Timeline for Double Series involving Gamma function
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2017 at 19:38 | comment | added | Nemo | There is a much simpler proof, just expand the function $\frac{1}{(3-(1-u)-(1-v))^{3x}}$ into the powers of $(1-u)^n(1-v)^m$ and integrate termwise. | |
| Jul 25, 2017 at 19:03 | vote | accept | maliesen | ||
| Jul 25, 2017 at 18:10 | comment | added | Nemo | It is easy to prove that $\int_0^\infty\int_0^\infty \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{\Gamma(3x)}$ | |
| Jul 25, 2017 at 16:01 | history | edited | Nemo | CC BY-SA 3.0 |
added 253 characters in body
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| Jul 25, 2017 at 15:58 | comment | added | Matt Young | I didn't check any details of this answer, but for fun I had Mathematica spot-check if the double integral equals $\Gamma^3(x)/(3 \Gamma(3x))$. For $x=1,2,3,4,5,6$, at least, Mathematica can verify both sides agree as rational numbers. A couple other random non-integral choices of $x$ also gives agreement up to the default accuracy. | |
| Jul 25, 2017 at 14:44 | history | answered | Nemo | CC BY-SA 3.0 |