<s>[INST] How to use the Fourier transform for show that $ \int_{0}^{\infty} \frac{sin^2(x)}{x^2} dx = \frac{\pi}{2} $? 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} $ ?
 [/INST] $$ \mathcal{F}(F(\alpha) = \frac{1-\cos(x)}{x^2} = \frac{4\sin^2(\frac{x}{2})}{\pi x^2} $$
Let $\alpha = 0$
$$ \int_{0}^{\infty} \frac{4\sin^2(\frac{x}{2})}{\pi x^2} \ cos(0) dx = 1 $$
$$ t=x/2 \Rightarrow x=2t \Rightarrow \frac{dx}{dt} = 2 $$
$$ 2 \int_{0}^{\infty} \frac{4\sin^2(t)}{4 \pi t^2} dt = 1 \iff \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = 1 \iff \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = \frac{\pi}{2} $$
</s>