Integral using Parseval's Theorem How would I integrate 
$$\int_{-\infty}^{+\infty} \frac{\sin^{2}(x)}{x^{2}}\,dx$$  using Fourier Transform methods, i.e. using Parseval's Theorem ? 
How would I then use that to calculate: 
$$\int_{-\infty}^{+\infty} \frac{\sin^{4}(x)}{x^{4}}\,dx$$? 
 A: Both integrals can be computed through integration by parts, since:
$$ \int_{\mathbb{R}}\frac{\sin^2 x}{x^2}\,dx = \int_{\mathbb{R}}\frac{\sin(2x)}{x}\,dx = \pi$$
and: 
$$ \int_{\mathbb{R}}\frac{\sin^4 x}{x^4}\,dx = \int_{\mathbb{R}}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{3x^3}\,dx=\int_{\mathbb{R}}\frac{\cos(2x)-\cos(4x)}{3x^2}\,dx=\frac{2\pi}{3}.$$
The general integral
$$\int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^n\,dx $$
can be computed from the pdf of the Irwin-Hall distribution.
A: Parseval in this case states that
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$
where $F$ and $G$ are the respective FTs of $f$ and $g$.  (Integrability conditions such as absolute integrability over the real line must be satisfied for both pairs of functions.)
When $f(x) = g(x) = \sin{x}/x$, then $F(k) = G(k) = \pi I_{[-1,1]}$ and we have
$$\int_{-\infty}^{\infty} dx \frac{\sin^2{x}}{x^2} = \frac1{2 \pi} \pi^2 \int_{-1}^1 dk  = \pi$$
When $f(x) = g(x) = \sin^2{x}/x^2$, then $F(k) = G(k) = \pi (1-|k|/2) I_{[-2,2]} $, and we have
$$\int_{-\infty}^{\infty} dx \frac{\sin^4{x}}{x^4} = \frac1{2 \pi} \pi^2 \int_{-2}^2 dk \left ( 1-\frac{|k|}{2} \right )^2 = \frac{2 \pi}{3}$$
