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Question Number 27909 by mondodotto@gmail.com last updated on 16/Jan/18

Commented by abdo imad last updated on 16/Jan/18

$$wehave\frac{d}{dx}ln(1-x)=\frac{-1}{1-x}=-\sum _{n=0}^{\propto }{x}^n$$$$\Rightarrow ln(1-x)=-\sum _{n=0}^{\propto }\frac{x^{n+1}}{n+1}=-\sum _{n=1}^{\propto }\frac{x^n}{n}+\lambda and\lambda =f\left(0\right)=0$$$$soln(1-x)=-\sum _{n=1}^{\propto }\frac{x^n}{n}and$$$${\int }_0^{1}ln\left(x\right)ln(1-x)dx=-{\int }_0^1\left(\sum _{n=1}^{\propto }\frac{x^n}{n}\right)lnxdx$$$$=-\sum _{n=1}^{\propto }\frac{1}{n}{\int }_0^1{x}^{n}lnxdxbutintegrationbypartsgive$$$${\int }_0^1{x}^{n}lnxdx={\left[\frac{1}{n+1}{x}^{n+1}ln\left(x\right)\right]}_0^1-{\int }_0^1\frac{1}{n+1}{x}^{n+1}\frac{dx}{x}$$$$=-\frac{1}{{(n+1)}^2}\Rightarrow I=\sum _{n=1}^{\propto }\frac{1}{n{(n+1)}^2}letdecompose$$$$F\left(x\right)=\frac{1}{x{(x+1)}^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$a={lim}_{x->0}xF\left(x\right)=1$$$$c={lim}_{x->-1}{(x+1)}^{2}F\left(x\right)=-1so$$$$F\left(x\right)=\frac{1}{x}+\frac{b}{x+1}-\frac{1}{{(x+1)}^2}$$$$F\left(1\right)=\frac{1}{4}=1+\frac{b}{2}-\frac{1}{4}\Rightarrow 1+\frac{b}{2}=\frac{1}{2}\Rightarrow b=-1$$$$F\left(x\right)=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}andI={lim}_{n->\propto }{S}_n/$$$$S_n=\sum _{k=1}^n\frac{1}{k{(k+1)}^2}=\sum _{k=1}^n\frac{1}{k}-\sum _{k=1}^n\frac{1}{k+1}-\sum _{k=1}^n\frac{1}{{(k+1)}^2}$$$$but\sum _{k=1}^n=H_n^{}and\sum _{k=1}^n\frac{1}{{(k+1)}^2}=\sum _{k=1}^{n+1}\frac{1}{k^2}-1$$$$\sum _{k=1}^n\frac{1}{k+1}=\sum _{k=2}^{n+1}\frac{1}{k}=H_{n+1}-1$$$$S_n=H_n-H_{n+1}+1-\sum _{k=1}^{n+1}\frac{1}{k^2}+1$$$${lim}_{n->\propto }{H}_n-H_{n+1}=0$$$${lim}_{n->\propto }\sum _{k=1}^{n+1}\frac{1}{k^2}=\frac{{\pi }^2}{6}{lim}_{n->\propto }{S}_n=2-\frac{{\pi }^2}{6}$$$$I=2-\frac{{\pi }^2}{6}.$$

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