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

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.

• Are you allowed to use residue theorem? – rtybase May 12 '19 at 9:41
• I am afraid no, we have never done that. – TwoStones May 12 '19 at 9:42
• What about Cauchy's integral formula? – rtybase May 12 '19 at 9:45
• Should not the integral by computation of $\tilde f(\omega)$ be from $-1$ to $1$? – user May 12 '19 at 9:50
• @user which integral do you mean? – TwoStones May 12 '19 at 9:54

Hint: Use Plancherel theorem $$\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int_{-\infty }^{\infty}|{\hat {f}}(\omega)|^{2}\,d\omega$$ then $$\int _{-1}^{1}dx=\int_{-\infty }^{\infty}\frac{4}{2\pi}\dfrac{\sin^2\omega}{\omega^2}\,d\omega$$
• But why disappeared the $t$ from the $\sin^2(\omega t)$ in your integral? – TwoStones May 12 '19 at 9:53
• @TwoStones $t$ is integration variable and doesn't appear in Fourier transform. – Nosrati May 12 '19 at 9:57
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$$\frac{\pi}{8}\int_{\Bbb R}|(f\ast f)(t)|^2dt=\frac{\pi}{4}\int_0^2(2-t)^2dt=\frac{2\pi}{3}.$$