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Question Number 94310 by mathmax by abdo last updated on 18/May/20

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

Answered by mathmax by abdo last updated on 19/May/20

$$\mathrm{another}\mathrm{way}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}=\frac{1}{{\mathrm{x}}^2}\mathrm{and}\mathrm{v}=\ln (1-{\mathrm{x}}^2)\Rightarrow$$$$\mathrm{I}={\left[(1-\frac{1}{\mathrm{x}})\ln (1-{\mathrm{x}}^2)\right]}_0^1-{\int }_0^1(1-\frac{1}{\mathrm{x}})\times \frac{-2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}$$$$=0+2{\int }_0^1\frac{\mathrm{x}-1}{1-{\mathrm{x}}^2}\mathrm{dx}=-2{\int }_0^1\frac{1-\mathrm{x}}{(1-\mathrm{x})(1+\mathrm{x})}\mathrm{dx}=-2{\int }_0^1\frac{\mathrm{dx}}{1+\mathrm{x}}$$$$=-2{[\ln \mid 1+\mathrm{x}\mid ]}_0^1=-2\ln \left(2\right)\Rightarrow ★\mathrm{I}=-2\ln \left(2\right)★$$

Answered by mathmax by abdo last updated on 19/May/20

$$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1-\mathrm{u})=\frac{-1}{1-\mathrm{u}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \ln (1-\mathrm{u})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$$$$(\mathrm{c}=0)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \ln (1-{\mathrm{x}}^2)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}}}{\mathrm{n}}\mathrm{and}$$$$\frac{\ln (1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}-2}}{\mathrm{n}}\Rightarrow {\int }_0^1\frac{\ln (1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{dx}$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}(2\mathrm{n}-1)}\mathrm{let}{\mathrm{s}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}(2\mathrm{k}-1)}\Rightarrow$$$$\frac{{\mathrm{s}}_{\mathrm{n}}}{2}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}(2\mathrm{k}-1)}=\sum _{\mathrm{k}=1}^{\mathrm{n}}(\frac{1}{2\mathrm{k}-1}-\frac{1}{2\mathrm{k}})$$$$=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}-1}-\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{2\mathrm{n}}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}$$$$={\mathrm{H}}_{2\mathrm{n}}-{\mathrm{H}}_{\mathrm{n}}\sim \ln \left(2\mathrm{n}\right)+\gamma +\mathrm{o}\left(\frac{1}{2\mathrm{n}}\right)-\ln \left(\mathrm{n}\right)-\gamma -{\mathrm{o}}^{{}^{\prime }}\left(\frac{1}{\mathrm{n}}\right)\to \ln \left(2\right)\Rightarrow$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{s}}_{\mathrm{n}}=2\ln \left(2\right)\Rightarrow {\int }_0^1\frac{\ln (1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{dx}=-2\ln \left(2\right)$$

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