Closed form for $\sum_{n=1}^\infty\frac{\cos(\pi \log n)}{n^2}$
Solution 1:
Of course there is. The fastest way to obtain it is to heuristically write $\cos(\pi\ln n)$ as $\frac12(n^{i\pi}+n^{-i\pi})$. The answer is then given by $$\sum_{n=1}^{\infty}\frac{\cos\left(\pi\ln n\right)}{n^2}=\frac{\zeta(2+i\pi)+\zeta(2-i\pi)}{2}.$$