Using the Cauchy Principal Value, I need to compute the following integral
$$\int_{-\infty}^\infty\frac{\cos(ax) – \cos(bx)}{x^2}dx$$
I have used the standard semi-circle contour with an indentation around the singularity at $x=0$. However integrating around the outer semicircle and smaller one around $0$, I find they have no contribution to the integral and the residue is also $0$. However I know the integral is not equal to $0$.
Where have I gone wrong?
Best Answer
The integral can be rewritten as a double integral
$$\int_{-\infty}^\infty \frac{\cos(ax)-\cos(bx)}{x^2}\:dx = \int_{-\infty}^\infty \int_a^b \frac{\sin(yx)}{x}\:dy\:dx$$
Swapping the order of integration gives us
$$\int_a^b \int_{-\infty}^\infty \frac{\sin z}{z} \:dz \:dy = \int_a^b \pi \:dy = \pi(b-a)$$
by using the substitution $z = yx$ on the inside. The sinc integral can be done by considering a rectangular contour instead of a semicircle. This insight means you could use a rectangular contour for the original integral if you wish.