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Sep 21, 2023 at 1:20 comment added Rafael You are right. I didn't need to look at the Poisson formula. I guess I was fixated on that because of the Hilbert transform I was also looking at. In the end I started going back at some definitions after the comments and realized, my confusion came from the fact that the papers I was reading were not careful defining the Hilbert transform and omitted the fact, that the integrals were principal valued.
Sep 20, 2023 at 22:57 comment added Anurag Sahay There's a few ways to prove the cotangent identity you are looking for that avoid Poisson summation, I think. Have a look at this link: math.stackexchange.com/questions/1849878/…
Sep 20, 2023 at 19:48 comment added Rafael I know this is a standard fact. My point is that I have found countless times (such as sciencedirect.com/topics/mathematics/hilbert-transform or sciencedirect.com/topics/psychology/hilbert-transform) that the Fourier transform of $1/x$ in the usual sense is being used as the sign function. My question is exactly why that is and I would have settled for a Cauchy Principal Value kind of integral, but many authors seem to imply otherwise.
Sep 20, 2023 at 19:38 comment added Christian Remling $1/x$ is not locally integrable and thus doesn't have a FT. What is true is that the FT of (the tempered distribution) $\textrm{PV}(1/x)$ (principal value) is $c\textrm{ sign}(x)$, but this is a standard fact and not a suitable question for this site. Please use math.stackexchange.com instead.
Sep 20, 2023 at 17:40 history asked Rafael CC BY-SA 4.0