Algebra Questions

Question Number 150404 by mathdanisur last updated on 12/Aug/21

$$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\$$

Answered by Ar Brandon last updated on 12/Aug/21

$$\Omega\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx},\:{x}={a}\mathrm{tan}\vartheta \\$$ $$\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left({a}\mathrm{tan}\vartheta\right)}{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \vartheta}\centerdot{a}\mathrm{sec}^{\mathrm{2}} \vartheta{d}\vartheta \\$$ $$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{ln}{a}+\mathrm{ln}\left(\mathrm{tan}\vartheta\right)\right){d}\vartheta=\frac{\pi\mathrm{ln}{a}}{\mathrm{2}{a}} \\$$ $$\Rightarrow\Omega'\left({a}\right)=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{ln}{a}}{{a}^{\mathrm{2}} }\right)=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{a}\mathrm{ln}{x}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\$$ $$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\pi}{\mathrm{4}{a}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{ln}{a}}{{a}^{\mathrm{2}} }\right) \\$$ $$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\pi}{\mathrm{4}} \\$$

Commented bymathdanisur last updated on 12/Aug/21

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\$$

Answered by ajfour last updated on 12/Aug/21

$$\Omega=\mathrm{ln}\:{x}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$ $$\:\:\:\:\:+\int\left\{\frac{\mathrm{1}}{{x}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}{dx} \\$$ $$\Omega={G}\mathrm{ln}\:{x}+\int\:\frac{{Gdx}}{{x}} \\$$ $${G}=\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{x}}\left(−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\$$ $${let}\:\:{x}=\mathrm{tan}\:\theta \\$$ $${G}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\$$ $$\:=\frac{\mathrm{2}\theta+\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}}=\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\$$ $$\mathrm{2}\left({G}\mathrm{ln}\:{x}\right)=\left(\mathrm{ln}\:{x}\right)\mathrm{tan}^{−\mathrm{1}} {x} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{x}\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$ $$\mathrm{2}\int\:\frac{{Gdx}}{{x}}=\int\left\{\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right\}{dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={H}+\mathrm{tan}^{−\mathrm{1}} {x}+{c} \\$$ $${H}=\mathrm{ln}\:{x}\mathrm{tan}^{−\mathrm{1}} {x}−\int\frac{\mathrm{ln}\:{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\$$ $$\\$$ $$\Omega=\frac{\mathrm{ln}\:{x}}{\mathrm{2}{x}}\left(−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)−\frac{\mathrm{ln}\:{x}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\$$ $$\:\:\:\:\:+\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{x}\mathrm{tan}^{−\mathrm{1}} {x} \\$$ $$\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}\:{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+{c} \\$$ $$... \\$$

Commented bymathdanisur last updated on 12/Aug/21

$$\mathrm{Thankhou}\:\mathrm{Ser},\:\mathrm{ans}.? \\$$

Answered by mnjuly1970 last updated on 12/Aug/21

$$\:\:−\frac{\pi}{\mathrm{4}}\:.... \\$$

Answered by Lordose last updated on 12/Aug/21

$$\\$$ $$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{tan}\theta} {=}\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{log}\left(\mathrm{tan}\boldsymbol{\theta}\right)}{\mathrm{sec}^{\mathrm{2}} \boldsymbol{\theta}} \\$$ $$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{log}\left(\mathrm{sin}\theta\right)\mathrm{d}\theta\:−\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{log}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta \\$$ $$\Omega\:=\:\frac{\partial}{\partial\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{sin}^{\mathrm{a}} \left(\theta\right)\mathrm{d}\theta\mid_{\mathrm{a}=\mathrm{0}} \:−\:\frac{\partial}{\partial\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{a}} \left(\theta\right)\mathrm{d}\theta\mid_{\mathrm{a}=\mathrm{2}} \\$$ $$\boldsymbol{\mathrm{B}}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2x}−\mathrm{1}} \left(\mathrm{t}\right)\mathrm{cos}^{\mathrm{2y}−\mathrm{1}} \left(\mathrm{t}\right)\mathrm{dt} \\$$ $$\Omega\:=\:\frac{\partial}{\partial\mathrm{a}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)\right)\mid_{\mathrm{a}=\mathrm{0}} \:−\:\frac{\partial}{\partial\mathrm{a}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)\right)\mid_{\mathrm{a}=\mathrm{2}} \\$$ $$\boldsymbol{\Omega}\:=\:−\frac{\boldsymbol{\pi}}{\mathrm{4}} \\$$