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Question Number 32477 by prof Abdo imad last updated on 25/Mar/18

$$calcilate{\int }_0^1\frac{ln(1-x^2)}{x^2}dx$$

Commented by prof Abdo imad last updated on 31/Mar/18

$$for\mid t\mid <1\frac{1}{1-t}=\sum _{n=0}^{\mathrm{\infty }}{t}^n\Rightarrow -ln\mid 1-t\mid =\sum _{n=0}^{\mathrm{\infty }}\frac{t^{n+1}}{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n}=-ln(1-t)and$$$$ln(1-x^2)=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n}\Rightarrow \frac{ln(1-x^2)}{x^2}=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n-2}}{n}$$$$\Rightarrow {\int }_0^1\frac{ln(1-x^2)}{x^2}dx=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(2n-1)}$$$$letput{S}_n=\sum _{k=1}^n\frac{1}{k(2k-1)}wehave$$$$\frac{1}{2}{S}_n=\sum _{k=1}^n\frac{1}{2k(2k-1)}=\sum _{k=1}^n(\frac{1}{2k-1}-\frac{1}{2k})$$$$=\sum _{k=1}^n\frac{1}{2k-1}-\frac{1}{2}{H}_{n}but$$$$\sum _{k=1}^n\frac{1}{2k-1}=1+\frac{1}{3}+....+\frac{1}{2n-1}$$$$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2n}$$$$=H_{2n}-\frac{1}{2}{H}_n$$$$\frac{1}{2}{S}_n=H_{2n}-H_n=ln\left(2n\right)+\gamma -ln\left(n\right)-\gamma +o\left(1\right)$$$$=ln\left(\frac{2n}{n}\right)+\gamma +o\left(1\right)\to ln\left(2\right)\Rightarrow {lim}_{n\to \mathrm{\infty }}{S}_n=2ln\left(2\right)$$$$\Rightarrow {\int }_0^1\frac{ln(1-x^2)}{x^2}=-2ln\left(2\right).$$

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