Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 139283 by qaz last updated on 25/Apr/21

$${\int }_0^{\mathrm{\infty }}\frac{{ln}^{2}x}{x^2+a^2}dx=\frac{\pi }{8a}({\pi }^2+4{ln}^{2}a),a>0$$

Answered by Ar Brandon last updated on 10/Aug/21

$$\mho ={\int }_0^{\mathrm{\infty }}\frac{{\ln }^2\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx},\mathrm{x}=\mathrm{atan}\theta$$$$=\frac{1}{\mathrm{a}}{\int }_0^{\frac{\pi }{2}}{\ln }^2\left(\mathrm{atan}\theta \right)\mathrm{d}\theta$$$$=\frac{1}{\mathrm{a}}{\int }_0^{\frac{\pi }{2}}({\ln }^2\mathrm{a}+2\ln \left(\mathrm{a}\right)\ln \left(\tan \theta \right)+{\ln }^2\left(\tan \theta \right))\mathrm{d}\theta$$$$=\frac{\pi }{2\mathrm{a}}{\ln }^2\mathrm{a}+\frac{1}{\mathrm{a}}{\int }_0^{\frac{\pi }{2}}{\ln }^2\left(\tan \theta \right)\mathrm{d}\theta$$$$\mathrm{f}\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\theta \mathrm{d}\theta =\frac{\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)}{2},\mathrm{f}\left(0\right)=\frac{\pi }{2}$$$$\mathrm{lnf}\left(\alpha \right)=\ln \mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)+\ln \mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)+\ln \left(\frac{1}{2}\right)$$$$\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{\mathrm{f}\left(\alpha \right)}=\frac{1}{2}(\psi \left(\frac{\alpha +1}{2}\right)-\psi \left(\frac{1-\alpha }{2}\right))$$$$\mathrm{f}{}^{\prime }\left(0\right)=\frac{\pi }{4}(\psi \left(\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right))=0$$$$\mathrm{f}{}^{″}\left(\alpha \right)=\frac{\mathrm{f}\left(\alpha \right)}{4}({\psi }^{\prime }\left(\frac{\alpha +1}{2}\right)+{\psi }^{\prime }\left(\frac{1-\alpha }{2}\right))+\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{2}(\psi \left(\frac{\alpha +1}{2}\right)+\psi \left(\frac{1-\alpha }{2}\right))$$$$\mathrm{f}{}^{″}\left(0\right)=\frac{\pi }{8}\left(\frac{{\pi }^2}{{\sin }^2\left(\frac{\pi }{2}\right)}\right)+0=\frac{{\pi }^3}{8}={\int }_0^{\frac{\pi }{2}}{\ln }^2\left(\tan \theta \right)\mathrm{d}\theta$$$$\mho =\frac{\pi }{2\mathrm{a}}{\ln }^2\mathrm{a}+\frac{{\pi }^3}{8\mathrm{a}}=\frac{\pi }{8\mathrm{a}}({\pi }^2+{4\ln }^2\mathrm{a})$$

