Suppose $$ \int_{0}^{\infty} f(x)\cos(\alpha x)dx = F(\alpha) = \left\{ \begin{array}{lcc} 1-\alpha & if & 0\le \alpha \le 1 \\ \\ \hspace{0.6cm} 0 & if & \alpha>1 \\ \end{array} \right. $$ $$\mathcal{F}(F(\alpha) = \frac{2}{\pi} \int_{0}^{\infty} F(\alpha)\cos(\alpha x) d\alpha = \int_{0}^{1} (1-\alpha)\cos(\alpha x) d\alpha = \frac{1-\cos(x)}{x^2} $$
How to use this for show that $ \int_{0}^{\infty} \frac{\sin^2(x)}{x^2} dx = \frac{\pi}{2} $ ?
$$ \mathcal{F}(F(\alpha) = \frac{1-\cos(x)}{x^2} = \frac{4\sin^2(\frac{x}{2})}{\pi x^2} $$
Let $\alpha = 0$
$$ \int_{0}^{\infty} \frac{4\sin^2(\frac{x}{2})}{\pi x^2} \ cos(0) dx = 1 $$
$$ t=x/2 \Rightarrow x=2t \Rightarrow \frac{dx}{dt} = 2 $$
$$ 2 \int_{0}^{\infty} \frac{4\sin^2(t)}{4 \pi t^2} dt = 1 \iff \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = 1 \iff \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = \frac{\pi}{2} $$