# Contour integral- Complex variables

## Homework Statement

evaluate $\int \frac{sinh(ax)}{sinh(\pi x)}$ where the integral runs from 0 to infinity

## 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

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Ray Vickson
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## Homework Statement

evaluate $\frac{sinh(ax)}{sinh(\pi x)}$

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

What is it about $\sinh(ax)/\sinh(\pi x)$ that you want to evaluate? You did not say!
Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks

Ray Vickson
Your function does not have poles at $z = i n \pi$ because $\sinh(i n \pi^2 ) \neq 0$ for integer $n$.
Your function does not have poles at $z = i n \pi$ because $\sinh(i n \pi^2 ) \neq 0$ for integer $n$.
Again, my mistake. I meant at $z=ni$ Im not great at copying the problem from my work to the computer :P