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I have the following sum that I wish to evaluate: $$ \sum_{n=0}^{\infty} \left(\int_0^1 du\,u^{1+n+\epsilon}\right) \frac{\Gamma(n+2) \Gamma(1+\epsilon)}{\Gamma(n+3+\epsilon)} {}_2F_1\left(1, n+2, n+3+\epsilon, -\frac{a}{b}\right) = \sum_{n=0}^{\infty} \frac{1}{n+2+\epsilon} \frac{\Gamma(n+2) \Gamma(1+\epsilon)}{\Gamma(n+3+\epsilon)} {}_2F_1\left(1, n+2, n+3+\epsilon, -\frac{a}{b}\right)\,\,(1)$$ Does anyone know whether such a sum can be computed or if it exists? The hypergeometric is indeed convergent in the region I am working in, $b>-a, b>0, a<0$. I see that to use the series expansion though that necessarily $c=n+3+\epsilon \in \mathbb{Z}_{>0}$. This would be the case if $\epsilon=0$ since (in general this is not the case). So maybe I could write $${}_2F_1\left(1, n+2, n+3+\epsilon, -\frac{a}{b}\right) = \text{lim}_{\epsilon \rightarrow 0} \frac{\Gamma(n+3+\epsilon)}{\Gamma(n+2)} \sum_{k=0}^{\infty} \frac{\Gamma(1+k) \Gamma(n+2+k)}{\Gamma(n+3+\epsilon+k)} \frac{z^k}{k!}$$ The gamma factors outside this sum at least cancel those in $(1)$ but I don't know if what I did is any helpful.

Thanks for any comments!

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  • $\begingroup$ Non trivial sum - Now, that's an understatement... :-$)$ $\endgroup$ – Lucian Nov 30 '15 at 22:37
  • $\begingroup$ @Lucian: :P, basically the integral I am trying to evaluate is the following $$\int_0^1 dz \int_0^1 du \frac{u^{1+\alpha} (1-z)^{\alpha}z}{(b+z)(1-uz)},$$ for $b$ and $\alpha$ real constants not necessarily integers. In the above I reexpressed the coupled term (1-uz) as a geometric series. I've tried multiple things, like substitutions and partial fractions but all seem doomed to fail. Would you have any further ideas? $\endgroup$ – CAF Dec 1 '15 at 8:47

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