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Question Number 117223 by mnjuly1970 last updated on 10/Oct/20

$$...nicemathematics...$$$$proof::$$$$\mathrm{\Omega }=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}{\left[{tan}^{-1}\left(\frac{1}{x}\right)\right]}^{2}dx=ln\left(2\right)$$$$...m.n.1970...$$

Answered by mathmax by abdo last updated on 10/Oct/20

$$=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}{\arctan }^2\left(\frac{1}{\mathrm{x}}\right)\mathrm{dx}\mathrm{by}\mathrm{parts}$$$$\pi \mathrm{I}={\left[\mathrm{x}{\arctan }^2\left(\frac{1}{\mathrm{x}}\right)\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\mathrm{x}.2\arctan \left(\frac{1}{\mathrm{x}}\right).\times \frac{-\frac{1}{{\mathrm{x}}^2}}{1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\mathrm{x}\arctan \left(\frac{1}{\mathrm{x}}\right)\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\arctan \left(\frac{1}{\mathrm{x}}\right)\mathrm{dx}$$$$={\left[\ln (1+{\mathrm{x}}^2)\arctan \left(\frac{1}{\mathrm{x}}\right)\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\ln (1+{\mathrm{x}}^2).\frac{-\frac{1}{{\mathrm{x}}^2}}{1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\tan \theta }{\int }_0^{\frac{\pi }{2}}\frac{\ln (1+{\tan }^2\theta )}{1+{\tan }^2\theta }(1+{\tan }^2\theta )\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\ln \left(\frac{1}{{\cos }^2\theta }\right)\mathrm{d}\theta =-2{\int }_0^{\frac{\pi }{2}}\ln \left(\cos \theta \right)\mathrm{d}\theta =-2(-\frac{\pi }{2}\ln 2)=\pi \ln \left(2\right)$$$$\Rightarrow \pi \mathrm{I}=\pi \ln \left(2\right)\Rightarrow \mathrm{I}=\ln \left(2\right)$$

Commented by mnjuly1970 last updated on 10/Oct/20

$$thankyouverymuchmr$$$$max.nicesolutionas$$$$always..$$

Commented by mathmax by abdo last updated on 10/Oct/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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