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\begin{document}

\title[Complete monotonicity of a power-exponential function]
{Logarithmically complete monotonicity of a power-exponential function involving the logarithmic and psi functions}

\author[B.-N. Guo]{Bai-Ni Guo}
\address[Guo]{School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China}
\email{\href{mailto: B.-N. Guo <bai.ni.guo@gmail.com>}{bai.ni.guo@gmail.com}, \href{mailto: B.-N. Guo <bai.ni.guo@hotmail.com>}{bai.ni.guo@hotmail.com}}
\urladdr{\url{http://www.researchgate.net/profile/Bai-Ni_Guo/}}

\author[F. Qi]{Feng Qi}
\address[Qi]{College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, 028043, China; Department of Mathematics, School of Science, Tianjin Polytechnic University, Tianjin City, 300387, China}
\email{\href{mailto: F. Qi <qifeng618@gmail.com>}{qifeng618@gmail.com}, \href{mailto: F. Qi <qifeng618@hotmail.com>}{qifeng618@hotmail.com}, \href{mailto: F. Qi <qifeng618@qq.com>}{qifeng618@qq.com}}
\urladdr{\url{http://qifeng618.wordpress.com}}

\begin{abstract}
Let $\Gamma$ and $\psi=\frac{\Gamma'}{\Gamma}$ be respectively the classical Euler gamma function and the psi function and let $\gamma=-\psi(1)=0.57721566\dotsc$ stand for the Euler-Mascheroni constant. In the paper, the authors simply confirm the logarithmically complete monotonicity of the power-exponential function $q(t)=t^{t[\psi(t)-\ln t]-\gamma}$ on the unit interval $(0,1)$, concisely deny that $q(t)$ is a Stieltjes function, surely point out fatal errors appeared in the paper [V. Krasniqi and A. Sh. Shabani, \emph{On a conjecture of a logarithmically completely monotonic function}, Aust. J. Math. Anal. Appl. \textbf{11} (2014), no.~1, Art.~5, 5~pages; Available online at \url{http://ajmaa.org/cgi-bin/paper.pl?string=v11n1/V11I1P5.tex}], and partially solve a conjecture posed in the article [B.-N. Guo, Y.-J. Zhang, and F. Qi, \textit{Refinements and sharpenings of some double inequalities for bounding the gamma function}, J. Inequal. Pure Appl. Math. \textbf{9} (2008), no.~1, Art.~17; Available online at \url{http://www.emis.de/journals/JIPAM/article953.html}].
\end{abstract}

\subjclass[2010]{Primary 33B15; Secondary 26A48, 26A51, 33B10, 33B99, 44A10}

\keywords{Logarithmically complete monotonicity; Stieltjes function; logarithmic function; psi function; power-exponential function; conjecture}

