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Topic: Help with identity
Replies: 15   Last Post: May 9, 2013 7:13 AM

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Mike Trainor

Posts: 28
Registered: 4/21/13
Re: Help with identity
Posted: May 1, 2013 7:34 PM
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On Fri, 26 Apr 2013 14:31:57 +0100, Robin Chapman
<> 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^{n-1}/(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^(n-1) 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

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.


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