Integration Questions

Question Number 139282 by qaz last updated on 25/Apr/21

$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\:{x}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}{a}}\centerdot{ln}\:{a}\:\:\:\:\:\:\:\:\:\:,{a}>\mathrm{0}\: \\$$

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

$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mathrm{dx},\:\mathrm{x}=\mathrm{atan}\theta \\$$ $$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{atan}\theta\right)\mathrm{d}\theta=\frac{\pi}{\mathrm{2a}}\mathrm{lna}+\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{tan}\theta\right)\mathrm{d}\theta \\$$ $$\:\:\:\:=\frac{\pi}{\mathrm{2a}}\mathrm{lna}\:\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\theta\right)\mathrm{d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta\right) \\$$