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Question Number 163682 by mnjuly1970 last updated on 09/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1{\left({\mathrm{Li}}_2^{}\left(x\right)\right)}^{2}dx=?◼m.n$$$$------$$$$$$

Answered by Kamel last updated on 09/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1{\left({\mathrm{Li}}_2^{}\left(x\right)\right)}^{2}dx=?◼m.n$$$$------$$$$\mathrm{\Omega }\stackrel{IBP}{=}\frac{{\pi }^4}{36}+{\int }_0^{1}Ln(1-x){Li}_2\left(x\right)dx$$$$\int Ln(1-x)dx=-(1-x)Ln(1-x)-x$$$$\therefore \mathrm{\Omega }\stackrel{IBP}{=}\frac{{\pi }^4}{36}-\frac{{\pi }^2}{3}-2{\int }_0^1\frac{(1-x){Ln}^2(1-x)}{x}dx-2{\int }_0^{1}Ln(1-x)dx$$$$=\frac{{\pi }^4}{36}-\frac{{\pi }^2}{3}-4\zeta \left(3\right)+6$$

Commented by mnjuly1970 last updated on 09/Jan/22

$$verynicesirkamel$$

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