■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。
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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
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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
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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
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ψ(1)=Σ(-1)^(n)/(2n+1)=arctan(1)=π/4
1/cosx=2Σψ(2m+1)(2/π)^(2m+1)・x^2m
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