\end{eqnarray}
By choosing $\beta$ is one of the
roots of following equation
\begin{eqnarray}
\label{alpha}
 B \beta^2 - 2C \beta + A =0, \quad \text{or} \quad \beta=
 \dfrac{C \pm \sqrt{C^2 - AB}}{B},
\end{eqnarray}
the integral $J_3$ is casted into
\begin{eqnarray}
&&\dfrac{J_3}{\Gamma\left (\frac{6-d}{2} \right)} =
\int \limits_0^1 dx\int \limits_{\beta x}^{1-(1-\beta)x} dy\;
\left\{ \Big[2(C-B\beta)y + D - E\beta \Big] x
+ By^2+ Ey+ F - i\rho   \right\}^{\frac{d-6}{2}} .
\nonumber\\
\end{eqnarray}
It is also important to note
that the final result will be independent of
$\beta$ in Eq.~(\ref{alpha}). It means that
we are free to choice one of roots $\beta$
in Eq.~(\ref{alpha}). As a result,
the integrand is linearized of $x$,
hence the $x$-integration can be evaluated
first. In order to work out the $x$-integration,
we split the integration as follows
\begin{eqnarray}
\int \limits_0^1 dx\; \int \limits_{\beta x}^{1-(1-\beta)x} dy
=\left\{  \int \limits_0^1 dy \; \int \limits_{0}^{1} dx
-  \int \limits_0^{\beta}dy\; \int \limits_{\frac{y}{\beta}}^{1} dx
-  \int \limits_{\beta}^1 dy\; \int \limits_{\frac{y-1}{\beta-1}}^{1} dx
\right\}.
\end{eqnarray}
\begin{figure}[ht]
\begin{center}
\begin{pspicture}(-3, -2.5)(3, 2.5)
\psset{linewidth=.1pt}
\psset{unit = 0.9}
\psline{->}(-4,-2.5)(-4,2.5)
\psline(-4, 2)(-1.5, 0)
\rput(-4.5, 2){$1$}
\rput(-1.5, -0.5){$1$}
\pspolygon[fillstyle=crosshatch, fillcolor=gray,
linestyle=none, hatchcolor=lightgray](-4,2)(-4,0)(-1.5,0)
\psline{->}(4,-2.5)(4,2.5)
\psline{->}(-7,0)(-1,0)
\psline{->}(1,0)(7,0)
\psline(4,2)(6.5,1)
\psline(4,0)(6.5,1)
\psline[linestyle=dashed](4,1)(6.5,1)
\psline[linestyle=dashed](4,2)(6.5,2)
\psline[linestyle=dashed](6.5,0)(6.5,2)
\pspolygon[fillstyle=crosshatch, fillcolor=gray,
linestyle=none, hatchcolor=lightgray](4,2)(4,0)(6.5,1)
\rput(-3.8, 2.6){$y$}
\rput(6.5, -0.5){$1$}
\rput(3.5, 2){$1$}
\rput(3.5, 1){$\beta$}
\rput(4.2, 2.6){$y$}
\rput(-1.2,-0.2){$x$}
\rput(7,-0.2){$x$}
\rput(0., 0.6){$y' = y+ \beta x$}
\psline{->}(-0.3,0)(0.3,0)
\end{pspicture}
\end{center}
\caption{\label{region} The integration
looks like after after shifting
$y \rightarrow y - \beta x $.}
\end{figure}
To archive a more symmetric
form we make a further transformations
$y \rightarrow  \beta (1 - y )$ for
the second integral and $y \rightarrow  1 -(1 - \beta )y$
for the third integral respectively.
This brings some order in the arguments of the
integrands. The denominators are all of
the linear form $\sim (y - y_i)$ for $i = 1, 2, 3$.
It is also easy to confirm that all $y_i$ follows
the equations
\begin{eqnarray}
 p_i^2 (y_k -x_{ij})^2 +M_{ij} -i\rho = M_3 -i\rho,
\end{eqnarray}
for $i,j,k =1,2,3$.

Finally, following an idea
in~\cite{'tHooft:1978xw},
we add extra terms which sum
all of them is to zero for cancelling
the residue of the pole at $y_i$.
The result reads
\begin{eqnarray}
&&\hspace{-0.5cm}
\dfrac{ \lambda^{1/2}(p_1^2, p_2^2, p_3^2) }
{\Gamma\left (\frac{4-d}{2} \right)}
\; J_3 =  \\
&&=
-\int\limits_0^1 dy \frac{\left[p_1^2 y^2
-(p_1^2 + m_1^2 -m_2^2)y +m_1^2 -i\rho \right]
^{\frac{d-4}{2} }}{y-y_3}
+  \int\limits_0^1 dy
\frac{(M_{3} -i\rho)^{\frac{d-4}{2} }}{y-y_3} \nonumber\\
&& \hspace{0cm} \quad
-\int\limits_0^1 dy \frac{\left[p_2^2 y^2
-(p_2^2 + m_2^2 -m_3^2)y +m_2^2 -i\rho \right]
^{\frac{d-4}{2} }}{y-y_1}
+  \int\limits_0^1 dy
\frac{(M_{3} -i\rho)^{\frac{d-4}{2} }}{y-y_1}
\nonumber \\
&& \hspace{0cm} \quad
-\int\limits_0^1 dy \frac{\left[p_3^2 y^2
-(p_3^2 + m_3^2 -m_1^2)y +m_3^2 -i\rho \right]
^{\frac{d-4}{2} }}{y-y_2}
+
\int\limits_0^1 dy
\frac{(M_{3} -i\rho)^{\frac{d-4}{2} }}{y-y_2}.
