Integrating $ \int \limits_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \operatorname d\!x $

Suppose that $\int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}$. By an integration by parts, $$\int_0^\infty \frac{\sin^2 x}{x^2} dx = -\frac{\sin^2 x}{x}|_0^\infty + \int_0^\infty \frac{\sin (2x)}{x} dx .$$

Check that $\lim_{x\to \infty}\frac{\sin^2 x}{x} = \lim_{x\to 0} \frac{\sin^2 x}{x}=0$. By a change of variable $t=2x$, $\int_0^\infty \frac{\sin (2x)}{x} dx = \int_0^\infty \frac{\sin t}{t} dt = \pi/2$ using our supposition. Hence $\int_0^\infty \frac{\sin^2 x}{x^2} dx = \frac{\pi}{2}$.

Therefore $\int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx = \pi$.