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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?

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Longeyes
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  • Where is it stated that one may not consider a limit of a countably infinite number of residues? In fact, this is common when using the Residue theorem to evaluate certain infinite sums. – Ron Gordon Jan 03 '13 at 11:33
  • @DavideGiraudo SORRY! really bad typo :S it is still supposed to be sinh – Longeyes Jan 03 '13 at 11:44
  • @rlgordonma That's the way we learned it (as far as I know) and also on wikipedia they state: "Suppose U is a simply connected open subset of the complex plane, and a1,...,an are finitely many points of U and f is a function which is defined and holomorphic on U \ {a1,...,an}." link I took that to mean, that the theorem only applied to a finite amount of residues, have I misunderstood this? – Longeyes Jan 03 '13 at 11:46
  • I think you are being too restrictive. Think about the limit of such a number of points as that number gets very large. – Ron Gordon Jan 03 '13 at 12:05
  • Also, the contour you should use should be in the lower half-plane when $\omega > 0$ and vice-versa. – Ron Gordon Jan 03 '13 at 12:06
  • @rlgordonma Oh, right forgot that the contour should be in the lower half-plane. Is the argument that because the zeros of sinh(x) don't have a limit they're countable and that's why I can use the theorem? – Longeyes Jan 03 '13 at 12:10

1 Answers1

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

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

$$ = \pi \frac{\sin{k \pi}}{\cos{k \pi} + \cosh{\omega \pi}} $$

Ron Gordon
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