Which meromorphic functions are logarithmic derivatives of other meromorphic functions? Let $f$ be a meromorphic function defined on the whole complex plane. 

Is there a characterization in terms of easier-to-test properties of $f$ whether or not $f=g'/g$ for some entire $g$?

The question is in the spirit of theorems like:


*

*If $f$ is entire and nonzero, then $f=e^g$ for entire $g$. (More generally, if $f$ is holomorphic and nonzero in a simply connected domain, then $f=e^g$ for $g$ holomorphic in that domain.)

*If $f$ is a rational function with no nonzero residue at any pole, then $f = g'$ for a rational $g$.
My main question is as above but here are some generalizations I would also be interested in if you happen to know anything about them: (1) what if I just required $g$ satisfying $f=g'/g$ to be meromorphic? (2) What if I just asked $f$ to be meromorphic in a simply connected domain?
 A: If $g$ is holomorphic in some domain $D$ then $f = g'/g$ has only simple poles, and the residue at each pole is a positive integer.
Conversely, if  $f$ is meromorphic in the simply-connected domain $D$ with
only simple poles, and the residue at each pole is a positive integer,
then there is a holomorphic function $g$ in $D$ such that $f = g'/g$.
Proof: Let $P \subset D$ be the set of (isolated) poles of $f$.
Fix a point $z_0 \in D \setminus P$ and define $g$ in $D \setminus P$
by 
$$
  g(z) = \exp \left( \int_\gamma f(t) \, dt \right )
$$
where $\gamma$ is any path in $D \setminus P$ connecting $z_0$ to $z$.
The definition is independent of the choice of $\gamma$: If
$\gamma_1$ and $\gamma_2$ are two paths from $z_0$ to $z$
then $\gamma_1 -\gamma_2$ is a closed path in $D \setminus P$,
and from the residue theorem it follows that
$$
 \int_{\gamma_1} f(t) \, dt - \int_{\gamma_2} f(t) \, dt
 = 2 \pi i \sum_{p \in P} \text{n} ( \gamma_1 -\gamma_2, p) \,
\text{Res} (f, p)
$$
and that is $2 \pi i$ times some integer. It follows that
$$
\exp \left( \int_{\gamma_1} f(t) \, dt \right ) = 
\exp \left( \int_{\gamma_2} f(t) \, dt \right ) \, .
$$ 
$g$ is holomorphic in $D \setminus P$ and satisfies $g'/g = f$.
It remains to show that $g$ can be continued analytically to $D$.
Let $z_1 \in P$ be a pole of $f$ with residue $k$. Then
$$
 f(z) = \frac{k}{z-z_1} + f_1(z)
$$
where $f_1$ is holomorphic in a neighborhood $U$ of $z_1$.
Using the same technique as above,
$$
g_1(z) = (z - z_1)^k \exp \left( \int_\gamma f_1(t) \, dt \right)
\quad \text{($\gamma$ is any path from $z_1$ to $z$ in $U$)}
$$
is holomorphic in $U$ and satisfies $g_1'/g_1 = f$.
If follows that $g$ and $g_1$ are constant multiples of each other
in $U\setminus \{ z_1 \}$, which means that $g$ has a holomorphic
continuation to $z_1$ (and a $k$-fold zero in that point).
Remarks: The condition "$D$ simply-connected" is used when applying
the residue theorem, and this condition cannot be dropped: For $a
\in \Bbb C \setminus \Bbb Z$, $f(z) = a/z$ is holomorphic in
$D = \{ z : 1 < |z| < 2 \}$, but not a logarithmic derivative.
If $f$ is holomorphic in $D$ with simple poles and
the residue at each pole is a (positive or negative) integer,
then the same technique shows that $f = g'/g$ for some
meromorphic function $g$ in $D$.
A: If $g$ is a nonconstant analytic function, the poles of $g'/g$ are at zeros of $g$.  At a zero of $g$ of multiplicity $m$, $g'/g$ has a simple pole with residue $m$.  Thus a necessary condition for $f$ to be representable as $g'/g$ is that
its poles are all simple with positive integer residues.
Conversely, suppose $f$ satisfies this condition.  Then we know the zeros of any possible candidate $g$ and their multiplicities.  Of course these zeros can't have an accumulation point.  The Weierstrass factorization theorem says that given any sequence
of complex numbers $a_j$ with no accumulation point and positive integers $m_j$, there is an entire function $g_0$ with that sequence  of zeros and multiplicities.  Then $h = g_0'/g_0 - f$ is an entire function.  Let $g_1 = \exp(H)$ where $H$ is an antiderivative of $h$.  This has no zeros.  Then we have
$g_1'/g_1 = h$.  And now $g = g_0/g_1$ satisfies $$ \dfrac{g'}{g} = \dfrac{g_0}{g_0} - \dfrac{g_1'}{g_1} = f$$ 
