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Question Number 206340 by mnjuly1970 last updated on 12/Apr/24

$$$$$${\int }_0^1\frac{ln(1-x)ln(1+x)}{x}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\mathrm{\Omega }}_n$$$$$$$$find:\underset{n=1}{\sum ^{\mathrm{\infty }}}n{\mathrm{\Omega }}_n=?$$

Answered by sniper237 last updated on 12/Apr/24

$$\frac{1}{1+x}=\sum _{n⩾0}{(-x)}^n\Rightarrow ln(1+x)=\sum _{n⩾1}\frac{{(-1)}^n{x}^n}{n}$$$$So{\mathrm{\Omega }}_n=\frac{{(-1)}^n}{n}{\int }_0^1{x}^{n-1}ln(1-x)dx$$$$V=ln(1-x)V{}^{\prime }=\frac{1}{x-1}U{}^{\prime }=x^{n-1}U={(-1)}^n\frac{x^n-1}{n}$$$$n{\mathrm{\Omega }}_n={\left[UV\right]}_0^{\lambda \to 1}-\frac{{(-1)}^n}{n}{\int }_0^1(1+x+...+x^{n-1})dx$$$$=\underset{k=1}{\sum ^n}\frac{{(-1)}^{n+1}}{nk}$$$$$$

Answered by Mathspace last updated on 12/Apr/24

$${ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n$$$$\Rightarrow ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{n+1}}{n+1}+c(c=0)$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^n$$$$\Rightarrow \frac{ln(1+x)}{x}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^{n-1}$$$$\Rightarrow {\int }_0^1\frac{ln(1-x)ln(1+x)}{x}dx$$$$={\int }_0^1\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^{n-1}\right)lnxdx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{\int }_0^1{x}^{n-1}lnxdx$$$$byparts{u}_n={\int }_0^1{x}^{n-1}lnx$$$$={\left[\frac{x^n}{n}lnx\right]}_0^1-{\int }_0^1\frac{x^n}{n}\frac{dx}{x}$$$$=-\frac{1}{n}{\int }_0^1{x}^{n-1}dx=-\frac{1}{n}{\left[\frac{x^n}{n}\right]}_0^1$$$$=-\frac{1}{n^2}\Rightarrow$$$$I=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}(-\frac{1}{n^2})$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^3}\Rightarrow {\mho }_n=\frac{{(-1)}^{n-1}}{n^3}$$$$and\sum _{n=1}^{\mathrm{\infty }}n{\mho }_n=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}$$$$=-\delta \left(2\right)=-(2^{1-2}-1)\xi \left(2\right)$$$$=\frac{1}{2}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}n{\mho }_n=\frac{{\pi }^2}{12}$$

Commented by sniper237 last updated on 13/Apr/24

$$u_n\stackrel{?}{=}{\int }_0^1{x}^{n-1}ln(1-x)dx\stackrel{\to 1-x}{=}{\int }_0^1{(1-x)}^{n-1}lnxdx$$

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