\thanks{This paper was typeset using \AmS-\LaTeX}

\maketitle

\section{Introduction}

For bounding the classical Euler gamma function
\begin{equation*}
\Gamma(z)=\int^\infty_0t^{z-1} e^{-t}\td t, \quad \Re(z)>0
\end{equation*}
and for bounding the ratio $\frac{\Gamma(x)}{\Gamma(y)}$, Guo, Zhang, and Qi established some inequalities in~\cite{note-on-li-chen.tex}. When comparing two inequalities, they created the function
\begin{equation}\label{q(t)-dfn}
q(t)=t^{t[\psi(t)-\ln t]-\gamma}, \quad t>0
\end{equation}
in~\cite[Remark~8]{note-on-li-chen.tex}, where $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ is called the psi function and $\gamma=0.577\dotsc$ stands for the Euler-Mascheroni constant. As said in~\cite[Remark~8]{note-on-li-chen.tex}, Guo, Zhang, and Qi demonstrated, by using the software \textsc{Mathematica} to plot, the decreasing monotonicity of $q(t)$ on $(0,\infty)$. Then they conjectured that the function $q(t)$ is logarithmically completely monotonic on $(0,\infty)$. A function $f$ is said to be logarithmically completely monotonic on an interval $I$ if it is infinitely differentiable and satisfies
\begin{equation*}
(-1)^k[\ln f(t)]^{(k)}\ge0
\end{equation*}
on $I$ for $k\in\mathbb{N}$, where $\mathbb{N}$ denotes the set of positive integers. See~\cite{Atanassov, CBerg, absolute-mon-simp.tex, compmon2, minus-one, auscm-rgmia, e-gam-rat-comp-mon, schur-complete, Schilling-Song-Vondracek-2nd} and plenty of closely related references in~\cite{JAAC384.tex}. For our own convenience, we denote the set of logarithmically completely monotonic functions on $I$ by $\mathcal{L}[I]$.
\par
Recall from~\cite[Chapter~XIII]{mpf-1993}, \cite[Chapter~1]{Schilling-Song-Vondracek-2nd}, and~\cite[Chapter~IV]{widder} that a function $f(x)$ is said to be completely monotonic on an interval $I$ if and only if $f$ has derivatives of all orders and satisfies
\begin{equation*}
(-1)^{k-1}f^{(k-1)}(x)\ge0
\end{equation*}
on $I$ for $k\in\mathbb{N}$. The class of completely monotonic functions on $(0,\infty)$ may be characterized by~\cite[p.~161, Theorem~12b]{widder} which reads that a necessary and sufficient condition that $f(x)$ should be completely monotonic for $0<x<\infty$ is that
\begin{equation*}
f(x)=\int_0^\infty e^{-xs}\td\mu(s),
\end{equation*}
where $\mu(s)$ is non-decreasing and the integral converges for $0<x<\infty$. We will use $\mathcal{C}[I]$ to denote the set of completely monotonic functions on $I$.
\par
Also recall from~\cite{CBerg, Schilling-Song-Vondracek-2nd, widder} that if $f(x)$ can be represented as
\begin{equation*}
f(x)=a+\int_0^\infty\frac1{x+s}\td\mu(s)
\end{equation*}
on $(0,\infty)$, where $a\ge0$, the measure $\mu$ is nonnegative on $[0,\infty)$, and
\begin{equation*}
  \int_0^\infty\frac1{1+s}\td\mu(s)<\infty,
\end{equation*}
then $f(x)$ is said to be a Stieltjes function. The set of Stieltjes functions will be denoted by $\mathcal{S}$.
\par
In~\cite{Krasniqi-Shabani-AJMAA-14}, Krasniqi and Shabani claimed that they confirmed the above conjecture. However, because $-\ln t\in\mathcal{C}[(0,1)]$, but $-\ln t\not\in\mathcal{C}[(0,\infty)]$, all the proofs of~\cite[Conjecture~2.1 and Theorem~2.2]{Krasniqi-Shabani-AJMAA-14} are wrong.
\par
Our main results in this paper may be stated as the following theorem.

\begin{thm}\label{note-on-li-chen-conj-thm}
The function $q(t)$ defined by~\eqref{q(t)-dfn} satisfies
\begin{gather}
q(t)\in\mathcal{L}[(0,1)], \quad q(t)\not\in\mathcal{S}, \label{q(t)2property}\\
\lim_{t\to0^+}q(t)=\infty,\quad \lim_{t\to\infty}q(t)=0.\label{2-limits}
\end{gather}
\end{thm}

It is clear that Theorem~\ref{note-on-li-chen-conj-thm} partially confirms the logarithmically complete monotonicity of the function $q(t)$ on $(0,\infty)$ and partially solves the above conjecture posed in~\cite[Remark~8]{note-on-li-chen.tex}.

