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Question Number 151230 by mnjuly1970 last updated on 19/Aug/21

$$$$$$prove:$$$${\int }_0^{\mathrm{\infty }}\frac{ln(1+x^2)}{x^2(1+x^2)}dx=\pi ln\left(\frac{e}{2}\right)..$$

Commented by Lordose last updated on 19/Aug/21

Commented by Lordose last updated on 19/Aug/21

Answered by Lordose last updated on 19/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}\stackrel{\mathbf{P}.\mathbf{F}}{=}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{dx}-{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{\Phi }-\mathrm{\Psi }$$$$\mathrm{\Phi }\stackrel{\mathbf{IBP}}{=}-\frac{\ln (1+{\mathrm{x}}^2)}{\mathrm{x}}{\mid }_{0^{+}}^1+{\int }_0^1\frac{2}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{\Phi }=-\log \left(2\right)+\frac{\mathit{\pi }}{2}$$$$\mathrm{\Psi }={\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}\stackrel{\mathrm{x}=\tan \left(\mathrm{y}\right)}{=}-2{\int }_0^{\frac{\mathit{\pi }}{4}}\ln \left(\cos \left(\mathrm{y}\right)\right)\mathrm{dy}$$$$\mathrm{\Psi }=2{\int }_0^{\frac{\mathit{\pi }}{4}}(\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\cos \left(2\mathrm{ky}\right)+\log \left(2\right))\mathrm{dy}=2\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}{\int }_0^{\frac{\mathit{\pi }}{4}}\cos \left(2\mathrm{ky}\right)\mathrm{dy}+2\log \left(2\right){\int }_0^{\frac{\mathit{\pi }}{4}}1\mathrm{dy}$$$$\mathrm{\Psi }=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{k}}\sin \left(\frac{\mathit{\pi }\mathrm{k}}{2}\right)}{{\mathrm{k}}^2}+\frac{\mathit{\pi }\log \left(2\right)}{2}$$$$\mathrm{\Psi }=\frac{\mathit{\pi }}{2}\log \left(2\right)-\mathbf{G}$$$$\mathrm{\Omega }=\frac{\mathit{\pi }}{2}-\log \left(2\right)-\frac{\mathit{\pi }}{2}\log \left(2\right)+\mathbf{G}$$$$\mathbf{\Omega }=\frac{\mathit{\pi }}{2}-\mathbf{log}\left(2\right)(1+\frac{\mathit{\pi }}{2})+\mathbf{G}$$$$\mathbf{G}={\mathrm{Catalan}}^{\prime }\mathrm{s}\mathrm{Constant}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{k}}\sin \left(\frac{\mathit{\pi }\mathrm{k}}{2}\right)}{{\mathrm{k}}^2}$$

Commented by mnjuly1970 last updated on 19/Aug/21

$$thankyousomuchmrlordose$$$$mymistake\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{ln(1+x^2)}{x^2(1+x^2)}dx$$$$grateful...$$

Answered by qaz last updated on 19/Aug/21

$${\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}$$$$={\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{dx}-{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-\frac{\ln (1+{\mathrm{x}}^2)}{\mathrm{x}}{\mid }_0^1+{\int }_0^1\frac{2}{1+{\mathrm{x}}^2}\mathrm{dx}+2{\int }_0^{\pi /4}\mathrm{lncos}\mathrm{xdx}$$$$=-\ln 2+\frac{\pi }{2}+{\int }_0^{\pi /2}\mathrm{lncos}\frac{\mathrm{x}}{2}\mathrm{dx}$$$$=-\ln 2+\frac{\pi }{2}+\frac{1}{2}{\int }_0^{\pi /2}\ln \frac{1+\cos \mathrm{x}}{2}\mathrm{dx}$$$$=-\ln 2+\frac{\pi }{2}+\frac{1}{2}{\int }_0^{\pi /2}\ln \frac{\sin \mathrm{x}}{\tan \frac{\mathrm{x}}{2}}\mathrm{dx}-\frac{1}{2}{\int }_0^{\pi /2}\ln 2\mathrm{dx}$$$$=-\ln 2+\frac{\pi }{2}-\frac{\pi }{2}\ln 2-{\int }_0^{\pi /4}\mathrm{lntan}\mathrm{xdx}$$$$=-\ln 2+\frac{\pi }{2}-\frac{\pi }{2}\ln 2-{\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-\ln 2+\frac{\pi }{2}-\frac{\pi }{2}\ln 2-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\mathrm{lnxdx}$$$$=-\ln 2+\frac{\pi }{2}-\frac{\pi }{2}\ln 2-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{(2\mathrm{n}+1)}^2}$$$$=-\ln 2+\frac{\pi }{2}-\frac{\pi }{2}\ln 2+\mathrm{G}$$

Answered by Lordose last updated on 19/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}\stackrel{\mathrm{x}=\tan \left(\mathrm{y}\right)}{=}-2{\int }_0^{\frac{\mathit{\pi }}{2}}{\cot }^2\mathrm{yln}\left(\cos \left(\mathrm{y}\right)\right)\mathrm{dy}$$$$\mathrm{\Omega }=-2\frac{\partial }{\partial \mathrm{a}}{\mid }_{\mathrm{a}=2}{\int }_0^{\frac{\mathit{\pi }}{2}}{\cos }^{\mathrm{a}}\left(\mathrm{y}\right){\sin }^{-2}\left(\mathrm{y}\right)\mathrm{dy}$$$$\mathrm{\Omega }=-\frac{\partial }{\partial \mathrm{a}}{\mid }_{\mathrm{a}=2}\left(\mathbf{B}(\frac{\mathrm{a}+1}{2},-\frac{1}{2})\right)=-\frac{1}{2}({\mathit{\psi }}^{\left(0\right)}\left(\frac{\mathrm{a}+1}{2}\right)-{\mathit{\psi }}^{\left(0\right)}\left(\frac{\mathrm{a}}{2}\right))\mathbf{B}(\frac{\mathrm{a}+1}{2},-\frac{1}{2})$$$$\mathbf{\Omega }=\frac{1}{2}\mathit{\pi }(\mathit{\gamma }+{\mathit{\psi }}^{\left(0\right)}\left(\frac{3}{2}\right))$$

Answered by qaz last updated on 19/Aug/21

$${\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}$$$$={\int }_0^{\pi /2}\frac{\mathrm{lnsec}{}^2\mathrm{x}}{\tan {}^2\mathrm{x}}\mathrm{dx}$$$$=-2{\int }_0^{\pi /2}\cot {}^2\mathrm{xlncos}\mathrm{xdx}$$$$=2{\int }_0^{\pi /2}(1-{\csc }^2\mathrm{x})\mathrm{lncos}\mathrm{xdx}$$$$=-\pi \ln 2-2{\int }_0^{\pi /2}{\csc }^2\mathrm{xlncos}\mathrm{xdx}$$$$=-\pi \ln 2+2\cot \mathrm{xlncos}\mathrm{x}{\mid }_0^{\pi /2}+2{\int }_0^{\pi /2}\mathrm{dx}$$$$=-\pi \ln 2+\pi$$

Commented by mnjuly1970 last updated on 19/Aug/21

$$thanksalot..$$

Commented by peter frank last updated on 19/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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