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Question Number 61979 by maxmathsup by imad last updated on 13/Jun/19

$$find{\int }_0^{1}ln\left(x\right)ln(1+x)dx$$

Commented by maxmathsup by imad last updated on 13/Jun/19

$$letA={\int }_0^{1}ln\left(x\right)ln(1+x)dxwehave{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n\Rightarrow$$$$ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{n+1}}{n+1}+c(c=0)\Rightarrow ln(1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{x}^n}{n}$$$$with\mid x\mid <1\Rightarrow A={\int }_0^1\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{x}^n}{n}\right)ln\left(x\right)dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{n}ln\left(x\right)dxlet{A}_n={\int }_0^1{x}^{n}ln\left(x\right)dxbyparts$$$$A_n={\left[\frac{1}{n+1}{x}^{n+1}ln\left(x\right)\right]}_{x\to 0}^1-{\int }_0^1\frac{1}{n+1}{x}^{n+1}\frac{dx}{x}=-\frac{1}{n+1}{\int }_0^1{x}^{n}dx=-\frac{1}{{(n+1)}^2}\Rightarrow$$$$A=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}\frac{1}{{(n+1)}^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n{(n+1)}^2}$$$$letdecomposeF\left(x\right)=\frac{1}{x{(x+1)}^2}\Rightarrow F\left(x\right)=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$a={lim}_{x\to 0}xF\left(x\right)=1$$$$c={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=-1\Rightarrow F\left(x\right)=\frac{1}{x}+\frac{b}{x+1}-\frac{1}{{(x+1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=1+b\Rightarrow b=-1\Rightarrow F\left(x\right)=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}\Rightarrow$$$$A=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}=-ln\left(2\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}-1=ln\left(2\right)-1$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}-1$$$$let\delta \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^x}and\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}(x>1)wehaveproved$$$$\delta \left(x\right)=(2^{1-x}-1)\xi \left(x\right)\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{2}\frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}=\frac{{\pi }^2}{12}-1\Rightarrow A=-ln\left(2\right)-ln\left(2\right)+1-\frac{{\pi }^2}{12}+1\Rightarrow$$$$A=2-2ln\left(2\right)-\frac{{\pi }^2}{12}.$$$$$$

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