Integral of the day XXII

Nnice integral representation of $\pi$

Today we show the proof of this integral posted by @integralsbot \[\int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = -\pi \] For the proof we will use the Lobachevsky integral formula.

Proof

Recall the Lobachevsky integral formula:

Proposition. Let $f(x)$ a complex-valued, $\pi$-periodic function that is absolutely integrable over a single period. Therefore \[ \int_{0}^{\infty} \frac{\sin^2 x}{x^2}f(x)dx = \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx \] Note that \[ \int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = \int_{0}^{\infty} \frac{\sin^2 x}{x^2}\frac{\ln \cos^2 x}{\sin^2 x} dx \] Since the function \[ f(x) = \frac{\ln \cos^2 x}{\sin^2 x} dx\] is $\pi$-periodic it follows that \[ I = \int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = \int_{0}^{\infty} \frac{\sin^2 x}{x^2}\frac{\ln \cos^2 x}{\sin^2 x} dx= \int_{0}^{\frac{\pi}{2}} \frac{\ln \cos^2 x}{\sin^2 x} dx\] Hence \begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\ln \cos^2 x}{\sin^2 x} dx =& 2\int_{0}^{\frac{\pi}{2}} \frac{\ln \cos x}{\sin^2 x} dx\\ =& 2\int_{0}^{1} \frac{\ln(w)}{(1-w^2)^{\frac{3}{2}}} dw \quad \left(w\mapsto \cos x \right)\\ \stackrel{IBP}{=}& \underbrace{2\frac{\ln(w)w}{\sqrt{1-w^2}}\Big|_{0}^{1}}_{=0}-2\int_{0}^{1} \frac{1}{\sqrt{1-w^2}}dw \\ =& -2\arcsin(1)\\ =& -\pi \end{align*} Therefore \[\boxed{\int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = -\pi }\]