I have the infinite series from a Fourier series problem:
$\sum_{n=1}^\infty \frac{1}{n^2}sin(n\pi x)sin(n\pi y)$
which is stated to be proportional to $x*(1-y)$ when $0 \leq x \leq 1$ and $x \leq y \leq 1$
However, when I'm evaluating that sum for generic $x$ and $y$ in Mathematica, etc., I end up with a mess of polylogarithms:
$\frac{1}{4}[Li(2,exp(i\pi (x-y))) + Li(2,exp(-i\pi (x-y))) - Li(2,exp(i\pi (x+y))) - Li(2,exp(-i\pi (x+y)))]$
Plotting this over the range of applicability of the analytical solution appears to be identical, but I have no idea what methods or approximations goes into simplifying these functions, or if I could avoid working with polylogarithms altogether.
Figured it out after a lot of searching.
Rewriting $\sin(n\pi x) \sin(n\pi y)$ as $\cos(n\pi (x-y)) - \cos(n\pi (x+y))$ so we now have two sums that go as $\sum_n \cos(n\theta)/n^2$
This turns out to be a standard Clausen function, which has an analytic solution from comparing its Fourier series to the Bernoulli series, and gives for $\theta < 2\pi$,
$\sum_n \cos(n\theta)/n^2 = \frac{\pi^2}{6} - \frac{\pi\theta}{2} + \frac{\theta^2}{4}$
From which point its easy to get the identity I was looking for