TITLE: Solve $\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$
QUESTION [6 upvotes]: I have this problem
$\displaystyle \int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$ 
I think it diverges, but I cannot prove it. (I'm in my first calculus course, so I don't know of any more advanced methods than 

u-substitution (including trig subs)
integration by parts

REPLY [2 votes]: Since $(\sin^2 x)/x$ is non-negative on $(0, \frac{\pi}{2})$, 
$$
\int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx
  \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx.
$$
Further, $\cos^2(x + \frac{\pi}{2}) = \sin^2 x$ for all real $x$, so substitution ($x = u + \frac{\pi}{2}$) gives
$$
\int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx
  = \int_0^\infty \frac{\sin^2 x}{x + \frac{\pi}{2}}\, dx
  \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx.
$$
Adding these inequalities,
$$
\int_{\frac{\pi}{2}}^\infty \frac{dx}{x}
  = \int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx
  + \int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx
  \leq 2\int_0^\infty \frac{\sin^2 x}{x}\, dx,
$$
and the lower bound is clearly divergent.