Very difficult sum of series Compute the following sum 
$$\sum_{n=1}^{+\infty}\frac{1}{n^3 \sin(n \pi \sqrt{2})}$$
Source :
Very difficult sum of series
Like jmerry on AoPS I have no idea how to compute the sum.
Any ideas ? Perhaps someone knows already the result..
Thank you in advance for your time.

Partial answer:
$$\sum_{n\ge 1}\frac{1}{n^3\sin(n\pi \sqrt{2})}=\sum_{n\ge 1}\sum_{k=0}^{\infty}\frac{e^{-jn\pi\sqrt{2}}}{2j n^3}e^{-2jkn\pi\sqrt{2}}=\frac{1}{2j}\sum_{k\ge 0}\mbox{Li}_{3}\left(e^{-j(2k+1)\pi \sqrt{2}}\right)$$ where $\mbox{Li}_s(\cdot)$ is the polylogarithm function. We need to see if there is a closed form for this kind of summation.
