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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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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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    If 0, 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. No need to seperate if |t|>2.2012-12-07
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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, 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.2012-12-07