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

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?

• Thank you for editing my typo! – Eric Auld Jul 13 '13 at 22:42

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