TITLE: How to prove that $\int_{-\infty}^{\infty} \frac{1-\cos x}{x^2} dx$ equal to $\pi $?
QUESTION [6 upvotes]: How to prove that $\int_{-\infty}^{\infty} \frac{1-\cos x}{x^2} dx$ equal to $\pi $?

Is there any simple approach that does not require knowledge in Fourier Analysis or Complex analysis?

REPLY [4 votes]: Note that $$I=\int_{-\infty}^{\infty}\frac{1-\cos\left(x\right)}{x^{2}}dx=2\int_{0}^{\infty}\frac{1-\cos\left(x\right)}{x^{2}}dx
 $$ $$\stackrel{IBP}{=}2\int_{0}^{\infty}\frac{\sin\left(x\right)}{x}dx=\frac{1}{i}\int_{0}^{\infty}\frac{e^{ix}-e^{-ix}}{x}dx=\frac{1}{i}\lim_{\epsilon\rightarrow0^{+}}\int_{0}^{\infty}\frac{e^{\epsilon+ix}-e^{\epsilon-ix}}{x}dx
 $$ and now using the Frullani's theorem with complex parameters we get $$I=\frac{1}{i}\lim_{\epsilon\rightarrow0^{+}}\log\left(\frac{\epsilon+i}{\epsilon-i}\right)=\color{red}{\pi}$$ as wanted.