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\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 3 (2009), no. 2, 9--15\\
$\frac{\rule{4.55in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
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$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]}


\title[New upper bounds for Mathieu--type series]{New upper bounds for Mathieu--type series}

\author[\v Z. Tomovski, T.K. Pog\'any]{\v Zivorad Tomovski$^1$$^{*}$ and Tibor K. Pog\'any$^2$}

\address{$^1$ Institute of Mathematics, St. Cyril and Methodius University, MK-1000 Skopje, Macedonia.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{tomovski@iunona.pmf.ukim.edu.mk}}

\address{$^2$ Faculty of Maritime Studies, University of Rijeka, 51000 Rijeka, Croatia.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{poganj@pfri.hr}}

\dedicatory{{\rm Communicated by Z. P\'ales}}

\subjclass[2000]{Primary: 26D15; Secondary 33E20.}

\keywords{Mathieu series, alternating Mathieu--series,
Hardy--Hilbert integral inequality, upper bound inequality.}

\date{Received: 26 September 2008; Accepted: 25 January 2009.
\newline \indent $^{*}$ Corresponding author}

\begin{abstract}
The Mathieu's series $S(r)$ was considered firstly by \'E.L. Mathieu
in 1890, its alternating variant $\widetilde{S}(r)$ has been
recently introduced by Pog\'any {\em et al.} [Appl. Math. Comput.
173 (2006), 69--108], where various bounds have been established for
$S, \widetilde{S}$. In this note we obtain new upper bounds over
$S(r), \widetilde{S}(r)$ with the help of Hardy--Hilbert double
integral inequality.
\end{abstract} \maketitle

\section{Introduction and preliminaries}

\noindent The series
   \begin{eqnarray*}
      S(r) = \sum_{n=1}^\infty \frac{2n}{(n^2+r^2)^2}
   \end{eqnarray*}
is named after \'Emile L\'eonard Mathieu (1835--1890), who investigated it in his 1890 book \cite{Mat} written on the
elasticity of solid bodies. Bounds for this series are needed for the solution of boundary value problems for the
biharmonic equations in a two--dimensional rectangular domain, see \cite[Eq. (54),  p. 258]{SCH}. The alternating
version of $S(r)$, that is
   \begin{eqnarray*}
      \widetilde{S}(r) = \sum_{n=1}^\infty (-1)^{n-1}\,\frac{2 n}{(n^2+r^2)^2}
   \end{eqnarray*}
was introduced following certain Tomovski's ideas and recently
discussed by Pog\'any \textit{et al.} in \cite{PST}. Applications of
alternating Mathieu series $\widetilde{S}(r)$ concerning ODE which
solution is the Butzer--Flocke--Hauss Omega function were studied in
\cite{BPS}, \cite{PS}. The integral representations of $S(r),
\widetilde{S}(r)$ \cite{E}, \cite{PST} respectively, reads as
follows:
   \begin{eqnarray} \label{A2}
      S(r) = \frac1{r}\int_0^\infty \frac{x\, \sin(rx)}{e^x-1}\, {\rm d}x, \qquad \quad
      \widetilde{S}(r) = \frac1{r}\int_0^\infty \frac{x\, \sin(rx)}{e^x+1}\, {\rm d}x\, .
   \end{eqnarray}
These integral expressions will be the starting points in our study.

\section{Results required}

\noindent Let us consider a H\"older pair $(p,q), p^{-1}+q^{-1}=1, p>1$, two non--negative functions
$f\in L^p(\mathbb R_+),\,g\in L^q(\mathbb R_+)$, and let us denote $\|\cdot\|_{L_s(\mathbb R_+)}:= \|\cdot\|_s$
the usual integral $L_s$--norm on the set of positive reals. The celebrated Hardy--Hilbert (or Hilbert) integral
inequality \cite{MPF} reads
   \begin{eqnarray} \label{B0}
      \int_0^\infty \int_0^\infty \frac{f(x)g(y)\, {\rm d}x{\rm d}y}{x+y} \le \frac{\pi}{\sin(\pi/p)}\|f\|_p\|g\|_q\, .
   \end{eqnarray}
The inequality is strict unless at least one of $f,g$ is zero, and the constant on the right in \eqref{B0} is
the best possible \cite{MPF}.

