# How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?

in a theoretical physics book, the author makes the following claim:

$$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$ and justifies this as

These series can be derived from a theorem which states that any meromorphic function may be expanded as a summation over its poles and residues at those poles

What's the name of that theorem? It's not really a Laurent series, since the Laurent series is for an expansion around one particular point only. I can see that the poles occur whenever $z = (2n+1)i\pi$ for $n \in \mathbb{N}$, but then where does that constant $1/2$ come from?

EDIT: Well, it appears that the general claim isn't valid, so now I'd be interested in a justification for the expansion in my particular example...

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Mittag-Leffler's theorem guarantees the existence of a meromorphic function $g(z)$ whose poles and principal parts are given by any values specified. Then, if $f(z)$ is a meromorphic function, then $f(z) - g(z)$ is holomorphic, and it remains to compute this difference. In practice this is probably nontrivial, because $g(z)$ is not uniquely determined, but for functions with nice poles and principal parts, this is possible.

Such a possibility applies in your case with $f(z) = 1/(e^z + 1)$. We can justify the formula you gave in your question by using an approach based on a discussion between me and one of my friends, so I do not claim the credit for these ideas.

In order to properly handle the convergence of the infinite sum, we should first symmetrize the infinite sum you gave, so instead let $$g(z) = \sum_{k > 0 \text{, odd}} \left( \frac{1}{k i \pi - z} - \frac{1}{k i \pi + z} \right) = -\sum_{k > 0 \text{, odd}} \frac{2z}{z^2 + k^2 \pi^2}$$

We can check that $g(z)$ is a meromorphic function whose poles and principal parts match that of $f(z) = 1/(e^z + 1)$, so it follows that $f(z)- g(z)$ is entire. It remains to compute this difference. First, notice that both $f(z)$ and $g(z)$ are $2 \pi i$ periodic. So to check the growth of $f(z), g(z)$, we need only check the behavior as $\mathrm{Re}(z) \rightarrow \pm \infty$. Notice that $f(z) \rightarrow 1,0$ as $\mathrm{Re}(z) \rightarrow -\infty, +\infty$, respectively. Thus it follows that $f(z)$ is in fact uniformly bounded away from its poles. To check $g(z)$, split the sum as $$g(z) = \sum_{0<k<2|z|/\pi, \text{ odd}} + \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} = S_1(z) + S_2(z)$$ Now, notice that for $\mathrm{Re}(z)$ sufficiently large, \begin{align} |S_1(z)| & = \left|\frac{-2}{z} \sum_{0<k<2|z|/\pi, \text{ odd}} \frac{1}{1 + k^2 \pi^2/z^2} \right| \le \frac{2}{|z|} \frac{2|z|}{\pi} = 4,\\ |S_2(z)| & = \left| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} \right| \\ & \le \frac{8}{3} |z| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{1}{ \pi^2 k^2} \\ & \le \frac{8}{3} |z| \int_{-1 + 2|z|/\pi}^{\infty} \frac{1}{\pi^2 s^2} \, ds \\ & \le \frac{8}{3} \frac{|z|}{-1 + 2|z|/\pi} \le C \end{align} for $C > 0$ a constant. Thus $g(z)$ is also uniformly bounded away from its poles. Then, it follows that the difference $f(z) - g(z)$ is uniformly bounded, and being entire, then by Louiville's theorem it must be constant. Now, we compute that $$f(0) - g(0) = \frac{1}{2} - 0 = \frac{1}{2}$$ and hence $f(z) - g(z) \equiv 1/2$.

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Awesome. Thank you very much. –  Lagerbaer May 19 '13 at 3:23

Taken literally, the claim is certainly not true. Meromorphic functions are not uniquely determined by their poles and corresponding residues. Take for example $$\frac{1}{z} \qquad\text{and}\qquad \frac{e^z}{z},$$ which have the same poles and same residues.

However, Mittag-Leffler's theorem as noted by Random Variable is related.

For your particular example, you can start from a well known series expansion of $\cot z$: $$\cot z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z+n\pi} = \frac1z + \sum_{n=0}^\infty \frac{2z}{n^2\pi^2-z^2},$$ see for example this for a derivation.

Let's interpret your series as a symmetric sum and rewrite it by summing the terms pairwise: $$(\operatorname{term} 0 + \operatorname{term} -1) + (\operatorname{term} 1 + \operatorname{term} -2) + \cdots$$

This will give us $$\sum_{n=-N-1}^N \frac{1}{(2n+1)i\pi - z} = -i\sum_{n=0}^N \frac{2z}{(2n+1)^2\pi^2 + z^2},$$ which is very similar to the series for $\cot z$, but where we have thrown away half the terms. Inspired by this observation, let's compute $$\cot z - \frac12\cot \frac z2 = \frac1z + \sum_{n=0}^\infty \frac{2z}{n^2\pi^2-z^2} - \frac1{z} - \sum_{n=0}^\infty \frac{2z}{4n^2\pi^2-z^2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2-z^2}.$$ Hence $$\cot iz - \frac12\cot \frac {iz}{2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2-(iz)^2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2+z^2}.$$ Consequently, your sum is \begin{align} \frac12 -i\sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2 + z^2} &= \frac12+\frac i2\cot \frac {iz}{2} - i\cot iz \\ &= \frac{1}{e^z+1}. \end{align}