\nonumber
\end{eqnarray}
The analytic result for $J_3$
can be written in a compact form
\begin{eqnarray}
\label{j3final}
\dfrac{J_3}{\Gamma\left (\frac{4-d}{2} \right)}
&=& \;\left\{
-\dfrac{
(M_{3} -i\rho)^{\frac{d-4}{2} }
} {\lambda^{1/2}(p_1^2, p_2^2, p_3^2)}
\cdot J^{(d=4)}_{123}  +
\dfrac{1} {\lambda^{1/2}(p_1^2, p_2^2, p_3^2)}
\cdot
J^{(d)}_{123}
\right\} \nonumber\\
 && \nonumber\\
 && +
\left\{
 (1,2,3) \leftrightarrow (2,3,1)
\right\}
+
\left\{
(1,2,3) \leftrightarrow (3,1,2)
\right\},
\end{eqnarray}
with
\begin{eqnarray}
J_{ijk}^{(d)}&=& - \int\limits_0^1 dy
\frac{\left[p_i^2 y^2 -(p_i^2 + m_i^2
-m_j^2)y +m_i^2 -i\rho \right]
^{\frac{d-4}{2} }}{y-y_k},
\end{eqnarray}
for $i,j,k =1,2,3.$ Where the integrand's
poles are given
\begin{eqnarray}
y_1 &=& 1-\dfrac{D - E\beta +2(C - B\beta)}{2(1-\beta)
(C - B\beta)},
\; y_2 =  1+ \dfrac{D - E\beta}{2(C - B\beta)},
\; y_3 = - \dfrac{D - E\beta }{2\beta(C - B\beta)}.
\end{eqnarray}
Applying the formula for master integral in
appendix $A$, we will present
the result of $J_{ijk}$ in terms of
Appell $F_1$ functions.
For instant, one takes $J_{123}$
for an example. This term  is expressed as follows
\begin{eqnarray}
\label{j123general1}
J_{123} &=&
-\left(\dfrac{\partial_3 S_3 }{\sqrt{8G_3 } } \right)
\Bigg[ \left(\dfrac{\partial_2 S_{12} }{G_{12} } \right)
\dfrac{ (m_1^2)^{\frac{d -4}{2} } }{M_{3}-m_1^2 }
F_1\left(1; 1,\frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{12} -m_1^2}{M_{3} -m_1^2},
1-\dfrac{M_{12} }{m_1^2} \right)\nonumber\\
&&\hspace{1.7cm}+
\left(\dfrac{\partial_1 S_{12} }{G_{12} } \right)
\dfrac{ (m_2^2)^{\frac{d-4}{2} }}{M_{3}-m_2^2 }
F_1\left(1; 1, \frac{4-d}{2}; \frac{3}{2};
\dfrac{M_{12} -m_2^2}{M_{3} -m_2^2},
1-\dfrac{M_{12} }{m_2^2} \right)  \Bigg] \nonumber\\
&&+\Bigg[ \dfrac{M_{12} -m_1^2}
{2(M_{3}-m_1^2)} (m_1^2)^{\frac{d-4}{2}}
F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_1^2}{M_{3} -m_1^2},
1-\dfrac{M_{12} }{m_1^2} \right)
\\
&& \hspace{1.4cm}
-\dfrac{M_{12} -m_2^2}{2(M_{3}-m_2^2)}
(m_2^2)^{\frac{d-4}{2} }
F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{M_{12} -m_2^2}{M_{3} -m_2^2},
1-\dfrac{M_{12} }{m_2^2} \right) \;\;\Bigg],
\nonumber
\end{eqnarray}
provided that $\left|\frac{M_{12} -m_{1;2}^2}
{M_{3} -m_{1;2}^2}\right|,
\;\left|1-\frac{M_{12} }{m_{1;2}^2}\right|<1$.
