‘Integral’ of a Weierstrass $ \wp $-function. I'm revising for my finals and I've seen a question which asks:
Is there a meromorphic function $f: \mathbb{C}/\Lambda \to \mathbb{P}^1$ such that $f' = \wp$? 
There is a hint which says consider the poles of such a function, and I'm pretty sure no such function can exist, but if anyone could supply me with a proof or further hint I would be really grateful.
 A: Let me introduce a function known as the $\zeta$ of Weierstrass ($\zeta: \mathbb{C}\backslash \Lambda \to \mathbb{P}^1$, see edit 2 for the case $f:\mathbb{C} / \Lambda \to \mathbb{P}^1$), let me prove to you all you need to know about this function.
DEFINITION: We define the function $\zeta$ of Weierstrass as follows
$$
\zeta(z)=\frac{1}{z} + {\sum \limits_{\omega \in \Lambda}} ' \left[\frac{1}{z-\omega}+\frac{1}{\omega} + \frac{z}{\omega^2}\right]
$$
Claim $\zeta'=-\wp$
PROOF
It is easily shown* (see the edit 1 below) that the series appearing on the definition of the function $\zeta$ converges uniformly on $\mathbb{C}\backslash \Lambda$, then we can differentiate it term by term, giving 
\begin{align*}
\frac{d}{dz}\zeta(z) & = -\frac{1}{z^2} + {\sum \limits_{\omega \in \Lambda}} ' \left[-\frac{1}{(z-\omega)^2} + \frac{1}{\omega^2}\right]\\
& = - \left( \frac{1}{z^2} + {\sum \limits_{\omega \in \Lambda}} ' \left[\frac{1}{(z-\omega)^2} - \frac{1}{\omega^2}\right]\right) = -\wp(z) \  \ \ \blacksquare
\end{align*}
Finally note that for all $\omega \in \Lambda $
$$
\lim_{z\to \omega} (z-\omega)\zeta(z)=1
$$
Then the $\zeta$ function is indeed meromorphic since it has a simple pole in each $\omega \in \Lambda$, with principal part 
$$
\frac{1}{z-\omega}
$$
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EDIT 1* To show that the series on $\zeta$ converges uniformly you only need the next result, which is sometimes presented as a definition of $\zeta$:
Lemma The $\zeta$ function of Weierstrass satisfies that
$$
\zeta(z)=\frac{1}{z} - \int_{0}^{z} \left( \wp(u)-\frac{1}{u^2} \right)du
$$
where $\int_0^z$ is the line integral over a rectifiable path $\gamma$, from $0$ to $z$ without passing by any points of $\Lambda$. 
PROOF By your previous knowledge on the $\wp$ function you know that the series $\wp(u)-1/u^2$ converges uniformly on $\mathbb{C}\backslash \Lambda$, then it can by integrated term by term
\begin{align*}
 \int\limits_{0}^{z} \left( \wp(u)-\frac{1}{u^2} \right)du & ={\sum \limits_{\omega \in \Lambda}} '   \int\limits_{0}^{z} \left[\frac{1}{(u-\omega)^2}-\frac{1}{\omega^2}\right]du \\
  & ={\sum \limits_{\omega \in \Lambda}} ' \left[\int\limits_{0}^{z} \frac{du}{(u-\omega)^2}-\int\limits_{0}^{z} \frac{du}{\omega^2}\right] \\
   & ={\sum \limits_{\omega \in \Lambda}} ' \left[\frac{-1}{(u-\omega)^2}\Big|_0^z-\frac{u}{\omega^2}\Big|_0^z\right]\\
   & ={\sum \limits_{\omega \in \Lambda}} ' \left[-\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}-\frac{z}{\omega^2}\right] = \frac{1}{z}-\zeta(z) \ \ \ \blacksquare 
\end{align*}
Then of course, since the series on $\zeta$ is obtained by integrating a uniformly convergent series, then it must converge uniformly. 
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EDIT 2 As pointed out in the comments the $\zeta$ function is not elliptic, since all the singularities are simple poles (also, it can be shown $\zeta(z+2\omega_j)=\zeta(z)+2\eta_j$, where $\omega_j$ are the half periods generating $\Lambda$, and the $\eta_j$ are obtained from the Legendre Relations). Then this also proves that there dose not exists a function $f: \mathbb{C}/\Lambda \to \mathbb{P}^1$ such that $f'=\wp$, since again $-\zeta$ is not an elliptic function. 

So the function $\ f$ you are looking for should be $\ f=-\zeta$, but since it is not an elliptic one, such $\ f$ dose not exist.

Obs This exercise shows that, although, the elliptic functions are closed under differentiation they are not under integration.
