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\vskip 1cm{\LARGE\bf Integral Representations and Binomial \\
\vskip .1in
Coefficients
}
\vskip 1cm
\large
Xiaoxia Wang\footnote{
This work is supported by Shanghai Leading
Academic Discipline Project, Project No.\ J50101,
and a post-doctoral fellowship from
the University of Salento, Department of Mathematics.} \\
Department of Mathematics\\
Shanghai University\\
Shanghai, China\\
\href{mailto:xiaoxiawang@shu.edu.cn}{\tt xiaoxiawang@shu.edu.cn} \\
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\begin{abstract}
In this article, we present
two extensions of Sofo's theorems on
integral representations of ratios of reciprocals of double
binomial coefficients.
From the two extensions,
we get several new relations between integral representations
and binomial coefficients.
\end{abstract}
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\section{Introduction}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Recently, Sofo \cito{sofo} extended the result in relation to the integral representations
of ratios of reciprocals of the double binomial coefficients with the help of Beta function
in integral form. In \cito{sofo}, Sofo investigated integral representations for
\[\sum_{n=0}^\infty f_n(a, b, j, k, t),\]
which is a function of the reciprocal double binomial coefficients and derivatives,
and then Sofo reproved many results in \cite{kn:cloitre,kn:gs,kn:sondow}.
For the completion of this article, we reproduce the $\Gamma$-function defined through Euler integral
\[\Gamma(x)=\int_0^\infty u^{x-1}e^{-u}\text{d}u
\xqdp \text{with}\xqdp \mathfrak{R}(x)>0,\]
and Beta function
\bnm
B(s, t)=\int_0^1z^{s-1}(1-z)^{t-1}dz=\frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}
\quad \text{for} \:\: \mathfrak{R}(s)>0\:\: \text{and}\:\: \mathfrak{R}(t)>0
\enm
which is very useful in the work of simplification and representation of binomial sums in closed
integral form. The reader may refer to \cite{kn:alm-kra,kn:alz-kar,kn:kra-rao,kn:sri-choi}.
Throughout this paper, $\mathbb{N}$ and $\mathbb{R}$ denote the
natural numbers and the real numbers, respectively.
In fact, Sofo first gave the following theorem in \cito{sofo20091} about the
relation between the
integral representations and double binomial coefficients.
\begin{thm}\label{sofo1}
For $t\in \mathbb{R}$ and $a,\: b,\: n, \:j$ and $k \in \mathbb{N}$ subject to $|t|\leq1$ and $j, \:k >0$, then
\bnm
\+\+\sum_{n=0}^\infty\frac{t^n}{{an+j\choose j}{bn+k\choose k}}
=jk\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}}{1-tx^ay^b}dxdy\\[2mm]
\+\+=_{j+k+1}F_{j+k}\ffnk{c}{t}
{1,\:\:\frac{1}{a},\:\:\frac{2}{a},\:\:\cdots,\:\:\frac{j}{a},\:\:\frac{1}{b},\:\:\frac{2}{b},\:\:\cdots,\:\:\frac{k}{b}}
{\frac{a+1}{a},\frac{a+2}{a}, \cdots, \frac{a+j}{a},\frac{b+1}{b},\frac{b+2}{b},\cdots, \frac{b+k}{b}}.
\enm
\end{thm}
The $_{j+k+1}F_{j+k}(t)$ in this theorem is the general
hypergeometric series, the readers can refer to
Bailey~\cito{bailey} and Slater~\cito{slater}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In recent work \cito{sofo}, Sofo extended Theorem \ref{sofo1} by the following theorem.
