# Fourier transform of $\frac{1}{\sinh(x+a)}$ for complex $a$

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?

• 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 at 14:05
• 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 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 $$\omega$$.