On Thu, 02 May 2013 09:41:43 +0100, Robin Chapman <R.J.Chapman@ex.ac.uk> wrote:
>> 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?
Finally dug out the books and saw an example of how to do these integrals -- suddenly everything came back. The book had a example of how to do the integral of
cos(3*x)/(a - cos(x))
They did have z^3 + z^(-3) in the denominator and messed with the 3rd order pole. But, in the margin, in my own hand was a note saying that z^3 is enough and that the residues add up to the same either way!!