Consider the scaled parametric integral
   \begin{eqnarray*}
      \mathcal I_p = \int_0^\infty\frac{|\sin{x}|^p}{x^p} \qquad \big( p>1\big)\, .
   \end{eqnarray*}
We point out that in \cite[p. 663]{AMM} the following estimate has been proved:
   \begin{eqnarray*}
      \mathcal I_p\le \frac{\pi}{2} \sqrt{\frac{2}{p}} \qquad \big(p\ge 2\big).
\end{eqnarray*}
However, we shall give another estimate over $\mathcal I_p$ when $p>1$.
\begin{lemma} For all $p>1$ the following estimate holds
   \begin{eqnarray} \label{B1}
      \mathcal I_p \le q
   \end{eqnarray}
where $q$ is the conjugate H\"older pair to $p$.
\end{lemma}
\begin{proof}
Let us write
   \[ \mathcal I_p :=  \int_0^1 \frac{|\sin x|^p}{x^p}\, {\rm d}x +
      \int_1^\infty \frac{|\sin x|^p}{x^p}\, {\rm d}x\, .\]
Then, by the estimate $\sin x\le x,\, x\in [0,1]$ and by the redundant $|\sin x|\le 1, x> 1$ respectively,
we easily deduce
   \begin{eqnarray*}
      \mathcal I_p \le \int_0^1 {\rm d}x + \int_1^\infty \frac{{\rm d}x}{x^p} = 1+ \frac1{p-1} = q \, .
   \end{eqnarray*}
This finishes the proof of the Lemma.
\end{proof}

\section{Main results}

\noindent At first we establish an upper bound for both $S(r), \widetilde{S}(r)$ of magnitude $O\big(r^{-1/2}\big)$.
\begin{theorem} Let $(p,q),\, p>1$ be a H\"older pair. Then we have
   \begin{eqnarray} \label{C-1}
      \widetilde{S}(r) \le S(r) \le \frac{16\sqrt{\pi}\, q^{1/(2p)}p^{1/(2q)}}{\sqrt{r}\,\sin^{1/2}(\pi/p)} =: C_p(r)\, .
   \end{eqnarray}
Moreover, the best/sharpest upper bound estimate
   \[ C_2(r) = \frac{16 \sqrt{2\pi}}{\sqrt{r}}\]
is obtained if $p=q=2$.
\end{theorem}
\begin{proof} It is sufficient to prove the inequality on the left in \eqref{C-1} since the right one
can be proved similarly. First, we give two elementary inequalities:
   \begin{align} \label{C-2}
      \frac x{e^x+1} \le \frac x{e^x-1} &\le \frac 2{e^{x/2}} \qquad \qquad \qquad \big( x \ge 0\big) \\
      \label{C-3}
      \frac{xy(x+y)}{64} \le \exp \Big\{ \frac x4+\frac y4 + \frac{x+y}4\Big\} &= \exp\Big\{ \frac{x+y}2\Big\}
      \qquad \big(x,y\ge 0\big)\, .
   \end{align}
Thus, we have
   \begin{align}
      \big({S}(r)\big)^2 &= \frac1{r^2} \int_0^\infty \int_0^\infty \frac{xy\sin(rx)\sin(ry)}{(e^x-1)(e^y-1)}\,
                                      {\rm d}x{\rm d}y \nonumber  \\
                                 &\le \frac4{r^2} \int_0^\infty \int_0^\infty |\sin(rx)\sin(ry)|e^{-(x+y)/2}
                                      {\rm d}x{\rm d}y
                                      \nonumber \tag{{\rm by}\,\,\eqref{C-2}} \\
                                 &\le \frac{256}{r^2} \int_0^\infty \int_0^\infty \frac{|\sin(rx)\sin(ry)|}{xy(x+y)}\,
                                      {\rm d}x{\rm d}y
                                      \nonumber \tag{{\rm by}\,\,\eqref{C-3}}\, .
   \end{align}
Taking $f(x) = x^{-1}|\sin(rx)|=g(x)$ we apply the Hardy--Hilbert inequality to the last expression, such that
one transforms into
   \begin{align}
      \big({S}(r)\big)^2 &\le \frac{256 \pi}{r^2 \sin(\pi/p)}\,
                              \Bigg( \int_0^\infty \frac{|\sin(rx)|^p}{x^p}\, {\rm d}x\Bigg)^{1/p}\,
                              \Bigg( \int_0^\infty \frac{|\sin(ry)|^q}{y^q}\, {\rm d}y\Bigg)^{1/q}\, \nonumber \\
                         &=   \frac{256 \pi\, r^{(p-1)/p + (q-1)/q}}{r^2\, \sin(\pi/p)}\,
                              (\mathcal I_p)^\frac{1}{p}(\mathcal I_q)^\frac{1}{q} \,\nonumber\\
                                     &\le \frac{256 \pi\, q^\frac1p \cdot p^\frac1q}{r\, \sin(\pi/p)}
                                     \tag{{\rm by}\,\,\eqref{B1}}
   \end{align}
This is equivalent to the asserted result \eqref{C-1}, since the termwise comparation of defining formul{\ae}
showes that $\widetilde{S}(r) \le S(r)$.

