find-0-x-2-2-x-4-x-2-1-dx-

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

find \int_{0}^{\infty} \frac{x^{2} - 2}{x^{4} + x^{2} + 1} 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 Q 117446

x^{2} - 2 = (x^{2} - 1) - 1

= (x^{2} - 1) - \frac{(x^{2} + 1) - (x^{2} - 1)}{2}

= \frac{3}{2} (x^{2} - 1) - \frac{1}{2} (x^{2} + 1)

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

I = \int_{0}^{\infty} \frac{x^{2} - 2}{x^{4} + x^{2} + 1} dx = \frac{3}{2} \int_{0}^{\infty} \frac{(x^{2} - 1)}{x^{4} + x^{2} + 1} dx - \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1)}{x^{4} + x^{2} + 1} dx

= \frac{3}{2} \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx - \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx

= \frac{3}{2} \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1} dx - \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} dx

= \frac{3}{2} \int_{+ \infty}^{+ \infty} \frac{du}{u^{2} - 1} - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dv}{v^{2} + 3} = - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dv}{v^{2} + 3}

= - {[\frac{1}{2 \sqrt{3}} Arctan (\frac{v}{\sqrt{3}})]}_{- \infty}^{+ \infty} = - \frac{1}{2 \sqrt{3}} (\frac{π}{2} + \frac{π}{2}) = - \frac{π}{2 \sqrt{3}}

Commented by Lordose last updated on 14/Oct/20

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}^{\infty} \frac{x^{2} - 2}{x^{4} + x^{2} + 1} d x \Rightarrow

I = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{x^{2} - 2}{x^{4} + x^{2} + 1} d x

l e t φ (z) = \frac{z^{2} - 2}{z^{4} + z^{2} + 1} p o l e s o f φ ?

z^{4} + z^{2} + 1 = 0 \Rightarrow u^{2} + u + 1 = 0 (u = z^{2)}

Δ = - 3 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}

u_{2} = e^{- \frac{i 2 π}{3}}

\Rightarrow φ (z) = \frac{z^{2} - 2}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})}

= \frac{z^{2} - 2}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}

R e s (φ, e^{\frac{i π}{3}}) = \frac{e^{\frac{2 i π}{3}} - 2}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{2 π}{3}))}

= \frac{1}{2 i \sqrt{3}} (e^{\frac{i π}{3}} - 2 e^{- \frac{i π}{3}})

R e s (φ, - e^{- \frac{i π}{3}}) = \frac{e^{- \frac{2 i π}{3}} - 2}{(- 2 e^{- \frac{i π}{3}}) (- 2 i s i n (\frac{2 π}{3})}

= \frac{e^{- \frac{i π}{3}} - 2 e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{2 i \sqrt{3}} {e^{\frac{i π}{3}} - 2 e^{- \frac{i π}{3}} + e^{- \frac{i π}{3}} - 2 e^{\frac{i π}{3}}}

= \frac{π}{\sqrt{3}} {- e^{\frac{i π}{3}} - e^{- \frac{i π}{3}}}

= - \frac{π}{\sqrt{3}} (2 c o s (\frac{π}{3})) = - \frac{π}{\sqrt{3}} \Rightarrow

I = - \frac{π}{2 \sqrt{3}}

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