Let $g$ be a meromorphic function on $\mathbb{C}$. Suppose $g$ has poles of order 1 with integer residue. Prove that exists $f$ meromophic such that
$$g(z)=\frac{f'(z)}{f(z)}.$$
I was able to prove it defining
$$f(z)=e^{\int_\gamma g(w)dw}$$
where $\gamma$ is a curve that joins $0$ with $z$ without passing through the poles of $g$. However, the exercise is expected to be solved using the theorem of Mittag-Leffler. I have been trying but I don't have a clue how to do it that way. Any suggestions or recomendations?
Best Answer
Sketch of the proof using Weierstrass Theorem - let the poles of $g$ be divided into two sets depending on whether their residues are positive or negative integers (either set can be empty of course), $a_k, n_k \ge 1$, $b_k, -m_k \le -1$.
Then by the Weiertrass factorization theorem we can get entire functions $h_1, h_2$ with zeroes only at $a_k,b_k$, of orders $n_k, m_k$ respectively (if either set is empty take the corresponding $h$ identically $1$).
Then if $h=\frac{h_1}{h_2}$ we have that $h$ is a meromorphic function st $h'/h$ has the poles of $g$ with residues precisely the same, since for $a_k$, one has $h=(z-a_k)^{n_k}q_k$ with $q_k$ having no zero or pole at $a_k$ and then $h'/h=n_k/(z-a_k)+q_k'/q_k$, with $q_k'/q_k$ analytic at $a_k$ and same for $b_k, -m_k$
But this implies that $g-h'/h=g_1$ entire, so if $G_1$ is an antiderivative of $g_1$, we can take $f=he^{G_1}$ and $f'/f=h'/h+g_1=g$ is the required represenation of $g$