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)](https://www.tinkutara.com/question/Q167397.png)
$${IBP}\Rightarrow\Omega=\left[{ln}\left(\mathrm{1}+{x}\right){Li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){ln}\left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${ln}\left(\mathrm{1}+{x}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}+\mathrm{1}}{x}^{{j}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{k}\geqslant\mathrm{0}} {\sum}{x}^{{k}} \\ $$$$\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}−{x}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\underset{{m}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}+\mathrm{1}}{x}^{{n}+\mathrm{1}} ={S} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}+\mathrm{1}}=−\overset{−} {{H}}_{{n}+\mathrm{1}} \\ $$$${S}=−\underset{{n}\geqslant\mathrm{0}} {\sum}\overset{−} {{H}}_{{n}+\mathrm{1}} {x}^{{n}+\mathrm{1}} =−\underset{{n}\geqslant\mathrm{1}} {\sum}{H}_{{n}} {x}^{{n}} \\ $$$$\Omega=−\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \overset{−} {{H}}_{{n}} {x}^{{n}} {ln}\left({x}\right){dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\overset{−} {{H}}_{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\overset{−} {{H}}_{{n}+\mathrm{1}} +\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\overset{−} {{H}}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\overset{−} {{H}}_{{n}} }{{n}^{\mathrm{2}} }−\eta\left(\mathrm{3}\right)=\Omega \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\overset{−} {{H}}_{{n}} }{{n}^{{q}+\mathrm{1}} }=\zeta\left({q}\right){ln}\left(\mathrm{2}\right)−\frac{{q}}{\mathrm{2}}\zeta\left({q}+\mathrm{1}\right)+\eta\left({q}+\mathrm{1}\right)+\underset{{k}=\mathrm{1}} {\overset{{q}} {\sum}}\eta\left({k}\right)\eta\left({q}+\mathrm{1}−{k}\right) \\ $$$${Euler}\:{formula}\:{For}\:{harominc}\:{sum} \\ $$$$\Omega=\zeta\left(\mathrm{2}\right){ln}\left(\mathrm{2}\right)−\zeta\left(\mathrm{3}\right)+\eta\left(\mathrm{3}\right)+\eta\left(\mathrm{1}\right)\eta\left(\mathrm{2}\right)−\eta\left(\mathrm{3}\right) \\ $$$$=\zeta\left(\mathrm{2}\right){ln}\left(\mathrm{2}\right)−\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}_{} }{ln}\left(\mathrm{2}\right).\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\zeta\left(\mathrm{3}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$