Q:

Proving that $\int_{1}^{\infty} \frac{\sin^{2}(3x)}{x} dx$ diverges

I have to prove that $$\int_{1}^{\infty} \frac{\sin^{2}(3x)}{x}\,dx$$ diverges.
Can anyone give a hand? I'm totally stuck.

A:

We have, by integration by parts:
$$ \int_{1}^{M}\frac{\sin^2(3x)}{x}\,dx = \left.\left(\frac{1}{2}-\frac{\sin(6x)}{12 x}\right)\right|_{1}^{M}+\int_{1}^{M}\left(1-\frac{\sin(6x)}{6x}\right)\,\frac{dx}{2x}$$
but $1-\frac{\sin 6x}{6x}$ is trivially greater than $\frac{5}{6}$ for any $x\geq 1$ and $$\frac{5}{12}\cdot\lim_{M\to +\infty}\int_{1}^{M}\frac{dx}{x} = +\infty.$$

Actually the integral diverges like $\frac{1}{2}\log(M)>\frac{5}{12}\log(M)$, since $\int_{1}^{+\infty}\frac{\cos(6x)}{x}\,dx$ is convergent by Dirichlet's test (integral version), but no matter the constant, the main term of the asymptotics is $\log(M)$ as expected.

