The question here is sparked by the discussion inside this question about indefinite sum(antidifference) of tan(x).
A proposed solution was a function $$f_1(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$ The function involved is the q-digamma function. Symbolically it resolves to be antidifference of tan(x), one can check it by following this link to Wolfran Alpha's result: http://tiny.cc/60mmf
But it turned out that neither Wolfram Alpha, nor any other software is able to evaluate the q-polygamma function with $q=e^{2 i}$ numerically. Attempting to evaluate it manually also turned out to be too difficult.
A second solution seemed to be easier to evaluate. It was a series that converged absolutely and also has been proven to be an antidifference of tan(x):
$$f_2(x)=-\sum _{k=1}^{\infty } \left(\psi left( \psi \left(k \pi -\frac{\pi }{2}+1-x\right)+\psi \left(k \pi -\frac{\pi }{2}+x\right)-\psi {2}+x\right) \right. $$ $$ \qquad \left. -\psi \left(k \pi -\frac{\pi }{2}+1\right)-\psi \left(k \pi -\frac{\pi }{2}\right)\right)+C$$
This function has a fancy graphic (see the initial discussion). Since it is known that a function can have several antidifferences which differ between each other by a 1-periodic function, it is interesting, whether the both functions $f_1(x)$ and $f_2(x)$ are equal or what is the difference between them (of course if the first function actually can be evaluated numerically).
