
Re: Help with identity
Posted:
May 1, 2013 8:50 PM


On Wed, 01 May 2013 19:34:16 0400, Mike Trainor <mtrainor@hotmail.com> wrote:
>On Fri, 26 Apr 2013 14:31:57 +0100, Robin Chapman ><R.J.Chapman@ex.ac.uk> wrote: > >>On 26/04/2013 13:10, Mike Trainor wrote: >> > >>> It comes down to doing the integral of >>> >>> cos(ny)/(cosh(x)  cos(y)) >>> >>> from 0 to 2 pi, for integer n, where x => 0. >> >>How about integrating z^{n1}/(cosh(x)  (z+1/z)/2) >>over the unit circle in C? > >Thanks, Robin, from bringing back 25+ year old >memories ... have not done this kind of work in >a while. Funny about the cosh(x) terms as it >simplifies the terms. > >I do have a question as my memory is shot and >I cannot figure it out. I see why you would have >z^(n1) and not z^n as the 'dy' becomes >dz/(i*z). Now, there are simple poles at >z = exp(+/ x), and only the z = exp(x) lies >within the contour as x > 0 in my case, at least. >That gives the cosech(x) term I need in the >answer. > >But, I have a question. What above the >z^n term in the numberator that comes due >to the numerator that should be there from >the cos(ny) part? That messes up things as >the residues now have exp(n*x) and, >unforturnately, exp(n*x). Other than that, >it all works out. > >I have yet to get to the basement, locate and >pull out my grad school textbooks .... I guess >the answer is there. > >Thanks for the pointer and would appreciate it >if you can give me a hint. > >tia >mt
And, I might as well add that I have not gotten near the nth order pole at z = 0 :)
mt

