Timeline for Why does the Gamma-function complete the Riemann Zeta function?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 14, 2021 at 9:31 | comment | added | Watson | @SimonHenry : see KConrad's answer below (from 2019). Essentially this is because we know the zeros/poles of $\Gamma(s)$. | |
| Oct 26, 2015 at 9:59 | comment | added | Simon Henry | But there is a lot of functions which are their own Fourier transform, isn't it ? why this one in particular ? what would happen with other functions with the same property ? | |
| Dec 4, 2009 at 22:32 | comment | added | Ben Webster♦ | Ah, Terry Tao's post is clearly the answer I was looking for. | |
| Dec 4, 2009 at 4:44 | history | answered | David E Speyer | CC BY-SA 2.5 |