
Re: Help with identity
Posted:
May 2, 2013 4:41 AM


On 02/05/2013 00:34, Mike Trainor 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.
No we don't. Put z = cos(y) + i sin(y) = exp(iy). What is z^n in terms of y?

