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Question Number 118030 by mathmax by abdo last updated on 14/Oct/20
find ∫_0 ^∞ ((x^2 −2)/(x^4 +x^2 +1))dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$
Commented by Ar Brandon last updated on 14/Oct/20
My idea is often to express the terms in the numerator like those of the square-roots of the first and last terms in the denominator. Just like in Q117446 x^2 −2=(x^2 −1)−1 =(x^2 −1)−(((x^2 +1)−(x^2 −1))/2) =(3/2)(x^2 −1)−(1/2)(x^2 +1)
$$\mathrm{My}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{often}\:\mathrm{to}\:\mathrm{express}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{numerator}\:\mathrm{like}\:\mathrm{those}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}-\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{last}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{denominator}.\:\mathrm{Just}\:\mathrm{like}\:\mathrm{in}\:\mathrm{Q117446} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)−\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$
Answered by Ar Brandon last updated on 14/Oct/20
I=∫_0 ^∞ ((x^2 −2)/(x^4 +x^2 +1))dx=(3/2)∫_0 ^∞ (((x^2 −1))/(x^4 +x^2 +1))dx−(1/2)∫_0 ^∞ (((x^2 +1))/(x^4 +x^2 +1))dx =(3/2)∫_0 ^∞ ((1−(1/x^2 ))/(x^2 +1+(1/x^2 )))dx−(1/2)∫_0 ^∞ ((1+(1/x^2 ))/(x^2 +1+(1/x^2 )))dx =(3/2)∫_0 ^∞ ((1−(1/x^2 ))/((x+(1/x))^2 −1))dx−(1/2)∫_0 ^∞ ((1+(1/x^2 ))/((x−(1/x))^2 +3))dx =(3/2)∫_(+∞) ^(+∞) (du/(u^2 −1))−(1/2)∫_(−∞) ^(+∞) (dv/(v^2 +3))=−(1/2)∫_(−∞) ^(+∞) (dv/(v^2 +3)) =−[(1/(2(√3)))Arctan((v/( (√3))))]_(−∞) ^(+∞) =−(1/(2(√3)))((π/2)+(π/2))=−(π/(2(√3)))
$$\mathcal{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{1}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{3}}\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{+\infty} ^{+\infty} \frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} +\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} +\mathrm{3}} \\ $$$$\:\:\:=−\left[\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\mathrm{Arctan}\left(\frac{\mathrm{v}}{\:\sqrt{\mathrm{3}}}\right)\right]_{−\infty} ^{+\infty} =−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$
Commented by Lordose last updated on 14/Oct/20
Excellente
$$\mathrm{Excellente} \\ $$
Commented by Ar Brandon last updated on 14/Oct/20
Merci. J'suis ravi !
Commented by Ar Brandon last updated on 14/Oct/20
S'il vous plaît, pouvez-vous jeter un oeil sur la Question 117972 ? Il semble avoir un problème avec votre réponse.
Commented by Lordose last updated on 15/Oct/20
Je pense que la meilleure façon de résoudre la question 117927 est de dessiner les cartes.��
Commented by Ar Brandon last updated on 15/Oct/20
Je n'arrive pas à le faire malheureusement
Commented by Lordose last updated on 15/Oct/20
D'accord. Je suis un peu occupé maintenant. quand je serai libre, je résoudrai.
Commented by Ar Brandon last updated on 15/Oct/20
D'accord, merci
Answered by Bird last updated on 15/Oct/20
I=∫_0 ^∞ ((x^2 −2)/(x^4 +x^2 +1))dx ⇒ I =(1/2)∫_(−∞) ^(+∞ ) ((x^2 −2)/(x^(4 ) +x^2 +1))dx let ϕ(z)=((z^2 −2)/(z^4 +z^2 +1)) poles ofϕ? z^4 +z^2 +1=0 ⇒u^2 +u+1=0 (u=z^(2)) Δ=−3 ⇒u_1 =((−1+i(√3))/2)=e^((i2π)/3) u_2 =e^(−((i2π)/3)) ⇒ϕ(z)=((z^2 −2)/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) =((z^2 −2)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/3) )+Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) )=((e^((2iπ)/3) −2)/(2e^((iπ)/3) (2isin(((2π)/3))))) =(1/(2i(√3))) (e^((iπ)/3) −2e^(−((iπ)/3)) ) Res(ϕ,−e^(−((iπ)/3)) )=((e^(−((2iπ)/3)) −2)/((−2e^(−((iπ)/3)) )(−2isin(((2π)/3)))) =((e^(−((iπ)/3)) −2e^((iπ)/3) )/(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(2i(√3))){e^((iπ)/3) −2e^(−((iπ)/3)) +e^(−((iπ)/3)) −2e^((iπ)/3) } =(π/( (√3))){ −e^((iπ)/3) −e^(−((iπ)/3)) } =−(π/( (√3)))(2cos((π/3))) =−(π/( (√3))) ⇒ I =−(π/(2(√3)))
$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty\:} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}\:} \:+{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${let}\:\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} −\mathrm{2}}{{z}^{\mathrm{4}} +{z}^{\mathrm{2}} \:+\mathrm{1}}\:{poles}\:{of}\varphi? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow{u}^{\mathrm{2}} +{u}+\mathrm{1}=\mathrm{0}\:\left({u}={z}^{\left.\mathrm{2}\right)} \right. \\ $$$$\Delta=−\mathrm{3}\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${u}_{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{z}^{\mathrm{2}} −\mathrm{2}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)=\frac{{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{2}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\left({e}^{\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right) \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)=\frac{{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{2}}{\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left(−\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right.} \\ $$$$=\frac{{e}^{−\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{{e}^{\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} +{e}^{−\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left\{\:−{e}^{\frac{{i}\pi}{\mathrm{3}}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=−\frac{\pi}{\:\sqrt{\mathrm{3}}}\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\right)\:=−\frac{\pi}{\:\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${I}\:=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$

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