\section{Proof of Theorem~\ref{note-on-li-chen-conj-thm}}
In~\cite{CBerg, Chen-Qi-TKJM-2005, absolute-mon-simp.tex, compmon2, minus-one, schur-complete}, the inclusions
\begin{equation}\label{S-L-C-relation}
\mathcal{S}\setminus\{0\}\subset\mathcal{L}[(0,\infty)] \quad \text{and}\quad \mathcal{L}[I]\subset\mathcal{C}[I]
\end{equation}
were discovered. For more information, see~\cite{JAAC384.tex} and plenty of closely related references therein.
In order to prove the first property in~\eqref{q(t)2property}, from the first inclusion in~\eqref{S-L-C-relation}, it suffices to show that the function $q(t)$ is a Stieltjes function. In~\cite{CBerg}, Berg remarked that $f\in\mathcal{S}$ and $f\ne0$ if and only if $\frac1{xf(x)}\in\mathcal{S}$. Therefore, if $q(t)\in\mathcal{S}$, then $\frac1{tq(t)}\in\mathcal{S}$, and so $\frac1{tq(t)}$ should be decreasing on $(0,\infty)$. However, an easy calculation leads to
$$
1=\frac1{tq(t)}\bigg|_{t=1}<\frac1{tq(t)}\bigg|_{t=2}=2^{3(\gamma-1)+2\ln2}=1.08518525\dotsc
$$
As a result, the function $q(t)$ is not a Stieltjes function, that is, the second property in~\eqref{q(t)2property} is true.
\par
In~\cite[pp.~374--375, Theorem~1]{psi-alzer} and~\cite[p.~105, Theorem~1]{theta-new-proof.tex-BKMS}, Alzer, Guo, and Qi proved that
\begin{equation*}
\theta_\alpha(x)=x^\alpha[\ln x-\psi(x)]\in\mathcal{C}[(0,\infty)]
\end{equation*}
if and only if $\alpha\le1$ and that
\begin{gather}
\lim_{x\to0^+}\theta_1(x)=1,\quad \lim_{x\to\infty}\theta_1(x)=\frac12, \label{theta1limit}\\
\lim_{x\to0^+}\theta_\alpha(x)=\infty,\quad \lim_{x\to\infty}\theta_\alpha(x)=0, \quad \alpha<1.\notag
\end{gather}
It is clear that
\begin{equation}\label{lnq(t)-dfn}
\ln q(t)=\{t[\psi(t)-\ln t]-\gamma\}\ln t=(-\ln t)[\theta_1(t)+\gamma].
\end{equation}
Since $-\ln t\in\mathcal{C}[(0,1)]$ and the product or sum of finitely many completely monotonic functions is still a completely monotonic function on their common interval, it is easy to see that $\ln q(t)\in\mathcal{C}[(0,1)]$ and $q(t)\in\mathcal{L}[(0,1)]$.
\par
The limits in~\eqref{2-limits} follows immediately from taking $t\to0^+$ and $t\to\infty$ on both sides of the equation~\eqref{lnq(t)-dfn} and making use of the limits in~\eqref{theta1limit}.
\par
The first limit in~\eqref{2-limits} may also be deduced directly from the argument
\begin{align*}
\lim_{t\to0^+}\ln q(t)&=\Bigl\{\lim_{t\to0^+}[t\psi(t)]-\lim_{t\to0^+}(t\ln t)-\gamma\Bigr\}\lim_{t\to0^+}\ln t\\
&=\Bigl\{\lim_{t\to0^+}[t\psi(t)+1-1]-0-\gamma\Bigr\}\lim_{t\to0^+}\ln t\\
&=\biggl\{\lim_{t\to0^+}\biggl(t\biggl[\psi(t)+\frac1t\biggr]\biggr)-1-\gamma\biggr\}\lim_{t\to0^+}\ln t\\
&=\biggl\{\lim_{t\to0^+}[t\psi(t+1)]-1-\gamma\biggr\}\lim_{t\to0^+}\ln t\\
&=-(1+\gamma)\lim_{t\to0^+}\ln t\\
&=\infty.
\end{align*}
The proof of Theorem~\ref{note-on-li-chen-conj-thm} is complete.

\section{Remarks}

\begin{rem}
On the interval $(1,\infty)$, it is obvious that the function $\ln t$ is positive and increasing and $\theta_1(t)+\gamma\in\mathcal{C}[(1,\infty)]$. Therefore, we can not make clear the monotonicity of the functions $(\ln t)[\theta_1(t)+\gamma]=-\ln q(t)$ and $(-\ln t)[\theta_1(t)+\gamma]=\ln q(t)$, say nothing of the logarithmically complete monotonicity of $q(t)$, on $(1,\infty)$. This is the key difficulty to completely solve the above conjecture.
\end{rem}

\begin{rem}
By a similar argument to the second property in~\eqref{q(t)2property}, we can show that $1-G(x)\not\in\mathcal{S}$, where
\begin{equation*}
G(x)=\biggl[1-\frac{\ln x}{\ln(x+1)}\biggr]x\ln x, \quad x>0.
\end{equation*}
However, Guo and Qi conjectured that $1-G(x)\in\mathcal{C}[(0,\infty)]$ in~\cite{unit-ball.tex-Optim.Lett} and its preprint~\cite{unit-ball.tex}. Later, Berg and Pedersen verified this conjecture in~\cite{berg-pedersen-Alzer}.
\end{rem}

\section{An open problem}

For $\beta\in\mathbb{R}$, let
\begin{equation*}
h_\beta(t)=t^{t[\psi(t)-\ln t]-\beta}, \quad t>0.
\end{equation*}
If $h_\beta(t)\in\mathcal{L}[(0,\infty)]$, then the first derivative of its logarithm
\begin{equation*}
[\ln h_\beta(t)]'=\frac{t^2\psi'(t)\ln t-t (\ln t)^2-2 t \ln t+t \psi(t)+t\psi(t)\ln t-b}t
\end{equation*}
should be non-positive on $(0,\infty)$, that is,
\begin{equation*}
b\ge t^2(\ln t)\psi'(t)+t(1+\ln t)\psi(t)-(\ln t+2)t \ln t\to-\frac12, \quad t\to\infty.
\end{equation*}
This means that $\beta\ge-\frac12$ is a necessary condition such that $h_\beta(t)\in\mathcal{L}[(0,\infty)]$.
\par
Is the condition $\beta\ge-\frac12$ sufficient for $h_\beta(t)\in\mathcal{L}[(0,\infty)]$?

\begin{open}
Determine the best constant $-\frac12\le\beta\le\gamma$ such that $h_\beta(t)\in\mathcal{L}[(0,\infty)]$.
\end{open}

\subsection*{Acknowledgements}
The authors thank Valmir Krasniqi for his comments on the original version of this paper through an e-mail on 2 March 2015.

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\end{document}
