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Oct 23, 2016 at 16:24 history closed Michael Renardy
Alex Degtyarev
Franz Lemmermeyer
R.P.
Daniel Loughran
Not suitable for this site
Oct 23, 2016 at 10:22 review Close votes
Oct 23, 2016 at 15:08
Oct 23, 2016 at 9:47 history edited Myshkin
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Oct 23, 2016 at 9:47 answer added Myshkin timeline score: 4
Feb 19, 2013 at 9:20 comment added Marc Palm I voted to close. Yes, with little knowledge of the Riemann zeta function it is obviously not analytically, but meromorphically continuable to the complex plane. This question is too elementary for this forum.
Feb 19, 2013 at 3:00 comment added Włodzimierz Holsztyński ... and a chance to reject it.
Feb 19, 2013 at 2:44 comment added Venkataramana If you can show that $\frac{1}{\zeta (s)}$, which is defined for $re(s)>1$, extends to an $analytic$ function for $re(s)>\frac{1}{2}$ , you will get at least a million dollars.
Feb 19, 2013 at 2:37 comment added S. Carnahan I'm not sure this answers the question you really want to ask, but meromorphic functions always have analytic continuations.
Feb 19, 2013 at 2:33 history asked pat porto CC BY-SA 3.0