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## Homework Statement

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

## Homework Equations

## 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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