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What part of the claim is not true? The series expansion I give above, or the general claim about meromorphic functions? –  Lagerbaer May 18 '13 at 22:04
The general claim. The expansion is valid, but the justification is incomplete at best. –  mrf May 18 '13 at 22:05
Actually, let me think about your particular expansion. Some care is needed, since the series doesn't converge absolutely. –  mrf May 18 '13 at 22:11

This seems to be some kind of application of the Residue theorem: https://en.wikipedia.org/wiki/Residue_theorem. Here, given $z_0$ at which $f$ is holomorphic, define $$g(z) = \frac{f(z)}{z - z_0}.$$ Then $g$ has poles where $f$ does, and of the same residue, but also has one extra pole, at $z_0$, whose residue we can compute via Cauchy's integral theorem to be exactly $f(z_0)$. The residue at the pole $p$ can be computed by taking $$\lim_{z\rightarrow p} zg(z) = \frac{\mbox{Res}(f,p)}{p - z_0}.$$

Now, choose a finite set of poles and your favorite point $z_0$ and a contour $\gamma$ around them so that the winding numbers are all 1 (a large box will do). Integrating around the contour will give, by the residue theorem and the above calculations:

$$f(z_0) = \frac{1}{2\pi i}\int_\gamma g(z) dz - \sum_{p \mbox{ inside } \gamma} \frac{Res(f,p)}{p - z_0}.$$

Now, the catch is to "let $\gamma$ go to infinity", to get an infinite sum. As already pointed out, there are some issues with that, so more careful estimates are needed, and these answer the question of "where did that constant come from"? I unfortunately don't have time to carry out those right now, but I will update this answer with those estimates once I have them.

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The formula you quote can be easily obtained from the more general formula for the pole (or Mittag-Leffler) expansion of a meromorphic function $f$: $$\tag 1 f(z) = f(0) + \sum_{n=1}^\infty \alpha_n \left( \frac{1}{z-z_n} + \frac{1}{z_n} \right),$$ where

1. $f$ is a meromorphic function with simple poles at $z_n, \, n \in \mathbb{N}$, holomorphic at $z=0$,
2. $\alpha_n$ is the residue of $f$ at the pole $z_n$,
3. $f$ is uniformly bounded on every circle $\mathcal C_n$ with radius $R_n$ containing all poles $z_k$ with $k \le n$, i.e. $$\tag 2\exists A > 0 | \,\, \forall (n \in \mathbb{N}, z \in \mathcal C_n), \,\, |f(z)| \le A$$

### Application of (1) to the problem at hand

Defining $$f(z) = \frac{1}{e^z+1},$$ we see that $f$ has simple poles at $z_n = (2n+1)i \pi$ for $n \in \mathbb{Z}$, with residue $\alpha_n = -1$ for each $n \in \mathbb{N}$. Direct application of (1), noting that $f(0)=1/2$, gives $$\frac{1}{e^z+1} = \frac{1}{2} - \sum_{n \in \mathbb{Z}} \left[ \frac{1}{z- i (2n+1) \pi} + \frac{1}{i(2n+1)\pi} \right].$$ The expansion given in the OP follows from the observation that $$\sum_{n \in \mathbb{Z}} \frac{1}{(2n+1)\pi} = 0.$$

### Proof of (1)

Suppose without loss of generality that the poles $z_n$ are ordered monotonically as in $$0 < |z_1| \le |z_2| \le \dots,$$ and fix some $z_0 \in \mathbb{C}$ such that $f$ is holomorphic at $z_0$.

Consider the integral $$\tag 3 I_n(z_0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z-z_0},$$ which applying Residue's theorem becomes $$\tag 4 I_n(z_0) = f(z_0) + \sum_{k=1}^n \frac{\alpha_k}{z_k-z_0}.$$ In particular for $z_0=0$ we have $$\tag 5 I_n(0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z} = f(0) + \sum_{k=1}^n \frac{\alpha_k}{z_k}.$$ Taking the difference between (3) and (5) (in the integral form) we see that: $$\tag 6 |I_n(z_0) - I_n(0) | \le \oint_{\mathcal C_n} \frac{|dz|}{2\pi} \left| \frac{z_0 f(z)}{(z-z_0) z} \right| \le \frac{Az_0}{R_n - |z_0|} \to 0, \text{when } n \to \infty.$$ However, we also know that $$\tag 7 I_n(z_0)-I_n(0) = f(z_0) - f(0) - \sum_{k=1}^n \left( \frac{1}{z_0-z_k} + \frac{1}{z_k} \right),$$ which taking the limit $n\to \infty$ gives the wanted result (1).

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