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The complex function $$f(z)=\frac{1}{e^{1/z}-1}$$ has an essential singularity at $z=0$, and an infinite quantity of poles inside every open neighborhood containing it.

Let $\mathbb{R}\ni\epsilon>0$. What is the value of $$\int_{|z|=\epsilon} \frac{dz}{e^{1/z}-1}=?$$

Dr Potato
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    What about $z = \frac{1}{2\pi i}$ ? There are an infinite number of these poles inside the unit disk. – Ninad Munshi Jun 08 '20 at 05:44
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    $f$ does not have an essential singularity at $z=0,$ Let $z_k:=\frac{1}{2k \pi i}$ for $k \in \mathbb Z \setminus {0}.$ Then each $z_k$ is a pole of $f$. Thus, $0$ is a limit point of poles of $f$. Therefore, $f$ does not have an isolated singularity at $0$. – Fred Jun 08 '20 at 05:54
  • @NinadMunshi Already corrected. Thanks! – Dr Potato Jun 13 '20 at 17:50
  • @Fred Why not being an isolated singularity makes it not an essential one? – Dr Potato Jun 13 '20 at 17:51

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Instead of giving ourselves a headache with residues, we can convert this into different line integral

$$\int_{|z|=1} \frac{dz}{e^{\frac{1}{z}}-1} = \int_0^{2\pi} \frac{ie^{it}\:dt}{e^{e^{-it}}-1} = \int_0^{2\pi} \frac{ie^{2it}e^{-it}\:dt}{e^{e^{-it}}-1} = \int_{|z|=1} \frac{dz}{z^2(e^z-1)}$$

The second integral only has a pole of order $3$ at $z=0$ inside the unit disk. Calculating the Laurent series gives us

$$z^{-3}\left(1+\frac{1}{2}z+\frac{1}{6}z^2+\cdots\right)^{-1} = z^{-3}\left(1-\frac{z}{2}-\frac{1}{6}z^2+\frac{1}{4}z^2+\cdots\right) = \frac{1}{z^3} - \frac{1}{2z^2} + \frac{1}{12z}+\cdots$$

by geometric series. The residue is $\frac{1}{12}$ thus the integral evaluates to

$$\int_{|z|=1} \frac{dz}{e^{\frac{1}{z}}-1} = \int_{|z|=1} \frac{dz}{z^2(e^z-1)} = \frac{\pi i}{6}$$

and since the original function did not have any poles outside the unit disk, the integral on $|z|=2$ will be exactly the same.

Ninad Munshi
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  • In the second line, the first expression of the integrand is not hard to obtain, but following the first equality I didn't get how did you ordered the geometric series. In order to be sure if this is the only term --in the series of series-- contributing to the $(-1)$-th coefficient of the Laurent series expression of the integrand, the generic term of the infinite sum must be explicit. please help us to write it down. – Dr Potato Jun 08 '20 at 18:26
  • @DrPotato the generic term is not necessary all we need are the lowest terms up to the second power and some logic to realize the product of higher order terms will never give a second power. – Ninad Munshi Jun 08 '20 at 21:13
  • How could the integral possible exists if the loop encloses an infinite quantity of poles? – Dr Potato Jun 10 '20 at 20:43
  • @DrPotato are you saying infinite sums cannot exist? That is absurd – Ninad Munshi Jun 10 '20 at 20:45
  • Yes. They exist. They always exists, but without known convergence for such sort of sums: seems to me that all its nearby poles have the same residue. – Dr Potato Jun 10 '20 at 20:51
  • @DrPotato they do not, the residues are a scalar multiple of the sum of $\frac{1}{k^2}$ – Ninad Munshi Jun 10 '20 at 21:07
  • I just don't get it. This Laurent series is centered at the essential singularity of the function, which has a pole in any open neighborhood of this center, thereby the radius of convergence for such power (Laurent) series would be zero! Please help me understand WTH. – Dr Potato Jun 10 '20 at 23:51
  • @DrPotato as many comments have explained to you already in this post and the other one, there is no essential singularity at $0$ because it is not isolated. – Ninad Munshi Jun 11 '20 at 00:01