# Compute the given integral using the residue theorem

Consider the integral $$\int_{-\infty}^{\infty} \frac{1}{x^2}e^{iax} \, dx.$$

I'd like to compute this using Cauchy's residue theorem. On examples like this that I've done in the past, the procedure has been to use the contour given by a closed semi-circle in the upper (or lower) half plane and then apply the residue theorem. However, here the singularity is at the origin so I can't do the same as before.

What else can be done here instead?

• Use a similar semi-circular contour, but with a semi-circular dent around the origin. Take the limit as the radius of the dent goes to zero. – bames Feb 18 '18 at 14:54
• This integral diverges for all $a$. – Fabian Feb 18 '18 at 14:55
• Hint: Try to use a contour which is a closed upper plane semi-circle for both internal and external bounds, so that $r_1=\epsilon$ and $r_2=R$ where $\epsilon\to0$ and $R\to\infty$. This is almost like the simple method but with solving the problem of the singularity of origin. Take a look at this: math.stackexchange.com/questions/980970/… – Mehrdad Zandigohar Feb 18 '18 at 15:07
• The sine part is odd and cancels out. The cosine part is equivalent to $\frac{1}{x^2}$ around zero, hence diverges. – Arnaud Mortier Feb 18 '18 at 15:10

There are various ways of making sense of some divergent integrals - principal value, summability methods, whatever. As far as I can see none of them work here. The problem is that there's no cancellation available in the singularity of $1/t^2$ at the origin.
People often say that $$\int_{-\infty}^\infty e^{iat}dt=\delta(a).$$ That's true in the sense of distributions. Saying it's true in the sense of distributions means that if $\phi$ is a test function then (in informal notation)$\newcommand{\ip}[2]{\langle #1,#2\rangle}$ $$\int_{-\infty}^\infty\ip{e^{iat}}{\phi(a)}=\ip{\delta}\phi,$$ and that last equation is actually true, since $$\int\hat\phi=\phi(0).$$
But if we try to "interpret" the integral in that sense here we run into $\int\hat\phi(\xi)/\xi^2$, and that integral doesn't exist either.