How would I show the result below using contour integration?

Question

How would I show the result below using contour integration? $$\int_{-\infty}^{\infty} \frac{\cos bx - \cos ax}{x^2} dx = \pi (a-b)$$ where a>b>0 using contour integration. Any help would be greatly appreciated, thanks!

Answer 1

See Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? on performing contour integration to get $\int_0^\infty \frac{\sin x}x=\frac\pi2$. Then

\begin{align} & \int_{-\infty}^{\infty} \frac{\cos bx - \cos ax}{x^2} dx \overset{IBP} =2\int_0^\infty \frac{a\sin ax - b\sin bx}x=\pi (a-b) \end{align}