Can this sum be written in simpler terms?
$$\sum_{k=0}^\infty \frac{1}{z-k} \cdot \frac{\gamma(k,-\log x)}{\Gamma(k)}$$
(where $\gamma(k,-\log x)$ is the lower incomplete gamma function)
I'm pretty sure another expression for this value is the integral
$$-\int_1^x \log^{z-1} t \cdot \gamma(1-z,\log t) \,dt$$
but I haven't had any more luck with that either.