Take the 2-minute tour ×
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.

I try to evaluate $$\int_{-\infty}^\infty \frac{\sin^2 x}{x^2}e^{itx}\,dx$$ ($t$ real) using contour integrals, but encounter some difficulty. Perhaps someone can provide a hint. (I do not want to use convolution.)

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Rewrite your integral as $$\int_{-\infty}^\infty \frac{\sin^2 x}{x^2}e^{itx}\,dx = \int_{-\infty}^\infty \frac{1-\cos(2x)}{2x^2}e^{itx}\,dx = \int_{-\infty}^\infty \frac{2-e^{2ix}-e^{-2ix}}{4 x^2 }e^{itx}\,dx,$$ seperate the integral in three (or two) independent integrals and then apply the method of contour integrals.

share|improve this answer
    
But the real part of (1-e^2ix)e^{itx} is not (1-cos(2x)e^{itx}. This is part of my problem. –  TCL Dec 7 '12 at 19:04
    
@TCL: sorry, my bad. I edited the answer... Hope now everything is fine. –  Fabian Dec 7 '12 at 19:36
    
If $0<t<2$, rewrite as $$\frac{2e^{itx}-e^{(t+2)ix}-1}{4x^2}+\frac{1-e^{(t-2)ix}}{4x^2}$$ and integrate seperately. This is to gurantee that the integral over the circular arc goes to 0 as $R\to \infty$. Similar if $-2<t<0$. No need to seperate if $|t|>2$. –  TCL Dec 7 '12 at 21:42

An idea, defining

$$f(z):=\frac{e^{itz}\sin^2z}{z^2}\,\,,\,\,C_R:=[-R-\epsilon]\cup(-\gamma_\epsilon)\cup[\epsilon,R]\cup\gamma_R$$

with

$$\gamma_k:=\{z\in\Bbb C\;;\;|z|=k\,,\,\arg z\geq 0\}=\{z\in\Bbb C\;;\;z=ke^{i\theta}\,\,,\,0\leq\theta\leq\pi\}$$

in the positive direction (check the minus sign in $\,\gamma_\epsilon\,$ above!).

This works assuming $\,0<t\in\Bbb R\,$, Jordan's lemma in the lemma and its corollaty in the answer here

share|improve this answer
    
But sin z/z does not tend to 0 as z-->infinity, so Jordan's lemma cannot be used. –  TCL Dec 7 '12 at 19:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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