I'm trying to evaluate this sum.
$$S = \sum^\infty_{r=1} \left(\left(\frac{e^{ir}}{r^2}\right)\left(\tan^{-1}(r\sqrt{\pi})\right)\right)$$
I used the exponential definitions of $\sin(\theta)$ and $\cos(\theta)$ to get it into the form...
$$S = \frac{-i}{2}\sum^\infty_{r=1}\left(\left(\frac{e^{ir}}{r^2}\right)\left(\ln\left(-\frac{r\sqrt{\pi}-i}{r\sqrt\pi + i}\right)\right)\right)$$
After this, I'd separated the -1 from the natural logarithm so I could get the polylogarithm and my new expression was
$$S = \frac{\pi}{2}\text{Li}_2(e^i) - \frac{i}{2}S_m$$
$$S_m = \sum^\infty_{r=1} \left(\left(\frac{e^{ir}}{r^2}\right)\left(\ln\left(\frac{r\sqrt\pi - i}{r\sqrt\pi + i}\right)\right)\right)$$
Now I'm stuck on this step. I have no clue where to go from here and am kind of looking for ideas... It could be that my initial approach to it all is wrong in itself, in which any other suggestions on how I could approach it would be appreciated.
Edit: left an extra bracket by mistake