Rewriting \eqref{C-1} with the notation $x:=1/p$, we deduce
   \[ C_{1/x}(r) = \frac{16 \sqrt{\pi}}{\sqrt{r(1-x)^xx^{1-x}\sin(\pi x)}}
    \qquad \big( 0<x<1 \big)\, .\]
An easy computation shows that the denominator is maximal and hence the upper bound constant is minimal if
$x=1/2$, that is, when $p=q=2$.
\end{proof}
Now, we will extend this result, scaling the exponent of $r$ in the upper bound \eqref{C-1}. The achieved
magnitude should be $O\big(r^{-1/(2p)}\big),\,p>1$.
\begin{theorem} Let $(p,q), p>1$ be a H\"older pair. Then for all $r>0, v>1$ we have
   \begin{eqnarray} \label{C6}
      \widetilde{S}(r) \le S(r) \le \frac{C(p,v)}{r^{1/(2p)}}
   \end{eqnarray}
where
   \begin{eqnarray*}
      C(p,v) := \frac{2^{(5q+1)/(2q)}\,\max\{2^{1/(2p)},2^{1/(2q)}\}\,\big( \pi p\big)^{1/(2p)}\,\big(\Gamma(q)\Gamma(2q)\big)^{1/(2q)}}
                     {q^{3/2}\,\big(\sin(\pi/p)\,(p-1/v)^{1/v}(p-1+1/v)^{1-1/v}\big)^{1/(2p)}} \, .
   \end{eqnarray*}
\end{theorem}
\begin{proof} For a given H\"older pair $(p,q), p>1$ and for some $r>0$ consider
   \begin{align*}
      \big(S(r)\big)^2 &= \frac1{r^2} \int_0^\infty \int_0^\infty \frac{xy\sin(rx)\sin(ry)}{(e^x-1)(e^y-1)}\,
                {\rm d}x{\rm d}y \\
             &= \frac1{r^6} \int_0^\infty \int_0^\infty \frac{\sin(x)\sin(y)}{xy(x+y)^{1/p}}\,
                \cdot\, \frac{x^2 y^2 (x+y)^{1/p}}{(e^{x/r}-1)(e^{y/r}-1)}\, {\rm d}x{\rm d}y  \, .
   \end{align*}
By the H\"older inequality we conclude
   \begin{align} \label{C2}
      \big(S(r)\big)^2 &\le \frac1{r^6} \Bigg( \int_0^\infty \int_0^\infty \frac{|\sin(x)\sin(y)|^p}{x^py^p(x+y)}\,
                                        {\rm d}x{\rm d}y \Bigg)^{1/p} \nonumber \\
                        & \qquad \qquad \times  \Bigg(\int_0^\infty \int_0^\infty \frac{x^{2q} y^{2q} (x+y)^{q-1}}
                                        {(e^{x/r}-1)^q(e^{y/r}- 1)^q}\,
                                        {\rm d}x{\rm d}y \Bigg)^{1/q}\, .
   \end{align}
Choosing this time $w$ as the H\"older conjugate pair to given $v>1$ and specifying
