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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.)

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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.

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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



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

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

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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

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