Timeline for Why does the Gamma-function complete the Riemann Zeta function?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2023 at 14:36 | comment | added | Steven Clark | Yes, but it always simplifies to $$\zeta(s)=\frac{\Gamma(\hat{f},1-s)}{\Gamma(f,s)}\,\zeta(1-s)=2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\,\zeta(1-s).$$ It also works with (at least some) distributions and non-Schwartz functions where perhaps the optimal choice is $f(x)=\delta(|x|-1)$ in which case $$\hat{f}(w)=\mathcal{F}_x[f(x)](w)=2 \cos (2 \pi w)$$ and $\Gamma(f,s)=2$. | |
| Sep 11, 2014 at 22:30 | comment | added | paul garrett | @VítTuček, :) ... | |
| Sep 11, 2014 at 22:28 | comment | added | Vít Tuček | Bingo! The second Hermite function yields an innocent factor of $(1-2s)$. Thank you for clarification. | |
| Sep 11, 2014 at 22:07 | comment | added | paul garrett | @VítTuček, ah, one little point is that the game is to integrate over the whole $\mathbb R^\times$, so using odd Schwartz function $xe^{-\pi x^2}$ simply produces $0$, avoiding the seeming paradox. (I'd wager that using the second Hermite polynomial produces no contradiction!) For quadratic imaginary fields, for example, the range of choices of equivariance under the circle action (thinking of Iwasawa-Tate set-up) gives many chances to cancel-and-be-zero, etc. | |
| Sep 11, 2014 at 22:00 | comment | added | Vít Tuček | Starting from $\Gamma(f,s)\zeta(s) = \Gamma(\widehat{f},1-s)\zeta(1-s)$ and multiplying both sides by $\Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{1-s}{2}\right)$ I obtain, after cancellation of the classical functional equation for $\xi(s)$, the following equation $$ \Gamma\left(\frac{1-s}{2}\right)\Gamma\left(\frac{1+s}{2}\right) = -\imath \frac{\pi}{\sin(\pi s)}, $$ which can't be true since for real $s$ the left hand side is real whereas the right hand side is imaginary. Where did I make mistakes? | |
| Sep 11, 2014 at 21:59 | comment | added | Vít Tuček | @paulgarrett: Right, but... Let's see what happens for $f(x) = xe^{-\pi x^2}$ which is an eigenfunction of $$\mathcal{F}(g)(s) = \int_\mathbb{R} g(x)e^{-2\pi \imath x s}$$ of eigenvalue $-\imath$. The Gamma factor is $$\Gamma(f,s) = \frac{1}{2\sqrt{\pi}} \pi^{-\frac{s}{2}} \Gamma\left(\frac{s+1}{2}\right).$$ | |
| Sep 11, 2014 at 0:17 | comment | added | paul garrett | @VítTuček, and, if one wants greater harmony with the normalization(s) of the Gaussian in common use for the various version of Hermite polynomials, and using then the fact that $H_{n+1}(x){\rm Gaussian}(x)=(d/dx\pm c\cdot x)(H_n(x){\rm Gaussian}(x))$, so the polynomial factor can get normalized to $(s-1)(s_2)...$ rather than a messier one due to a mismatch of normalizations... if one wants. | |
| Sep 10, 2014 at 22:46 | comment | added | Vít Tuček | For the record, the Hermite functions yield just a polynomial multiple of the standard Gamma factor. For example $H_{16}$ leads to $2027025 + 16 (-1 + s) s (274455 + 2 (-1 + s) s (27051 + 8 (-1 + s) s (127 + (-1 + s) s)))$ | |
| Sep 10, 2014 at 16:53 | comment | added | Vít Tuček | I guess that would be the Dirac comb. See mathoverflow.net/a/39187/6818 | |
| Sep 10, 2014 at 16:47 | comment | added | David E Speyer | If I'm thinking correctly, $\zeta(s)$ is formally $\Gamma(g,1-s)$ where $g$ is a sum of $\delta$ functions at the positive integers. Is there some sense in which $g = \hat{g}$ here? | |
| Sep 10, 2014 at 16:41 | comment | added | Vít Tuček | So one can use any fixed point of the Fourier transform for $f$. Any idea what do we get for the appropriate Hermite functions? | |
| Sep 10, 2014 at 16:31 | history | answered | paul garrett | CC BY-SA 3.0 |