The poles are at $x=(2n+1)\pi\mathrm i$, and the corresponding residues are $-2\mathrm e^{-(2n+1)\pi k}$. If we ignore issues of convergence, this suggests that the integral is
$$ -4\pi\mathrm i\sum_{n=0}^\infty\mathrm e^{-(2n+1)\pi k}=\frac{-4\pi\mathrm i\mathrm e^{-\pi k}}{1-\mathrm e^{-2\pi k}}=\frac{-2\pi\mathrm i}{\sinh\pi k}\;. $$
To make this rigorous in the context of tempered distributions, you can take the derivative of the function being Fourier-transformed, calculate the resulting convergent integral, and divide by $-\mathrm ik$ (see Tempered distributions and Fourier transform at Wikipedia). The residues are slightly more complicated to evaluate, but as one might expect the result is the same.