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Contour integral- Complex variables

  1. Jun 11, 2016 #1
    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. jcsd
  3. Jun 11, 2016 #2

    Ray Vickson

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    What is it about ##\sinh(ax)/\sinh(\pi x)## that you want to evaluate? You did not say!
     
  4. Jun 11, 2016 #3
    Oops- My mistake I forgot the integral and to state that it runs from 0 to infinity- I will fix that. Thanks
     
  5. Jun 11, 2016 #4

    Ray Vickson

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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##.
     
  6. Jun 11, 2016 #5
    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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