\tan(\pi\alpha\cdot i)=i\cdot\tanh(\pi\alpha)=i\cdot\sum_{k=0}^{\infty}{\frac{8\pi\alpha}{(2k+1)^2\pi^2+4\pi^2\alpha^2}}=\frac{2\alpha i}{\pi}\cdot\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}