I need to find the value of the integral:
$\int_{-\infty}^{\infty} \frac{sin^2x}{x^2}dx $
Right now progress:
Because the value of $\frac{sin^2x}{x^2}$ is convergent, the integral will be equal to its principal value. So $\int_{-\infty}^{\infty} \frac{sin^2z}{z^2}dz = P\int_{-\infty}^{\infty}\frac{sin^2x}{x^2}dx $
We can write $sin^2 x = \frac{(e^{ix} - e^{-ix})^2}{(2i)^2x^2}= \frac{e^{2ix} - 2 e^{ix} e^{-ix} + e^{-2ix}}{(2i)^2x^2}$ however, now $e^{2ix} - 2+ e^{-2ix} = cosh(2ix)/2 -2$
For cosh, we know that the right contour to use is rectangular contour however, I'm kind of stuck and would really appreciate some insight. Thanks.