Evaluate $\displaystyle \int_0^\infty\left(\frac{\sin x}x\right)^2\,dx$
$\displaystyle
\int_0^\infty {\left( {\frac{{\sin x}}{x}} \right)^2 \,dx} = \int_0^\infty {\frac{{1 - \cos 2x}}{{2x^2 }}\,dx} .$
Use the following parameter: $\displaystyle \frac{1}{{x^2 }} = \int_0^\infty {ue^{ - ux} \,du} .$
(The only reason that I'd ever create a double integral is that I can reverse the integration order.)
From there you can solve the rest, without problems.
Or we can define
$\displaystyle I(\varphi ) = \int_0^\infty {\frac{{\sin ^2 (\varphi x)}}
{{x^2 }}\,dx} ,\,\varphi > 0.$
Then you can apply the Leibniz's Rule for differentiation under the integral sign. You'd get
$\displaystyle I(\varphi ) = \frac{\pi}
{2}\varphi .$
The conclusion follows.
$\displaystyle \int_{0}^{\infty} \frac{\cos ax - \cos bx}{x^2} dx = \int_{0}^{\infty} \int_a^b \frac{\sin tx}{x}dt~dx = \int_a^b \int_0^{\infty}\frac{\sin tx}{x}dx~dt$.
But $\displaystyle \int_0^{\infty} \frac{\sin tx}{x} dx = \int_a^b \frac{\pi}{2} dt = \frac{\pi}{2}(b-a)$.
That means,
$\displaystyle \int_0^{\infty} \frac{\sin^2 x}{x^2} dx = \frac{1}{2}\int_0^{\infty}\frac{1-\cos 2x}{x^2} dx = \frac{1}{2} \int_0^{\infty} \frac{\cos 0x - \cos 2x}{x^2} dx = \frac{1}{2}(2\pi ) = \pi$.
Where did I make a mistake? Anyways, you get the idea I am too lazy to check this right now.