Quick question. Could somebody please explain to me why it is that
$$\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \ dx = \pi$$
for every positive integer $n$? This integral showed up when I was computing a certain normalization constant. I was planning on just labeling it $I_n$ and moving on with my life but then Wolfram Alpha informed me it always equals $\pi$. Thanks!
How to Show Integral of Sine Functions Equals Pi
calculusdefinite integralsintegration
Best Answer
It depends how rigorous you want to be. For $n=1$ this is a classic integral, that I'll assume you have seen before/can easily find. For $n>1$ we have the following generalization if we let $a>b\geqslant 0$
$$\begin{aligned}2\int_0^{\infty}\frac{\sin(ax)\sin(bx)}{x^2}\;dx &=\int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\;dx \\ &= \int_0^{\infty}\int_{a-b}^{a+b}\frac{\sin(xy)}{x}\;dy \;dx\\ &=\int_{a-b}^{a+b}\int_0^{\infty}\frac{\sin(xy)}{x}\;dx \;dy \\ &=\int_{a-b}^{a+b}\frac{\pi}{2}\;dy \\ &=\pi b\end{aligned}$$