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.

  • $\begingroup$ Do you mean the lower incomplete gamma function $\gamma(\alpha, \beta)$ or the upper incomplete gamma function $\Gamma(\alpha, \beta)$? $\endgroup$ – njuffa Sep 20 '16 at 0:03
  • $\begingroup$ The lower incomplete gamma function. Thanks for spotting that - I've edited the question to fix that. $\endgroup$ – Nathan McKenzie Sep 20 '16 at 0:28

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