■量子化とラマヌジャンの和(その25)

integral(0,∞)x^n/[e^x-1]=Γ(n+1)ζ(n+1)

ζ(z)=Σ1/n^z

φ(z)=Σ(-1)^(n-1)/n^z

ξ(z)=Σ1/(2n-1)^z

ψ(z)=Σ(-1)^(n-1)/(2n-1)^z

と定義します。

===================================

xcotx=1-2Σζ(2m)x^2m/π^2m

xcotx=Σ(0,∞)(-1)^mB2m・(2x)^2m/(2m)!ですから

ζ(2m)=1/2・(2π)^2m/(2m)!・(-1)^m-1B2m

ζ(2)=π^2/6,ζ(4)=π^4/90,ζ(6)=π^6/645,ζ(8)=π^8/9450,ζ(10)=π^10/93555

===================================

φ(z)=(1-2^(1-z))ζ(z)

φ(1)=log2

x/sinx=1+2Σφ(2m)x^2m/π^2m

x/sinx=1+2Σ(0,∞)(1-2^2m-1)(-1)^mB2m(x)^2m/(2m)!ですから

φ(2m)=(1-2^(1-2m)(2π)^2m/2(2m)!・(-1)^m-1B2m

φ(0)=1/2,ζ(0)=-1/2

===================================

ξ(z)=(1-2^(-z))ζ(z)

xtanx=2Σξ(2m)(2x)^2m/π^2m

xtanx=Σ(0,∞)(1-2^2m)(-1)^mB2m(2x)^2m/(2m)!ですから

ξ(2m)=(1-2^(-2m))(2π)^2m/2(2m)!・(-1)^m-1B2m

ξ(0)=0

===================================

ψ(1)=Σ(-1)^(n)/(2n+1)=arctan(1)=π/4

1/cosx=2Σψ(2m+1)(2/π)^(2m+1)・x^2m

===================================