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This is what I saw in a paper. It never occurred to me that a diverging function like $|x|$ can have a meaningful Fourier transform.

It should have something to do with the so-called distribution theory, right?

ps. Here $k$ is real.

poisson
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    Without further context, it's hard to tell what sort of convergence or equality is implied here. At the very least, the formula in the title holds as it stands for $k$ in the upper half-plane, and that function extends via analytic continuation to the whole plane (even if the integral no longer converges there). Various regularizations (this sort of integral is common in physics, for example) give a similar result. But the integral itself certainly doesn't converge for $\text{Im}, k \leq 0$. – anomaly Sep 16 '22 at 12:34
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    This is indeed a Fourier transform in the sense of distributions that is written formally as an integral. The $1/k^2$ on the right-hand-side can be interpreted as the https://en.wikipedia.org/wiki/Hadamard_regularization of $1/k^2$. This can also be viewed as the second derivative in the sense of distributions of $\ln(k)$. See for example https://math.stackexchange.com/questions/3723136/the-fourier-transform-of-1-p3/3724502#3724502 for another similar case where (with $d=1$) the finite part of $1/|k|$ is defined as the derivative of $\ln(k)$. – LL 3.14 Sep 16 '22 at 13:00
  • See THIS ANSWER. – Mark Viola Sep 16 '22 at 22:10

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