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I am trying to understand how to generally compute the Fourier transform of the function $\frac{1}{\sinh(x+a)}$, where $a$ is a general complex number. Plugging the equation into Wolfram Alpha gives a definite answer for real a. However, if I make the coefficient imaginary, Wolfram Alpha is unable to give an answer. Naively, I would expect the original answer to hold for general $a$, for two reasons

  • In doing the Fourier transform, one can just perform a change of variables to $u = x + a$, and compute the integral in a way that is seemingly insensitive to the details of $a$.
  • The Fourier transform can be done analytically using the method of residues (example here). In this case, the coefficient $a$ shifts the location of the poles, but not in a way that alters whether or not they are picked up by the contour integration.

Is this reasoning sound? Or is Wolfram Alpha picking up on some subtlety that I'm missing?

Henry Shackleton
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  • How do you define the fourier transform of $\operatorname{csch}(x+a)$, since it's not an $L^1$ function? Did you try to compute it with that definition? – Botond Aug 20 '19 at 14:05
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    Concretely you don't know how to prove the residue theorem for that particular function ? It is not hard, let $C_R$ be the rectangle enclosing $[-R,R] + i[0,R]$ then your FT is $\lim_{R \to \infty} \int_{C_ R}\frac{e^{-i\omega z}}{\sinh(z+a)}dz$, can you show the inner integral is the sum of residues of the simple poles ? – reuns Aug 20 '19 at 19:13

1 Answers1

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The transform that you have is for the distribution defined as $(\operatorname{csch}, \phi) = \operatorname{v.\!p.} \int_{\mathbb R} \operatorname{csch}(x) \phi(x) dx$, the Fourier integral doesn't exist when $\operatorname{Im} a/\pi \in \mathbb Z$.

Assume now $\operatorname{Im} a/\pi \not \in \mathbb Z$. If $\omega > 0$, then $$\operatorname*{Res}_{x = -a + i \pi k} \operatorname{csch}(x + a) = (-1)^k, \\ \int_{\mathbb R} \operatorname{csch}(x + a) e^{i \omega x} dx = 2 \pi i \sum_{\pi k > \operatorname{Im} a} (-1)^k e^{i \omega (-a + i \pi k)} = \frac {2 \pi i e^{\pi \lceil \operatorname{Im} a/\pi \rceil (i - \omega) - i a \omega}} {1 + e^{-\pi \omega}}.$$ The integral and the expression on the rhs are analytic in the strip $-1 < \operatorname{Im} \omega < 1$, therefore they coincide for all real $\omega$.

Maxim
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