Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 119157 by mathdave last updated on 22/Oct/20

Answered by mindispower last updated on 23/Oct/20

$$x=\frac{1}{t}\Rightarrow$$$$={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(\frac{1}{t}\right)}{1+\frac{1}{t}}.\frac{dt}{t^2.\frac{1}{\sqrt{t}}}={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(\frac{1}{t}\right)}{t+1}.\frac{dt}{\sqrt{t}}$$$$2{\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(x\right)}{1+x}.\frac{dx}{\sqrt{x}}={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(x\right)+{tan}^{-}\left(\frac{1}{x}\right)}{x+1}.\frac{dx}{\sqrt{x}}$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}2.\frac{d\sqrt{x}}{(1+{\left(\sqrt{x}\right)}^2)}=\pi {\left[arctan\left(\sqrt{x}\right)\right]}_0^{\mathrm{\infty }}=\frac{{\pi }^2}{2}$$$${\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(x\right)}{1+x}.\frac{dx}{\sqrt{x}}=\frac{{\pi }^2}{4}$$$$icompleteanswer$$$$2{\int }_0^{\mathrm{\infty }}\frac{{tan}^{-}\left(x\right)}{1+x}.\frac{dx}{\sqrt{x}}=\frac{\pi }{4}inprevious$$

Commented by mathdave last updated on 22/Oct/20

Commented by mathmax by abdo last updated on 22/Oct/20

$$\mathrm{let}\mathrm{try}\mathrm{parametric}\mathrm{method}\mathrm{we}\mathrm{have}$$$$\mathrm{I}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{\mathrm{t}(1+{\mathrm{t}}^2)}\left(2\mathrm{t}\right)\mathrm{dt}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{at}}^2\right)}{{\mathrm{t}}^2+1}\mathrm{dt}(\mathrm{a}>0)\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{t}}^2}{(1+{\mathrm{a}}^2{\mathrm{t}}^4)({\mathrm{t}}^2+1)}\mathrm{dt}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{({\mathrm{a}}^2{\mathrm{z}}^4+1)({\mathrm{z}}^2+1)}$$$$\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2({\mathrm{z}}^4+\frac{1}{{\mathrm{a}}^2})({\mathrm{z}}^2+1)}=\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2({\mathrm{z}}^2-\frac{\mathrm{i}}{\mathrm{a}})({\mathrm{z}}^2+\frac{\mathrm{i}}{\mathrm{a}})({\mathrm{z}}^2+1)}$$$$=\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2(\mathrm{z}-\sqrt{\frac{\mathrm{i}}{\mathrm{a}}})(\mathrm{z}+\sqrt{\frac{\mathrm{i}}{\mathrm{a}}})(\mathrm{z}-\sqrt{\frac{-\mathrm{i}}{\mathrm{a}}})(\mathrm{z}+\sqrt{\frac{-\mathrm{i}}{\mathrm{a}}})(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$=\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2(\mathrm{z}-\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$\mathrm{residus}\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,\mathrm{i})\}$$$$\mathrm{Res}(\phi ,\mathrm{i})=\frac{-1}{2\mathrm{i}({\mathrm{a}}^2+1)}$$$$\mathrm{Res}(\phi ,\frac{1}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{1}{\mathrm{a}}\frac{\mathrm{i}}{{\mathrm{a}}^2\left(\frac{2}{\sqrt{\mathrm{a}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\right)\left(\frac{2\mathrm{i}}{\mathrm{a}}\right)(\frac{\mathrm{i}}{\mathrm{a}}+1)}=\frac{{\mathrm{a}}^2\sqrt{\mathrm{a}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}}{{4\mathrm{a}}^3(\mathrm{a}+\mathrm{i})}$$$$...\mathrm{be}\mathrm{continued}...$$

Commented by mindispower last updated on 23/Oct/20

$$yesthankyousir$$

Commented by mathmax by abdo last updated on 23/Oct/20

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

Answered by mathmax by abdo last updated on 22/Oct/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{\sqrt{\mathrm{x}}(1+\mathrm{x})}\mathrm{dx}\mathrm{changement}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{\mathrm{t}(1+{\mathrm{t}}^2)}\left(2\mathrm{t}\right)\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2}\mathrm{dt}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{\arctan \left({\mathrm{z}}^2\right)}{{\mathrm{z}}^2+1}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})=2\mathrm{i}\pi \times \frac{\mid \arctan (-1)\mid }{2\mathrm{i}}$$$$=\pi \times \frac{\pi }{4}=\frac{{\pi }^2}{4}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com