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?

  • $\begingroup$ 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? $\endgroup$ – Botond Aug 20 at 14:05
  • 1
    $\begingroup$ 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 ? $\endgroup$ – reuns Aug 20 at 19:13

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 $\omega$.


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