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Question Number 150062 by 0731619 last updated on 09/Aug/21

Answered by Ar Brandon last updated on 09/Aug/21

$${li}_2\left(x\right)\underset{function}{\stackrel{dilogarithm}{=}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n^2}=-{\int }_0^x\frac{ln(1-t)}{t}dt$$$$\mathrm{\Omega }={\int }_0^1{\mathrm{Li}}_2\left(x\right)dx={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n^2}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1})$$$$=\zeta \left(2\right)-H_n+H_n-1=\frac{{\pi }^2}{6}-1$$

Answered by mindispower last updated on 09/Aug/21

$${li}_2\left(x\right)=-{\int }_0^x\frac{ln(1-t)}{t}dt$$$$\int {li}_2\left(x\right)dxbypart={xli}_2\left(x\right)+\int ln(1-x)dx={xli}_2\left(x\right)$$$$+(x-1)ln(1-x)+x+c$$

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