I want to solve $$\int_{-\infty}^{\infty} \frac{e^{x/7}}{1+e^{x}}dx$$ using Cauchy's Residue Theorem, but according to Wolfram Alpha there's a total of 7 simple poles for this equation. I rewrite this as $$\int_{-\infty}^{\infty} \frac{e^{z/7}}{1+e^{z}}dz$$ and want to find residues using $$Res(f, c) = \lim_{z\to c}(z - c)f(z)$$ along with L'Hopital's Rule but I'm not getting the correct numbers. Also, is it really necessary to use all the residues from all the poles in the end for Cauchy's Residue Theorem? I apologize if this is trivial but I'm unable to find the correct answer. I'd really appreciate a simple walk through of the process. Thank you!
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You have a pole whenever $e^z = -1$. There are infinitely many, not just $7$. – Robert Israel Apr 13 '20 at 23:27
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More specifically, the poles have the form $z = i\pi (2n+1)$ where $n \in \Bbb Z$. – PrincessEev Apr 13 '20 at 23:28
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But I think you want to use a rectangular contour with horizontal sides on $\text{Im}(z) = 0$ and $\text{Im}(z) = 2 \pi$. Then there's only one pole to worry about. – Robert Israel Apr 14 '20 at 00:54
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4Does this answer your question [put $z=e^{x/7}$, then it's the case $a=7$]? Improper integration involving complex analytic arguments – Integrand Oct 14 '20 at 19:30