Definite integral involving logarithm of cosine Does anyone know the provenance of or the answer to
the following integral
$$\int_0^\infty\ \frac{\ln|\cos(x)|}{x^2} dx $$
Thanks.
 A: Lucian's answer is just fine (as always), but from
$$ \sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}=\frac{1}{\sin^2 x}\tag{1}$$
for any $x\in(-\pi,\pi)$ it also follows that:
$$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{\log\cos^2 x}{x^2}\,dx = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\log\cos x}{\sin^2 x}\,dx=-\frac{1}{2}\int_{0}^{+\infty}\frac{\log(1+t^2)}{t^2}\,dt$$
by replacing $x$ with $\arctan t$ in the last step. Integration by parts now leads to:
$$ I = -\int_{0}^{+\infty}\frac{dt}{1+t^2} = \color{red}{-\frac{\pi}{2}}.\tag{2}$$
Someone may ask now: How to prove $(1)$? 
Well, for such a purpose, start from the Weierstrass product for the sine function:
$$\frac{\sin z}{z}=\prod_{n\geq 1}\left(1-\frac{z^2}{\pi^2 n^2}\right)$$
then consider the logarithm of both sides and differentiate it twice with respect to $z$.
A: Hint: Let $I(n)=\displaystyle\int_0^\infty\frac{1-\cos^{2n}x}{x^2}~dx.~$ Prove first that $I(n)=n\pi~\dfrac{\displaystyle{2n\choose n}}{4^n}~,~$ then evaluate $I'(0)$.
A: This integral is equal to 
$$
\frac{1}{2} \int_0^\infty \frac{\ln (\cos^2 x)}{x^2} dx = \frac{1}{2}(-\pi) = -\frac{\pi}2$$
The easiest place to remember seeing this is Gradshteyn and Ryzhik, where it appears as definite integral 4.322.6.  The source quoted there is Fikhtebgik'ts, G. M. (http://en.wikipedia.org/wiki/Grigorii_Fichtenholz on Wikipedia) in the book Kurs differntsial'nogo i integral'ogo ischizdat, Vloume 2, page 686. 
The book is pictured on the WP page. Quoting the Wiki description, "Fichtenholz's books about analysis are widely used in Eastern European and Chinese universities due to its exceptionality of detailed and well-ordered presentation of material about mathematical analysis. " 
If this is an example of content in an introductory class on calculus, I think I am glad it has not been translated into English for me to have read as an undergraduatei!
