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.
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):
$$f_2(x)=-\sum _{k=1}^{\infty } \left( \psi \left(k \pi -\frac{\pi }{2}+1-x\right)+\psi \left(k \pi -\frac{\pi }{2}+x\right) \right. $$ $$ \qquad \left. -\psi \left(k \pi -\frac{\pi }{2}+1\right)-\psi \left(k \pi -\frac{\pi }{2}\right)\right)+C$$
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).
