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Proof for an integral involving sinc function
How do I show that $\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \ dx = \pi$?

$\int_{-\infty}^{\infty}\sin^2(x)/x^2=\pi$ according to wolfram alpha. That is such a beautiful result! But how do I calculate this integral by hand?

Thanks in advance.

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    If you are familiar with Fourier Analysis, you can calculate the inverse of $\frac{\sin x}{x}$, then use Parseval-Plancherel equality to get the result.2012-05-06

5 Answers 5

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The answer may be found by using complex analysis, specifically the residue theorem. A full deriviation may be found here.

I know of an easy way to derive this result using real analysis alone.

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    While the link looks nice at first sight, I think there is a problem with the integral of 2/z^2. Actually it.should go.to.infinity for.countour C0. Something is swept under the rug.2018-04-07
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An easy way:

$I(a)=\int_{-\infty}^{\infty}\frac{\sin^2(ax)dx}{x^2};=>$

$=>\frac{dI}{da}=\int_{-\infty}^{\infty}\frac{\sin(2ax)dx}{x}=\pi;=>$

$=>I(a)=\pi a+const;=>$

$I(a)=\pi a$ because $I(0)=0$

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    This is cute. How'd you think of it?2016-11-28
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First we split $\sin^2(x)=\frac{(1-e^{2ix})+(1-e^{-2ix})}{4}$. To avoid the pole at $x=0$, drop the path of integration a bit below the real line (this function has no poles and it vanishes at infinity, so this is okay).

Next, let $\gamma^+$ be the path below the real axis, then circling back in a semi-circular path counterclockwise around the upper half-plane; and let $\gamma^-$ be the path below the real axis, then circling back in a semi-circular path clockwise around the lower half-plane.

$\hspace{4cm}$enter image description here

Note that $\gamma^+$ circles the pole at $x=0$ of $\frac{(1-e^{2ix})}{4x^2}$ and $\gamma^-$ misses the pole at $x=0$ of $\frac{(1-e^{-2ix})}{4x^2}$.

Therefore, $ \begin{align} \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x &=\int_{-\infty-i}^{\infty-i}\frac{1-\cos(2x)}{2x^2}\mathrm{d}x\\ &=\int_{-\infty-i}^{\infty-i}\frac{(1-e^{2ix})+(1-e^{-2ix})}{4x^2}\mathrm{d}x\\ &=\color{green}{\int_{\gamma^+}\frac{(1-e^{2ix})}{4x^2}\mathrm{d}x}+\color{red}{\int_{\gamma^-}\frac{(1-e^{-2ix})}{4x^2}\mathrm{d}x}\\ &=\color{green}{2\pi i\frac{-2i}{4}}+\color{red}{0}\\ &=\pi \end{align} $

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    Would the downvoter care to explain if something is wrong? I don't see anything wrong, so I would appreciate knowing if there is some error.2017-12-04
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Use Parseval theorem $\int_{-\infty}^{\infty}dx |f(x)|^{2}= \int_{-\infty}^{\infty}du|F(u)|^{2} $

the Fourier inverse transform of $ \frac{sin(x)}{x} $ is an step function (window function )

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This is, for example, Exercise 2 of chapter 11 of the book Complex analysis by Bak and Newman. The hint is: integrate $\frac{e^{2iz}-1-2iz}{z^2}$ around a large semi-circle.