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\markboth{\footnotesize \emph{\emph{Global Journal of Mathematical Analysis}}}{\footnotesize \emph{\emph{Global Journal of Mathematical Analysis}}}
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\begin{center} \huge{\bf An integral representation of the Catalan numbers}
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\centerline{\bf Xiao-Ting Shi$^*$, Fang-Fang Liu, Feng Qi}

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{\small
\begin{center}
\emph{Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China}
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\centerline{\emph{*Corresponding author e-mail: xiao-ting.shi@hotmail.com, xiao-ting.shi@qq.com}}

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\noindent \hspace{-3 pt}{\scriptsize \textbf{ Copyright \copyright 2015 Xiao-Ting Shi, Fang-Fang Liu, Feng Qi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.}}
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\noindent \textbf{Abstract}\\
\centerline{}
In the paper, the authors establish an integral representation of the Catalan numbers, connect the Catalan numbers with the (logarithmically) complete monotonicity, and pose an open problems on the logarithmically complete monotonicity of a function involving ratio of gamma functions.\\
\centerline{}
\noindent {\footnotesize \emph{\textbf{Keywords}}:  \emph{Catalan number; integral representation; complete monotonicity; logarithmically complete monotonicity; open problem}}

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\noindent\hrulefill


\section{Introduction}

It is known~\cite{wolfram-Catalan} that, in combinatorics, the Catalan numbers $C_n$ for $n\ge0$ form a sequence of natural numbers that occur in tree enumeration problems of the type, ``In how many ways can a regular $n$-gon be divided into $n-2$ triangles if different orientations are counted separately?'' The solution is the Catalan number $C_{n-2}$. They are named after the Belgian mathematician Eug\`ene Charles Catalan.
The first few Catalan numbers $C_n$ for $0\le n\le11$ are
\begin{equation*}
1,\quad 1,\quad 2,\quad 5,\quad 14,\quad 42,\quad 132,\quad 429,\quad 1430,\quad 4862,\quad 16796,\quad 58786.
\end{equation*}
Explicit formulas of $C_n$ for $n\ge0$ include
\begin{equation*}
C_n=\frac1{n+1}\binom{2n}{n}
=\frac{(2n)!}{n!(n+1)!}
=\frac{2^n(2n-1)!!}{(n+1)!}
=(-1)^n2^{2n+1}\binom{\frac12}{n+1}
=\frac1n\binom{2n}{n-1}
={}_2F_1(1-n,-n;2;1)
\end{equation*}
and
\begin{equation}\label{CatalanN-Gamma}
C_n=\frac{4^n\Gamma(n+1/2)}{\sqrt\pi\,\Gamma(n+2)},
\end{equation}
where
\begin{equation*}
\Gamma(z)=\int^\infty_0t^{z-1} e^{-t}\td t, \quad \Re(z)>0
\end{equation*}
is the classical Euler gamma function and
\begin{equation*}%\label{hypergeom-f}
{}_pF_q(a_1,\dotsc,a_p;b_1,\dotsc,b_q;z)=\sum_{n=0}^\infty\frac{(a_1)_n\dotsm(a_p)_n} {(b_1)_n\dotsm(b_q)_n}\frac{z^n}{n!}
\end{equation*}
is the generalized hypergeometric series defined for complex numbers $a_i\in\mathbb{C}$ and $b_i\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}$, for positive integers $p,q\in\mathbb{N}$, and in terms of the rising factorials
\begin{equation*}
(x)_n=
\begin{cases}
x(x+1)(x+2)\dotsm(x+n-1), & n\ge1,\\
1, & n=0.
\end{cases}
\end{equation*}
The asymptotic form for the Catalan numbers is
\begin{equation*}
C_x\sim\frac{4^x}{\sqrt\pi\,}\biggl(x^{-3/2}-\frac98x^{-5/2}+\frac{145}{128}x^{-7/2}+\dotsm\biggr).
\end{equation*}
For more information on the Catalan numbers $C_n$, please also refer to the monographs~\cite{Comtet-Combinatorics-74, GKP-Concrete-Math-2nd} and the website \url{https://en.wikipedia.org/wiki/Catalan_number} and references therein.
\par
In this paper, motivated by the explicit expression~\eqref{CatalanN-Gamma} and by virtue of an integral representation of the gamma function $\Gamma(x)$, we establish an integral representation of the Catalan numbers $C_x$ for $x\ge0$.
\par
Our main result can be stated as the following theorem.

