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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 2, 2013 9:39 AM
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On Thu, 02 May 2013 09:41:43 +0100, Robin Chapman
<> wrote:

>On 02/05/2013 00:34, Mike Trainor wrote:
>> 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
>> 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*

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 n-th order pole is
definitely bad news :-)

thanks again

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