Test for divergence of $int_{0}^{infty} frac{sin^2(x)}{x}dx$ without evaluating the integral

I would like to prove that $$\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$$ diverges without actually evaluating the integral. Is there a convergence test from calculus or real analysis that can show that this integral diverges?

Thanks.

Edit: Someone pointed out that this is a possible duplicate. However, the question put forth as a possible duplicate asks about $\sin(x^2)$, not about $\sin^2(x)$.

Answer

It is a divergent integral by Kronecker's lemma, since $\sin^2(x)$ is a non-negative function with mean value $\frac{1}{2}$. In more explicit terms, by integration by parts we have

$$\int_{\pi}^{N\pi}\frac{\sin^2(x)}{x}\,dx =\color{blue}{\left[\frac{1}{2}-\frac{\sin(2x)}{4x}\right]_{\pi}^{N\pi}}+\color{red}{\frac{1}{2}\int_{\pi}^{N\pi}\frac{dx}{x}}+\color{blue}{O(1)}$$
where the blue terms are bounded, but the red term equals $\frac{1}{2}\log N$.