   \[ f(x)=g(x) = x^{-p}|\sin(x)|^p\,, \]
we evaluate by the Hardy--Hilbert inequality \eqref{B0} the first integral from above:
   \begin{align} \label{C3}
      \mathcal J &=  \int_0^\infty \int_0^\infty \frac{|\sin(x)\sin(y)|^p}{x^py^p(x+y)}\, {\rm d}x{\rm d}y \nonumber \\
                 &\le  \frac{\pi}{\sin(\pi/p)}\,\Bigg( \int_0^\infty \frac{|\sin(x)|^{pv}}{x^{pv}}\,
                       {\rm d}x\Bigg)^{1/v}\,
                       \Bigg( \int_0^\infty \frac{|\sin(y)|^{pw}}{y^{pw}}\, {\rm d}y\Bigg)^{1/w}\, .
   \end{align}
Estimating \eqref{C3} by \eqref{B1} we deduce
   \begin{eqnarray*}
      \mathcal J \le \frac{\pi}{\sin(\pi/p)}\,\frac{pv}{(pv-1)^{1/v}((p-1)v+1)^{1-1/v}}\, .
   \end{eqnarray*}
The second integral in \eqref{C2} we evaluate in the following way:
   \begin{align} \label{C5}
      \mathcal K   &=   \int_0^\infty \int_0^\infty \frac{x^{2q} y^{2q} (x+y)^{q-1}}
                        {(e^{x/r}-1)^q(e^{y/r}-1)^q}\, {\rm d}x{\rm d}y \nonumber \\
                   &=   r^{5q+1}\int_0^\infty \frac{x^{2q} y^{2q} (x+y)^{q-1}}
                        {(e^{x}-1)^q(e^{y}-1)^q}\, {\rm d}x{\rm d}y \nonumber \\
                   &\le r^{5q+1}\, \max \{2,\,2^{q-1}\}\, \int_0^\infty \frac{x^{3q-1}\, {\rm d}x}{(e^{x}-1)^q}\,
                        \int_0^\infty \frac{y^{2q}\, {\rm d}y}{(e^{y}-1)^q}\\ \label{C51}
                   &\le (2r)^{5q+1} q^{-3q} \max \{2,\, 2^{q-1}\}\, \Gamma(q)\Gamma(2q)\, \nonumber.
   \end{align}
where in \eqref{C5} we make use of the estimate (such that follows by \eqref{C-2}):
   \begin{eqnarray*}
      \int_0^\infty \frac{x^\alpha}{(e^x-1)^q}\, {\rm d}x \le 2^q \int_0^\infty x^{\alpha -q}e^{-qx/2}\, {\rm d}x
            = \frac{2^{\alpha+1}}{q^{\alpha-q+1}}\, \Gamma(\alpha-q+1) \, ,
   \end{eqnarray*}
specified for $\alpha=3q-1,\, 2q$ respectively. So, the upper bound over $S(r)$ in \eqref{C6} is proved.
Repeating the termwise comparation procedure for $S(r), \widetilde{S}(r)$, we clearly deduce \eqref{C6}.
\end{proof}

