Timeline for zeta-function regularized integrals

7 events

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Dec 5, 2013 at 22:56	vote	accept	user6818
Dec 5, 2013 at 22:54	comment	added	user6818		- I guess a similar proof you have given for $\xi(3)$ will also give this above.
Dec 5, 2013 at 22:53	comment	added	user6818		Thanks for the efforts. Now I remember that there is this general statement that for $Re(q)>0$ one can write, $\xi(q) = \frac{\pi^q}{2^{1-2q}(2^q - 2) \Gamma(q)} \int _0 ^\infty dx \frac{ e^{-2\pi\sqrt{x}} x^{\frac{q}{2} -1 } }{1 + e^{-2\pi\sqrt{x}} }$ - now for $q=3$ this seems to match the first integral equality I wrote down - and now if I power-series expand the denominator of this integrand and integrate term-by-term I get, $\xi(q) = \frac{2^q}{2-2^q}\sum_{s=1}^{\infty} \frac{ (-1)^s }{s^q } $ - which matches your expression for $\xi(3)$
Dec 5, 2013 at 22:27	history	edited	user11000	CC BY-SA 3.0	added 577 characters in body
Dec 5, 2013 at 22:23	comment	added	user11000		@user6818 What is the first integration equality you are talking about?
Dec 5, 2013 at 22:12	comment	added	user6818		I am a bit confused about what you are saying - (1) Why is $\xi(3) = - \frac{4}{3} \sum_{k=1}^\infty (-1)^k \frac{1}{k^3}$ ? (...that doesn't naively seem to be the usual definition of the Riemann zeta function...) (2) Are you proving the first integration equality also somehow?
Dec 5, 2013 at 0:05	history	answered	user11000	CC BY-SA 3.0