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: $$\int_{-\infty}^{+\infty} \frac{\sin^{4}(x)}{x^{4}}\,dx$$?

Solution 1:

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.

Solution 2:

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

$$\int_{-\infty}^{\infty} dx \frac{\sin^4{x}}{x^4} = \frac1{2 \pi} \pi^2 \int_{-2}^2 dk \left ( 1-\frac{|k|}{2} \right )^2 = \frac{2 \pi}{3}$$