How to prove this formula I want to know how to prove this:
$$
\sum_{k=1}^\infty \arctan\left( \frac{x}{k^2+a} \right)
 = \pi \left\lfloor \frac{b}{\sqrt{2}} \right\rfloor
    + \arctan\left( \frac{x}{b^2} \right)
    - \arctan\left[ \tanh\left( \frac{\pi}{\sqrt{2}} \frac{x}{b} \right)
                    \cot\left( \frac{\pi}{\sqrt{2}} b \right) \right]
$$
for $0 \le a, b = \sqrt{\sqrt{x^2 + a^2} - a}$.
At least in the special case $x=1$,$a=0$
Thank you ..

Let $\;\displaystyle u + iv = \pi\sqrt{a+ix} = \frac{\pi}{\sqrt{2}}\left(\frac{x}{b} + ib\right)$ and follow nearly exactly the same step as in this answer, you get
$$\begin{align}
\sum_{k=1}\tan^{-1}\left(\frac{x}{k^2+a}\right) 
&= \tan^{-1}\left(\frac{\tan v}{\tanh u}\right) - \tan^{-1}\left(\frac{v}{u}\right) + \pi N\\
&= \tan^{-1}\left(\frac{x}{b^2}\right) - \tan^{-1}\left(\tanh\left(\frac{\pi x}{\sqrt{2}{b}}\right)\cot\left(\frac{\pi b}{\sqrt{2}}\right)\right) + \pi N
\end{align}$$
for some integer $N$. The only thing left is to figure out what $N$ is...
