I am working on a problem where the following sum appears: $$F(s, t)=\frac{1}{\Gamma(1+2\alpha)}\sum_{n=0}^{\infty}{\frac{s^{n} t^{n}}{\left[(s+1)(t+1)\right]^{n+1+\alpha}}\frac{\Gamma(n+1+2\alpha)}{\Gamma(n+1)}}$$ where $s, t$ are positive real numbers and $\alpha>-1/2$. If I let Wolfram Mathematica calculate this seemingly awful expression, I get the following: $$F(s, t)=\frac{(s+1)^{\alpha}(t+1)^{\alpha}}{(s+t+1)^{1+2\alpha}}$$ and it seems it doesn't take that long to compute it too! So I assume it is a "sort of" well-known formula. Does anybody know if this formula is well-known, and if it is, could they tell me where I could find it in the literature? Otherwise, could anybody point an approach for proving this? Thanks in advance for any help!

This identity crops up in the study of some linear operators I have, whose kernel is a product of Gamma functions.

Sorry for the edits, but I realized too late I had adjust a tiny mistake.

  • $\begingroup$ For Mathematica all such functions are just series. So you can consider both this functions as series of $s$ and compare $k$th coefficients in these series. $\endgroup$ – Alexey Ustinov Oct 31 '15 at 2:02
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    $\begingroup$ $\frac{\Gamma(n+2\alpha+1)}{\Gamma(n+1)\Gamma(2\alpha+1)} = \binom{n+2\alpha}{n}=(-1)^n\binom{-2\alpha-1}{n}$ and thus the sum represents a simple binomial expansion. $\endgroup$ – Max Alekseyev Oct 31 '15 at 2:07

Expanding my comment.

Notice that $\frac{\Gamma(n+2\alpha+1)}{\Gamma(n+1)\Gamma(2\alpha+1)}=\binom{n+2\alpha}{n}=(-1)^n\binom{-2\alpha-1}{n}$ and thus the sum can be rewritten as $$\frac{1}{[(s+1)(t+1)]^{\alpha+1}}\sum_{n=0}^{\infty} \binom{-2\alpha-1}{n} \left(-\frac{st}{(s+1)(t+1)}\right)^n$$ $$ = \frac{1}{[(s+1)(t+1)]^{\alpha+1}}\left(1-\frac{st}{(s+1)(t+1)}\right)^{-2\alpha-1} = \frac{[(s+1)(t+1)]^{\alpha}}{(s+t+1)^{2\alpha+1}}$$ as expected.


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