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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?

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  • $\begingroup$ 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. $\endgroup$ – bames Feb 18 '18 at 14:54
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    $\begingroup$ This integral diverges for all $a$. $\endgroup$ – Fabian Feb 18 '18 at 14:55
  • $\begingroup$ 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/… $\endgroup$ – Mehrdad Zandigohar Feb 18 '18 at 15:07
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    $\begingroup$ The sine part is odd and cancels out. The cosine part is equivalent to $\frac{1}{x^2}$ around zero, hence diverges. $\endgroup$ – Arnaud Mortier Feb 18 '18 at 15:10
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The integral doesn't exist.

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.

One of the strongest methods is claiming that the integral exists in the sense of distributions. That doesn't work here either. To explain why not, first we should give an example where that does work, and explain why it works and what it means.

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.

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