A Fourier transform using contour integral

Question

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.)

Answer 1

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.

Answer 2

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