Matematiikassa Riemannin zeeta-funktio on kompleksitason kuvaus, joka liittyy alkulukujen jakaumaan ja on siksi mielenkiintoinen mm. lukuteorian kannalta. [1]
Riemannin zeeta-funktio
on määritelty kompleksiluvuille
, joiden reaaliosa
, summaksi
.
Alueessa
tämä sarja suppenee ja zeeta-funktio on analyyttinen. Bernhard Riemann keksi, että zeeta-funktiota voidaan analyyttisesti jatkaa meromorfiseksi funktioksi, joka on määritelty koko kompleksitasossa lukuun ottamatta pistettä
. Tämä funktio on kyseessä Riemannin hypoteesissa.
Jos
pätevät kaavat
![{\displaystyle \zeta (s)={\frac {2^{s-1}}{s-1}}-2^{s}\int \limits _{0}^{\infty }{\frac {\sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\ (\mathrm {e} ^{\pi \,t}+1)}}\,\mathrm {d} t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0cbb9ff0194e009e2a219e857deed725aaefcdc)
ja
![{\displaystyle \zeta (s)={\frac {1}{s-1}}+{\frac {1}{2}}+2\int \limits _{0}^{\infty }{\frac {\sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\ (\mathrm {e} ^{2\,\pi \,t}-1)}}\,\mathrm {d} t.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43a426d89a5efce7dbc2eb8fc7e8ac4316b6eb8e)
Jos
on
.
Integraali zeetafunktion derivaatalle on
![{\displaystyle {\zeta '(s)=2^{s-1}\left({\frac {\log 2}{s-1}}-{\frac {1}{(s-1)^{2}}}+\int \limits _{0}^{\infty }{\frac {2\arctan t\cdot \cos(s\arctan t)+\log {\frac {4}{1+t^{2}}}\cdot \sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\cdot (e^{\pi t}+1)}}\mathrm {d} t\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5877092a40d05df2e89cdba542c4a55cc3723c1e)
joka pätee kaikille kompleksiluvuille paitsi kun s=1.
![{\displaystyle \sum _{k=2}^{\infty }\zeta (k)x^{k-1}=-\psi _{0}(1-x)-\gamma }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6766c608bf100aa433b34559bbe6cdda32f78aea)
missä ψ0 on digammafunktio.
![{\displaystyle \sum _{n=2}^{\infty }(\zeta (n)-1)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1707919eef6cd1e84cc44f3cde817a1f2f082fd0)
![{\displaystyle \sum _{n=1}^{\infty }(\zeta (2n)-1)={\frac {3}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c33eb669ae68e3f2ca3bed394b6c3a27134a5bba)
![{\displaystyle \sum _{n=1}^{\infty }(\zeta (2n+1)-1)={\frac {1}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c12a9b1cf865cc2a8d6d0247659e7beac898f73a)
![{\displaystyle \sum _{n=2}^{\infty }(-1)^{n}(\zeta (n)-1)={\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e56f079686e8256ee19fb038816f9e1182264941)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{2^{2n}}}={\frac {1}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf639c5f9a6c3516a12fd35ec8d43f46388ecc21)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{4^{2n}}}={\frac {13}{30}}-{\frac {\pi }{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fad75cf20bf950c92a2556e4467be2c88c5a4e5b)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{8^{2n}}}={\frac {61}{126}}-{\frac {\pi }{16}}({\sqrt {2}}+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/866a92965d3cadd8a9b43fd6d930ef07f1443150)
![{\displaystyle \sum _{n=1}^{\infty }(\zeta (4n)-1)={\frac {7}{8}}-{\frac {\pi }{4}}\left({\frac {e^{2\pi }+1}{e^{2\pi }-1}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/609e68e4e15abab35ccfc7aa674a69f34e7ecc3d)
![{\displaystyle \log 2=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34c77c062c125a9e521088c5c26f433c90deadb1)
![{\displaystyle \log \pi =\sum _{n=2}^{\infty }{\frac {(2({\tfrac {3}{2}})^{n}-3)(\zeta (n)-1)}{n}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b50d25d562e20a47d1fbb5017dd93b4fd60e33c)
Sarjoja Eulerin vakiolle:
![{\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)}{n}}=\gamma }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ea679fcfca1f65496b3da256cb1f03fde8a279f)
![{\displaystyle \sum _{n=2}^{\infty }{\frac {\zeta (n)-1}{n}}=1-\gamma }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a08241be47efa01f9a27c070380118fcffab37d)
![{\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)-1}{n}}=\ln 2+\gamma -1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96ce5056c7e5b7098a3a35c85f69d8c403b71877)
![{\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)}{n2^{n-1}}}=\gamma -\log {\frac {4}{\pi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c6013096c851fff1f7cc6458d287f11e61b9e79)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n+1)-1}{(2n+1)\ 2^{2n}}}=1+\log {\frac {2}{3}}-\gamma .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcf2a3d0a9ffe503004930939e34443f5548b263)
Sarja Catalanin vakiolle:
![{\displaystyle {\frac {1}{16}}\sum _{n=1}^{\infty }(n+1)\ {\frac {3^{n}-1}{4^{n}}}\ \zeta (n+2)=G.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e146954c0f0b1e020070355e39f675036b23698a)
![{\displaystyle \sum _{n=1}^{\infty }(-1)^{n}t^{2n}\left[\zeta (2n)-1\right]={\frac {t^{2}}{1+t^{2}}}+{\frac {1-\pi t}{2}}-{\frac {\pi t}{e^{2\pi t}-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88eec406cb0d9554da544b9614fe5581fb6c0786)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {\zeta (k+n+2)-1}{2^{k}}}{{n+k+1} \choose {n+1}}=\left(2^{n+2}-1\right)\zeta (n+2)-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9c901f6e88fa236721f966618af96e5a6dd0fe9)
![{\displaystyle \sum _{k=0}^{\infty }{k+\nu +1 \choose k}\left[\zeta (k+\nu +2)-1\right]=\zeta (\nu +2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80e6bbc24b257d55d40c9355ca7a84304071001c)
![{\displaystyle \sum _{k=0}^{\infty }{k+\nu +1 \choose k+1}\left[\zeta (k+\nu +2)-1\right]=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16f2fb2ee7e0568c9a2b8badbb0932865d12809d)
![{\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k+1}\left[\zeta (k+\nu +2)-1\right]=2^{-(\nu +1)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a5d3c969127771d6343b29fda8c68e1ef01fefa)
![{\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k+2}\left[\zeta (k+\nu +2)-1\right]=\nu \left[\zeta (\nu +1)-1\right]-2^{-\nu }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f083804addcac40e42af2534b693d62a606210d)
![{\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k}\left[\zeta (k+\nu +2)-1\right]=\zeta (\nu +2)-1-2^{-(\nu +2)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02e48542bc883438b81ad6d57f5ba7c62ec35d8b)
- ↑ Thompson, Jan & Martinsson, Thomas: Matematiikan käsikirja, s. 341–342. Helsinki: Tammi, 1994. ISBN 951-31-0471-0.