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Question Number 45232 by maxmathsup by imad last updated on 10/Oct/18

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

Commented by maxmathsup by imad last updated on 11/Oct/18

$$letI={\int }_0^{1}ln\left(x\right)ln(1+x)dxwehavefor\mid x\mid <1$$$${ln}^{{}^{\prime }}(1+x)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n\Rightarrow ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^{n+1}+\lambda$$$$x=0\Rightarrow \lambda =0\Rightarrow ln(1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^n\Rightarrow$$$$I=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{n}ln\left(x\right)dxbyparts$$$$A_n={\int }_0^1{x}^{n}ln\left(x\right)dx={\left[\frac{1}{n+1}{x}^{n+1}ln\left(x\right)\right]}_{x\to 0}^1-{\int }_0^1\frac{1}{(n+1)}{x}^{n}dx$$$$=-\frac{1}{{(n+1)}^2}\Rightarrow I=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n{(n+1)}^2}letdecomposeF\left(x\right)=\frac{1}{x{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$a={lim}_{x\to 0}xF\left(x\right)=1$$$$c={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=-1\Rightarrow F\left(x\right)=\frac{1}{x}+\frac{b}{x+1}-\frac{1}{{(x+1)}^2}$$$$F\left(1\right)=\frac{1}{4}=1+\frac{b}{2}-\frac{1}{4}\Rightarrow 1=4+2b-1=3+2b\Rightarrow 2b=-2\Rightarrow b=-1\Rightarrow$$$$F\left(x\right)=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}\Rightarrow$$$$I=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}but$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}=-ln\left(2\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}=ln\left(2\right)-1$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(n+1)}^2}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}+1letfind\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}$$$$wehave\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^2}-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$=\frac{1}{4}\frac{{\pi }^2}{6}-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}but\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{3}{4}\mathrm{\Sigma }\frac{1}{n^2}=\frac{3}{4}\frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=\frac{{\pi }^2}{24}-\frac{{\pi }^2}{8}=-\frac{{\pi }^2}{12}\Rightarrow$$$${\int }_0^{1}ln\left(x\right)ln(1+x)dx=-ln\left(2\right)-ln\left(2\right)+1-(-\frac{{\pi }^2}{12}+1)$$$$=\frac{{\pi }^2}{12}-2ln\left(2\right).$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

$${\int }_0^{1}ln(1+x)\times ln\left(x\right)dx\approx -0.21\left(approx\right)$$

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