What is the Nontrivial Residue Integral for the Function sin(bx)/(e^pi x-1)?

[rpf_rr]

I found this integral in a book ("A course of modern analysis", Whittaker):
\int_{0}^{\infty} \frac{sin(bx)}{e^{\pi x}-1} dx
I tried to use residue theorem in the rectangular domain [0,R]x[0,i], with R-> \infty , but i couldn't do the integral in [0,i]

 

[zetafunction]

expand the sine function into a Taylor series, and use the representation

\int_{0}^{\infty}dx (exp(x)-1)^{-1}x^{s-1} = \Gamma (s) \zeta (s)

this should work

 

[Gib Z]

Instead of the rectangle with base [0,R], height i, use height 2i (looking at the term in the denominator, you want to move up by a full period, you'll see why). Also, don't bother trying to explicitly find the integral along the upper side.

The integral you are interested in, call it I, is the lower side of the rectangle. Express the integral along the upper side in terms of I. That way, the integral along the two horizontal sides will add to (something)I.

Alternatively, you could just expand it as a geometric series:
\frac{ \sin bx}{e^{\pi x} - 1} = \frac{ e^{-\pi x} \sin bx}{1- e^{-\pi x}} = e^{-\pi x} \sin bx + e^{-2\pi x} \sin bx + e^{-3\pi x} \sin bx + \cdots

Those integrals are simple to do with integration by parts.

 

[rpf_rr]

yes, you are right, 2i, I wrote wrong, but the problem for me is the integral "from 0 to 2i", i can't find a simple expression, while Mathematica do (only a cotangent and something like 1/p, or similar). The other methods, for me (maybe I'm not too experienced) gives too "complex" results. thanks for the answers however.