TITLE: Fourier transform of $ \frac{\sinh(kx)}{\sinh(x)}$
QUESTION [5 upvotes]: Possible Duplicate:
Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$ 

Im trying to compute the integral of

$$I = \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx$$

for $0 < k < 1$ over the contour of the half circle in the upper plane.
I know I have residues for $x= i\pi n, n\in \mathbb N $. The second part of the contour where $$ x= R~e^{it} , t\in[0,\pi]$$ goes to 0, so I would be left with
$$I = \lim_{R\rightarrow \infty} \int_{-R}^{R} \frac{\sinh(kt)}{\sinh(t)}e^{-i\omega t}dt = \lim_{R\rightarrow \infty} 2\pi i \sum_{\mid z \mid=R}Res_{z}\left( \frac{\sinh(kt)}{\sinh(t)}e^{-i\omega t}\right)$$
My problem now is though, that I thought I'm not allowed to use the method of residues because I have an infinite amount of them?

REPLY [3 votes]: $$I = \int_{-\infty}^{\infty} dx \:  \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} $$
By taking a semicircle in the lower half-plane for $\omega > 0$ or the upper half-plane for $\omega < 0$ and using the Residue theorem:
$$ = i 2 \pi (-i) \sum_{n=1}^{\infty} \sin{k \pi n} \exp{(-\omega \pi n)} $$
which is:
$$ = 2 \pi \Im{\frac{1}{\exp{[\pi (w - i k)]} - 1}}$$
or
$$ = \pi \frac{\sin{k \pi}}{\cos{k \pi} + \cosh{\omega \pi}} $$