One finds another representation for $J_{123}$ is as
(by applying Eq.~(\ref{K1}) in appendix $A$)
\begin{eqnarray}
\label{j123general2}
J_{123} &=&
-\left(\dfrac{\partial_3 S_3 }{\sqrt{8G_3 } } \right)
\Bigg[ \left(\dfrac{\partial_2 S_{12} }{G_{12} } \right)
\dfrac{ (M_{12} -i\rho)^{\frac{d-4}{2} } }{M_{3}-M_{12} }
F_1\left(\frac{1}{2}; 1,\frac{4-d}{2};
\frac{3}{2}; \dfrac{M_{12} -m_1^2}{M_{12} -M_{3} },
1-\dfrac{m_1^2}{M_{12} } \right)\nonumber\\
&&\hspace{1.5cm}+
\left(\dfrac{\partial_1 S_{12} }{G_{12} } \right)
\dfrac{ (M_{12} -i\rho)^{\frac{d-4}{2} }}{M_{3}-M_{12} }
F_1\left(\frac{1}{2}; 1, \frac{4-d}{2};
\frac{3}{2}; \dfrac{M_{12} -m_2^2}{M_{12} -M_{3} },
1-\dfrac{m_2^2}{M_{12} } \right)
\Bigg]
\nonumber\\
&&+\Bigg[ \dfrac{M_{12} -m_1^2}{2(M_{12}-M_{3}) }
(M_{12} -i\rho)^{\frac{d-4}{2} }
F_1\left(1; 1,\frac{4-d}{2};
2; \dfrac{M_{12} -m_1^2}{M_{12} -M_{3} },
1-\dfrac{m_1^2}{M_{12} } \right)
\\
&& \hspace{1.0cm}
-\dfrac{M_{12} -m_2^2}{2(M_{12}-M_{3}) }
(M_{12} -i\rho)^{\frac{d-4}{2} }
F_1\left(1; 1,\frac{4-d}{2};
2; \dfrac{M_{12} -m_2^2}{ M_{12} -M_{3} },
1-\dfrac{m_2^2}{M_{12} } \right)
\;\;\Bigg], \nonumber
\end{eqnarray}
provided that
$\left|\frac{M_{12} -m_{1;2}^2}{ M_{12} -M_{3} } \right|,\;
\left|1-\frac{m_{1;2}^2}{M_{12} }\right|<1$.

It is important to note that the results
in (\ref{j123general1}, \ref{j123general2})
are only valid if the absolute value of
the arguments of {the Appell $F_1$ functions}
in these representations are less
than $1$. If these kinematic variables do not
satisfy this condition.
One has to perform analytic {continuations},
we refer the work of~\cite{olsson}
for Appell $F_1$.

The results shown in
Eqs.~(\ref{j123general1}, \ref{j123general2})
are also new hypergeometric representations
for scalar one-loop three-point
functions in general space-time dimension.
From these representations, one can perform
analytic continuation
of the result in the cases
of $M_{ij}=0, m_i^2, m_j^2$, $M_3=0$
(for $i,j,k=1,2,3$), $G_3=0$ and
massless case, etc. This can be done
by applying transformations for
Appell $F_1$ functions
(seen appendix $B$).

For a example, we consider
the case of $G_3=0$. In the case, $\beta = c/b$,
repeating the calculation, we arrive at
\begin{eqnarray}
\label{g123J30}
J_3
= -\sum\limits_{k=1}^3
  \left(\dfrac{\partial_k S_3}{2S_3} \right)
  {\bf k^{-} }  J_3(d; \{p_i^2\}, \{m_i^2\} ).
\end{eqnarray}
The operator ${\bf k^{-} }$ is defined in
such a way that
it will reduce the three-point integrals
to two-point integrals by shrinking
an propagator in the integrand of $J_3$.
This equation equivalents with Eq.~($46$)
in Ref.~\cite{Devaraj:1997es}. Noting that
we have already arrived this relation
in (\ref{j3gram}) of previous subsection.
\subsection{Tensor one-loop three-point integrals}
Following tensor reduction method
in Ref.~\cite{Davydychev:1991va}, tensor
one-loop three-point integrals with
rank $M$ can be presented in terms
of scalar ones with the shifted space-time
dimension:
\begin{eqnarray}
&& \hspace{-1cm}
J^{(3)}_{\mu_1\mu_2\cdots\mu_M} (d;\{\nu_1,\nu_2,\nu_3\}) =
\\
&=& \int \frac{d^d k}{i\pi^{d/2}}
\dfrac{k_{\mu_1}k_{\mu_2}\cdots k_{\mu_M}}
{[(k+q_1)^2 -m_2^2 + i\rho]^{\nu_1}
[(k+q_2)^2 -m_3^2 + i\rho]^{\nu_2}
[(k+q_3)^2 -m_1^2 + i\rho]^{\nu_3}} \nonumber\\
&=&\sum\limits_{\lambda, \kappa_1, \cdots, \kappa_3}
\left(-\frac{1}{2}\right)^{\lambda}
\Big\{[g]^\lambda [q_1]^{\kappa_1}
[q_2]^{\kappa_2} [q_3]^{\kappa_3} \Big\}_
{\mu_1\mu_2\cdots\mu_M}    \\
&&\hspace{1.4cm}
\times (\nu_1)_{\kappa_1}(\nu_2)_{\kappa_2}(\nu_3)_{\kappa_3}
\;\;
J_3\Big(d+2(M-\lambda);
\{\nu_1+\kappa_1,\nu_2+\kappa_2,\nu_3+\kappa_3\} \Big).
\nonumber
\end{eqnarray}
In this formula, the condition for the indices
$\lambda, \kappa_1, \kappa_2$ and
$\kappa_3$ is $2\lambda +\kappa_1+\kappa_2+\kappa_3=M$.
Moreover, these indices also follow the constrain
$0 \leq \kappa_1, \kappa_2, \kappa_3 \leq M$
and $0 \leq \lambda \leq [M/2]$
(integer of $M/2$). The notation
$(a)_\kappa = \Gamma(a+\kappa)/\Gamma(a)$
is the Pochhammer symbol. The structure of
tensor $\{[g]^\lambda [q_1]^{\kappa_1} [q_2]^{\kappa_2}
[q_3]^{\kappa_3} \}_{\mu_1\mu_2\cdots \mu_M} $
is symmetric with respect to
$\mu_1, \mu_2,\cdots, \mu_M$.
