0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.