\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \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} \\ \end{center} \vskip .2 in \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} \newtheorem{theorem}{Theorem} \newtheorem{thm}[theorem]{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \newenvironment{remk}{\begin{rema}\normalfont\quad}{\end{rema}} %\usepackage{mathrsfs} %\usepackage{latexsym,rotate,eucal,cite}%%%%%%%%%%%%%%% %\usepackage{amsmath,amsthm,amssymb,amsxtra}%%%%%%%%%%% \newcommand{\qdp}{\hspace*{1.5mm}}% \newcommand{\qqdp}{\hspace*{2.5mm}} \newcommand{\xqdp}{\hspace*{5.0mm}} 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in equation%% \newcommand{\xgatherx}[1]{\begin{gather}#1%%%%%%%%%%%% \end{gather}} %%%%%gather senza & %%%%% \newcommand{\xgatherz}[1]{\begin{gather*}#1%%%%%%%%%%% \end{gather*}} %%%%%gather senza & %%%%%% \newcommand{\xgatheredx}[1]{\begin{gathered}#1%%%%%%%% \end{gathered}}%gathered=no& in equation%% \newcommand{\mult}[2]{\begin{array}{#1}#2\end{array}}% \newcommand{\xmultx}[1]{\begin{multline}#1%%%%%%%%%%%% \end{multline}} %%%multline senza & %%%%%% \newcommand{\xmultz}[1]{\begin{multline*}#1%%%%%%%%%%% \end{multline*}}%%%multline senza & %%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF% %%%%%%%%%%%%%%%%%%%%% positions %%%%%%%%%%%%%%%%%%%%%% \newcommand{\centro}[1] {\begin{center}#1\end{center}} \newcommand{\sinistra}[1] {\begin{flushleft}#1\end{flushleft}} \newcommand{\destra}[1] {\begin{flushright}#1\end{flushright}} \newcommand{\centerbox}[2]{\centro{\begin{tabular}{|c|}\hline \parbox{#1}{\mbox{}\\[2.5mm]#2\mbox{}\\}\\\hline\end{tabular}}} \newcommand{\centrobmp}[3]{\vspace*{1mm} \centerbmp{#1}{#2}{#3}\vspace*{1mm}} \newcommand{\sav}[1]{\newsavebox{#1}% \sbox{#1}{\theequation}\\[-2mm]}%=\newline \newcommand{\usa}[1]{(\usebox{#1})}%{#1}={\some-nome} \newcommand{\summary}[2]{\begin{center}\parbox{#1} {{\sc\bf Summary}:\,{\small\it#2}}\end{center}} \newcommand{\riassunto}[2]{\begin{center}\parbox{#1} {{\sc\bf Riassunto}:\,{\small\it#2}}\end{center}} \newcommand{\grazia}[2]{\begin{center}\parbox{#1} {{\sc\bf Ringraziamento}:\,{\small\it#2}}\end{center}} \newcommand{\thank}[2]{\begin{center}\parbox{#1} {{\sc\bf Acknowledgement}:\,{\small\it#2}}\end{center}} \newcommand{\bbtm}[5]{\bibitem{kn:#1}{#2,}~{#3,}~\emph{#4}~{#5.}} \newcommand{\bbtmn}[4]{\bibitem{kn:#1}{#2,}~\emph{#3,}~{#4.}} \newcommand{\bbtmm}[4]{\bibitem{kn:#1}{#2,}~{#4.}} \newcommand{\cito}[1]{\cite{kn:#1}} \newcommand{\citu}[2]{\cite[#2]{kn:#1}} \newcommand{\mr}{\textbf{MR}\:} \newcommand{\zbl}{\textbf{Zbl}\:} \newcommand{\nota}[2]{\centerbox{#1}{\textbf{Achtung}\quad#2}} \newcommand{\graph}[3]{\begin{picture}(0,0) \put(#1,#2){#3}\end{picture}} \newcommand{\fwd}{\mbox{$\bigtriangleup\qdn\!\cdot\,\:$}} \newcommand{\bwd}{\bigtriangledown} \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 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} \bbtm{alm-kra}{G. 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Schroeppel} {} {Hakmem, Artificial Intelligence Memo No.\ 239, MIT, February 29 1972; available at \\ \href{http://adamo.decsystem.org/hakmem/hakmem/series.html} {\tt http://adamo.decsystem.org/hakmem/hakmem/series.html}} \bbtm{kra-rao}{C. Krattenthaler and K. S. Rao} {Automatic generation of hypergeometric identities by the beta integral method} {J. Comp. Appl. Math. } {$\mathbf{160}$ (2003), 159--173} \bbtm{sofo-triple}{A. Sofo} {Sums of reciprocals of triple binomial coefficients} {Int. J. Math. Math. Sci.} {2008, Art.\ ID 794181} \bbtm{sofo20091}{A. Sofo} {Some properties of reciprocals of double binomial coefficients} {Tamsui Oxf. J. Math. Sci.} {$\mathbf{25}$ (2) (2009), 141--151} \bbtm{sofo}{A. Sofo} {Harmonic numbers and double binomial coefficient} {Integral Transforms Spec. Funct.} {$\mathbf{20}$ (11) (2009), 847--857} \bbtmn{slater}{L. J. Slater} {Generalized Hypergeometric Functions} {Cambridge University Press, 1966} \bbtm{sondow}{J. Sondow} {Double integrals for Euler's constant and $\ln \frac{4}{\pi}$ and an analog of Hadjicosta's formula} {Amer. Math. Monthly} {$\mathbf{112}$ (2005), 61--65} \bbtmn{sri-choi}{H. M. Srivastava, J. Choi} {Series Associated with the Zeta and Related Functions} {Kluwer Academic Publishers, 2001} \end{thebibliography} \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}