Mittag-Leffler's Expansion Theorem

Theorem
Let $f$ be a meromorphic function with only simple poles continuous, or with a removable singularity, at $0$.

Let $X$ be the set of poles of $f$.

For $N \in \N$, let $C_N$ be a circle, centred at the origin, of radius $R_N$, where $R_N \to \infty$ as $N \to \infty$, such that $\partial C_N$ contains no poles of $f$ for any $N$.

Let $M > 0$ be a real number independent of $N$ such that for all $z \in \partial C_N$, $\cmod {\map f z} < M$, for all $N \in \N$.

Then:
 * $\displaystyle \map f z = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {z - n} + \frac 1 n}$

where:
 * $\Res f n$ is the residue of $f$ at $n$
 * $z$ is not a pole of $f$
 * $\displaystyle \map f 0 = \lim_{z \mathop \to 0} \map f z$ if $f$ has a removable singularity at $0$.

Proof
Let $\zeta \in \C \setminus X$.

Then:


 * $\displaystyle \frac {\map f z} {z - \zeta}$

has simple poles for $z \in X \cup \set \zeta$.

Let $X_N$ be the set of poles contained within $C_N$.

Then:

Setting $\zeta = 0$, we obtain:


 * $\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$

So:


 * $\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$

It remains to show that the integral on the left hand side vanishes as $N \to \infty$.

We have:

Letting $N \to \infty$ gives:


 * $\displaystyle 0 = \map f \zeta + \sum_{n \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$

Giving:


 * $\displaystyle \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$