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Question Number 150404 by mathdanisur last updated on 12/Aug/21
Ω =∫_( 0) ^( ∞) ((log(x))/((1 + x^2 )^2 )) dx = ?
$$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$
Answered by Ar Brandon last updated on 12/Aug/21
Ω(a)=∫_0 ^∞ ((lnx)/(a^2 +x^2 ))dx, x=atanϑ =∫_0 ^(π/2) ((ln(atanϑ))/(a^2 +a^2 tan^2 ϑ))∙asec^2 ϑdϑ =(1/a)∫_0 ^(π/2) (lna+ln(tanϑ))dϑ=((πlna)/(2a)) ⇒Ω′(a)=(π/2)((1/a^2 )−((lna)/a^2 ))=−∫_0 ^∞ ((2alnx)/((a^2 +x^2 )^2 ))dx ⇒∫_0 ^∞ ((lnx)/((a^2 +x^2 )^2 ))dx=−(π/(4a))((1/a^2 )−((lna)/a^2 )) ⇒∫_0 ^∞ ((lnx)/((1+x^2 )^2 ))dx=−(π/4)
$$\Omega\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx},\:{x}={a}\mathrm{tan}\vartheta \\ $$$$\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left({a}\mathrm{tan}\vartheta\right)}{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \vartheta}\centerdot{a}\mathrm{sec}^{\mathrm{2}} \vartheta{d}\vartheta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{ln}{a}+\mathrm{ln}\left(\mathrm{tan}\vartheta\right)\right){d}\vartheta=\frac{\pi\mathrm{ln}{a}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\Omega'\left({a}\right)=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{ln}{a}}{{a}^{\mathrm{2}} }\right)=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{a}\mathrm{ln}{x}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\pi}{\mathrm{4}{a}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{ln}{a}}{{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\pi}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 12/Aug/21
Thank You Ser
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$
Answered by ajfour last updated on 12/Aug/21
Ω=ln x∫(dx/((1+x^2 )^2 )) +∫{(1/x)∫(dx/((1+x^2 )^2 ))}dx Ω=Gln x+∫ ((Gdx)/x) G=∫(dx/((1+x^2 )^2 ))=(1/(2x))(−(1/(1+x^2 ))) −(1/2)∫(dx/(x^2 (1+x^2 ))) let x=tan θ G=(1/2)∫(1+cos 2θ)dθ =((2θ+sin 2θ)/4)=((tan^(−1) x)/2)+(x/(2(1+x^2 ))) 2(Gln x)=(ln x)tan^(−1) x +((xln x)/(1+x^2 )) 2∫ ((Gdx)/x)=∫{((tan^(−1) x)/x)+(1/(1+x^2 ))}dx =H+tan^(−1) x+c H=ln xtan^(−1) x−∫((ln xdx)/((1+x^2 ))) Ω=((ln x)/(2x))(−(1/(1+x^2 )))−((ln x)/2)∫(dx/(x^2 (1+x^2 ))) +((tan^(−1) x)/2)+(1/2)ln xtan^(−1) x −(1/2)∫((ln xdx)/((1+x^2 )))+c ...
$$\Omega=\mathrm{ln}\:{x}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:+\int\left\{\frac{\mathrm{1}}{{x}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}{dx} \\ $$$$\Omega={G}\mathrm{ln}\:{x}+\int\:\frac{{Gdx}}{{x}} \\ $$$${G}=\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{x}}\left(−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${let}\:\:{x}=\mathrm{tan}\:\theta \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:=\frac{\mathrm{2}\theta+\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}}=\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}\left({G}\mathrm{ln}\:{x}\right)=\left(\mathrm{ln}\:{x}\right)\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{x}\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}\int\:\frac{{Gdx}}{{x}}=\int\left\{\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={H}+\mathrm{tan}^{−\mathrm{1}} {x}+{c} \\ $$$${H}=\mathrm{ln}\:{x}\mathrm{tan}^{−\mathrm{1}} {x}−\int\frac{\mathrm{ln}\:{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\Omega=\frac{\mathrm{ln}\:{x}}{\mathrm{2}{x}}\left(−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)−\frac{\mathrm{ln}\:{x}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:+\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{x}\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}\:{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+{c} \\ $$$$… \\ $$
Commented by mathdanisur last updated on 12/Aug/21
Thankhou Ser, ans.?
$$\mathrm{Thankhou}\:\mathrm{Ser},\:\mathrm{ans}.? \\ $$
Answered by mnjuly1970 last updated on 12/Aug/21
−(π/4) ....
$$\:\:−\frac{\pi}{\mathrm{4}}\:…. \\ $$
Answered by Lordose last updated on 12/Aug/21
Ω = ∫_0 ^( ∞) ((log(x))/((1+x^2 )^2 ))dx =^(x=tanθ) ∫_0 ^( (𝛑/2)) ((log(tan𝛉))/(sec^2 𝛉)) Ω = ∫_0 ^(𝛑/2) cos^2 θlog(sinθ)dθ − ∫_0 ^(𝛑/2) cos^2 θlog(cosθ)dθ Ω = (∂/∂a)∫_0 ^(𝛑/2) cos^2 θsin^a (θ)dθ∣_(a=0) − (∂/∂a)∫_0 ^(𝛑/2) cos^a (θ)dθ∣_(a=2) B(x,y) = 2∫_0 ^(𝛑/2) sin^(2x−1) (t)cos^(2y−1) (t)dt Ω = (∂/∂a)((1/2)B(((a+1)/2),(3/2)))∣_(a=0) − (∂/∂a)((1/2)B((1/2),((a+1)/2)))∣_(a=2) 𝛀 = −(𝛑/4)
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{tan}\theta} {=}\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{log}\left(\mathrm{tan}\boldsymbol{\theta}\right)}{\mathrm{sec}^{\mathrm{2}} \boldsymbol{\theta}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{log}\left(\mathrm{sin}\theta\right)\mathrm{d}\theta\:−\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{log}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$$$\Omega\:=\:\frac{\partial}{\partial\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{sin}^{\mathrm{a}} \left(\theta\right)\mathrm{d}\theta\mid_{\mathrm{a}=\mathrm{0}} \:−\:\frac{\partial}{\partial\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{a}} \left(\theta\right)\mathrm{d}\theta\mid_{\mathrm{a}=\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{B}}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2x}−\mathrm{1}} \left(\mathrm{t}\right)\mathrm{cos}^{\mathrm{2y}−\mathrm{1}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\Omega\:=\:\frac{\partial}{\partial\mathrm{a}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)\right)\mid_{\mathrm{a}=\mathrm{0}} \:−\:\frac{\partial}{\partial\mathrm{a}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)\right)\mid_{\mathrm{a}=\mathrm{2}} \\ $$$$\boldsymbol{\Omega}\:=\:−\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$

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