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

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

A: 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:
$$
\sum_{n=1}^\infty \frac{\sin\pi n x\sin\pi n y}{\pi^2n^2}=\frac{(1-|x-y|)^2-(1-x-y)^2}{8}=\frac{(x+y)-|x-y|-2xy}{4}=\frac{\min(x,y)-xy}{2}.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
