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Contour integral- Complex variables

  • Thread starter Physgeek64
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  • #1
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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
 
Last edited:

Answers and Replies

  • #2
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!
 
  • #3
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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
 
  • #4
Ray Vickson
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Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks
Your function does not have poles at ##z = i n \pi ## because ##\sinh(i n \pi^2 ) \neq 0## for integer ##n##.
 
  • #5
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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
 

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