Weierstrass $\wp$ function question Given the Weierstrass $\wp$ function with periods $1,\tau$ and $\wp(z) := \sum\limits_{n^2+m^2\ne 0} \frac{1}{(z+m+n\tau)^2}-\frac{1}{(m+n\tau)^2}$, I am trying to show $\wp = (\pi^2 \sum\limits^\infty_{n=-\infty} \frac{1}{\sin^2(\pi(z+n\tau))})+K$ for some constant $K$.  Note I am not trying to prove what $K$ is as I know what it is but it is unimportant here.  I am just trying to show this must be true for some constant.  The way I am attempting to solve is the following.
I know $\wp^\prime(z)=-2\sum\limits_{n,m\in \mathbb{Z}} \frac{1}{(z+m+n\tau)^3}$.  Next I want to integrate both sides. Thus, if I can take the integration inside the sum I will be able to complete the problem easily from there.  However, that is my problem.  I do not know why I can take the integration inside.  I either never learned those rules or more likely have forgotten them.  So essentially my larger problem boils down to a small one.  Thanks for the help.
Prove $-2\int \sum\limits_{n,m\in \mathbb{Z}} \frac{1}{(z+m+n\tau)^3} = \sum\limits_{n,m\in \mathbb{Z}} \frac{1}{(z+m+n\tau)^2}$.
 A: The sum
$$\sum_{m,n\in\mathbb{Z}} \frac{1}{(z+m+n\tau)^2}$$
does not converge absolutely, so working with that is not easy, you have to explicitly prescribe the order of summation to get a well-defined sum, and need to justify each manipulation of the sum accordingly. That can be done here, but I think it's easier to prove
$$\wp(z) = \sum_{n=-\infty}^\infty \frac{\pi^2}{\sin^2 (\pi(z+n\tau))} + K$$
by considering the function
$$h(z) = \sum_{n=-\infty}^\infty \frac{\pi^2}{\sin^2 (\pi(z+n\tau))},$$
and either reach the conclusion by differentiating it, or arguing that it is an elliptic function with poles only in the lattice points, and whose principal parts coincide with that of the poles of $\wp$, whence $h-\wp$ is an entire elliptic function, hence constant.
First, one has to see that $h(z)$ is a meromorphic function. Since each term in the sum is evidently holomorphic in $\mathbb{C}\setminus \Omega$, where $\Omega$ is the lattice spanned by $1$ and $\tau$, it suffices to see that the sum converges locally uniformly.
For real $x,y$, we have $\lvert \sin (x+iy)\rvert^2 = \sin^2 x + \sinh^2 y$, so the terms in the sum of $h$ decay exponentially, which shows the locally uniform convergence of the sum. To make it precise, consider $A(M) := \{ z : \lvert\operatorname{Im} z\rvert \leqslant M\}$. For a given $M > 0$, choose $N \in \mathbb{N}$ such that $N\cdot \lvert \operatorname{Im}\tau\rvert > 2M$. Then for $\lvert n\rvert \geqslant N$ we have $\lvert \operatorname{Im} (z+n\tau)\rvert \geqslant \lvert n\rvert \cdot\lvert \operatorname{Im}\tau\rvert - M \geqslant \frac12\lvert n\rvert\cdot \lvert \operatorname{Im}\tau\rvert$, and hence
$$
\lvert \sin^2 (\pi(z+n\tau))\rvert = \lvert \sin^2 (\pi(z+n\tau))\rvert
\geqslant \sinh^2 (\pi n\operatorname{Im}\tau/2) \geqslant Ce^{\pi\lvert n \operatorname{Im}\tau\rvert},
$$
thus $$\sum_{\lvert n\rvert \geqslant N} \frac{\pi^2}{\sin^2 (\pi(z+n\tau))}$$ converges uniformly on $A(M)$ by the Weierstraß $M$-test.
Hence $h$ is a holomorphic function on $\mathbb{C}\setminus\Omega$. Since the sum converges compactly, termwise differentiation is legitimate, and we have
$$h'(z) = \sum_{n=-\infty}^\infty \frac{d}{dz}\left(\frac{\pi^2}{\sin^2 (\pi(z+n\tau))}\right).$$
Now we can use the partial fraction decomposition of $\dfrac{\pi^2}{\sin^2 \pi w}$ to obtain
$$h'(z) = \sum_{n=-\infty}^\infty \frac{d}{dz}\left(\sum_{m\in \mathbb{Z}} \frac{1}{(z+m+n\tau)^2}\right).$$
The partial fraction decomposition of $\dfrac{\pi^2}{\sin^2 \pi w}$ converges compactly, hence can be differentiated term by term, and thus
$$h'(z) = \sum_{n=-\infty}^\infty \sum_{m\in\mathbb{Z}} \frac{-2}{(z+m+n\tau)^3} = \wp'(z).$$
The nested sum here converges absolutely and compactly, thus here we can rearrange the sum as we please.
