Homework Help: Contour integral- Complex variables

1. Jun 11, 2016

Physgeek64

1. The problem statement, all variables and given/known data
evaluate $\int \frac{sinh(ax)}{sinh(\pi x)}$ where the integral runs from 0 to infinity

2. Relevant equations

3. The attempt at a solution
consider $\frac{sinh(az)}{sinh(\pi z)}$
Poles are at $z= n \pi i$
So I'm considering the contour integral around the closed contour from $-R$ to $- \epsilon$ then around a small semi-circle in the upper half plane of radius $\epsilon$, then along $\epsilon$ to $R$ then from $R$ to $R+i$, back along $R$ to $\epsilon$ along x+i, then along another semicircle, enclosing the pole at $z=n \pi i$ and then similarly on the other side of the integral

I have shown that the two end contributions are zero, and that over the semicircle at $z=0$ the contribution is zero to end up with

$2(1+cosa) \int \frac{sinh(ax)}{sinh( \pi x)}$ where this runs from 0 to infinity

and using Cauchy's reside theorem the only enclosed pole is the one at $z=i$ so this will be equal to $2 \pi i (sinh(ai))$

hence I get

that $\int \frac{sinh(ax)}{sinh( \pi x)}= \frac{- \pi sin(an)}{1+cos(a)}$

but the quoted answer is $\frac{1}{2}tan(\frac{a}{2})$

I feel like I may have over complicated my contour

Many thanks

Last edited: Jun 11, 2016
2. Jun 11, 2016

Ray Vickson

What is it about $\sinh(ax)/\sinh(\pi x)$ that you want to evaluate? You did not say!

3. Jun 11, 2016

Physgeek64

Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks

4. Jun 11, 2016

Ray Vickson

Your function does not have poles at $z = i n \pi$ because $\sinh(i n \pi^2 ) \neq 0$ for integer $n$.

5. Jun 11, 2016

Physgeek64

Again, my mistake. I meant at $z=ni$ Im not great at copying the problem from my work to the computer :P