This tensor is constructed from $\lambda$
of metric $g_{\mu\nu}$,
$\kappa_1$ of momentum $q_1$, $\cdots$,
$\kappa_3$ of momentum $q_3$.
The $J_3(d+2(M-\lambda);\{\nu_1+\kappa_1,
\nu_2+\kappa_2,\nu_3+\kappa_3\})$ are scalar
one-loop three-point functions with the shifted
space-time dimension
$d+ 2(M-\lambda)$,
raising powers of propagators
$\{\nu_i+\kappa_i\}$ for $i=1,2,3$.

For examples, we first take the simplest case $M=1$.
In this case, one has $\lambda =0$.
Subsequently, we get
\begin{eqnarray}
J^{(3)}_{\mu} (d;\{\nu_1,\nu_2,\nu_3\})
&=& \sum\limits_{k=1}^{3} \nu_k\; q_{k\mu} \;
J_3(d+2;\{\nu_1 +\delta_{1k},
\nu_2+ \delta_{2k},\nu_3+ \delta_{3k}\}),
\end{eqnarray}
with $\delta_{jk} $ is the Kronecker symbol.
We next consider $M=2$. One has
\begin{eqnarray}
J^{(3)}_{\mu_1\mu_2} (d; \{\nu_1,\nu_2,\nu_3\})
&=& -\frac{1}{2} g_{\mu_1\mu_2}\;
J_3 (d+2;\{\nu_1, \nu_2, \nu_3\})
\\
&&
\hspace{-2cm}
+ \sum\limits_{k=1}^{3} q_{k\mu_1} \; q_{k\mu_2} \; (\nu_k)_2
~
J_3 (d+4; \{\nu_1 +2\delta_{1k},
\nu_2 +2\delta_{2k}, \nu_3 +2\delta_{3k}\})
\nonumber\\
&&\hspace{-2cm}
+ \sum\limits_{k=1}^{3}\sum\limits_{k'>k}^{3}
(q_{k\mu_1} q_{k'\mu_2} + q_{k\mu_2}
q_{k'\mu_1}) \; (\nu_k)_1 \;(\nu_{k'})_1
\nonumber\\
&&\hspace{-2cm}\times
J_3 (d+4;\{\nu_1 +\delta_{1k}+\delta_{1k'},
\nu_2 +\delta_{2k}+\delta_{2k'},
\nu_3 +\delta_{3k}+\delta_{3k'}\} ). \nonumber
\end{eqnarray}
In the next step, the scalar integrals
$J_3\Big(d+2(M-\lambda);
\{\nu'_1,\nu'_2, \nu'_3\} \Big)$ will be
reduced to subset of master integrals by
using integration-by-part method
(IBP)~\cite{IBP}. By applying the
operator $\frac{\partial}{\partial k} \cdot k$
to the integrand of $J_3\left(d; \{\nu_1, \nu_2, \nu_3 \} \right)$
and choosing $k$ to be the momentum of three
internal lines ($k\equiv\{k+q_1, k+q_2, k+q_3\}$).
One arrives at the following system equations:
\begin{eqnarray}
\label{j30}
\begin{cases}
2 \nu_1 m_1^2 \mathbf{1^+} - \nu_2 Y_{12}
\mathbf{2^+} - \nu_3 Y_{13} \mathbf{3^+}
= (d- 2 \nu_1 -\nu_2-\nu_3) \mathbf{1} -\nu_2
\mathbf{1^-}\mathbf{2^+}
-\nu_3 \mathbf{1^-}\mathbf{3^+},      \\
- \nu_1 Y_{12} \mathbf{1^+} + 2 \nu_2 m_2^2
\mathbf{2^+} - \nu_3 Y_{23} \mathbf{3^+}
= (d-  \nu_1 - 2 \nu_2-\nu_3) \mathbf{1} - \nu_1
\mathbf{1^+}\mathbf{2^-}
-\nu_3 \mathbf{2^-}\mathbf{3^+},      \\
- \nu_1 Y_{13} \mathbf{1^+} - \nu_2 Y_{23}
\mathbf{2^+} +2 \nu_3 m_3^2 \mathbf{3^+}
= (d- \nu_1 -\nu_2-2 \nu_3) \mathbf{1} -\nu_1
\mathbf{1^+}\mathbf{3^-}
-\nu_2 \mathbf{2^+}\mathbf{3^-}.
\end{cases}
\end{eqnarray}
Where we used the following notation:
\begin{eqnarray}
\begin{cases}
\mathbf{1}     = J_3 (d; \{\nu_1,\nu_2,\nu_3\}), \\
\mathbf{1^\pm }= J_3 (d; \{\nu_1\pm 1,\nu_2,\nu_3\}), \\
\mathbf{2^\pm }= J_3 (d; \{\nu_1,\nu_2\pm 1,\nu_3\}), \\
\mathbf{3^\pm }= J_3 (d; \{\nu_1,\nu_2,\nu_3\pm 1\}).
