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Question Number 130781 by EDWIN88 last updated on 28/Jan/21

$${\int }_0^1\frac{\ln (1+x+x^2+x^3+...+x^n)}{x}dx?$$

Answered by mathmax by abdo last updated on 29/Jan/21

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^1\frac{\ln (1+\mathrm{x}+{\mathrm{x}}^2+....+{\mathrm{x}}^{\mathrm{n}})}{\mathrm{x}}\mathrm{dx}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^1\frac{\ln \left(\frac{1-{\mathrm{x}}^{\mathrm{n}+1}}{1-\mathrm{x}}\right)}{\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{\ln (1-{\mathrm{x}}^{\mathrm{n}+1})}{\mathrm{x}}\mathrm{dx}-{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$${\ln }^{{}^{\prime }}(1-\mathrm{x})=-\frac{1}{1-\mathrm{x}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \ln (1-\mathrm{x})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow {\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\mathrm{dx}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}$$$$=-\frac{{\pi }^2}{6}\mathrm{and}\ln (1-{\mathrm{x}}^{\mathrm{n}+1})=-\sum _{\mathrm{p}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{x}}^{\mathrm{n}+1}\right)}^{\mathrm{p}}}{\mathrm{p}}=-\sum _{\mathrm{p}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{(\mathrm{n}+1)\mathrm{p}}}{\mathrm{p}}\Rightarrow$$$${\int }_0^1\frac{\ln (1-{\mathrm{x}}^{\mathrm{n}+1})}{\mathrm{x}}\mathrm{dx}=-\sum _{\mathrm{p}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{p}}{\int }_0^1{\mathrm{x}}^{(\mathrm{n}+1)\mathrm{p}-1}\mathrm{dx}$$$$=-\sum _{\mathrm{p}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{p}(\mathrm{n}+1)\mathrm{p}}=-\sum _{\mathrm{p}=1}^{\mathrm{\infty }}\frac{1}{(\mathrm{n}+1){\mathrm{p}}^2}=-\frac{{\pi }^2}{6(\mathrm{n}+1)}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=-\frac{1}{\mathrm{n}+1}.\frac{{\pi }^2}{6}+\frac{{\pi }^2}{6}=\frac{{\pi }^2}{6}(1-\frac{1}{\mathrm{n}+1})=\frac{\mathrm{n}{\pi }^2}{6(\mathrm{n}+1)}$$

Commented by mathmax by abdo last updated on 29/Jan/21

$$\mathrm{remark}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=\frac{{\pi }^2}{6}$$

Commented by Lordose last updated on 29/Jan/21

$$\mathrm{Very}\mathrm{nice}\mathrm{sir}$$

Answered by mnjuly1970 last updated on 29/Jan/21

$${\varphi }_n={\int }_0^1\frac{1}{x}ln(1-x^{n+1})+{li}_2\left(1\right)$$$$=-\sum _{m⩾1}\frac{1}{m}{\int }_0^1\frac{x^{nm+m}}{x}dx+\frac{{\pi }^2}{6}$$$$=-\sum _{m⩾1}\frac{1}{m^2(n+1)}+\frac{{\pi }^2}{6}$$$$=-\frac{1}{n+1}\zeta \left(2\right)+\frac{{\pi }^2}{6}$$$${\varphi }_n=-\frac{{\pi }^2}{6(n+1)}+\frac{{\pi }^2}{6}$$$$ifn\to \mathrm{\infty }then{\varphi }_n\to \frac{{\pi }^2}{6}$$

Answered by Dwaipayan Shikari last updated on 29/Jan/21

$${\int }_0^1\frac{log(1+x+x+x^3+..+x^n)}{x}dx$$$$={\int }_0^1\frac{log(1-x^{n+1})}{x}-\frac{log(1-x)}{x}dx$$$$=-\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{(n+1)k-1}}{k}+\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{k-1}}{k}dx$$$$=-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)k^2}+\frac{{\pi }^2}{6}=\frac{{\pi }^2}{6}\left(\frac{n}{n+1}\right)$$

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