How to determine that the improper integral $\int_{\pi}^{\infty}\frac{\sin^2(x)}{x}dx$ diverges?

Question

How is the divergence of the improper integral $\int_{\pi}^{\infty}\frac{\sin^2(x)}{x}dx$ shown and determined with the improper integral comparison test or other methods?

Answer 1

We know that $|\sin(x)| \geq \frac12$ for $x \in [2\pi n + \pi/6, 2\pi n + 5\pi/6]$ (this holds for any $n \in \mathbb{N}$; to see why it's true just visualize the unit circle). Therefore we can estimate $$ \int_\pi^\infty \frac{\sin^2x}{x} \,dx \geq \sum_{n = 1}^{\infty} \int_{2\pi n + \pi/6}^{2 \pi n + 5\pi/6} \frac{\sin^2x}{x} \,dx \geq \sum_{n=1}^{\infty} \frac{1/4}{2\pi n + 5\pi/6}, $$ and this sum can be seen to diverge by comparison with the harmonic series $\sum_{n=1}^{\infty} \frac1n$.