\end{cases}
\end{eqnarray}
In order to solve system equations (\ref{j30}),
one first considers the following matrix:
\begin{eqnarray}
Y_{3} =
\begin{bmatrix}
\nu_1 Y_{11} &&& -\nu_2 Y_{12}  &&& -\nu_3 Y_{13}   \\
-\nu_1 Y_{12} &&& \nu_2 Y_{22}  &&& -\nu_3 Y_{23}   \\
-\nu_1 Y_{13} &&& -\nu_2 Y_{23}  &&& \nu_3 Y_{33}   \\
\end{bmatrix}
\text{with}\;
Y_{ij}&=&-(q_i-q_j)^2+m_i^2+m_j^2.
\end{eqnarray}
If det$(Y_{3}) \neq 0$, one then can present
$J_3 (d; \nu_{123}+1)$ in term of $J_3 (d; \nu_{123})$
with $\nu_{123}=\nu_1+\nu_2+\nu_3$. In this recurrence way
\cite{Laporta:2001dd},
we can arrive at the following integrals:
$J_2(d; \{\nu_1, \nu_2\})$
and $J_3(d; \{1,1,1\})$. By applying IPB once again
for
the former integrals, we will arrive at master
integrals such as:
$J_1(d; \{\nu\})$, $J_2(d; \{1,1\})$
which can be found in
\cite{Fleischer:2003rm}
and $J_3(d;\{1,1,1\})$ in this paper.
\section{Conclusions}
\noindent
New analytic formulas for
one-loop three-point
Feynman integrals in general space-time dimension
have presented in this paper. The results are expressed
in terms of Appell $F_1$ functions, considered all cases
of internal mass and external momentum assignments. We
have also cross-checked the analytic results in this
work with other paper in several special cases.
\\

\noindent
{\bf Acknowledgment:}~This research is funded by
Vietnam National University (VNU-HCM), Ho Chi Minh City
under grant number C$2019$-$18$-$06$???.
\section*{Appendix $A$: Evaluatin
the master integral
We are considering master integral
\begin{eqnarray}
 \mathcal{K} = \int\limits_0^1 dx\;
 \dfrac{\left[ p^2_ix^2
 - (p_i^2 +m_i^2-m_j^2)x +m_i^2 -i\rho
 \right]^{\frac{d-4}{2} }}{x - x_k},
\end{eqnarray}
with $p_i^2 \neq 0$. Where $|x_k|>1$
or $-1 \leq x_k <0$ and it follows
the below equation
\begin{eqnarray}
p^2_ix_k^2
 - (p_i^2 +m_i^2-m_j^2)x_k +m_i^2 -i\rho = M_3 -i\rho,
\end{eqnarray}
for $i,j,k=1,2,3$.

We will discuss on the method to evaluate this integral
under the above conditions. In the case of $0<x_k<1$,
one will perform analytic continuation the result for
all master integrals which are expected to appear in  the
general formula of $J_3$. This will be devoted in
concrete in section $2$.

In order to work out the master integral
one should write the polynomial of $x$ which appears in
numerator of the $\mathcal{K}$'s integrand as
follows
\begin{eqnarray}
 p^2_ix^2- (p_i^2 +m_i^2-m_j^2)x +m_i^2 -i\rho
 = p^2_i(x-x_{ij})^2 +M_{ij} -i\rho.
\end{eqnarray}
We have introduced the kinematic variables
\begin{eqnarray}
 x_{ij} &=& \dfrac{p_i^2 +m_i^2-m_j^2 }{2p_i^2},                \\
 x_{ji} &=&1-x_{ij}=\dfrac{p_i^2 -m_i^2+m_j^2 }{2p_i^2},
 \end{eqnarray}
for $p_i^2 \neq 0$. From the conventions,
we subsequently verify the below relations
\begin{eqnarray}
 \label{rij-rijk}
  p_i^2  x_{ij}^2 &=& m_i^2 - M_{ij},\\
  p_i^2  x_{ji}^2 &=& m_j^2 - M_{ij},\\
  p_i^2 (x_k-x_{ij})^2 &=& M_3 -M_{ij}, \\
  p_i^2 (1-x_k-x_{ji})^2 &=& M_3 -M_{ij},
 \end{eqnarray}
for $i,j =1,2,3$.

Using Mellin-Barnes relation~\cite{MB}
we then decompose the integrand as
\begin{eqnarray}
\label{MB1}
\mathcal{K} =
\frac{1}{2\pi i} \int\limits_{-i\infty}^{i\infty} ds\;
\dfrac{\Gamma(-s) \Gamma\left(\frac{4-d}{2} +s\right)}
{\Gamma\left(\frac{4-d}{2}\right)}
\left( \dfrac{1}{M_{ij} -i\rho }\right)^{\frac{4-d}{2} }
\int\limits_0^1
\dfrac{dx}{x-x_{k}}
\left[\dfrac{p_i^2(x-x_{ij})^2}{M_{ij} -i\rho} \right]^s.
