TITLE: How to compute $\int_{2/\pi}^{+\infty }\ln\cos(1/x)\,dx$?
QUESTION [5 upvotes]: What it says in the title.
 If $$I=\int_{2/\pi}^{+\infty }\ln\left({\cos\left({\frac{1}{x} }\right) }\right) \, \mathrm dx,$$ how could I proceed in order to find the value of $I$?

REPLY [4 votes]: Using a change of variable and a well-known fact:
$$I=\int_{0}^{\pi/2}\frac{\log\cos x}{x^2}\,dx=2\sum_{k\geq 1}\frac{(-1)^k}{k}\int_{0}^{\pi/2}\left(\frac{\sin(kx)}{x}\right)^2\,dx\tag{1}$$
but since:
$$ \cos x = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{2}$$
we have:
$$ \log \cos x = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m}\pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{m\,\pi^{2m}}x^{2m}\tag{3}$$
so the RHS of $(1)$ also equals:
$$ -\frac{2}{\pi} \sum_{m\geq 1}\frac{(1-4^{-m})\zeta(2m)}{ m(2m-1)}\tag{4}$$
that is a fast convergent series.