I have stumbled upon an infinite sum that from what I can see appears to be very well-behaved (I'm a physicist, not a mathematician), but I cannot find a good way of evaluating it in closed form. Does anyone here have any ideas of how I should proceed or happen to recognise this sum?
$$\sum _{n=1}^{\infty } \frac{ (-z)^n }{n} \frac{ \Gamma \left(\frac{n}{2}\right)^2 \Gamma \left(\frac{1}{2} n (1-i x)\right) \Gamma \left(\frac{1}{2} n (1+i x)\right) }{ \, \Gamma (n)^2} $$
where $x\in \mathbb{R}$? It is clear that since $\overline{\Gamma(z)}=\Gamma(\bar{z})$, that the summand is real, and I can evaluate it as a sum of hypergeometric functions when x=0, in which case it takes the form
$$\sum _{n=1}^{\infty } \frac{ (-z)^n }{n} \frac{ \Gamma \left(\frac{n}{2}\right)^4 }{ \, \Gamma (n)^2} = -\pi^2z \,_3F_2(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};\frac{z^2}{16}) -\frac{z^2}{2}\, _4F_3(1,1,1,1;\frac{3}{2},\frac{3}{2},2;\frac{z^2}{16}) $$
but is there a general form when $x>0$?
I have tried using the relation $ \Gamma \left(\frac{n}{2} (1-i x)\right) \Gamma \left(\frac{n}{2} (1+i x)\right) = \Gamma \left(\frac{n}{2}\right)^2 \prod _{k=0}^{\infty } \frac{1}{\frac{n^2 x^2}{(2 k+ x)^2}+1} $
but I don't see any obvious way this will help me. It could possibly work to try a series expansion for small values of $x$, but I would really like a closed form for all values of $x$ if possible...
Any help would be appreciated, thanks!