On the series $\sum \limits_{n=1}^{\infty} \frac{\sin (n \pi y) \sin \left ( n \pi x \right )}{n^2 \pi^2}$

Question

A friend of mine asked me to help him evaluate the series

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\sin (n \pi y) \sin \left ( n \pi x \right )}{n^2 \pi^2} \quad , \quad x , y \in (0, 1)$$

It does not ring any bells as to what it could be behind. The only thing I see is Fourier series and probably a dilogarithm function. But this is as far as I can see.. I cannot see to collect the pieces together.

I would like to help him and I am asking your help. Is there any closed form in terms of special function for this series?

Answer 1

This sum can be evaluated explicitly using Parseval's theorem: given

$$A(w) = \sum_{n=-\infty}^{\infty} a_n \, e^{i n w} $$ $$B(w) = \sum_{n=-\infty}^{\infty} b_n \, e^{i n w} $$

Then

$$\sum_{n=-\infty}^{\infty} a_n \bar{b}_n = \frac1{2 \pi} \int_{-\pi}^{\pi} dw \, A(w) \bar{B}(w) $$

To illustrate, I prove here that

$$\sum_{n=-\infty}^{\infty} \frac{\sin{n \pi x}}{n \pi} e^{i n w} = \begin{cases}1 & |w| \lt \pi x \\ 0 & |w| \gt \pi x \end{cases}$$

when $x \in [0,1)$. Accordingly,

$$\sum_{n=-\infty}^{\infty} \frac{\sin{n \pi x}}{n \pi} \frac{\sin{n \pi y}}{n \pi} = \frac1{2 \pi} \operatorname{min}{(2\pi x,2\pi y)} = \operatorname{min}{(x,y)}$$

Thus,

$$\sum_{n=1}^{\infty} \frac{\sin{n \pi x}}{n \pi} \frac{\sin{n \pi y}}{n \pi} = \frac12 \left (\operatorname{min}{(x,y)} - xy \right ) $$

Answer 2

Let us start with the well-known Fourier series for a repeated parabola: $$ x^2=\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos{nx};\quad -\pi\le x\le\pi, $$ which upon substitution $x=\pi(1-t)$ transforms to: $$ \pi^2(1-t)^2=\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{\cos\pi n t}{n^2} \Rightarrow (1-t)^2=\frac{1}{3}+4\sum_{n=1}^\infty \frac{\cos\pi n t}{\pi^2n^2}; \quad 0\le t\le 2. $$ It follows that $$ (1-t_1)^2-(1-t_2)^2=4\sum_{n=1}^\infty \frac{\cos\pi n t_1-\cos\pi n t_2}{\pi^2n^2}=8\sum_{n=1}^\infty \frac{\sin\pi n\frac{t_1+t_2}{2}\sin\pi n\frac{t_2-t_1}{2} }{\pi^2n^2}. $$ Finally substituting $t_1=|x-y|$, $t_2=x+y$, where the absolute value was taken to ensure $t_1\ge0$, one obtains: $$ \sum_{n=1}^\infty \frac{\sin\pi n x\sin\pi n y}{\pi^2n^2}=\frac{(1-|x-y|)^2-(1-x-y)^2}{8}=\frac{(x+y)-|x-y|-2xy}{4}=\frac{\min(x,y)-xy}{2}. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mathcal{S} & \equiv \sum_{n = 1}^{\infty}{\sin\pars{n\pi y}\sin\pars{n\pi x} \over n^{2}\pi^{2}} = \sum_{n = 1}^{\infty}xy\,\mrm{sinc}\pars{n\pi\verts{y}}\,\mrm{sinc}\pars{n\pi\verts{x}} \\[5mm] & = -xy + xy\sum_{n = 0}^{\infty} \mrm{sinc}\pars{n\pi\verts{y}}\,\mrm{sinc}\pars{n\pi\verts{x}} \end{align}

The series is evaluated with the Abel-Plana Formula when $\ds{\verts{x} + \verts{y} < 2}$ as I'll explain below. In any other case, we can use the periodic properties of the $\ds{\sin}$-function to render the arguments "inside the convergent region". The Abel-Plana formula can be used when the following expression vanishes out as $\ds{\Im\pars{z} \to \pm\infty}$:

\begin{align} &xy\,\mrm{sinc}\pars{z\pi\verts{y}}\,\mrm{sinc}\pars{z\pi\verts{x}} \expo{-2\pi\verts{\Im\pars{z}}} \\[5mm] \stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\sim}\,\,\,& \pm\,{\mrm{sgn}\pars{xy} \over 4\pi^{2}} \exp\pars{\rule{0pt}{4mm} -\bracks{\rule{0pt}{6mm}2 - \verts{x} - \verts{y}}\pi\verts{\Im\pars{z}}} \,\,\,\stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\to}\,\,\, {\Large 0} \\[2mm] &\ \bbx{\mbox{when}\ \verts{x} + \verts{y} < 2} \end{align} Then, \begin{align} \mathcal{S} & \equiv \sum_{n = 1}^{\infty}{\sin\pars{n\pi y}\sin\pars{n\pi x} \over n^{2}\pi^{2}} \\[5mm] & = -xy + xy\int_{0}^{\infty}\mrm{sinc}\pars{n\pi\verts{y}} \,\mrm{sinc}\pars{n\pi\verts{x}}\dd n + xy\bracks{{1 \over 2}\mrm{sinc}\pars{n\pi\verts{y}}\,\mrm{sinc}\pars{n\pi\verts{x}}} _{\ n\ =\ 0} \\[5mm] & = -\,{1 \over 2}\,xy + {xy \over \pi}\int_{0}^{\infty}\mrm{sinc}\pars{n\verts{y}} \,\mrm{sinc}\pars{n\verts{x}}\dd n \\[5mm] & = -\,{1 \over 2}\,xy + {xy \over \pi}\braces{\pi\,{\verts{x} + \verts{y} - \verts{\rule{0pt}{5mm}\verts{x} - \verts{y}} \over 4\verts{x}\verts{y}}} \\[5mm] & = \bbx{\mrm{sgn}\pars{x}\mrm{sgn}\pars{y} \min\braces{\verts{x},\verts{y}} - xy \over 2} \end{align}