\begin{thm}\label{sofo2}
For $t\in \mathbb{R}$ and $a,\: b, \:c, \:d,\: n, \:j$ and $k \in \mathbb{N}$ subject to
$a\geq b$, $c\geq d$, $j, \:k >0$ and
\[\big| t\frac{b^b(a-b)^{a-b}d^d(c-d)^{c-d}}{a^ac^c}\big|\leq1,\] then
\bnm
\+\+S(a,b,c,d,j,k,t)=\sum_{n=0}^\infty\frac{t^n}{{an+j\choose b n}{bn+k\choose d n}}\\[2mm]
\+\+=(a-b)(c-d)\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}U(1+U)}{(1-U)^3}\:dxdy\\[2mm]
\+\++\Big[(a-b)k+(c-d)j\Big]\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}U}{(1-U)^2}\:dxdy\\[2mm]
\+\++jk\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}}{1-U}\:dxdy,
\enm
where
\[U:=tx^b(1-x)^{a-b}y^d(1-y)^{c-d}.\]
\end{thm}
The purpose of this paper is to present two extensions of Sofo's Theorems \ref{sofo1} and
\ref{sofo2}, and then get some new results on the relations between integral
representations and binomial coefficients.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The first extension of Sofo's theorems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section, we prove an extension of Sofo's theorems,
which will lead to several new relations between double integral representations and
binomial coefficients.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{thm}[The first extension]\label{wxx}
For $t\in \mathbb{R}$ and $a,\: b, \:c, \:d,\: n, \:i,\: j, \:k $ and $l \in \mathbb{N}$ subject to
$a \geq b$, $c \geq d$, $j,\: k >0$, $i,\: l \geq 0$ and
\[\big| t\frac{b^b(a-b)^{a-b}d^d(c-d)^{c-d}}{a^ac^c}\big|\leq1,\]
then
\bnm
\+\+Q(a,b,c,d,i,j,k,l,t)=\sum_{n=0}^\infty\frac{t^n}{{an+j\choose b n+i}{cn+k\choose d n+l}}\\[3mm]
\+\+=(a-b)(c-d)\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}U(1+U)}{(1-U)^3}\:dxdy\\[2mm]
\+\++\big[(a-b)(k-l)+(c-d)(j-i)\big]\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}U}{(1-U)^2}\:dxdy\\[2mm]
\+\++(j-i)(k-l)\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}}{1-U}\:dxdy,
\enm
where
\[U:=tx^b(1-x)^{a-b}y^d(1-y)^{c-d}.\]
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Obviously, when $a=b, \:c=d$ and $i=l=0$, Theorem \ref{wxx} reduces to Theorem \ref{sofo1}
which is due to Sofo \cito{sofo20091}. Letting $i=l=0, \:a\neq b$ and $c\neq d$ in Theorem \ref{wxx},
we obtain Theorem \ref{sofo2} which is given by Sofo \cito{sofo}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{proof}
The summation of double binomial coefficients in this theorem can be expressed as follows:
\bnm
Q(a,b,c,d,i,j,k,l,t)
\+=\+\sum_{n=0}^\infty\frac{\Gamma(bn+i+1)\big[(a-b)n+j-i\big]\Gamma\big((a-b)n+j-i\big)}{\Gamma(an+j+1)}\\[2mm]
\+\times\+\frac{\Gamma(dn+l+1)\big[(c-d)n+k-l\big]\Gamma\big((c-d)n+k-l\big)}{\Gamma(cn+k+1)}t^n\\[2mm]
\+=\+\sum_{n=0}^\infty\Big[(a-b)n+j-i\Big]\Big[(c-d)n+k-l\Big]t^n\\[2mm]
\+\times\+ B\big(bn+i+1, (a-b)n+j-i\big)B\big(dn+l+1, (c-d)n+k-l\big).
\enm
Expanding the beta functions by the integral function, we express the $Q(a,b,c,d,i,j,k,l,t)$ as follows.
