
Re: Help with identity
Posted:
May 1, 2013 7:34 PM


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