Commented by qaz last updated on 25/Apr/21

$$iexpectsomeonewhocouldshowmeuseResidueTheorem,butstillthankyouSir.$$

Commented by Ar Brandon last updated on 25/Apr/21

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{welcome},\mathrm{Sir}$$

Answered by mathmax by abdo last updated on 26/Apr/21

$$\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{t}}\mathrm{lnx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\Rightarrow \mathrm{f}\left(\mathrm{t}\right){=}_{\mathrm{x}=\mathrm{ay}}{\int }_0^{\mathrm{\infty }}\frac{{\left(\mathrm{ay}\right)}^{\mathrm{t}}\ln \left(\mathrm{ay}\right)}{{\mathrm{a}}^2({\mathrm{y}}^2+1)}\mathrm{ady}$$$$={\mathrm{a}}^{\mathrm{t}-1}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\mathrm{t}}(\mathrm{lna}+\mathrm{lny})}{{\mathrm{y}}^2+1}\mathrm{dy}={\mathrm{a}}^{\mathrm{t}-1}\mathrm{lna}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\mathrm{t}}}{{\mathrm{y}}^2+1}\mathrm{dy}+{\mathrm{a}}^{\mathrm{t}-1}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\mathrm{t}}}{{\mathrm{y}}^2+1}\mathrm{lny}\mathrm{dy}$$$$\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\mathrm{t}}}{{\mathrm{y}}^2+1}\mathrm{dy}{=}_{{\mathrm{y}}^2=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\mathrm{t}}{2}}}{\mathrm{u}+1}\frac{\mathrm{du}}{2\sqrt{\mathrm{u}}}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\mathrm{t}}{2}+\frac{1}{2}-1}}{\mathrm{u}+1}\mathrm{du}=\frac{1}{2}\times \frac{\pi }{\sin \left(\pi \left(\frac{\mathrm{t}+1}{2}\right)\right)}=\frac{\pi }{2\sin (\frac{\pi \mathrm{t}}{2}+\frac{\pi }{2})}$$$$=\frac{\pi }{2\cos \left(\frac{\pi \mathrm{t}}{2}\right)}\mathrm{let}\mathrm{find}\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{\mathrm{t}}}{{\mathrm{y}}^2+1}\mathrm{lny}\mathrm{dy}$$$$\mathrm{\Phi }=-\frac{1}{2}\mathrm{Re}\left(\mathrm{\Sigma }\mathrm{Res}\left(\mathrm{w}{\mathrm{z}}_{\mathrm{i}}\right)\right)\mathrm{with}\mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^{\mathrm{t}}}{{\mathrm{z}}^2+1}{\ln }^2\mathrm{z}=\frac{{\mathrm{z}}^{\mathrm{t}}{\ln }^2\mathrm{z}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$\mathrm{Res}(\mathrm{w},\mathrm{i})=\frac{{\mathrm{i}}^{\mathrm{t}}{\ln }^2\left(\mathrm{i}\right)}{2\mathrm{i}}=\frac{{\mathrm{e}}^{\frac{\mathrm{i}\pi \mathrm{t}}{2}}{\left(\frac{\mathrm{i}\pi }{2}\right)}^2}{2\mathrm{i}}=-\frac{{\pi }^2}{8\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi \mathrm{t}}{2}}=\frac{\mathrm{i}{\pi }^2}{8}{\mathrm{e}}^{\frac{\mathrm{i}\pi \mathrm{t}}{2}}$$$$\mathrm{Res}(\mathrm{w},-\mathrm{i})=-\frac{1}{2\mathrm{i}}{(-\mathrm{i})}^{\mathrm{t}}{\ln }^2(-\mathrm{i})$$$$=-\frac{1}{2\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi \mathrm{t}}{2}}{(-\frac{\mathrm{i}\pi }{2})}^2=\frac{{\pi }^2}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}=-\frac{\mathrm{i}{\pi }^2}{8}{\mathrm{e}}^{-\frac{\mathrm{i}\pi \mathrm{t}}{2}}\Rightarrow$$$$\mathrm{\Sigma }\mathrm{Res}=\frac{\mathrm{i}{\pi }^2}{8}({\mathrm{e}}^{\frac{\mathrm{i}\pi \mathrm{t}}{2}}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi \mathrm{t}}{2}})=\frac{\mathrm{i}{\pi }^2}{8}\left(2\mathrm{i}\sin \left(\frac{\pi \mathrm{t}}{2}\right)\right)$$$$=-\frac{{\pi }^2}{4}\sin \left(\frac{\pi \mathrm{t}}{2}\right)\Rightarrow \mathrm{f}\left(\mathrm{t}\right)={\mathrm{a}}^{\mathrm{t}-1}\mathrm{lna}.\frac{\pi }{2\cos \left(\frac{\pi \mathrm{t}}{2}\right)}$$$$-{\mathrm{a}}^{\mathrm{t}-1}\frac{{\pi }^2}{4}\sin \left(\frac{\pi \mathrm{t}}{2}\right)\mathrm{but}\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{tlogx}}\mathrm{logx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{t}}{\log }^2\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{o}\right)={\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}$$$$\mathrm{rest}\mathrm{calculus}\mathrm{of}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)\mathrm{and}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{o}\right)....\mathrm{be}\mathrm{continued}...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com