\bnm
Q(a,b,c,d,i,j,k,l,t)
\+=\+\sum_{n=0}^\infty\Big[(a-b)n+j-i\Big]\Big[(c-d)n+k-l\Big]t^n\\[2.5mm]
\+\times\+\int_0^1x^{bn+i}(1-x)^{(a-b)n+j-i-1}dx\int_0^1y^{dn+l}(1-y)^{(c-d)n+k-l-1}dy\\[2.5mm]
\+=\+\sum_{n=0}^\infty\Big\{(a-b)(c-d)n^2+\big[(a-b)(k-l)+(c-d)(j-i)\big]n+(j-i)(k-l)\Big\}\\[2.5mm]
\+\times\+\int_0^1\int_0^1x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}
\Big[tx^b(1-x)^{a-b}y^d(1-y)^{c-d}\Big]^ndxdy
\enm
Exchanging the double integral function and summation, we get
\bnm
Q(a,b,c,d,i,j,k,l,t)
\+=\+\int_0^1\int_0^1x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}\Big\{\sum_{n=0}^\infty(a-b)(c-d)n^2U^n\\[2.5mm]
\++\+\big[(a-b)(k-l)+(c-d)(j-i)\big]nU^n+(j-i)(k-l)U^n\Big\}\:dxdy\\[2.5mm]
\+=\+(a-b)(c-d)\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}U(1+U)}{(1-U)^3}\:dxdy\\[2.5mm]
\++\+\big[(a-b)(k-l)+(c-d)(j-i)\big]\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}U}{(1-U)^2}\:dxdy\\[2.5mm]
\++\+(j-i)(k-l)\int_0^1\int_0^1\frac{x^i(1-x)^{j-i-1}y^l(1-y)^{k-l-1}}{1-U}\:dxdy,
\enm
where we have applied the derivation operator in the last equality to evaluate the summations.
Here the requirement $|tb^b(a-b)^{a-b}d^d(c-d)^{c-d}/a^ac^c|\leq1$ is for convergence.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=c=j=2, \:b=d=i=k=t=1$ and $l=0$ in Theorem \ref{wxx}, we have the following
result:
\bnm
\+Q\+(2,1,2,1,1,2,1,0,1)=\sum_{n=0}^\infty\frac{1}{{2n+2\choose n+1}{2n+1\choose n}}\\[2mm]
\+=\+\int_0^1\int_0^1\frac{x^2y(1-x)(1-y)\big[1+xy(1-x)(1-y)\big]}{\big[1-xy(1-x)(1-y)\big]^3}\:dxdy\\[2mm]
\++\+2\int_0^1\int_0^1\frac{x^2y(1-x)(1-y)}{\big[1-xy(1-x)(1-y)\big]^2}\:dxdy\\[2mm]
\++\+\int_0^1\int_0^1\frac{x}{1-xy(1-x)(1-y)}\:dxdy \\[2mm]
\+=\+\int_0^1\int_0^1\frac{x(1+xy-x^2y-xy^2+x^2y^2)}{\big[1-xy(1-x)(1-y)\big]^3}\:dxdy.
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=3,\: c=k=2$ and $b=d=i=j=l=t=1$ in Theorem \ref{wxx}, we get
\bnm
\+Q\+(3,1,2,1,1,1,2,1,1)=\sum_{n=0}^\infty\frac{1}{{3n+1\choose n+1}{2n+2\choose n+1}}\\[2mm]
\+=\+2\int_0^1\int_0^1\frac{x^2y^2(1-x)(1-y)\big[1+x y(1-x)^2(1-y)\big]}{\big[1-x y(1-x)^2(1-y)\big]^3}\:dxdy\\[2mm]
\++\+2\int_0^1\int_0^1\frac{x^2y^2(1-x)(1-y)}{\big[1-x y(1-x)^2(1-y)\big]^2}\:dxdy\\[2mm]
\+=\+4\int_0^1\int_0^1\frac{x^2y^2(1-x)(1-y)}{\big[1-x y(1-x)(1-y)^2\big]^3}\:dxdy.
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=j=3,\: b=c=i=k=2$ and $d=l=t=1$ in Theorem \ref{wxx},
we have
\bnm
\+Q\+(3,2,2,1,2,3,2,1,1)=\sum_{n=0}^\infty\frac{1}{{3n+3\choose 2n+2}{2n+2\choose n+1}}\\[3mm]
\+=\+\int_0^1\int_0^1\frac{x^4y^2(1-x)(1-y)\big[1+x^2y(1-x)(1-y)\big]}{\big[1-x^2y(1-x)(1-y)\big]^3}\:dxdy\\[3mm]
\++\+2\int_0^1\int_0^1\frac{x^4y^2(1-x)(1-y)}{\big[1-x^2y(1-x)(1-y)\big]^2}\:dxdy\\[3mm]
\++\+\int_0^1\int_0^1\frac{x^2y}{1-x^2y(1-x)(1-y)}\:dxdy\\[3mm]
\+=\+\int_0^1\int_0^1\frac{x^2y\big(1+x^2y-x^3y-x^2y^2+x^3y^2\big)}{\big[1-x^2y(1-x)(1-y)\big]^3}\:dxdy.
