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Question Number 136425 by Ar Brandon last updated on 21/Mar/21
If α>0 and β>0, prove ∫_0 ^∞ ((ln(αx))/(β^2 +x^2 ))dx=(π/(2β))ln(αβ)
$$\mathrm{If}\:\alpha>\mathrm{0}\:\mathrm{and}\:\beta>\mathrm{0},\:\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\alpha\mathrm{x}\right)}{\beta^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\pi}{\mathrm{2}\beta}\mathrm{ln}\left(\alpha\beta\right) \\ $$
Commented by Dwaipayan Shikari last updated on 21/Mar/21
Φ=∫_0 ^∞ ((log(α))/(β^2 +x^2 ))+∫_0 ^∞ ((log(x))/(x^2 +β^2 ))dx =((πlog(α))/(2β))+τ′(0) τ(η)=∫_0 ^∞ (x^η /(β^2 +x^2 ))dx=β^(η−1) ∫_0 ^∞ (u^η /(u^2 +1))du x=βu =(1/2)β^(η−1) ∫_0 ^∞ (t^(((η+1)/2)−1) /((t+1)^((η/2)+(1/2)−(η/2)+(1/2)) ))du=(β^(η−1) /2).((Γ(((η+1)/2))Γ((1/2)−(η/2)))/(Γ(2))) =(β^(η−1) /2).(π/(cos(((πη)/2)))) τ′(η)=((β^(η−1) log(β))/2).(π/(cos(((πη)/2))))+((β^(n−1) π^2 )/4)sec(((πη)/2))tan(((πη)/2)) τ′(0)=((log(β)π)/(2β)) Φ=((πlog(α))/(2β))+((πlog(β))/(2β))=((πlog(αβ))/(2β))
$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{{log}\left(\alpha\right)}{\beta^{\mathrm{2}} +{x}^{\mathrm{2}} }+\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({x}\right)}{{x}^{\mathrm{2}} +\beta^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi{log}\left(\alpha\right)}{\mathrm{2}\beta}+\tau'\left(\mathrm{0}\right) \\ $$$$\tau\left(\eta\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\eta} }{\beta^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}=\beta^{\eta−\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\eta} }{{u}^{\mathrm{2}} +\mathrm{1}}{du}\:\:\:\:\:{x}=\beta{u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\beta^{\eta−\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{{t}^{\frac{\eta+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left({t}+\mathrm{1}\right)^{\frac{\eta}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\eta}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} }{du}=\frac{\beta^{\eta−\mathrm{1}} }{\mathrm{2}}.\frac{\Gamma\left(\frac{\eta+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\eta}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\frac{\beta^{\eta−\mathrm{1}} }{\mathrm{2}}.\frac{\pi}{{cos}\left(\frac{\pi\eta}{\mathrm{2}}\right)} \\ $$$$\tau'\left(\eta\right)=\frac{\beta^{\eta−\mathrm{1}} {log}\left(\beta\right)}{\mathrm{2}}.\frac{\pi}{{cos}\left(\frac{\pi\eta}{\mathrm{2}}\right)}+\frac{\beta^{{n}−\mathrm{1}} \pi^{\mathrm{2}} }{\mathrm{4}}{sec}\left(\frac{\pi\eta}{\mathrm{2}}\right){tan}\left(\frac{\pi\eta}{\mathrm{2}}\right) \\ $$$$\tau'\left(\mathrm{0}\right)=\frac{{log}\left(\beta\right)\pi}{\mathrm{2}\beta} \\ $$$$\Phi=\frac{\pi{log}\left(\alpha\right)}{\mathrm{2}\beta}+\frac{\pi{log}\left(\beta\right)}{\mathrm{2}\beta}=\frac{\pi{log}\left(\alpha\beta\right)}{\mathrm{2}\beta} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 21/Mar/21
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Commented by Dwaipayan Shikari last updated on 21/Mar/21
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$$ \\ $$😃
Answered by mathmax by abdo last updated on 21/Mar/21
Φ=∫_0 ^∞ ((log(αx))/(x^2 +β^2 ))dx ⇒Φ=∫_0 ^∞ ((logα+logx)/(x^2 +β^2 ))dx =logα∫_0 ^∞ (dx/(x^2 +β^2 ))(→x=βu) +∫_0 ^∞ ((logx)/(x^2 +β^2 ))dx(→x=βu) =logα ∫_0 ^∞ ((βdu)/(β^2 (1+u^2 ))) +∫_0 ^∞ ((log(βu))/(β^2 (1+u^2 )))βdu =((logα)/β).(π/2) +(1/β)∫_0 ^∞ ((logβ +logu)/(1+u^2 ))du =((πlogα)/(2β)) +((logβ)/β).(π/2) +(1/β)∫_0 ^∞ ((logu)/(1+u^2 ))du(→0) =((π(logα+logβ))/(2β)) =(π/(2β))log(αβ)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}\left(\alpha\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}\alpha+\mathrm{logx}}{\mathrm{x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{log}\alpha\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\left(\rightarrow\mathrm{x}=\beta\mathrm{u}\right)\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logx}}{\mathrm{x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\mathrm{dx}\left(\rightarrow\mathrm{x}=\beta\mathrm{u}\right) \\ $$$$=\mathrm{log}\alpha\:\int_{\mathrm{0}} ^{\infty} \:\frac{\beta\mathrm{du}}{\beta^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}\left(\beta\mathrm{u}\right)}{\beta^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\beta\mathrm{du} \\ $$$$=\frac{\mathrm{log}\alpha}{\beta}.\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\beta}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{log}\beta\:+\mathrm{logu}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du} \\ $$$$=\frac{\pi\mathrm{log}\alpha}{\mathrm{2}\beta}\:+\frac{\mathrm{log}\beta}{\beta}.\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\beta}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logu}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\left(\rightarrow\mathrm{0}\right) \\ $$$$=\frac{\pi\left(\mathrm{log}\alpha+\mathrm{log}\beta\right)}{\mathrm{2}\beta}\:=\frac{\pi}{\mathrm{2}\beta}\mathrm{log}\left(\alpha\beta\right) \\ $$
Commented by Ar Brandon last updated on 21/Mar/21
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