■もうひとつのバーゼル問題(その20)

[0]∫(0,1)logx/(x−1)dx=ζ(2)=π^2/6

[1]∫(0,1)(x−1)/logxdx=log2

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[2]∫(0,1){(n−k)x^m+(k−m)x^n+(m−n)x^k}/(logx)^2dx=(m+1)(n−k)log(m+1)+(n+1)(k−m)log(n+1)+(k+1)(m−n)log(k+1)

[3]m=2,n=1,k=0

  ∫(0,1)(x−1)^2/(logx)^2dx=log(27/16)

[4]m=3,n=1,k=0

  ∫(0,1)(x−1)^2(x+2)/(logx)^2dx=log4

[5]m=3,n=2,k=0

  ∫(0,1)(x−1)^2(2x+1)/(logx)^2dx=log(2^16/3^2)

[6]m=3,n=2,k=1

  ∫(0,1)x(x−1)^2/(logx)^2dx=log(27/16)

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 一般に,

[7]f(x)=Σa(k)x^k,f(1)=0,f’(0)=1に対して

  ∫(0,1)f(x)/(logx)^2dx=log(Π(k+1)^(k+1)a(k))

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