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Oct 20, 2014 at 11:54 comment added GH from MO @VítTuček: Thank you, that makes sense.
Oct 17, 2014 at 0:29 comment added Vít Tuček But then you get a factor with $\pi$. This expression just gets rid of it and it lets you use Gelfond-Schneider theorem directly.
Oct 16, 2014 at 15:23 comment added GH from MO I think the same argument works with the simpler expression $\Gamma(z)\Gamma(1-z)$.
Oct 16, 2014 at 12:24 vote accept Vít Tuček
Oct 15, 2014 at 23:33 review First posts
Oct 15, 2014 at 23:39
Oct 15, 2014 at 23:25 history answered user60548 CC BY-SA 3.0