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Use fourier transform and convolution to show that $$\int_{-\infty}^{\infty}\frac{(\sin(t))^3}{t^3}dt=\frac{3\pi}{4}$$

convolution is defined by $$(f * g)(t) = \int_{-\infty}^{\infty}(f(s)*g(t-s)ds $$. Fourier transform defined by $$\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$

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Come on, are you listening? Give some context, please! –  Hans Lundmark Oct 22 '10 at 12:50
    
The function $\sin x/x$ is basically the Fourier transform of the indicator function of unit interval centred at $0$. Hence $\sin^3 x/x^3$ is the FT of the 3-fold convolution of that indicator function.... –  Robin Chapman Oct 22 '10 at 18:13
    
alvoutila, did you find the answer given to be sufficient? If so, please consider clicking the "check mark" to accept this answer. And please consider doing this for all of your questions. –  The Chaz 2.0 Sep 19 '11 at 19:16

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up vote 4 down vote accepted

Let $\chi$ be the indicate function of $[-1,1]$. Then the Fourier transform of $\chi$ is $$g(\xi)=\frac{\sin(2\pi\xi)}{\pi\xi}.$$

Compute the triple convolution $h(x)$ of $\chi$ and evaluate it at 0. That will be 3. From Fourier inverse formula one gets $$h(x)=\int_{-\infty}^\infty (g(\xi))^3e^{2\pi ix\xi}d\xi.$$ Evaluating at $x=0$ yields $$ 3=\int_{-\infty}^\infty \frac{\sin^3(2\pi\xi)}{\pi^3\xi^3}d\xi,$$ from which the result follows by substitution.

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