Edited by Valentin Fadeev, Thursday, 27 Mar 2014, 10:11
As a follow up thought I realized just how much easier it would have been to calculate the residue by definition, i.e. expanding the integrand in the Laurent series to get the coefficient . Let where is small:
Of course, care is needed when choosing the value of the root. It depends on the value of the argument set on the upper bound of the cut. Since I chose it to be 0, the correct value of the root is .
Now
Therefore, near the integrand has the following expansion:
where is the regular part of the expansion which is of no interest in this problem.
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