\end{eqnarray}
With the help of this relation, the Feynman
parameter integral will be casted
into the simpler form. It will be calculated
in terms of Gauss hypergeometric
functions. In particular, we have
\begin{eqnarray}
\mathcal{L} = \int\limits_0^1\dfrac{dx}{x-x_{k}}
\left[\dfrac{p_i^2(x-x_{ij})^2}{M_{ij} -i\rho} \right]^s
= \left\{ \int\limits_0^{x_{ij}} dx
+   \int\limits_{x_{ij}}^1 dx  \right\}
\dfrac{1}{x-x_{k}}
\left[\dfrac{p_i^2(x-x_{ij})^2}{M_{ij} -i\rho} \right]^s.
\end{eqnarray}
One makes a shift $x = x_{ij}\; x'$
(and $x = 1- x_{ji}\; x'$) for the first integral
(second integral) respectively. The result reads
\begin{eqnarray}
\mathcal{L}
&=& \left(- \dfrac{x_{ij}}{x_k} \right)
\left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
\int\limits_0^1 dx \;
\dfrac{(1-x)^{2s}}
{1-\left(  \dfrac{x_{ij}}{x_k}\right) x}
- \Big\{x_{ij} \leftrightarrow
x_{ji};\;  x_k \leftrightarrow 1-x_k\Big\}-\mathrm{term}.
\nonumber\\
\end{eqnarray}
We apply a further transformation
like $\xi = (1-x)^2 \geqslant 0$.
The Jacobian of the shift is
$dx = -\dfrac{d\xi}{2\sqrt{\xi}}$
and the integration domain is now $[1,0]$.
As a result of this shift, we arrive at
 \begin{eqnarray}
 \mathcal{L} &=&  -\left(\dfrac{x_{ij}}{2} \right)
 \left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
 \int\limits_0^1 d\xi \; \dfrac{\xi^{s}}{\sqrt{\xi}
 \left(x_k -x_{ij} +x_{ij}\sqrt{\xi} \right) } \\
 && \nonumber\\
 && - \Big\{x_{ij} \leftrightarrow x_{ji};\;  x_k
 \leftrightarrow 1-x_k\Big\}-\mathrm{term} \nonumber\\
 &=& -\dfrac{x_{ij}}{2(x_k -x_{ij} )}\;
 \left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
 \int\limits_0^1 d\xi \;
 \dfrac{\xi^{s-\frac{1}{2}}}{1-
 \dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2} \xi }\nonumber\\
 &&\nonumber\\
&& + \dfrac{x_{ij}^2 }{2(x_k -x_{ij} )^2}\;
\left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
 \int\limits_0^1 d\xi \; \dfrac{\xi^{s} }{1-
 \dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2} \xi }\\
 &&\nonumber\\
 && - \Big\{x_{ij} \leftrightarrow x_{ji};\;
 x_k \leftrightarrow 1-x_k\Big\}-\mathrm{term}. \nonumber\\
 \nonumber
 \end{eqnarray}
Following Eq.~(\ref{gauss-int}) in appendix $B$,
we can present this integral in terms of
Gauss hypergeometric functions
 \begin{eqnarray}\quad
 \mathcal{L} &=&  -\dfrac{x_{ij}}{2(x_k -x_{ij} )}
 \;\left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
 \dfrac{\Gamma\left(s+\frac{1}{2} \right)}
 { \Gamma\left(s+\frac{3}{2} \right)}
 \Fh21\Fz{s + \frac{1}{2}, 1}{s+ \frac{3}{2}}
 { \dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2} } \nonumber\\
 && +\dfrac{x_{ij}^2 }{2(x_k -x_{ij} )^2}
 \; \left(\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho} \right)^s
 \dfrac{\Gamma\left(s+1 \right)}{ \Gamma\left(s+2 \right)}
 \Fh21\Fz{s + 1, 1}{s+ 2}{\dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2} }
 \\
 &&\nonumber\\
 &&-\Big\{x_{ij}
 \leftrightarrow x_{ji};\;
 x_k \leftrightarrow 1-x_k\Big\}-\mathrm{term},
 \nonumber
 \end{eqnarray}
provided that
$\left|\frac{x_{ij}^2 }{ (x_k - x_{ij})^2}\right|<1$
for $i,j,k=1,2,3$
and $\mathcal{R}$e$\left(s+\frac{1}{2}\right)>0$.

Putting this result into Eq.~(\ref{MB1}), we are going
to evaluate the following Mellin-Barnes integral
\begin{eqnarray}
\label{MB2}
\dfrac{1}{2\pi i} \int\limits_{-i\infty}^{+i\infty}ds \;
\dfrac{\Gamma(-s)\;\Gamma(a +s) \;\Gamma(b+s)}{\Gamma(c+s) }
(-x)^s \; \Fh21\Fz{a+s,b'}{c+s}{y},
\end{eqnarray}
with $|\mathrm{Arg}(-x)|<\pi$ and $|x|<1$ and $|y|<1$.
