TITLE: Solve $ \int_{0}^{\infty} \sin^2 \left(\frac{1}{x}\right)\mathrm{d}x$
QUESTION [5 upvotes]: I think this integral does not converge. I want to  estimate downward the integral, but don't know how to.

REPLY [2 votes]: Set $u=1/x$, then $du=-dx/x^2$, and
$$
\int_0^\infty \sin^2\left(\frac{1}{x}\right)\,dx=\int_0^\infty\frac{\sin^2u\,du}{u^2},
$$
which converges, as 
$$
\frac{\sin^2u}{u^2}\le \max\left\{\frac{1}{u^2},1\right\}.
$$
Note. Using residue calculus one could obtain that
$$
\int_0^\infty\frac{\sin^2u\,du}{u^2}=\frac{\pi}{2}.
$$