Question Number 159268 by HongKing last updated on 14/Nov/21
$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{Li}_{\mathrm{3}} \left(\mathrm{1}\:-\:{xy}\right){dxdy} \\ $$$$ \\ $$
Answered by mindispower last updated on 15/Nov/21
![∫li_3 (x)=xli_3 (x)−∫li_2 (x)dx =xli_3 (x)−xli_2 (x)−∫ln(1−x)dx x(li_3 +(x)−li_2 (x))+(x−1)ln(1−x)−x+c Ω=Σ_(n≥1) ∫_0 ^1 ∫_0 ^1 (((1−xy)^n )/n^3 )dxdy =Σ_(n≥1) ∫_0 ^1 [−(((1−xy)^(n+1) )/((n+1)n^3 y))]_0 ^1 dy =Σ(1/(n^3 (n+1)))∫_0 ^1 (1/y)(1−(1−y)^(n+1) )dy =Σ_(n≥1) (H_(n+1) /(n^3 (n+1)))=Σ_(n≥1) (H_(n+1) /n^3 )−(H_(n+1) /n^2 )+(H_(n+1) /n)−(H_(n+1) /(n+1)) use euler harmonic sm Σ(H_n /n^p )=S(p)](https://www.tinkutara.com/question/Q159303.png)
$$\int{li}_{\mathrm{3}} \left({x}\right)={xli}_{\mathrm{3}} \left({x}\right)−\int{li}_{\mathrm{2}} \left({x}\right){dx} \\ $$$$={xli}_{\mathrm{3}} \left({x}\right)−{xli}_{\mathrm{2}} \left({x}\right)−\int{ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$${x}\left({li}_{\mathrm{3}} +\left({x}\right)−{li}_{\mathrm{2}} \left({x}\right)\right)+\left({x}−\mathrm{1}\right){ln}\left(\mathrm{1}−{x}\right)−{x}+{c} \\ $$$$\Omega=\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{xy}\right)^{{n}} }{{n}^{\mathrm{3}} }{dxdy} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left[−\frac{\left(\mathrm{1}−{xy}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right){n}^{\mathrm{3}} {y}}\right]_{\mathrm{0}} ^{\mathrm{1}} {dy} \\ $$$$=\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{y}}\left(\mathrm{1}−\left(\mathrm{1}−{y}\right)^{{n}+\mathrm{1}} \right){dy} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} }−\frac{{H}_{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }+\frac{{H}_{{n}+\mathrm{1}} }{{n}}−\frac{{H}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$ \\ $$$${use}\:{euler}\:{harmonic}\:{sm} \\ $$$$\Sigma\frac{{H}_{{n}} }{{n}^{{p}} }={S}\left({p}\right) \\ $$
Commented by HongKing last updated on 15/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{but}\:\mathrm{answer}.? \\ $$