Under these conditions,
one could close the integration contour to
the right side of the imaginary axis in the
$s$-complex plane. Subsequently, we take
into account the residua of the
sequence poles of $\Gamma(-s)$. The result
is presented as a series of Appell $F_1$
functions~\cite{Slater}
\begin{eqnarray}
\label{MB3}
\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(c)} \sum\limits_{m=0}^{\infty}
\dfrac{(a)_m (b)_m}{(c)_m} \dfrac{x^m}{m!} \Fh21\Fz{a+m,b'}{c+m}{y}
=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(c)}\; F_1 (a; b, b';c; x,y),
\nonumber
\end{eqnarray}
provided that $|\mathrm{Arg}(-x)|<\pi$
and $|x|<1$ and $|y|<1$. Finally, the result
for $\mathcal{K}$ reads
\begin{eqnarray}
\label{K1}
\mathcal{K} &=&  -\dfrac{x_{ij}}{(x_k -x_{ij} )}\;
\left(M_{ij} -i\rho\right)^{\frac{d-4}{2} }
\; F_1\left(\frac{1}{2}; 1, \frac{4-d}{2};
\frac{3}{2}; \dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2},
-\dfrac{p_i^2 x_{ij}^2}{M_{ij} -i\rho}  \right)\nonumber\\
&&+ \dfrac{x^2_{ij}}{2 (x_k -x_{ij} )^2}\;
\left(M_{ij} -i\rho\right)^{\frac{d-4}{2} }
\; F_1\left(1; 1, \frac{4-d}{2}; 2;
\dfrac{x_{ij}^2 }{ (x_k - x_{ij})^2},
-\dfrac{p_i^2 x_{ij}^2}{M_{ij}-i\rho }  \right)
\nonumber   \\
&&\nonumber \\
&&-\Big\{x_{ij} \leftrightarrow x_{ji};
\; x_k \leftrightarrow 1-x_k\Big\}-\mathrm{term},
\end{eqnarray}
provided that $\left|\frac{x_{ij}^2 }{ (x_k - x_{ij})^2} \right| <1$
and $ \left|-\frac{p_i^2 x_{ij}^2}{M_{ij}} \right| <1$ for $i,j,k=1,2,3$.
It is confirmed that
Arg$\left( \frac{x_{ij}^2 }{ (x_k - x_{ij} )^2 }\right)<\pi$
and Arg$\left(-\frac{p_i^2 x_{ij}^2}{M_{ij}}\right)<\pi$
due to the $i\rho$-term.
In other words, these kinematic variables are
never on the negative real axis.

Applying the transformation for Appell $F_1$ functions
(see Eq.~(\ref{f1relation1}) in appendix $B$, one finds
another representation for integral $\mathcal{K}$ as
\begin{eqnarray}
\label{K2}
 \mathcal{K} &=&
 \dfrac{x_{ij} (x_{ij}-x_k)
 \left(m_i^2 \right)^{\frac{d-4}{2} }
 }{ (x_k -x_{ij} )^2 - x_{ij}^2 }
 \; F_1\left(1; 1, \frac{4-d}{2};
 \frac{3}{2}; \dfrac{-x_{ij}^2}{ (x_k -x_{ij} )^2 -x_{ij}^2 },
\frac{p_i^2 x_{ij}^2 }{p_i^2 x_{ij}^2+M_{ij} -i\rho}  \right)
\nonumber\\
 && + \dfrac{x^2_{ij} \left(m^2_{i} \right)^{\frac{d-4}{2}}
 }{2 \left[ (x_k -x_{ij} )^2 -x_{ij}^2\right] }\;
 F_1\left(1; 1, \frac{4-d}{2}; 2;
 \dfrac{-x_{ij}^2}{ (x_k -x_{ij} )^2 -x_{ij}^2 },
\frac{p_i^2 x_{ij}^2 }{p_i^2 x_{ij}^2   +M_{ij}-i\rho }
\right)         \nonumber \\
 &&\nonumber\\
 &&-\Big\{x_{ij} \leftrightarrow x_{ji};\;
 x_k \leftrightarrow 1-x_k\Big\}-\mathrm{term},
\end{eqnarray}
provided that $ \left|-\frac{x_{ij}^2}
{ (x_k -x_{ij} )^2 -x_{ij}^2 } \right| <1$ and
$ \left| \frac{p_i^2 x_{ij}^2 }
{p_i^2 x_{ij}^2   +M_{ij} } \right| <1$ for $i,j,k=1,2,3$.
\section*{Appendix $B$:
Generalized hypergeometric functions}
The Gauss hypergeometric series are given
(see Eq.~($1.1.1.4$) in Ref.~\cite{Slater})
\begin{eqnarray}
\label{gauss-series}
\Fh21\Fz{a,b}{c}{z}
= \sum\limits_{n=0}^{\infty}
\dfrac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!},
\end{eqnarray}
provided that $|z|<1$. Here, the pochhammer symbol,
\begin{eqnarray}
(a)_n =\dfrac{\Gamma(a+n)}{\Gamma(a)},
\end{eqnarray}
is taken into account.
The integral representation for Gauss hypergeometric
functions is (see Eq.~(1.6.6) in Ref.~\cite{Slater})
\begin{eqnarray}
\label{gauss-int}
\Fh21\Fz{a,b}{c}{z}
=\dfrac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\limits_0^1 du
\; u^{b-1} (1-u)^{c-b-1} (1-zu)^{-a},
\end{eqnarray}
provided that  $|z|<1$ and Re$(c)>$Re$(b)>0$.