\enm
In fact, $Q(3,2,2,1,2,3,2,1,1)$ can be expressed as the Hakmem series \cito{hakmem} as follows:
\bnm
Q(3,2,2,1,2,3,2,1,1)\+=\+\sum_{n=0}^\infty\frac{1}{{3n+3\choose 2n+2}{2n+2\choose n+1}}\\[2mm]
\+=\+\sum_{n=0}^\infty\frac{(n+1)!(n+1)!(n+1)!}{(3n+3)!}=\sum_{n=0}^\infty\frac{n!n!n!}{(3n)!}-1\\[2mm]
\+=\+-1+\int_0^1\Big\{\frac{2(8+7t^2-7t^3)}{(4-t^2+t^3)^2}\\[2mm]
\++\+\frac{4t(1-t)(5+t^2-t^3)}{(4-t^2+t^3)^2\sqrt{(1-t)(4-t^2+t^3)}}
\arccos\big(\frac{2-t^2+t^3}{2}\big)\Big\}dt
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=4,\: b=c=j=k=2$ and $d=i=l=t=1$ in Theorem \ref{wxx}, we derive the following relation.
\bnm
\+Q\+(4,2,2,1,1,2,2,1,1)=\sum_{n=0}^\infty\frac{1}{{4n+2\choose 2n+1}{2n+2\choose n+1}}\\[2mm]
\+=\+2\int_0^1\int_0^1\frac{x^3y^2(1-x)^2(1-y)\big[1+x^2y(1-x)^2(1-y)\big]}{\big[1-x^2y(1-x)^2(1-y)\big]^3}\:dxdy\\[2mm]
\++\+3\int_0^1\int_0^1\frac{x^3y^2(1-x)^2(1-y)}{\big[1-x^2y(1-x)^2(1-y)\big]^2}\:dxdy\\[2mm]
\++\+\int_0^1\int_0^1\frac{x y}{1-x^2y(1-x)^2(1-y)}\:dxdy\\[2mm]
\+=\+\int_0^1\int_0^1\frac{xy\big(1+3x^2y-6x^3y+3x^4y-3x^2y^2+6x^3y^2-3x^4y^2\big)}{\big[1-xy^2(1-x)(1-y)^2\big]^3}\:dxdy.
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The second extension of Sofo's theorems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section, we give another extension of Sofo's Theorems \ref{sofo1} and \ref{sofo2}.
The following theorem is about the relation between three binomial coefficients and triple integral representations.
