Definite Integral using Residue Thm

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Homework Statement



Calculate the integral [ z^4/(1 + z^8) ] over negative infinity to positive infinity.

Homework Equations



Residue Theorem. Specifically for real-valued rational functions (on the real axis) where the denominator exceeds the degree of the numerator by at least two or more. The denominator has only complex poles, and the integral is given by 2πi(sum of all residues in the upper half plane U).

The Attempt at a Solution


I found the eight poles, only four of which are in the upper plane. So the residues in U we are looking for are for poles exp(πi/8), exp(3πi/8), exp(5πi/8), exp(7πi/8).

I posted this question before but unfortunately the help I was given didn't clear my confusion. The answer in the back of the book is π/4 * arcsin(3π/8) and I can't trace the steps to get this. I would greatly appreciate if somebody could walk me through finding the residue of one of these poles so that I can complete the rest of the question myself. I need to be able to get the answer in terms of arcsin which is part of my problem. When I multiplied the denominator all out for a pole then I get a huge expression of various exponents of e.

Thank you!
 
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Huge protip: If you have f(x)/g(x), and you have a simple pole, the residue at the pole is just f(x)/g'(x) evaluated at the point. This will save you hours of anguish trying to sum up different combinations of roots of unity
 
I think the answer shouldn't contain arcsin. It must be 1/sin.
 
Ok, yeah I foolishly wrote the answer is arcsine but it is actually π/4 * 1/sin(3π/8).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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