تابع بتا دیریکله

${\displaystyle \beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{s}}},}$

${\displaystyle \beta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1}e^{-x}}{1+e^{-2x}}}\,dx}$

${\displaystyle \beta (s)=4^{-s}\left(\zeta \left(s,{1 \over 4}\right)-\zeta \left(s,{3 \over 4}\right)\right).}$

${\displaystyle \beta (s)=2^{-s}\Phi \left(-1,s,{{1} \over {2}}\right),}$

${\displaystyle \beta (s)={\frac {1}{2^{s}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{\left(n+{\frac {1}{2}}\right)^{s}}}={\frac {1}{(-2)^{2s}(s-1)!}}\left[\psi ^{(s-1)}\left({\frac {1}{4}}\right)-\psi ^{(s-1)}\left({\frac {3}{4}}\right)\right]}$

${\displaystyle \beta (1-s)=\left({\frac {\pi }{2}}\right)^{-s}\sin \left({\frac {\pi }{2}}s\right)\Gamma (s)\beta (s)}$

${\displaystyle \beta (0)={\frac {1}{2}},}$

${\displaystyle \beta (1)\;=\;\arctan(1)\;=\;{\frac {\pi }{4}},}$

${\displaystyle \beta (2)\;=\;G,}$

${\displaystyle \beta (3)\;=\;{\frac {\pi ^{3}}{32}},}$

${\displaystyle \beta (4)\;=\;{\frac {1}{768}}\left(\psi _{3}\left({\frac {1}{4}}\right)-8\pi ^{4}\right),}$

${\displaystyle \beta (5)\;=\;{\frac {5\pi ^{5}}{1536}},}$

${\displaystyle \beta (7)\;=\;{\frac {61\pi ^{7}}{184320}},}$

${\displaystyle \beta (2k+1)={{({-1})^{k}}{E_{2k}}{\pi ^{2k+1}} \over {4^{k+1}}(2k)!},}$${\displaystyle \beta (2k+1)={{({-1})^{k}}{E_{2k}}{\pi ^{2k+1}} \over {4^{k+1}}(2k)!},}$

${\displaystyle \beta (-k)={{E{k}} \over {2}}.}$

${\displaystyle \beta (2k)={\frac {1}{2(2k-1)!}}\sum _{m=0}^{\infty }\left(\left(\sum _{l=0}^{k-1}{\binom {2k-1}{2l}}{\frac {(-1)^{l}A_{2k-2l-1}}{2l+2m+1}}\right)-{\frac {(-1)^{k-1}}{2m+2k}}\right){\frac {A_{2m}}{(2m)!}}{\left({\frac {\pi }{2}}\right)}^{2m+2k},}$

${\displaystyle \beta '(1)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {\ln(2n+1)}{2n+1}}\,=\,{\frac {\pi }{4}}{\big (}\gamma -\ln \pi )+\pi \ln \Gamma \left({\frac {3}{4}}\right)}$