\begin{thm}\label{CatalanN-Gamma-Int-thm}
For $x\ge0$, we have
\begin{equation}\label{CatalanN-Gamma-Int-Eq}
C_x=\frac{e^{3/2}4^{x}(x+1/2)^{x}}{\sqrt\pi\,(x+2)^{x+3/2}}\exp{\biggl[\int_0^\infty\beta(t)
\bigl(e^{-t/2}-e^{-2t}\bigr)e^{-xt}\td t\biggr]},
\end{equation}
where
\begin{equation*}
\beta(t)=\frac1t\biggl(\frac1{e^t-1}-\frac1t+\frac12\biggr).
\end{equation*}
\end{thm}

\section{A remark and an open problem}

Before proving Theorem~\ref{CatalanN-Gamma-Int-thm}, we give a remark on the formula~\eqref{CatalanN-Gamma-Int-Eq} and pose an open problem as follows.
\par
Recall from~\cite[Chapter~XIII]{mpf-1993}, \cite[Chapter~1]{Schilling-Song-Vondracek-2nd}, and~\cite[Chapter~IV]{widder} that an infinitely differentiable function $f$ is said to be completely monotonic on an interval $I$ if it satisfies\begin{equation*}
0\le(-1)^kf^{(k)}(x)<\infty
\end{equation*}
on $I$ for all $k\ge0$. Recall from~\cite{compmon2, minus-one} that an infinitely differentiable and positive function $f$ is said to be logarithmically completely monotonic on an interval $I$ if
\begin{equation*}
0\le(-1)^k[\ln f(x)]^{(k)}<\infty
\end{equation*}
hold on $I$ for all $k\in\mathbb{N}$. For more information on logarithmically completely monotonic functions, please refer to~\cite{subadditive-qi-guo-jcam.tex, JAAC384.tex, SCM-2012-0142.tex, Open-TJM-2003-Banach.tex}.
\par
The formula~\eqref{CatalanN-Gamma-Int-Eq} can be rearranged as
\begin{equation}\label{CatalanN-Gamma-Int-Rew}
\ln\biggl[\frac{\sqrt\pi\,(x+2)^{x+3/2}}{e^{3/2}4^{x}(x+1/2)^{x}}C_x\biggr]
=\int_0^\infty\beta(t)\bigl(e^{-t/2}-e^{-2t}\bigr)e^{-xt}\td t.
\end{equation}
Since the function $\beta(t)$ is positive on $(0,\infty)$, see~\cite{best-constant-one.tex, Qi-Springer-2012-Srivastava.tex, best-constant-one-simple-real.tex} and references therein, the right hand side of~\eqref{CatalanN-Gamma-Int-Rew} is a completely monotonic function on $(0,\infty)$. This means that the function
\begin{equation}\label{CatalanN-Gamma-times}
\frac{(x+2)^{x+3/2}}{4^{x}(x+1/2)^{x}}C_x, \quad x>0
\end{equation}
is logarithmically completely monotonic on $(0,\infty)$. Because any logarithmically completely monotonic function must be completely monotonic, see~\cite{JAAC384.tex} and references therein, the function~\eqref{CatalanN-Gamma-times} is also completely monotonic on $(0,\infty)$.
\par
The function~\eqref{CatalanN-Gamma-times} can be rewritten as
\begin{equation}\label{CatalanN-times}
\frac{(x+2)^{x+3/2} \Gamma(x+1/2)}{(x+1/2)^x \Gamma(x+2)}, \quad x>0.
\end{equation}
Hence, the logarithmically complete monotonicity of~\eqref{CatalanN-Gamma-times} implies the logarithmically complete monotonicity of~\eqref{CatalanN-times}. The function~\eqref{CatalanN-times} is a special case of the function
\begin{equation}\label{CatalanN-Gen-F}
\frac{\Gamma(x+a)}{(x+a)^x}
\frac{(x+b)^{x+b-a}}{\Gamma(x+b)}
\end{equation}
for $a,b\in\mathbb{R}$, $a\ne b$, and $x\in(-\min\{a,b\},\infty)$. It seems that the function~\eqref{CatalanN-Gen-F} does not appear in the expository and survey articles~\cite{bounds-two-gammas.tex, Gautschi-Kershaw-TJANT.tex, JAAC384.tex, Wendel-Gautschi-type-ineq-Banach.tex, Wendel2Elezovic.tex-JIA} and plenty of references therein. Therefore, we naturally pose an open problem below.

