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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2 Answers
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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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$\begingroup$ But the real part of (1-e^2ix)e^{itx} is not (1-cos(2x)e^{itx}. This is part of my problem. $\endgroup$– TCLDec 7, 2012 at 19:04
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$\begingroup$ @TCL: sorry, my bad. I edited the answer... Hope now everything is fine. $\endgroup$– FabianDec 7, 2012 at 19:36
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$\begingroup$ 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 $|t|>2$. $\endgroup$– TCLDec 7, 2012 at 21:42
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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
answered Dec 7, 2012 at 18:52
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$\begingroup$ But sin z/z does not tend to 0 as z-->infinity, so Jordan's lemma cannot be used. $\endgroup$– TCLDec 7, 2012 at 19:24