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Question Number 139282 by qaz last updated on 25/Apr/21
∫_0 ^∞ ((ln x)/(x^2 +a^2 ))dx=(π/(2a))∙ln a          ,a>0
$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\:{x}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}{a}}\centerdot{ln}\:{a}\:\:\:\:\:\:\:\:\:\:,{a}>\mathrm{0}\: \\ $$
Answered by Ar Brandon last updated on 25/Apr/21
Ω=∫_0 ^∞ ((lnx)/(x^2 +a^2 ))dx, x=atanθ      =(1/a)∫_0 ^(π/2) ln(atanθ)dθ=(π/(2a))lna+(1/a)∫_0 ^(π/2) ln(tanθ)dθ      =(π/(2a))lna (∫_0 ^(π/2) ln(sinθ)dθ=∫_0 ^(π/2) ln(cosθ)dθ)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mathrm{dx},\:\mathrm{x}=\mathrm{atan}\theta \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{atan}\theta\right)\mathrm{d}\theta=\frac{\pi}{\mathrm{2a}}\mathrm{lna}+\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{tan}\theta\right)\mathrm{d}\theta \\ $$$$\:\:\:\:=\frac{\pi}{\mathrm{2a}}\mathrm{lna}\:\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\theta\right)\mathrm{d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta\right) \\ $$

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