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Question Number 113190 by Dwaipayan Shikari last updated on 11/Sep/20

$${\int }_0^1\frac{logx}{x-1}dx$$

Commented by Dwaipayan Shikari last updated on 11/Sep/20

$$-{\int }_0^1\frac{logx}{1-x}dx=-{\int }_0^1\frac{log(1-x)}{x}dx={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n-1}}{n}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{n-1}}{n}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

Commented by Dwaipayan Shikari last updated on 11/Sep/20

$$Isitright?$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by mathdave last updated on 12/Sep/20

$$solution$$$$letI=-{\int }_0^1\frac{\ln x}{1-x}dx=-{\int }_0^1\underset{k=0}{\sum ^{\mathrm{\infty }}}{x}^k\ln xdx$$$$\frac{\partial }{\partial a}{\mid }_{a=1}I\left(a\right)=-\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\int }_0^1{x}^k.x^{a-1}dx=-\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\int }_0^1{x}^{k+a-1}dx$$$$I\left(a\right)=-\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}\left[\frac{1}{k+a}\right]$$$$I\left(1\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k+1)}^2}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}=\frac{{\pi }^2}{6}$$$$\because {\int }_0^1\frac{\ln x}{x-1}dx=\frac{{\pi }^2}{6}$$$$$$

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