Show that $\int_{0}^\infty \frac{1-\cos x}{x^2}dx=\pi/2$.

Question

I am trying to show that $$\int_{0}^\infty \frac{1-\cos x}{x^2}dx=\pi/2.$$The hint is "try simple substitution", and not incidentally, the previous problem has shown that $\int_0^\infty \frac{\sin^2(xu)}{u^2}du=\frac{\pi}{2}|x|$. This looks an awful lot like we'd like to reduce it to the earlier case, for $x=1$.

What shall we try to substitute for? I think we'd have some problems subbing for cosine, since it does not approach a limit at infinity (correct me if there is a way to make this substitution). Subbing for $x^2$ hasn't gotten me anywhere.

We might want to try to split it up, and see if anything better comes out of trying to integrate $\int_0^\infty \frac{\cos x}{x^2}dx$. No luck there so far.

Any ideas?

Answer 1

Hint: Use the identity

$$ \cos(x)=1-2\sin^2(x/2). $$

Answer 2

Double integral as an alternative $$ \begin{aligned} \int_0^{\infty} \frac{1-\cos x}{x^2} d x = & \int_0^{\infty}\left( \int_0^1 \frac{\sin (x y)}{x} d y \right)d x \\ = & \int_0^1\left(\int_0^{\infty} \frac{\sin (x y)}{x} d x\right) d y \\ = & \int_0^1 \frac{\pi}{2} d y \\ = & \frac{\pi}{2} \end{aligned} $$

Note: $\int_0^{\infty} \frac{\sin (x y)}{x} d x=\frac{\pi}{2}$ for all $y>0$.