Showing $\int_{-\infty}^{\infty}\frac{\sin^2(\omega)}{\omega^2}d\omega=\pi$

Question

Given the function $f$ with $f(t)=1$ for $|t|<1$ and $f(t)=0$ otherwise, I have to calculate its Fourier-transform, the convolution of $f$ with itself and from that I have to show that $$\int_{-\infty}^{\infty}\frac{\sin^2(\omega)}{\omega^2}d\omega=\pi$$ and $$\int_{-\infty}^{\infty}\frac{\sin^4(\omega)}{\omega^4}d\omega =\frac{2\pi}{3}$$

( $\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}$$\int_{-t}^{t}1\cdot e^{-i\omega t}dt$ be the Fourier-transform of $f$ )

For the first two parts I have:

$\tilde{f}(\omega)=\frac{2}{\sqrt{2\pi}}\frac{\sin(\omega t)}{\omega}$ and $(f*f)(\omega)=\frac{2}{\pi}\frac{\sin^2(\omega t)}{\omega^2}$.

But from here I dont know how to compute the integrals. My idea for the first one was using Fourier-Inversion of $(f*f)$ and then putting $t=1$. But that gives me $0$ for the integral.

Does someone has another idea? I would be grateful for any hint or advice!

Thank you.

Answer 1

Hint: Use Plancherel theorem $$\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int_{-\infty }^{\infty}|{\hat {f}}(\omega)|^{2}\,d\omega$$ then $$\int _{-1}^{1}dx=\int_{-\infty }^{\infty}\frac{4}{2\pi}\dfrac{\sin^2\omega}{\omega^2}\,d\omega$$

Answer 2

Your expression for $\tilde{f}\left(\omega\right)$ is incorrect; you should have $$\tilde{f}\left(\omega\right)=\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-i\omega t}dt=\sqrt{\frac{2}{\pi}}\frac{\sin\omega}{\omega}.$$The integrals you seek are then $$\int_{\Bbb R}\frac{\pi}{2}\left|\tilde{f}\left(\omega\right)\right|^{2}d\omega,\,\int_{\Bbb R}\frac{\pi^{2}}{4}\left|\tilde{f}\left(\omega\right)\right|^{4}d\omega.$$By Plancherel, the first integral is $\frac{\pi}{2}\int_{-1}^1dt=\pi$, while the second is $\frac{1}{2\pi}\frac{\pi^2}{4}\int_{\Bbb R}|(f\ast f)(t)|^2dt$. In terms of Iverson brackets, $$(f\ast f)(t)=\int_{\Bbb R}[u\in[-1,\,1]][t-u\in[-1,\,1]]du.$$I'll leave it as an exercise to verify $(f\ast f)(t)=(2-|t|)[t\in[-2,\,2]]$, so$$\frac{\pi}{8}\int_{\Bbb R}|(f\ast f)(t)|^2dt=\frac{\pi}{4}\int_0^2(2-t)^2dt=\frac{2\pi}{3}.$$

Answer 3

I will solve your second question. I am using Stein's convention that:

$$\mathcal{F}(g)(w)=\int_{-\infty}^\infty g(x)e^{-2\pi i w x}dx \quad \quad \mathcal{F}^{-1}(g)(x)=\int_{-\infty}^\infty g(w)e^{2\pi i w x}dw$$

$\mathcal{F}(\chi_{(-1,1)}/2)=\sin(2\pi w)/(2\pi w)$. From this it follows that:

$$ \mathcal{F}\left(\frac{1}{4}(2+x)\chi_{(-2,0)}(x)+\frac{1}{4}(2-x)\chi_{(0,2)}(x)\right)=\mathcal{F}(\chi_{(-1,1)}/2*\chi_{(-1,1)}/2)=\mathcal{F}(\chi_{(-1,1)}/2)\mathcal{F}(\chi_{(-1,1)}/2)=\left(\frac{\sin(2\pi w)}{2\pi w}\right)^4$$

Using the Inverse and a change of variables yields:

$$\frac{1}{4}(2+x)\chi_{(-2,0)}(x)+\frac{1}{4}(2-x)\chi_{(0,2)}(x)=\mathcal{F}^{-1}\left(\left(\frac{\sin(2\pi w)}{2\pi w}\right)^4\right)(x)=\frac{1}{2\pi}\mathcal{F}\left(\left(\frac{\sin( w)}{ w}\right)^4\right)(x/2\pi)$$ $$\therefore \mathcal{F}\left(\left(\frac{\sin( w)}{ w}\right)^4\right)(x)=2\pi (\frac{1}{4}(2+2\pi x)\chi_{(-2,0)}(x)+\frac{1}{4}(2-2\pi x)\chi_{(0,2)}(x))$$ By Plancharel's identity:

$$\int_{-\infty}^\infty \left(\frac{\sin(t)}{t}\right)^4dt=\int_{-\infty}^\infty \left(2\pi (\frac{1}{4}(2+2\pi x)\chi_{(-2,0)}+\frac{1}{4}(2-2\pi x)\chi_{(0,2)})\right)^2 dx= \frac{2 \pi}{3}$$