Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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.

share|cite|improve this question

marked as duplicate by Pedro Tamaroff, Martin Sleziak, Nate Eldredge, Zev Chonoles Jun 13 '12 at 6:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I added an answer there using Siminore's suggestion below to integrate by parts. – joriki May 6 '12 at 11:58
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. – y zh May 6 '12 at 14:02

5 Answers 5

up vote 3 down vote accepted

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.

share|cite|improve this answer
Just integrate by parts and use $\int_{-\infty}^{+\infty} \frac{\sin x}{x}dx$. – Siminore May 6 '12 at 11:24

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

share|cite|improve this answer
If by $\int _{-\infty -i}^{\infty -i}$ you mean the integral over the line $y=-1,$ how do we know the integral over this line is equal to the integral over $\mathbb{R}$? – The Substitute Apr 15 at 8:18
Because $\frac{\sin^2(x)}{x^2}$ has no singularities and tends to $0$ along the ends of the long rectangle $[-R,R]\cup\color{#C00000}{[R,R-i]}\cup[R-i,-R-i]\cup\color{#C00000}{[-R-i,-R]}$‌​. The integral along that contour tends to the difference of the two integrals as $R\to\infty$. (That was the intended meaning of the parenthetical comment.) – robjohn Apr 15 at 8:24

An easy way:



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

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

share|cite|improve this answer

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 )

share|cite|improve this answer

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.