%------------------------------------------------------------------------------ % Which version of paper: Here please write first, second, .... % Here please write the corresponding author and his/her e-mail. % Here please write the date of submission of paper or its revisions. %------------------------------------------------------------------------------ % \documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks, plainpages]{hyperref} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \begin{document} \setcounter{page}{9} \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)\\ \textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\ $\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.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( 01$. \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 $S(r)0$. 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}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( 0r_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}. \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{Berg} L. Berg, \textit{\"Uber eine absch\"atzung von Mathieu}, Math. Nachr. \textbf{7} (1952), 257–-259. \bibitem{AMM} M. Berger, \textit{Convexity}, Amer. Math. Monthly \textbf{97} (1990), no. 8, 650--678. \bibitem{BPS} P.L. Butzer, T.K. Pog\'any and H.M. Srivastava, \textit{A linear ODE for the Omega function associated with the Euler function$E_\alpha(z)$ and the Bernoulli function $B_\alpha(z)$}, Appl. Math. Lett. \textbf{19} (2006), 1073--1077. \bibitem{CH} J.G. van der Corput and L.O. 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