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Question Number 172764 by Mathematification last updated on 01/Jul/22

$${\int }_0^{\mathrm{\infty }}{\left(\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\right)}^2\mathrm{dx}=?$$

Answered by Mathspace last updated on 01/Jul/22

$$\mathrm{{\rm Y}}={\int }_0^{\mathrm{\infty }}\frac{{ln}^2(1-x)}{x^2}dx(u^{{}^{\prime }}=\frac{1}{x^2}andv={ln}^2(1+x))$$$$\mathrm{{\rm Y}}={[-\frac{1}{x}{ln}^2(1-x)]}_0^{+\mathrm{\infty }}$$$$+{\int }_0^{\mathrm{\infty }}\frac{1}{x}(-2)ln(1-x)dx$$$$=-2{\int }_0^{\mathrm{\infty }}\frac{ln(1-x)}{x}dx$$$${ln}^{{}^{\prime }}(1-x)=-\frac{1}{1-x}=-\sum _{n=0}^{\mathrm{\infty }}{x}^n\Rightarrow$$$$ln(1-x)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}+c(c=0)$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}\Rightarrow -\frac{ln(1-x)}{x}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}}{n}\Rightarrow$$$$\mathrm{{\rm Y}}=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^1{x}^{n-1}dx$$$$=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=2.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{3}$$$$\mathrm{{\rm Y}}=\frac{{\pi }^2}{3}$$

Commented by Mathematification last updated on 08/Jul/22

@Mathspace Please check the third and fourth line again sir.

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