Convergence of a Series involving $\cos$ and $\log$ Does the following series converge?
$$\sum_{k=1}^{\infty}\int_0^{\pi}\int_0^{\pi}\cos(2k(x-y))\log\big(\sin|\frac{x-y}{2}|\big)\,dx\,dy$$
 A: Today it is a fashion argument: see this question and this other one, too. 
We have that $\log(2\sin u)$ has a nice Fourier series:
$$\log(2\sin u)=-\sum_{n=1}^{+\infty}\frac{\cos(2nu)}{n}\tag{1}$$
hence:
$$\begin{eqnarray*} I_k &=& \int_{0}^{\pi}\int_{0}^{\pi}\cos(2k(x-y))\log\sin\left|\frac{x-y}{2}\right|\,dx\,dy\\ &=& 2\int_{0}^{\pi}\int_{0}^{x}\cos(2k(x-y))\log\sin\frac{x-y}{2}\,dy\,dx\\&=&2\int_{0}^{\pi}\int_{0}^{1}x\cos(2k(x-xt))\log\sin\frac{x-xt}{2}\,dt\,dx\\&=&2\int_{0}^{\pi}\int_{0}^{1}x\cos(2kxz)\log\left(2\sin\frac{xz}{2}\right)\,dz\,dx\\&=&-2\sum_{n=1}^{+\infty}\frac{1}{n}\int_{0}^{\pi}\int_{0}^{1}x\cos(2kxz)\cos(nxz)\,dz\,dx\end{eqnarray*}$$
Now the last double integral equals $\frac{\pi^2}{4}$ if $n=2k$,
$$4\frac{n^2+4k^2}{(n^2-4k^2)^2}$$
if $n$ is odd, zero otherwise, hence the contribute given by $n=2k$ makes the original double integral not summable over $k$:
$$\sum_{k=1}^{K}\int_{0}^{\pi}\int_{0}^{\pi}\cos(2kx)\log\sin\left|\frac{x-y}{2}\right|\,dx\,dy< -\frac{\pi^2}{4}\sum_{k=1}^{K}\frac{1}{k}<-\frac{\pi^2}{4}\log K.$$
