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\begin{center} \huge{\bf A logarithmically completely monotonic function involving the gamma function and originating from the Catalan numbers and function}
\end{center}

\centerline{}

\centerline{\bf Fang-Fang Liu$^*$, Xiao-Ting Shi, Feng Qi}

\centerline{}
{\small
\begin{center}
\emph{Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China}
\end{center}

\centerline{\emph{*Corresponding author e-mail: fang-liu@qq.com}}

\centerline{}
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\noindent \hspace{-3 pt}{\scriptsize \textbf{ Copyright \copyright 2015 Fang-Fang Liu, Xiao-Ting Shi, 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 find necessary conditions and sufficient conditions for a function involving the gamma function and originating from investigation of properties of the Catalan numbers and function in combinatorics to be logarithmically completely monotonic.\\
\centerline{}
\noindent {\footnotesize \emph{\textbf{Keywords}}:  \emph{necessary condition; sufficient condition; logarithmically completely monotonic function; gamma function; Catalan number; Catalan function}}

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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 such as ``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 $C_{n-2}$.
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 function $C_x$ 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 detailed information on the Catalan numbers $C_n$, please refer to the monographs~\cite{Comtet-Combinatorics-74, GKP-Concrete-Math-2nd} and references therein.
\par
In the paper~\cite{Catalan-Number-One.tex}, motivated by the explicit expression~\eqref{CatalanN-Gamma} and by virtue of an integral representation of the gamma function $\Gamma(x)$, the authors established an integral representation of the Catalan function $C_x$ for $x\ge0$.

\begin{thm}[{\cite[p.~131, Theorem~1]{Catalan-Number-One.tex}}]
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}

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
\end{equation}
is logarithmically completely monotonic on $(0,\infty)$. Because any logarithmically completely monotonic function must be completely monotonic, see~\cite[Eq.~(1.4)]{JAAC384.tex} and references therein, the function~\eqref{CatalanN-Gamma-times} is also completely monotonic on $(0,\infty)$.
\par
By virtue of~\eqref{CatalanN-Gamma}, 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 $F_{1/2,2}(x)$ of the general function
\begin{equation}\label{CatalanN-Gen-F}
F_{a,b}(x)=\frac{\Gamma(x+a)}{(x+a)^x}\frac{(x+b)^{x+b-a}}{\Gamma(x+b)}, \quad a,b\in\mathbb{R}, \quad a\ne b \quad x>-\min\{a,b\}.
\end{equation}
It is clear that
\begin{equation*}
F_{a,b}(x)F_{b,a}(x)=\biggl(\frac{x+b}{x+a}\biggr)^{b-a}
=\biggl(1+\frac{b-a}{x+a}\biggr)^{b-a}.
\end{equation*}
We notice that the function $F_{a,b}(x)$ 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, it is significant to naturally pose an open problem below.

\begin{open}[{\cite[Open Problem~1]{Catalan-Number-One.tex}}]
What are the necessary and sufficient conditions on $a,b\in\mathbb{R}$ such that the function $F_{a,b}(x)$ defined by~\eqref{CatalanN-Gen-F} is \textup{(}logarithmically\textup{)} completely monotonic in $x\in(-\min\{a,b\},\infty)$?
\end{open}

The aim of this paper is to give solutions to the above open problem.

\begin{thm}\label{NSCondition-Thm}
The sufficient conditions on $a,b$ for the function $[F_{a,b}(x)]^{\pm1}$ defined by~\eqref{CatalanN-Gen-F} to be logarithmically completely monotonic in $x\in(-\min\{a,b\},\infty)$ are $(a,b)\in D_{\pm}(a,b)$, where
\begin{equation*}
D_+(a,b)=\{(a,b):a\ge1,a>b\}\cup\biggl\{(a,b):a\le\frac12,a<b\biggr\}
\end{equation*}
and
\begin{equation*}
D_-(a,b)=\{(a,b):a\ge1,b>a\}\cup\biggl\{(a,b):a\le\frac12,b<a\biggr\}.
\end{equation*}
The necessary conditions on $a,b$ for the function $[F_{a,b}(x)]^{\pm1}$ to be logarithmically completely monotonic in $x\in(-\min\{a,b\},\infty)$ are
\begin{equation*}
a(a-b)\gtreqless\frac{a-b}2.
\end{equation*}
\end{thm}

