リーマンゼータ関数の特殊値
ζ ( 5 ) = 2 π 4 3 ( 2 5 − 1 ) [ log  2 + ∑ n = 1 ∞ ( 2 n + 5 ) ζ ( 2 n ) 2 2 n ( n + 2 ) ( 2 n + 3 ) ] {\displaystyle \zeta (5)={\frac {2\pi ^{4}}{3\,(2^{5}-1)}}{\Biggl [}\log 2+\sum _{n=1}^{\infty }{\frac {(2n+5)\,\zeta (2n)}{2^{2n}\,(n+2)(2n+3)}}{\Biggr ]}}
ζ ( 5 ) = 2 π 4 31 [ ∑ n = 1 ∞ ζ ( 2 n ) 2 2 n ( n + 2 ) ( 2 n + 1 ) − 3 ∑ n = 1 ∞ ζ ( 3 ) 2 2 n ( n + 1 ) ( 2 n + 3 ) ] {\displaystyle \zeta (5)={\frac {2\pi ^{4}}{31}}{\Biggl [}\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}(n+2)(2n+1)}}-3\sum _{n=1}^{\infty }{\frac {\zeta (3)}{2^{2n}\,(n+1)(2n+3)}}{\Biggr ]}}
ζ ( 5 ) = 1 24 ∫ 0 ∞ x 4 e x − 1 d x {\displaystyle \zeta (5)={\frac {1}{24}}\int _{0}^{\infty }\!{\frac {x^{4}}{e^{x}-1}}\,{\rm {d}}x}
ζ ( 3 ) = 2 π 2 7 [ 3 2 − log  π + ∑ n = 1 ∞ ζ ( 2 n ) 2 2 n − 1 n ( 2 n + 1 ) ( 2 n + 2 ) ] {\displaystyle \zeta (3)={\frac {2\pi ^{2}}{7}}{\Biggl [}{\frac {3}{2}}-\log \pi +\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n-1}\,n\,(2n+1)(2n+2)}}{\Biggr ]}}
ζ ( 3 ) = π 2 7 [ 1 − 4 ∑ n = 1 ∞ ζ ( 2 n ) 2 2 n ( 2 n + 1 ) ( 2 n + 2 ) ] {\displaystyle \zeta (3)={\frac {\pi ^{2}}{7}}{\Biggl [}1-4\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}\,(2n+1)(2n+2)}}{\Biggr ]}}
ζ ( 3 ) = 7 π 3 180 − 2 ∑ n = 1 ∞ 1 n 3 ( e 2 n π − 1 ) {\displaystyle \zeta (3)={\frac {7\pi ^{3}}{180}}-2\sum _{n=1}^{\infty }{\frac {1}{n^{3}\,(e^{2n\pi }-1)}}}
ゼータ関数の負の偶数での微分係数は、
d ζ d s | s = − 2 n = ( − 1 ) n ( 2 n ) ! ζ ( 2 n + 1 ) 2 ( 2 π ) 2 n {\displaystyle {\frac {{\rm {d}}\zeta }{{\rm {d}}s}}{\Bigg \vert }_{s=-2n}={\frac {(-1)^{n}\,(2n)!\,\zeta (2n+1)}{2\,(2\pi )^{2n}}}}
である。これはゼータ関数の一様収束性から項別に微分すれば簡単に示すことができる。この公式を用いて特殊値を計算すると、
d ζ d s | s = − 2 = − ζ ( 3 ) 4 π 2 {\displaystyle {\frac {{\rm {d}}\zeta }{{\rm {d}}s}}{\Bigg \vert }_{s=-2}=-{\frac {\zeta (3)}{4\pi ^{2}}}}
d ζ d s | s = − 4 = 3 ζ ( 5 ) 4 π 4 {\displaystyle {\frac {{\rm {d}}\zeta }{{\rm {d}}s}}{\Bigg \vert }_{s=-4}={\frac {3\,\zeta (5)}{4\pi ^{4}}}}
d ζ d s | s = − 6 = − 45 ζ ( 7 ) 8 π 6 {\displaystyle {\frac {{\rm {d}}\zeta }{{\rm {d}}s}}{\Bigg \vert }_{s=-6}=-{\frac {45\,\zeta (7)}{8\pi ^{6}}}}
ζ ( 3 ) = 5 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ( n ! ) 2 n 3 ( 2 n ) ! {\displaystyle \zeta (3)={\frac {5}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}\,(n!)^{2}}{n^{3}\,(2n)!}}}
ζ ( 3 ) = 1 4 ∑ n = 1 ∞ ( n ! ) 4 ( 30 n − 11 ) n 3 [ ( 2 n ) ! ] 2 ( 2 n − 1 ) {\displaystyle \zeta (3)={\frac {1}{4}}\sum _{n=1}^{\infty }{\frac {(n!)^{4}\,(30n-11)}{n^{3}\,[(2n)!]^{2}\,(2n-1)}}}
ζ ( 3 ) = 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ( n ! ) 10 ( 205 n 2 − 160 n + 32 ) n 5 [ ( 2 n ) ! ] 5 {\displaystyle \zeta (3)={\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}\,(n!)^{10}\,(205n^{2}-160n+32)}{n^{5}\,[(2n)!]^{5}}}}