The problem is short: Find the summation of following infinite summation:
$$a(s, t) = \sum_{k = 1}^{\infty} \frac{2 \sin(\pi k t) \sin(\pi k s)}{(\pi k)^2}$$
where $t, s$ is two constant number here. I guess Parseval's theorem could work here, but I cannot find suitable $f$ till now. Could anyone give some hint or idea?
$$\begin{align*} a(u,\,v) &=\frac{2}{\pi^2}\sum_{k=1}^\infty\frac{\sin(\pi k u)\sin(\pi k v)}{k^2} \\ &= \frac{1}{\pi^2}\sum_{k=1}^\infty\frac{\cos\left(\pi k(u-v) \right )-\cos\left(\pi k(u+v) \right )}{k^2}.\\ \end{align*}$$
So now we just need to investigate the sum
$$S(x)=\sum_{k=1}^\infty\frac{\cos(\pi k x)}{k^2}.$$
Fortunately this is a much easier Fourier series to deal with, and one may verify by evaluating the Fourier integrals that
$$S(x)=\frac{\pi^2}{6}-\frac{\pi^2}{2}x+\frac{\pi^2}{4}x^2,\qquad\text{for }0\leq x\leq2.$$
From this you can compute your sum, as
$$\begin{align*} a(u,\,v) &=\frac{1}{\pi^2}\left(S(|u-v|)-S(|u+v|)\right) \\ &=\frac{|u+v|}{2}-\frac{|u-v|}{2}-uv,\qquad\text{for }0\leq u-v \leq 2,\,0\leq u+v\leq2. \end{align*}$$
As a note, I couldn't find the closed form of $S(x)$ off the top of my head. But I was able to get an idea of what it should be by taking the Mellin transform of $\cos(\pi kx)$ and then summing over the residues in $s$ of
$$\frac{\Gamma(s)\cos(\pi s/2)\zeta(s+2)}{\pi^s}x^{-s}.$$