\section{Proof of Theorem~\ref{NSCondition-Thm}}

Taking the logarithm of $F_{a,b}(x)$ gives
\begin{equation*}
\ln F_{a,b}(x)=\ln\Gamma(x+a)-x\ln(x+a)-[\ln\Gamma(x+b)-(x+b-a)\ln(x+b)]
\triangleq f_a(x)-f_a(x+b-a).
\end{equation*}
Differentiating with respect to the variable $x$ of $f_a(x)$ yields
\begin{equation*}
f_a'(x)=\psi(x+a)-\ln(x+a)+\frac{a}{x+a}-1
\end{equation*}
and
\begin{equation}\label{psi-der-poly-eq}
f_a''(x)=\psi'(x+a)-\frac1{x+a}-\frac{a}{(x+a)^2}.
\end{equation}
\par
In~\cite[Theorem~1.3]{sandor-gamma-2-ITSF.tex}, it was found that
\begin{enumerate}
\item
the function $\psi(x)-\ln x+\frac{\alpha}x$ is completely monotonic on $(0,\infty)$ if and only if $\alpha\ge1$,
\item
the function $\ln x-\frac{\alpha}x-\psi(x)$ is completely monotonic on $(0,\infty)$ if and only if $\alpha\le\frac12$.
\end{enumerate}
This means that
\begin{enumerate}
\item
if $\alpha\ge1$, the function $\frac1x+\frac{\alpha}{x^2}-\psi'(x)$ is completely monotonic on $(0,\infty)$;
\item
if $\alpha\le\frac12$, the function $\psi'(x)-\frac1x-\frac{\alpha}{x^2}$ is completely monotonic on $(0,\infty)$.
\end{enumerate}
Equivalently,
\begin{enumerate}
\item
if $a\ge1$, the function $-f_a''(x-a)$ is completely monotonic on $(0,\infty)$;
\item
if $a\le\frac12$, the function $f_a''(x-a)$ is completely monotonic on $(0,\infty)$.
\end{enumerate}
Consequently,
\begin{enumerate}
\item
if $a\ge1$, the function $-f_a''(x)$ is completely monotonic on $(-a,\infty)$;
\item
if $a\le\frac12$, the function $f_a''(x)$ is completely monotonic on $(-a,\infty)$.
\end{enumerate}
As a result,
\begin{enumerate}
\item
when $a\ge1$ and $b-a>0$ the negativity of the function
\begin{equation*}
[\ln F_{a,b}(x)]''=f_a''(x)-f_a''(x+b-a)
\end{equation*}
is completely monotonic on $(-a,\infty)$;
\item
when $a\ge1$ and $b-a<0$, the function $[\ln F_{a,b}(x)]''$ is completely monotonic on $(-b,\infty)$;
\item
when $a\le\frac12$ and $b-a>0$, the function $[\ln F_{a,b}(x)]''$ is completely monotonic on $(-a,\infty)$;
\item
when $a\le\frac12$ and $b-a<0$, the function $-[\ln F_{a,b}(x)]''$ is completely monotonic on $(-b,\infty)$.
\end{enumerate}
In conclusion,
\begin{enumerate}
\item
when $(a,b)\in D_+(a,b)$, the function $[\ln F_{a,b}(x)]''$ is completely monotonic on $(-\min\{a,b\},\infty)$;
\item
when $(a,b)\in D_-(a,b)$, the function $-[\ln F_{a,b}(x)]''$ is completely monotonic on $(-\min\{a,b\},\infty)$.
\end{enumerate}
\par
By straightforward computation, we see that
\begin{align*}
\lim_{x\to\infty}[\ln F_{a,b}(x)]'&=\lim_{x\to\infty}\biggl[\psi(x+a)-\ln(x+a)+\frac{a}{x+a}-1 -\psi(x+b)+\ln(x+b)-\frac{a}{x+b}+1\biggr]\\
&=\lim_{x\to\infty}\biggl[\psi(x+a)-\psi(x+b)+\ln\frac{x+b}{x+a} +\frac{a(b-a)}{(x+a)(x+b)}\biggr]\\
&=0.
\end{align*}
This implies that, when $(a,b)\in D_\pm(a,b)$, the first logarithmic derivative satisfies $[\ln F_{a,b}(x)]'\lessgtr0$. By the definition of logarithmically completely monotonic functions, we conclude that, when $(a,b)\in D_\pm(a,b)$, the function $[F_{a,b}(x)]^{\pm1}$ is logarithmically completely monotonic on $(-\min\{a,b\},\infty)$.
\par
Conversely, if the function $[F_{a,b}(x)]^{\pm1}$ is logarithmically completely monotonic on $(-\min\{a,b\},\infty)$, then $\pm[\ln F_{a,b}(x)]'\le0$ which is equivalent to
\begin{equation*}
\psi(x+a)-\psi(x+b)+\ln\frac{x+b}{x+a} +\frac{a(b-a)}{(x+a)(x+b)}\lesseqgtr0.
\end{equation*}
This can be rearranged as
\begin{equation*}
a(a-b)\gtreqless(x+a)(x+b)\biggl[\psi(x+a)-\psi(x+b)+\ln\frac{x+b}{x+a}\biggr]\to\frac{a-b}2
\end{equation*}
as $x\to\infty$. Therefore, the necessary conditions are $a(a-b)\gtreqless\frac{a-b}2$.
The proof of Theorem~\ref{NSCondition-Thm} is complete.

\begin{rem}
About recent development on investigation of the function~\eqref{psi-der-poly-eq}, please refer to the manuscript~\cite{Remark-Qi-Zhao.tex} and plenty of references therein.
\end{rem}

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

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