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Question Number 164103 by mnjuly1970 last updated on 14/Jan/22

$$$$$$provethat$$$$$$$$\mathrm{\Omega }={\int }_0^1\ln \left(\frac{1+x}{1-x}\right).\frac{dx}{x\sqrt{1-x^2}}=\frac{{\pi }^2}{2}$$$$--m.n--$$$$$$

Answered by Lordose last updated on 14/Jan/22

$$\mathrm{\Omega }\stackrel{\mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}}}{=}2{\int }_0^1\frac{\ln \left(\frac{1}{\mathrm{x}}\right)}{{(1+\mathrm{x})}^2(\frac{1-\mathrm{x}}{1+\mathrm{x}}\sqrt{1-{\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)}^2}}\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^1\frac{\ln \left(\frac{1}{\mathrm{x}}\right)}{\sqrt{\mathrm{x}}(1-\mathrm{x})}\mathrm{dx}\stackrel{\mathrm{x}={\mathrm{x}}^2}{=}4{\int }_0^1\frac{\ln \left(\frac{1}{\mathrm{x}}\right)}{(1-{\mathrm{x}}^2)}\mathrm{dx}$$$$\mathrm{\Omega }=-4\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{2\mathrm{k}}\ln \left(\mathrm{x}\right)\mathrm{dx}\stackrel{\mathbf{IBP}}{=}4\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$\mathrm{\Omega }=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{k}+\frac{1}{2})}^2}\mathrm{dx}={\mathit{\psi }}^{\left(1\right)}\left(\frac{1}{2}\right)$$$$\mathbf{\Omega }=\frac{{\mathit{\pi }}^2}{2}▴▴▴$$

Commented by mnjuly1970 last updated on 14/Jan/22

$$mercey$$

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