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$$f\left ( z \right )=\frac{e^{az}}{1+e^{z}} ,\left ( a\in\left ( 0,1 \right ) \right )$$ The point $z=i\pi$ is one of the nonremovable singularities of this function. In order to expand it about that point I introduced the change of variables $z=\omega +i\pi$ and expanded the function $f\left(\omega \right)=e^{ai\pi}\frac{e^{a\omega}}{1-e^{\omega}}$ about $\omega=0$. Here is what I've come up with. $$e^{a\omega}=1+a\omega+\frac{a^{2}\omega^{2}}{2}+\frac{a^{3}\omega^{3}}{6}+...$$ Then I tried to use this well-known expansion $$\frac{1}{1-x}= \sum_{k= 0}^{\infty}x^{n}$$ and just use $e^{\omega}$ instead of $x$, however $\left | e^{\omega} \right |$ need not be $<1$ here, so the series might not converge.

Emir Šemšić
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  • Have you heard of Bernoulli numbers (or even Bernoulli polynomials, they appear here)? – Daniel Fischer Dec 02 '15 at 14:21
  • @DanielFischer Unfortunately that is way out of my scope right now – Emir Šemšić Dec 02 '15 at 14:25
  • Well. Point is, the Bernoulli numbers are defined via $$\frac{z}{e^z - 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n.$$ If you had already met them, that would probably make you a bit more comfortable, that's all. So the Bernoulli numbers naturally occur here. If you multiply that expansion with the series of $e^{az}$, you get $$\frac{z e^{az}}{e^z-1} = \sum_{n = 0}^{\infty} \frac{B_n(a)}{n!}z^n,$$ with polynomials $B_n(,\cdot,)$ [which, unsurprisingly, are closely related to the Bernoulli numbers $B_n$]. There's no (known) closed form for the Bernoulli numbers, so that's about as explicit … – Daniel Fischer Dec 02 '15 at 14:32
  • … as it gets. The Bernoulli numbers satisfy a relatively simple recurrence, however, so it's not too difficult to compute the first few $B_n$ explicitly. – Daniel Fischer Dec 02 '15 at 14:34
  • @DanielFischer Can we somehow obtain the Laurent expansion without ever having to know about Bernoulli numbers? – Emir Šemšić Dec 02 '15 at 15:29
  • Yes and no. Of course one doesn't need to know about Bernoulli numbers per se. But without a closed form, I don't see how you can get anywhere without an ansatz with unknown coefficients and then getting the recurrence. – Daniel Fischer Dec 02 '15 at 15:40

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