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25 Zeta and Related FunctionsRiemann Zeta Function

§25.6 Integer Arguments

Contents

§25.6(i) Function Values

25.6.1 ζ(0) =-12,
ζ(2) =π26,
ζ(4) =π490,
ζ(6) =π6945.
25.6.2 ζ(2n) =(2π)2n2(2n)!|B2n|,
n=1,2,3,.
25.6.3 ζ(-n) =-Bn+1n+1,
n=1,2,3,.
25.6.4 ζ(-2n) =0,
n=1,2,3,.
25.6.5 ζ(k+1)=1k!n1=1nk=11n1nk(n1++nk),
k=1,2,3,.
25.6.6 ζ(2k+1)=(-1)k+1(2π)2k+12(2k+1)!01B2k+1(t)cot(πt)dt,
k=1,2,3,.
25.6.7 ζ(2) =010111-xydxdy.
25.6.8 ζ(2) =3k=11k2(2kk).
25.6.9 ζ(3) =52k=1(-1)k-1k3(2kk).
25.6.10 ζ(4) =3617k=11k4(2kk).

§25.6(ii) Derivative Values

25.6.11 ζ(0)=-12ln(2π).
25.6.12 ζ′′(0)=-12(ln(2π))2+12γ2-124π2+γ1,

where γ1 is given by (25.2.5).

With c defined by (25.4.6) and n=1,2,3,,

25.6.13 (-1)kζ(k)(-2n) =2(-1)n(2π)2n+1m=0kr=0m(km)(mr)(ck-m)Γ(r)(2n+1)ζ(m-r)(2n+1),
25.6.14 (-1)kζ(k)(1-2n) =2(-1)n(2π)2nm=0kr=0m(km)(mr)(ck-m)Γ(r)(2n)ζ(m-r)(2n),
25.6.15 ζ(2n) =(-1)n+1(2π)2n2(2n)!(2nζ(1-2n)-(ψ(2n)-ln(2π))B2n).

§25.6(iii) Recursion Formulas

25.6.16 (n+12)ζ(2n)=k=1n-1ζ(2k)ζ(2n-2k),
n2.
25.6.17 (n+34)ζ(4n+2)=k=1nζ(2k)ζ(4n+2-2k),
n1.
25.6.18 (n+14)ζ(4n)+12(ζ(2n))2=k=1nζ(2k)ζ(4n-2k),
n1.
25.6.19 (m+n+32)ζ(2m+2n+2)=(k=1m+k=1n)ζ(2k)ζ(2m+2n+2-2k),
m0, n0, m+n1.
25.6.20 12(22n-1)ζ(2n)=k=1n-1(22n-2k-1)ζ(2n-2k)ζ(2k),
n2.

For related results see Basu and Apostol (2000).