\begin{open}
What are the necessary and sufficient conditions on $a,b\in\mathbb{R}$ such that the function~\eqref{CatalanN-Gen-F} is \textup{(}logarithmically\textup{)} completely monotonic in $x\in(-\min\{a,b\},\infty)$?
\end{open}

\section{Proof of Theorem~\ref{CatalanN-Gamma-Int-thm}}

Now we are in a position to give a proof of Theorem~\ref{CatalanN-Gamma-Int-thm}.
\par
Let
\begin{equation}
h(x)=(2\ln2)x-\ln\sqrt\pi\,+\ln\Gamma\biggl(x+\frac12\biggr)-\ln\Gamma(x+2),\quad x>0.
\end{equation}
Employing the formula~\cite[(3.22)]{Temme-96-book}
\begin{equation*}
\ln\Gamma(z)=\ln\Bigl(\sqrt{2\pi}\,z^{z-1/2}e^{-z}\Bigr)+\int_0^\infty\beta(t)e^{-zt}\td t
\end{equation*}
gives
\begin{align*}
h(x)&=(2\ln2)x-\ln\sqrt\pi\,+\ln\biggl[\sqrt{2\pi}\,\biggl(x+\frac12\biggr)^{x}e^{-(x+1/2)}\biggr] +\int_0^\infty\beta(t)e^{-(x+1/2)t}\td t \\
&\quad-\ln\Bigl[\sqrt{2\pi}\,(x+2)^{x+3/2}e^{-(x+2)}\Bigr]-\int_0^\infty\beta(t)e^{-(x+2)t}\td t\\
&=(2\ln2)x-\ln\sqrt\pi\,+\frac32+\ln\frac{(x+1/2)^{x}}{(x+2)^{x+3/2}} +\int_0^\infty\beta(t)\bigl(e^{-t/2}-e^{-2t}\bigr)e^{-xt}\td t.
\end{align*}
As a result, we acquire
\begin{align*}
C_x&=\exp\biggl\{(2\ln2)x-\ln\sqrt\pi\,+\frac32+\ln\frac{(x+1/2)^{x}}{(x+2)^{x+3/2}} +\int_0^\infty\beta(t)\bigl(e^{-t/2}-e^{-2t}\bigr)e^{-xt}\td t\biggr\}\\
&=\frac{e^{3/2}4^{x}(x+1/2)^{x}}{\sqrt\pi\,(x+2)^{x+3/2}}\exp{\biggl[\int_0^\infty\beta(t)
\bigl(e^{-t/2}-e^{-2t}\bigr)e^{-xt}\td t\biggr]}.
\end{align*}
The proof of Theorem~\ref{CatalanN-Gamma-Int-thm} is complete.

\begin{rem}
This paper is a companion of the articles~\cite{LCMF-Catalan.tex, LCMF-Catalan-NS.tex, Catalan-Number.tex, Cylinder-Circle-org.tex} and a slightly revised version of the preprint~\cite{Catalan-Number-One.tex}.
\end{rem}

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