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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.

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  • $\begingroup$ Are you allowed to use residue theorem? $\endgroup$ – rtybase May 12 '19 at 9:41
  • $\begingroup$ I am afraid no, we have never done that. $\endgroup$ – TwoStones May 12 '19 at 9:42
  • $\begingroup$ What about Cauchy's integral formula? $\endgroup$ – rtybase May 12 '19 at 9:45
  • $\begingroup$ Should not the integral by computation of $\tilde f(\omega)$ be from $-1$ to $1$? $\endgroup$ – user May 12 '19 at 9:50
  • $\begingroup$ @user which integral do you mean? $\endgroup$ – TwoStones May 12 '19 at 9:54
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Hint: Use Plancherel theorem $$\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int_{-\infty }^{\infty}|{\hat {f}}(\omega)|^{2}\,d\omega$$ then $$\int _{-1}^{1}dx=\int_{-\infty }^{\infty}\frac{4}{2\pi}\dfrac{\sin^2\omega}{\omega^2}\,d\omega$$

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  • $\begingroup$ But why disappeared the $t$ from the $\sin^2(\omega t)$ in your integral? $\endgroup$ – TwoStones May 12 '19 at 9:53
  • $\begingroup$ @TwoStones $t$ is integration variable and doesn't appear in Fourier transform. $\endgroup$ – Nosrati May 12 '19 at 9:57
  • $\begingroup$ Ah, I see. Do you have any tip for the second integral? $\endgroup$ – TwoStones May 12 '19 at 10:05
  • $\begingroup$ For the second use convolution Fourier transform. $\endgroup$ – Nosrati May 12 '19 at 10:13
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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$$\frac{\pi}{8}\int_{\Bbb R}|(f\ast f)(t)|^2dt=\frac{\pi}{4}\int_0^2(2-t)^2dt=\frac{2\pi}{3}.$$

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