\section{Discussion}

\hspace{-4mm}{\sf A.} In this research note we derive upper bounds for $S(r), \widetilde{S}(r)$, such that
possess the form
   \[ S(r) \le \frac{\Phi(\theta)}{r^\alpha} \qquad \big( \alpha>0\big)\, .\]
Here $\Phi(\theta)$ is an absolute constant and $\theta$ denotes the vector of scaling parameters.
We obtain our main results \eqref{C-1} and \eqref{C6} {\em via} the Hardy--Hilbert integral inequality.

At first, we recall some ancestor results such that will be compared to our bounds for small $r$.
In \cite{Mat} Mathieu posed his famous conjecture $S(r)<r^{-2},\, r>0$. The conjecture was proved after more
then 60 years by Berg \cite{Berg} and by Makai \cite{Makai}. Actually they showed more:
   \begin{eqnarray*}
       \frac1{r^2 + 1/2}<S(r)<\frac1{r^2} \qquad \big( r>0\big)\, .
   \end{eqnarray*}
Another proof of this upper bound has been given by van der Corput and Heflinger \cite{CH}. Diananda \cite{Dia}
improved Mathieu's bound to
   \begin{eqnarray} \label{Dian}
      S(r) \le \frac1{r^2} - \frac1{(2r^2+2r+1)(8r^2+3r+3)} \qquad \big( r>0\big)\, .
   \end{eqnarray}
Here has to be mentioned Guo's bound of magnitude $O(r^{-2})$, \cite[Eq. (10)]{Guo}.
\medskip

\hspace{-4mm}{\sf B.} We obtain easily an upper bound, such that is superior to Mathieu's bound $r^{-2}$ for small $r$.
Indeed, starting with the integral expressions for $S(r)$ and $\widetilde{S}(r)$ in \eqref{A2} we have
   \begin{eqnarray*}
      {S}(r) \le \frac 1r\, \int_0^\infty \frac{x\,{\rm d}x}{e^x-1} = \frac{\pi^2}{6r} =: S^\star(r) \quad
                 \text{and} \quad \widetilde{S}(r) \le \frac 1r\, \int_0^\infty \frac{x\,{\rm d}x}{e^x+1}
             = \frac{\pi^2}{12r}\,.
   \end{eqnarray*}
So, when $r\in (0, 6/\pi]$, it follows $S^\star(r)\le r^{-2}$.
\medskip

\hspace{-4mm}{\sf C.} Let us denote $S_1(r), S_2(r)$ the upper bounds in Theorems 3.1, 3.2 respectively.
Comparing Mathieu's bound with $S_1(r)$, solving the equation $S_1(r)=r^{-2}$ we find that
  \[ S_1(r) \le \frac1{r^2} \qquad \quad \Big( 0<r \le \frac{\sin^{2/3}(\pi/p)}
                       {4\sqrt[3]{4\pi}\, p^{1/(3q)}q^{1/(3p)}}:= r_1(p)<1\, \Big)\, . \]
Therefore, $S_1(r)$ is obviously superior to bounds with magnitude
$O(r^{-2}),\, r$ small. Similar comparisons involving $S_2(r)$
and/or Diananda's \eqref{Dian} and Guo's bounds one leaves to the
interested reader. These analyses show that our bounds
\eqref{C-1}, \eqref{C6} mainly improve the earlier ones.
\medskip

\hspace{-4mm}{\sf D.} Let us compare $S_1(r)$ and $S_2(r)$. It is not hard to see that
   \[ r_0 := r_0(p,v) = \frac{2^{3q-1}\pi\, p^{2-q}q^{4q-1}\big( (p-1/v)^{1/v}(p-1+1/v)^{1-1/v}\big)^{q-1}}
                       {\sin(\pi/p)\, \max\{ 2, 2^{q-1}\}\,\Gamma(q)\Gamma(2q)}\]
is the unique positive solution of $S_1(r)=S_2(r)$. Accordingly, it follows that
   \[ S_2(r) < S_1(r) \qquad \big( r\in (0,r_0)\big)\, , \]
while for $r>r_0$ the reversed conclusion holds. We point out that $r_0$ can easily skip $1$;
for instance $r_0(2,2) = 512\, \pi$.
\medskip

\hspace{-4mm}{\sf E.} Because the alternating Mathieu series has been introduced recently in \cite{PST},
the here established bounds are unique until now. However, for $r$ large the bounding inequalities presented
also in \cite{PST} are sharper than the here presented ones.

\medskip
\noindent \textbf{Acknowledgements:} The present investigation was
supported partially by the Research Project No. 05-437/1 of
\textit{Ministry of Education and Sciences of Macedonia} and by the
Research Project No. 112-2352818-2814 of the \textit{Ministry of
Sciences, Education and Sports of Croatia}.

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\end{document}

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