\begin{thm}[The second extension]\label{wxx-2}
For $t\in \mathbb{R}$ and $a,\: b, \:c, \:d,\: e,\: n,\: j, \:k $ and $m \in \mathbb{N}$ subject to
$a\geq b$, $c\geq d$, $j, \:k, \: m >0$ and
\[\big| t\frac{b^b(a-b)^{a-b}d^d(c-d)^{c-d}}{a^ac^c}\big|\leq1,\]
then
\bnm
\+\+T(a,b,c,d,e,j, k, m,t)=\sum_{n=0}^\infty\frac{t^n}{{an+j\choose b n}{cn+k\choose d n}{en+m\choose en}}\\[3mm]
\+\+=m(a-b)(c-d)\int_0^1\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}(1-z)^{m-1}W(1+W)}{(1-W)^3}\:dxdydz\\[2mm]
\+\++m\big[k(a-b)+j(c-d)\big]\int_0^1\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}(1-z)^{m-1}W}{(1-W)^2}\:dxdydz\\[2mm]
\+\++mkj\int_0^1\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}(1-z)^{m-1}}{1-W}\:dxdydz,
\enm
where
\[W:=tx^b(1-x)^{a-b}y^d(1-y)^{c-d}z^e.\]
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Clearly, when $a=b, \:c=d$ and $e=0$, Theorem \ref{wxx-2} reduces to
Sofo's \cito{sofo20091} Theorem \ref{sofo1}. Letting
$e=0, \:a\neq b$ and $c\neq d$ in Theorem \ref{wxx-2}, we obtain
Theorem \ref{sofo2} which is presented by Sofo \cito{sofo}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The proof of this theorem is as same as we have given for
Theorem \ref{wxx}. Now we present some examples
of this theorem.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=b,\:c=d$ and $t=\pm1$ in Theorem \ref{wxx-2}, we obtain the following
results:
\bnm
\+T\+(a,a,c,c,e,j, k, m,\pm1)=\sum_{n=0}^\infty\frac{(\pm1)^n}{{an+j\choose an}{cn+k\choose cn}{en+m\choose en}}\\[3mm]
\+=\+j k m\int_0^1\int_0^1\int_0^1\frac{(1-x)^{j-1}(1-y)^{k-1}(1-z)^{m-1}}{1\mp x^ay^cz^e}\:dxdydz,
\enm
which is due to Sofo \cito{sofo-triple}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=c=2,\: b=d=e=j=k=m=t=1$ in Theorem \ref{wxx-2}, we have the following
result:
\bnm
\+T\+(2,1,2,1,1,1,1,1,1)=\sum_{n=0}^\infty\frac{1}{{2n+1\choose n}{2n+1\choose n}{n+1\choose n}}\\[2mm]
\+=\+\int_0^1\int_0^1\int_0^1\frac{xyz(1-x)(1-y)\big[1+xyz(1-x)(1-y)\big]}{\big[1-xyz(1-x)(1-y)\big]^3}\:dxdydz\\[2mm]
\++\+2\int_0^1\int_0^1\int_0^1\frac{xyz(1-x)(1-y)}{\big[1-xyz(1-x)(1-y)\big]^2}\:dxdydz\\[2mm]
\++\+\int_0^1\int_0^1\int_0^1\frac{1}{1-xyz(1-x)(1-y)}\:dxdydz,\\[2mm]
\+=\+\int_0^1\int_0^1\int_0^1\frac{1+xyz-x^2yz-xy^2z+x^2y^2z}{\big[1-xyz(1-x)(1-y)\big]^3}\:dxdydz.
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Letting $a=3,\: b=c=2$ and $d=e=k=j=m=t=1$ in Theorem \ref{wxx-2}, we establish the following
result:
\bnm
\+T\+(3,2,2,1,1,1,1,1,1)=\sum_{n=0}^\infty\frac{1}{{3n+1\choose 2n}{2n+1\choose n}{n+1\choose n}}\\[2mm]
\+=\+\int_0^1\int_0^1\int_0^1\frac{x^2y z(1-x)(1-y)\big[1+x^2y z(1-x)(1-y)\big]}{[1-x^2y z(1-x)(1-y)]^3}\:dxdydz\\[2mm]
\++\+2\int_0^1\int_0^1\int_0^1\frac{x^2y z(1-x)(1-y)}{\big[1-x^2y z(1-x)(1-y)\big]^2}\:dxdydz\\[2mm]
\++\+\int_0^1\int_0^1\int_0^1\frac{1}{1-x^2y z(1-x)(1-y)}\:dxdydz,\\[2mm]
\+=\+\int_0^1\int_0^1\int_0^1\frac{1+x^2yz-x^3yz-x^2y^2z+x^3y^2z}{\big[1-x^2y z(1-x)(1-y)\big]^3}\:dxdydz.
\enm
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B65; Secondary 05A10, 33C20.
\noindent \emph{Keywords: }
beta function, integral representations, binomial coefficients.
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received April 12 2010;
revised version received June 7 2010.
Published in {\it Journal of Integer Sequences}, June 9 2010.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in
\end{document}