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Question Number 114330 by mnjuly1970 last updated on 18/Sep/20

Commented by mnjuly1970 last updated on 18/Sep/20

$$easyquestion\uparrow \uparrow \uparrow$$

Answered by Dwaipayan Shikari last updated on 18/Sep/20

$${\lambda }_1={\int }_0^{\mathrm{\infty }}\frac{log(1+x^2)}{1+x^2}dx$$$${\int }_0^{\frac{\pi }{2}}\frac{log(1+{tan}^2\theta )}{{sec}^2\theta }.{sec}^2\theta d\theta$$$$-2{\int }_0^{\frac{\pi }{2}}log\left(cos\theta \right)d\theta =\pi log\left(2\right)$$$${\lambda }_2={\int }_0^{\frac{\pi }{4}}\frac{log(1+tan\theta )}{{sec}^2\theta }.{sec}^2\theta d\theta$$$$={\int }_0^{\frac{\pi }{4}}log(cosx+sinx)-log\left(cosx\right)={\int }_0^{\frac{\pi }{4}}\frac{1}{2}log2+log\left(cos(\frac{\pi }{4}-x)\right)-log\left(cosx\right)=I$$$$=\frac{\pi }{8}log\left(2\right)+0=\frac{\pi }{8}log\left(2\right)$$$$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\pi log\left(2\right)}{\frac{\pi }{8}log\left(2\right)}=8$$$$$$$$$$

Commented by mnjuly1970 last updated on 18/Sep/20

$$thankyousir.verynice.$$

Answered by mathmax by abdo last updated on 18/Sep/20

$$\mathrm{we}\mathrm{have}{\lambda }_1={\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\tan \theta }{\int }_0^{\frac{\pi }{2}}\frac{\ln (1+{\tan }^2\theta )}{1+{\tan }^2\theta }(1+{\tan }^2\theta )\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\ln \left(\frac{1}{{\cos }^2\theta }\right)\mathrm{d}\theta =-2{\int }_0^{\frac{\pi }{2}}\ln \left(\cos \theta \right)\mathrm{d}\theta =-2(-\frac{\pi }{2}\ln \left(2\right))=\pi \ln \left(2\right)$$$${\lambda }_2={\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\mathrm{tant}}{\int }_0^{\frac{\pi }{4}}\frac{\ln (1+\mathrm{tant})}{1+{\tan }^2\mathrm{t}}(1+{\tan }^2\mathrm{t})\mathrm{dt}$$$$={\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tant})\mathrm{dt}{=}_{\mathrm{t}=\frac{\pi }{4}-\mathrm{u}}{\int }_0^{\frac{\pi }{4}}\ln (1+\tan (\frac{\pi }{4}-\mathrm{u})\mathrm{du}$$$$={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\mathrm{tanu}}{1+\mathrm{tanu}})\mathrm{du}={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{2}{1+\mathrm{tanu}}\right)\mathrm{du}$$$$=\frac{\pi }{4}\ln \left(2\right)-{\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tanu})\mathrm{du}=\frac{\pi }{4}\ln \left(2\right)-{\lambda }_2\Rightarrow$$$$2{\lambda }_2=\frac{\pi }{4}\ln \left(2\right)\Rightarrow {\lambda }_2=\frac{\pi }{8}\ln \left(2\right)\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{\pi \ln \left(2\right)}{\frac{\pi }{8}\ln \left(2\right)}\Rightarrow ★\frac{{\lambda }_1}{{\lambda }_2}=8★$$

Commented by mnjuly1970 last updated on 18/Sep/20

$$thankyoumaster\left(ostad\right)in$$$$ourlanguage.verynice.grateful$$$$$$$$teful$$

Commented by mathmax by abdo last updated on 18/Sep/20

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

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