Riemannin zeeta-funktio

Matematiikassa Riemannin zeeta-funktio on kompleksitason kuvaus, joka liittyy alkulukujen jakaumaan ja on siksi mielenkiintoinen mm. lukuteorian kannalta. ^[1]

Määritelmä

Riemannin zeeta-funktio ${\displaystyle \zeta (s)}$ on määritelty kompleksiluvuille ${\displaystyle s}$, joiden reaaliosa ${\displaystyle >1}$, summaksi

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}$.

Alueessa ${\displaystyle \{s\in \mathbb {C} :\mathrm {Re~} s>1\}}$ 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 ${\displaystyle 1}$. Tämä funktio on kyseessä Riemannin hypoteesissa.

Integraaleja

Jos ${\displaystyle s\in \mathbb {C} \setminus \{1\}}$ pätevät kaavat

${\displaystyle \zeta (s)={\frac {2^{s-1}}{s-1}}-2^{s}\int \limits _{0}^{\infty }{\frac {\sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\ (\mathrm {e} ^{\pi \,t}+1)}}\,\mathrm {d} t}$

ja

${\displaystyle \zeta (s)={\frac {1}{s-1}}+{\frac {1}{2}}+2\int \limits _{0}^{\infty }{\frac {\sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\ (\mathrm {e} ^{2\,\pi \,t}-1)}}\,\mathrm {d} t.}$

Jos ${\displaystyle 0<Re(s)<1\!}$ on

${\displaystyle \int _{0}^{\infty }x^{s-1}{\frac {\gamma x+\log \Gamma (1+x)}{x^{2}}}\,dx={\frac {\pi }{\sin(\pi s)}}{\frac {\zeta (2-s)}{2-s}}\!}$.

Integraali zeetafunktion derivaatalle on

${\displaystyle {\zeta '(s)=2^{s-1}\left({\frac {\log 2}{s-1}}-{\frac {1}{(s-1)^{2}}}+\int \limits _{0}^{\infty }{\frac {2\arctan t\cdot \cos(s\arctan t)+\log {\frac {4}{1+t^{2}}}\cdot \sin(s\arctan t)}{(1+t^{2})^{\frac {s}{2}}\cdot (e^{\pi t}+1)}}\mathrm {d} t\right)}}$

joka pätee kaikille kompleksiluvuille paitsi kun s=1.

Kaavoja jotka sisältävät zeetafunktion

${\displaystyle \sum _{k=2}^{\infty }\zeta (k)x^{k-1}=-\psi _{0}(1-x)-\gamma }$

missä ψ₀ on digammafunktio.

${\displaystyle \sum _{n=2}^{\infty }(\zeta (n)-1)=1}$

${\displaystyle \sum _{n=1}^{\infty }(\zeta (2n)-1)={\frac {3}{4}}}$

${\displaystyle \sum _{n=1}^{\infty }(\zeta (2n+1)-1)={\frac {1}{4}}}$

${\displaystyle \sum _{n=2}^{\infty }(-1)^{n}(\zeta (n)-1)={\frac {1}{2}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{2^{2n}}}={\frac {1}{6}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{4^{2n}}}={\frac {13}{30}}-{\frac {\pi }{8}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{8^{2n}}}={\frac {61}{126}}-{\frac {\pi }{16}}({\sqrt {2}}+1)}$

${\displaystyle \sum _{n=1}^{\infty }(\zeta (4n)-1)={\frac {7}{8}}-{\frac {\pi }{4}}\left({\frac {e^{2\pi }+1}{e^{2\pi }-1}}\right)}$

${\displaystyle \log 2=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}}$

${\displaystyle \log \pi =\sum _{n=2}^{\infty }{\frac {(2({\tfrac {3}{2}})^{n}-3)(\zeta (n)-1)}{n}}.}$

Sarjoja Eulerin vakiolle:

${\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)}{n}}=\gamma }$

${\displaystyle \sum _{n=2}^{\infty }{\frac {\zeta (n)-1}{n}}=1-\gamma }$

${\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)-1}{n}}=\ln 2+\gamma -1}$

${\displaystyle \sum _{n=2}^{\infty }(-1)^{n}{\frac {\zeta (n)}{n2^{n-1}}}=\gamma -\log {\frac {4}{\pi }}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {\zeta (2n+1)-1}{(2n+1)\ 2^{2n}}}=1+\log {\frac {2}{3}}-\gamma .}$

Sarja Catalanin vakiolle:

${\displaystyle {\frac {1}{16}}\sum _{n=1}^{\infty }(n+1)\ {\frac {3^{n}-1}{4^{n}}}\ \zeta (n+2)=G.}$

${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}t^{2n}\left[\zeta (2n)-1\right]={\frac {t^{2}}{1+t^{2}}}+{\frac {1-\pi t}{2}}-{\frac {\pi t}{e^{2\pi t}-1}}}$

${\displaystyle \sum _{k=0}^{\infty }{\frac {\zeta (k+n+2)-1}{2^{k}}}{{n+k+1} \choose {n+1}}=\left(2^{n+2}-1\right)\zeta (n+2)-1}$

${\displaystyle \sum _{k=0}^{\infty }{k+\nu +1 \choose k}\left[\zeta (k+\nu +2)-1\right]=\zeta (\nu +2)}$

${\displaystyle \sum _{k=0}^{\infty }{k+\nu +1 \choose k+1}\left[\zeta (k+\nu +2)-1\right]=1}$

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k+1}\left[\zeta (k+\nu +2)-1\right]=2^{-(\nu +1)}}$

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k+2}\left[\zeta (k+\nu +2)-1\right]=\nu \left[\zeta (\nu +1)-1\right]-2^{-\nu }}$

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}{k+\nu +1 \choose k}\left[\zeta (k+\nu +2)-1\right]=\zeta (\nu +2)-1-2^{-(\nu +2)}}$

Katso myös

• Zeeta-funktio

Lähteet

Kirjallisuutta

• Thompson, Jan & Martinsson, Thomas: Matematiikan käsikirja. Helsinki: Tammi, 1994. ISBN 951-31-0471-0.