The series of Appell $F_1$ functions are  given
(see Eq.~(8.13) in Ref.~\cite{Slater})
\begin{eqnarray}
\label{appell-series}
F_1(a; b, b'; c; x, y)
= \sum\limits_{m=0}^{\infty}\sum\limits_{n=0}^{\infty}
  \dfrac{(a)_{m+n} (b)_m (b')_n}{(c)_{m+n}\; m! n!} x^m y^n,
\end{eqnarray}
provided that $|x|<1$ and $|y|<1$.
The single integral representation for $F_1$ is
(see Eq.~(8.25) in Ref.~\cite{Slater})
\begin{eqnarray}
\label{appell-int}
F_1(a; b, b'; c; x, y)
=  \dfrac{\Gamma(c)}{\Gamma(c-a)\Gamma(a)} \int\limits_0^1 du\;
   u^{a-1} (1-u)^{c-a-1} (1-xu)^{-b}(1-yu)^{-b'},
\end{eqnarray}
provided that Re$(c)$ $>$ Re$(a)>0$ and $|x|<1$, $|y|<1$.
\subsection*{Transformations for Gauss
$_2F_1$ hypergeometric functions}
Basic linear transformation formulas for
Gauss $_2F_1$ hypergeometric functions
collected from Ref.~\cite{Slater}
are listed as follows
\begin{eqnarray}
\Fh21\Fz{a,b}{c}{z} &=& \Fh21\Fz{b,a}{c}{z}
\label{tran2F1a}   \\
&=& (1-z)^{c-a-b} \Fh21\Fz{c-a,c-b}{c}{z}
\label{tran2F1b} \\
&=& (1-z)^{-a} \Fh21\Fz{ a,c-b}{c}{\frac{z}{z-1} }
\label{tran2F1c} \\
&=& (1-z)^{-b} \Fh21\Fz{ b,c-a}{c}{\frac{z}{z-1} }
\label{tran2F1d}\\
&=& \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
\Fh21\Fz{a,b}{a+b-c+1}{1-z} \nonumber \\
&+& (1-z)^{c-a-b} \frac{\Gamma(c) \Gamma(a+b-c)}{\Gamma(a) \Gamma(b)}
\Fh21\Fz{c-a,c-b}{c-a-b+1}{1-z}\\
&=& \frac{\Gamma(c) \Gamma(b-a)}{\Gamma(b) \Gamma(c-a)} (-z)^{-a}
\Fh21\Fz{a,1-c+a}{1-b+a}{\frac{1}{z} } \nonumber \\
&+& \frac{\Gamma(c) \Gamma(a-b)}{\Gamma(a) \Gamma(c-b)} (-z)^{-b}
\Fh21\Fz{b,1-c+b}{1-a+b}{\frac{1}{z} }\label{z-}.
\end{eqnarray}
\subsection*{Transformations for Appell
$F_1$ hypergeometric functions}
We collect all transformations for Appell
$F_1$ functions from
Refs.~\cite{Slater}. The first
relation for $F_1$ is mentioned,
\begin{eqnarray}
\label{f1relation1}
F_1\Big(a;b,b';c;x,y\Big)=(1-x)^{-b}(1-y)^{-b'}
F_1\Big(c-a;b,b';c;\frac x{x-1},\frac y{y-1}\Big).
\end{eqnarray}
If $b'=0$, we arrive at the well-known
Pfaff--Kummer transformation for the $_2F_1$.
In detail, one has
\begin{eqnarray}
F_1\Big(a;b,b';c;x,y\Big)=(1-x)^{-a}
F_1\!\left(a;-b-b'+c,b';c;\frac x{x-1},\frac{y-x}{1-x}\right).
\end{eqnarray}
Furthermore, if $c=b+b'$, one then obtains
\begin{eqnarray}
F_1\Big(a;b,b';b+b';x,y\big)&=&(1-x)^{-a}
\Fh21\Fz{a,b'}{b+b'}{\frac{y-x}{1-x}}
\\
&=&
(1-y)^{-a}\Fh21\Fz{a,b}{b+b'}{\frac{x-y}{1-y}}.
\end{eqnarray}
Similarly,
\begin{eqnarray}
 F_1\left(a;b,b';c;x,y\right)&=&(1-y)^{-a}
 F_1\!\left(a;b,c-b-b';c;\frac{x-y}{1-y},\frac y{y-1}\right),
 \\
 F_1\left(a;b,b';c;x,y\right) &=&
 {\mbox{\small $(1-x)^{c-a-b}(1-y)^{-b'}$}}
 F_1\!\left(c-a;c-b-b',b';c;x,\frac{x-y}{1-y}\right),
\\
 F_1\left(a;b,b';c;x,y\right) &=&
 {\mbox{\small $(1-x)^{-b}(1-y)^{c-a-b'}$}}
 F_1\!\left(c-a;b,c-b-b';c;\frac{y-x}{1-x},y\right).
 \end{eqnarray}


\section{\@startsection{section}{1}{\z@}{3.5ex plus 1ex minus
   .2ex}{2.3ex plus .2ex}{\bf}}
\def\subsection{\@startsection{subsection}{2}{\z@}{3.25ex plus 1ex minus
   .2ex}{1.5ex plus .2ex}{\bf}}
\def\bf Appendix B: {\Roman{section}.}
\def\Alph{subsection}.{\Alph{subsection}.}
\def\arabic{subsubsection}.{\arabic{subsubsection}.}
\def