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Question Number 137518 by Lordose last updated on 03/Apr/21

Ω = \int_{0}^{\infty} \frac{u^{2} + 2}{u^{4} + {2 u}^{2} + 2} du

Answered by mathmax by abdo last updated on 03/Apr/21

Φ = \int_{0}^{\infty} \frac{x^{2} + 2}{x^{4} + {2 x}^{2} + 2} dx \Rightarrow Φ = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{x^{2} + 2}{x^{4} + {2 x}^{2} + 2} dx let

φ (z) = \frac{z^{2} + 2}{z^{4} + {2 z}^{2} + 2} poles of φ ?

z^{4} + {2 z}^{2} + 2 = 0 \Rightarrow t^{2} + 2 t + 2 = 0 (t = x^{2})

Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow t_{1} = - 1 + i and t_{2} = - 1 - i

t_{1} = \sqrt{2} (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \sqrt{2} e^{\frac{i 3 π}{4}} and t_{2} = \sqrt{2} e^{- \frac{i 3 π}{4}}

\Rightarrow φ (z) = \frac{z^{2} + 2}{(z^{2} - \sqrt{2} e^{\frac{i 3 π}{4}}) (z^{2} - \sqrt{2} e^{- \frac{i 3 π}{4}})} let α = \sqrt{\sqrt{2}} \Rightarrow

φ (z) = \frac{z^{2} + 2}{(z - α e^{\frac{i 3 π}{8}}) (z + α e^{\frac{i 3 π}{8}}) (z - α e^{- \frac{i 3 π}{8}}) (z + α e^{- i \frac{3 π}{8}})}

\Rightarrow \int_{R} φ (z) dz = 2 i π {Res (φ, α e^{\frac{i 3 π}{8}}) + Res (φ, - α e^{- \frac{i 3 π}{8}})}

Res (φ, α e^{\frac{i 3 π}{8}}) = \frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{2 α e^{\frac{i 3 π}{8}} \sqrt{2} (2 isin (\frac{3 π}{4}))}

= \frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{4 i α} e^{- \frac{i 3 π}{8}}

Res (φ, - α e^{- \frac{i 3 π}{8}}) = \frac{α^{2} e^{- \frac{3 π i}{4}} + 2}{- 2 α e^{- \frac{i 3 π}{8}} (- \sqrt{2}) (2 isin (\frac{3 π}{4})}

= \frac{α^{2} e^{- i \frac{3 π}{4}} + 2}{4 i α} e^{\frac{i 3 π}{8}} \Rightarrow

\int_{R} φ (z) dz = 2 i π {\frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{4 i α} e^{- \frac{i 3 π}{8}} + \frac{α^{2} e^{- \frac{i 3 π}{4}} + 2}{4 i α} e^{\frac{i 3 π}{8}}}

= \frac{π}{2 α} {α^{2} e^{\frac{i 3 π}{8}} + {2 e}^{- \frac{i 3 π}{8}} + α^{2} e^{- \frac{i 3 π}{8}} + {2 e}^{\frac{i 3 π}{8}}}

= \frac{π}{2 α} {2 α^{2} \cos (\frac{3 π}{8}) + 4 \cos (\frac{3 π}{8})}

= \frac{π}{2 α} (2 α^{2} + 4) \cos (\frac{3 π}{8}) = \frac{π}{α} (α^{2} + 2) \cos (\frac{3 π}{8})

= \frac{π}{\sqrt{\sqrt{2}}} (\sqrt{2} + 2) \cos (\frac{3 π}{8}) = 2 Φ

\cos (\frac{3 π}{8}) = \cos (\frac{π}{8} + \frac{π}{4}) = \cos (\frac{π}{8}) \cos (\frac{π}{4}) - \sin (\frac{π}{8}) \sin (\frac{π}{4})

= \frac{\sqrt{2 + \sqrt{2}}}{2} \times \frac{1}{\sqrt{2}} - \frac{\sqrt{2 - \sqrt{2}}}{2} \times \frac{1}{\sqrt{2}}

and Φ = \frac{π}{2 \sqrt{\sqrt{2}}} (2 + \sqrt{2}) \cos (\frac{3 π}{8})