
Re: Help with identity
Posted:
May 2, 2013 9:39 AM


On Thu, 02 May 2013 09:41:43 +0100, Robin Chapman <R.J.Chapman@ex.ac.uk> wrote:
>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?
Yes, I see that  and we can take the real part. This also avoids a pole at z = 0.
I was writing cos(n x) as (z^n + z^(n))/2, just as in the denominator. I see we do not *have* to.
Thanks a million .... I think I have got it, but I will pull out the books and refresh my knowledge as I am sure there is something like Jordan's lemma that says why the z^(n) term is bad news. The nth order pole is definitely bad news :)
thanks again mt

