A series sum in a potential flow problem A point source is located at the origin in a complex plane and two parallel walls are held at $\pm ib$. The complex potential should be solved by the method of image. The sum takes
$$
w(z)=\frac{Q}{2\pi}\ln z + \frac{Q}{2\pi}\sum_{n=1}^{\infty}\left[\ln(z-i\cdot2nb)\right]-\frac{Q}{2\pi}\sum_{n=1}^{\infty}\left[\ln(z+i\cdot2nb)\right]
$$
After some trivial manipulation,
$$
w(z)=\frac{Q}{2\pi}\sum_{-\infty}^{\infty}\ln(z+i\cdot 2nb)
$$
After a google search, I find this solution which confirms my steps and implies that the series above is equal to
$$
\sum_{-\infty}^{\infty}\ln(z+i\cdot 2nb)=\ln \sinh\left(\frac{\pi z}{2b}\right)
$$
However, I still don't know how to perform the summation by hand (I tried Maple without any answer).
 A: I have made a mistake during the model problem. The correct potential is
\begin{align}
w(z)&=\frac{Q}{2\pi}\log z+\frac{Q}{2\pi}\sum_{n=1}^{\infty}\left(\log(z-i\cdot 2nb)+\log(z-i\cdot 2nb)\right)\\
&=\frac{Q}{2\pi}\log z+\frac{Q}{2\pi}\sum_{n=1}^{\infty}\log(z^2+4n^2b^2)
\end{align}
Taking the exponential function, the infinite product is
\begin{aligned}
   \mathrm{e}^{w(z)}&=\left(z\prod_{n=1}^{\infty}(z^2+4n^2b^2)\right)^{\frac{Q}{2\pi}}\\
        &=\left(\prod_{n=1}^{\infty}n^2\cdot 4b^2z\prod_{n=1}^{\infty}\left(1+\frac{z^2}{4n^2b^2} \right)\right)^{\frac{Q}{2\pi}}\\
        &=\left(\frac{8b^3}{\pi}\sinh \left(\frac{\pi z}{2b} \right) \prod_{n=1}^{\infty}n^2\right)^{\frac{Q}{2\pi}}
  \end{aligned}
Though
$$
\prod_{n=1}^{\infty}n^2
$$
diverges, it does not have an influence on the velocity field. Therefore, the potential can be written as
$$
w(z)=\frac{Q}{2\pi}\log\left(\sinh\left(\frac{\pi z}{2b}\right) \right)
$$
A: $$Exp(\frac{2 \pi \: w(z)}{Q}) =z  \prod_{n=1}^{\infty}\frac{(z-i\cdot2nb)}{(z+i\cdot2nb)} = z  \prod_{n=1}^{\infty}\frac{(-\frac{z \: i}{2nb}-1)}{(1-\frac{z \: i}{2nb})} \\ =  z  \prod_{n=1}^{\infty}\frac{(1+\frac{z^2}{4n^2b^2})}{(-1+\frac{z \: i}{nb} + \frac{z^2}{4n^2b^2})} = \frac{2 b \sinh \left(\frac{\pi  z}{2 b}\right)}{\pi } \prod_{n=1}^{\infty}\frac{1}{(-1+\frac{z \: i}{nb} + \frac{z^2}{4n^2b^2})} $$
where the final product diverges.
