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Let $f$ a function $f:\mathbb{C}\to\mathbb{C},$ $$f(z)=\frac{1}{(e^{\frac{1}{z}}-1)(e^z-1)}.$$ Trying to integrate this function in a closed contour around $0$ has been impossible to me. Let $\epsilon>0\in\mathbb{R}$. $$ \int_{|z|=\epsilon}\frac{dz}{(e^{\frac{1}{z}}-1)(e^z-1)}=? $$ The function $f$ turns out to be the same after applying a otherwise useful contour integration around an essential singularity technique.

Is this "integrand" $f(z)=\frac{1}{(e^{\frac{1}{z}}-1)(e^z-1)}$, evaluated at some loop around its essential singularity at $z=0$, an integrable function? In case yes, How to find it?

Dr Potato
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1 Answers1

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1. When $\frac{1}{2\pi} < \epsilon < 2\pi$, we may use the Taylor series

$$ \frac{z}{e^z - 1} = -\frac{z}{2} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n}, \qquad |z| < 2\pi, $$

where $B_k$'s are the Bernoulli numbers, to give the following Laurent expansion:

$$ \frac{1}{(e^{1/z}-1)(e^z - 1)} = \biggl( -\frac{z}{2} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n} \biggr)\biggl( -\frac{1}{2z} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!z^{2n}} \biggr) $$

for $\frac{1}{2\pi} < |z| < 2\pi$. So by the Residue Theorem and using $B_0 = 1$ and $B_2=\frac{1}{6}$, we get

$$ \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{\mathrm{d}z}{(e^{1/z}-1)(e^z - 1)} = -\frac{1}{2} \biggl( B_0 + \frac{B_2}{2!} \biggr) = -\frac{13}{24}. $$

This indeed confirms the conjecture in the comment.

2. As $\epsilon > 1$ increases, it will cross the poles at $\pm 2\pi i k$. So, if $2n\pi < \epsilon < 2(n+1)\pi$, then

\begin{align*} &\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{\mathrm{d}z}{(e^{1/z}-1)(e^z - 1)} \\ &=-\frac{13}{24}+\sum_{k=1}^{n} \biggl( \underset{z=2\pi ki}{\mathrm{Res}}\,\frac{1}{(e^{1/z}-1)(e^z - 1)} + \underset{z=-2\pi ki}{\mathrm{Res}}\,\frac{1}{(e^{1/z}-1)(e^z - 1)} \biggr) \\ &=-\frac{13}{24}+\sum_{k=1}^{n} \underbrace{\biggl( \frac{1}{e^{-i/(2\pi k)} - 1} + \frac{1}{e^{i/(2\pi k)} - 1} \biggr)}_{=-1} = \boxed{-\frac{13}{24} - n}. \end{align*}

Sangchul Lee
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  • Nice!... and for $2\pi<\epsilon$? – Dr Potato Jun 11 '20 at 20:09
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    @DrPotato, I added a computation for that regime. – Sangchul Lee Jun 11 '20 at 20:24
  • Why is $\operatorname{Res}_{z=2\pi ki}\frac{1}{(e^{1/z}-1)(e^z-1)} = \frac{1}{e^{-i/(2\pi k)}-1 }$? – Dr Potato Jun 11 '20 at 23:10
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    @DrPotato, It relies on the following simple observation: If $g(a)\neq 0$ and $g(z)/h(z)$ has a simple pole at $a$, then $$\underset{z=a}{\mathrm{Res}},\frac{g(z)}{h(z)}=\frac{g(a)}{h'(a)}.$$ This is easily proved by applying the Cauchy Integration Formula to $$\int_{|z-a|=\epsilon}\frac{g(z)}{h(z)},\mathrm{d}z=\int_{|z-a|=\epsilon}\frac{g(z)}{h(z)/(z-a)}\cdot\frac{1}{z-a},\mathrm{d}z,$$ where now $\frac{g(z)}{h(z)/(z-a)}$ extends to a holomorphic function near $z=a$ with the value $\frac{g(a)}{h'(a)}$ at $z=a$. – Sangchul Lee Jun 12 '20 at 00:18
  • Awesome! Where did you get this? – Dr Potato Jun 12 '20 at 01:29
  • I thought it was the answer, but just noticed: @SangchulLee You start showing the power series expression in terms of Bernoulli numbers. One of them is evaluated at $z$ and the other one at $w=1/z$. Both indeed, as you affirm, converge only when its argument, $z$ and $w$ respectively, is inside the open disk. However, the multiplication of both series, evaluated one in $z$ and the other in $1/z$, have radius of convergence equal to $0$ in the $z$-complex plane, thus the remaining part of the argument is not valid. – Dr Potato Jun 15 '20 at 18:59
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    @DrPotato, Both of the Laurent series $$ \frac{z}{e^z-1} = -\frac{z}{2} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n} \qquad\text{and}\qquad \frac{1/z}{e^{1/z}-1} = -\frac{1}{2z} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!z^{2n}} $$ converge absolutely over the annulus $\frac{1}{2\pi}<|z|<2\pi$. So, not only one may freely rearrange the order of the terms in the double sum arising from the product, but also can distribute the integral symbol term-wise (thanks to the locally uniform convergence of the resulting double sum). – Sangchul Lee Jun 15 '20 at 19:05
  • @DrPotato, In case you have not been exposed to the topic of Laurent series, here is a very brief idea about what it is: A holomorphic function $f$ over an annulus $r<|z-z_0|<R$ may be expanded as a Laurent series: $$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n.$$ The theory of Laurent series has many parallels to that of power series, which makes it extremely useful. (You may also check the Wikipedia article about this.) – Sangchul Lee Jun 15 '20 at 19:19