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Question Number 167267 by mnjuly1970 last updated on 11/Mar/22

Answered by mindispower last updated on 15/Mar/22

IBP⇒Ω=[ln(1+x)Li_2 (1−x)]_0 ^1 −∫_0 ^1 ((ln(1+x)ln(x))/(1−x))dx =−∫_0 ^1 ((ln(x)ln(1+x))/(1−x))dx ln(1+x)=Σ_(j≥0) (((−1)^j )/(j+1))x^(j+1) (1/(1−x))=Σ_(k≥0) x^k ((ln(1+x))/(1−x))=Σ_(n≥0) Σ_(m=0) ^n (((−1)^m )/(m+1))x^(n+1) =S Σ_(m=0) ^n (((−1)^m )/(m+1))=−H_(n+1) ^− S=−Σ_(n≥0) H_(n+1) ^− x^(n+1) =−Σ_(n≥1) H_n x^n Ω=−Σ_(n≥1) ∫_0 ^1 H_n ^− x^n ln(x)dx =Σ_(n≥1) (H_n ^− /((n+1)^2 ))=Σ_(n≥1) ((H_(n+1) ^− +(((−1)^n )/(n+1)))/((n+1)^2 )) =Σ_(n≥1) (H_(n+1) ^− /((n+1)^2 ))+Σ_(n≥1) (((−1)^n )/((n+1)^3 )) =Σ_(n≥1) (H_n ^− /n^2 )−η(3)=Ω Σ_(n≥1) (H_n ^− /n^(q+1) )=ζ(q)ln(2)−(q/2)ζ(q+1)+η(q+1)+Σ_(k=1) ^q η(k)η(q+1−k) Euler formula For harominc sum Ω=ζ(2)ln(2)−ζ(3)+η(3)+η(1)η(2)−η(3) =ζ(2)ln(2)−ζ(3)+(1/2)ln(2)ζ(2) =(3/2_ )ln(2).(π^2 /6)−ζ(3) =(π^2 /4)ln(2)−ζ(3)

IBP⇒Ω=[ln(1+x)Li2(1−x)]01−∫01ln(1+x)ln(x)1−xdx=−∫01ln(x)ln(1+x)1−xdxln(1+x)=∑j⩾0(−1)jj+1xj+111−x=∑k⩾0xkln(1+x)1−x=∑n⩾0∑nm=0(−1)mm+1xn+1=S∑nm=0(−1)mm+1=−H−n+1S=−∑n⩾0H−n+1xn+1=−∑n⩾1HnxnΩ=−∑n⩾1∫01H−nxnln(x)dx=∑n⩾1H−n(n+1)2=∑n⩾1H−n+1+(−1)nn+1(n+1)2=∑n⩾1H−n+1(n+1)2+∑n⩾1(−1)n(n+1)3=∑n⩾1H−nn2−η(3)=Ω∑n⩾1H−nnq+1=ζ(q)ln(2)−q2ζ(q+1)+η(q+1)+∑qk=1η(k)η(q+1−k)EulerformulaForharomincsumΩ=ζ(2)ln(2)−ζ(3)+η(3)+η(1)η(2)−η(3)=ζ(2)ln(2)−ζ(3)+12ln(2)ζ(2)=32ln(2).π26−ζ(3)=π24ln(2)−ζ(3)

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