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Question Number 118030 by mathmax by abdo last updated on 14/Oct/20

$$\mathrm{find}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2-2}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}$$

Commented by Ar Brandon last updated on 14/Oct/20

$$\mathrm{My}\mathrm{idea}\mathrm{is}\mathrm{often}\mathrm{to}\mathrm{express}\mathrm{the}\mathrm{terms}\mathrm{in}\mathrm{the}$$$$\mathrm{numerator}\mathrm{like}\mathrm{those}\mathrm{of}\mathrm{the}\mathrm{square}-\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{first}$$$$\mathrm{and}\mathrm{last}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{denominator}.\mathrm{Just}\mathrm{like}\mathrm{in}\mathrm{Q}117446$$$${\mathrm{x}}^2-2=({\mathrm{x}}^2-1)-1$$$$=({\mathrm{x}}^2-1)-\frac{({\mathrm{x}}^2+1)-({\mathrm{x}}^2-1)}{2}$$$$=\frac{3}{2}({\mathrm{x}}^2-1)-\frac{1}{2}({\mathrm{x}}^2+1)$$

Answered by Ar Brandon last updated on 14/Oct/20

$$\mathcal{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2-2}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}=\frac{3}{2}{\int }_0^{\mathrm{\infty }}\frac{({\mathrm{x}}^2-1)}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{({\mathrm{x}}^2+1)}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\frac{3}{2}{\int }_0^{\mathrm{\infty }}\frac{1-\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$$$=\frac{3}{2}{\int }_0^{\mathrm{\infty }}\frac{1-\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-1}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+3}\mathrm{dx}$$$$=\frac{3}{2}{\int }_{+\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2-1}-\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2+3}=-\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2+3}$$$$=-{\left[\frac{1}{2\sqrt{3}}\mathrm{Arctan}\left(\frac{\mathrm{v}}{\sqrt{3}}\right)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=-\frac{1}{2\sqrt{3}}(\frac{\pi }{2}+\frac{\pi }{2})=-\frac{\pi }{2\sqrt{3}}$$

Commented by Lordose last updated on 14/Oct/20

$$\mathrm{Excellente}$$

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

$$I={\int }_0^{\mathrm{\infty }}\frac{x^2-2}{x^4+x^2+1}dx\Rightarrow$$$$I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-2}{x^4+x^2+1}dx$$$$let\phi \left(z\right)=\frac{z^2-2}{z^4+z^2+1}polesof\phi ?$$$$z^4+z^2+1=0\Rightarrow u^2+u+1=0(u=z^{2)}$$$$\mathrm{\Delta }=-3\Rightarrow u_1=\frac{-1+i\sqrt{3}}{2}=e^{\frac{i2\pi }{3}}$$$$u_2=e^{-\frac{i2\pi }{3}}$$$$\Rightarrow \phi \left(z\right)=\frac{z^2-2}{(z^2-e^{\frac{i2\pi }{3}})(z^2-e^{-\frac{i2\pi }{3}})}$$$$=\frac{z^2-2}{(z-e^{\frac{i\pi }{3}})(z+e^{\frac{i\pi }{3}})(z-e^{-\frac{i\pi }{3}})(z+e^{-\frac{i\pi }{3}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,e^{\frac{i\pi }{3}})+Res(\phi ,-e^{-\frac{i\pi }{3}})\}$$$$Res(\phi ,e^{\frac{i\pi }{3}})=\frac{e^{\frac{2i\pi }{3}}-2}{2e^{\frac{i\pi }{3}}\left(2isin\left(\frac{2\pi }{3}\right)\right)}$$$$=\frac{1}{2i\sqrt{3}}(e^{\frac{i\pi }{3}}-2e^{-\frac{i\pi }{3}})$$$$Res(\phi ,-e^{-\frac{i\pi }{3}})=\frac{e^{-\frac{2i\pi }{3}}-2}{(-2e^{-\frac{i\pi }{3}})(-2isin\left(\frac{2\pi }{3}\right)}$$$$=\frac{e^{-\frac{i\pi }{3}}-2e^{\frac{i\pi }{3}}}{2i\sqrt{3}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{2i\pi }{2i\sqrt{3}}\{e^{\frac{i\pi }{3}}-2e^{-\frac{i\pi }{3}}+e^{-\frac{i\pi }{3}}-2e^{\frac{i\pi }{3}}\}$$$$=\frac{\pi }{\sqrt{3}}\{-e^{\frac{i\pi }{3}}-e^{-\frac{i\pi }{3}}\}$$$$=-\frac{\pi }{\sqrt{3}}\left(2cos\left(\frac{\pi }{3}\right)\right)=-\frac{\pi }{\sqrt{3}}\Rightarrow$$$$I=-\frac{\pi }{2\sqrt{3}}$$

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