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As a follow up to this integral, I must ask as to how I would evaluate it:

EDIT: Okay, let me simplify some things:

I started with: $\frac{e^{-sa}}{2\pi i}\int_{|z|=1} \frac{z^ne^{\frac{-sb}{2}(z-\frac{1}{z})}}{z(a+\frac{b}{2}(z-\frac{1}{z}))}dz$ (after some fixes to make it a bit better to understand).

Because of Cauchy's Integral Formula( $\frac{1}{2\pi i}\int_Cf(z)dz=Res(f(z))$), I need to find the residue of that whole mess of a function.

Problem is, how do I do that?

Do I multiply the bottom part out, move the exponential into the denominator, and then calculate residue that way, or is it something different?

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AstroFox
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  • Have you seen Cauchy's Integral formula? https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula – David P Aug 28 '16 at 07:26
  • You attempted what, exactly? In order to apply the residue theorem, you have to locate the zeroes of $a+(z-iz)iz/2$ lying in $|z|<1$ and compute the residues of $f(z)$ at these points. Then you have to deal with the singularity at $z=0$. – Jack D'Aurizio Aug 28 '16 at 09:06
  • Since my original equation set $a=1$, I had wolfram factor out the bottom portion. That's where I became confused - I did not know as to what I was supposed to do from there: do I separate the entire residue into two residues, compute them, and add them up? – AstroFox Aug 28 '16 at 16:48
  • Huge Changes have been made. I remodified the initial question so that it made more sense, and could possibly be solved in a better manner. – AstroFox Aug 28 '16 at 19:48
  • I think $z - \frac1{z}$ should be inside the exponent in the numerator. This significantly complicates the computation of the residue at $z=0$. What are you looking for? It is unlikely there is a closed form for this integral. – sometempname Aug 28 '16 at 20:34

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