Ω = ∫_0 ^( ∞) ((u^2 +2)/(u^4 +2u^2 +2))du


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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
	$Φ =∫_0 ^∞ ((x^2 +2)/(x^4 +2x^2 +2))dx ⇒Φ =(1/2)∫_(−∞) ^(+∞) ((x^2 +2)/(x^4 +2x^2 +2))dx let ϕ(z)=((z^2 +2)/(z^4 +2z^2 +2)) poles of ϕ? z^4 +2z^2 +2=0 ⇒t^2 +2t+2=0 (t=x^2 ) Δ^′ =1−2=−1 ⇒t_1 =−1+i and t_2 =−1−i t_1 =(√2)(−(1/( (√2)))+(i/( (√2)))) =(√2)e^((i3π)/4) and t_2 =(√2)e^(−((i3π)/4)) ⇒ϕ(z)=((z^2 +2)/((z^2 −(√2)e^((i3π)/4) )(z^2 −(√2)e^(−((i3π)/4)) ))) let α =(√(√2)) ⇒ ϕ(z)=((z^2 +2)/((z−αe^((i3π)/8) )(z+α e^((i3π)/8) )(z−αe^(−((i3π)/8)) )(z+α e^(−i((3π)/8)) ))) ⇒∫_R ϕ(z)dz =2iπ{ Res(ϕ,αe^((i3π)/8) ) +Res(ϕ,−αe^(−((i3π)/8)) )} Res(ϕ,αe^((i3π)/8) ) =((α^2 e^((i3π)/4) +2)/(2αe^((i3π)/8) (√2)(2isin(((3π)/4))))) =((α^2 e^((i3π)/4) +2)/(4iα)) e^(−((i3π)/8)) Res(ϕ ,−α e^(−((i3π)/8)) ) =((α^2 e^(−((3πi)/4)) +2)/(−2α e^(−((i3π)/8)) (−(√2))(2isin(((3π)/4)))) =((α^2 e^(−i((3π)/4)) +2)/(4iα)) e^((i3π)/8) ⇒ ∫_R ϕ(z)dz =2iπ{((α^2 e^((i3π)/4) +2)/(4iα))e^(−((i3π)/8)) +((α^2 e^(−((i3π)/4)) +2)/(4iα)) e^((i3π)/8) } =(π/(2α)){ α^2 e^((i3π)/8) +2e^(−((i3π)/8)) +α^2 e^(−((i3π)/8)) +2e^((i3π)/8) } =(π/(2α)){2α^2 cos(((3π)/8)) +4cos(((3π)/8))} =(π/(2α))(2α^2 +4)cos(((3π)/8)) =(π/α)(α^2 +2)cos(((3π)/8)) =(π/( (√(√2))))((√2)+2)cos(((3π)/8)) =2Φ cos(((3π)/8))=cos((π/8)+(π/4))=cos((π/8))cos((π/4))−sin((π/8))sin((π/4)) =((√(2+(√2)))/2)×(1/( (√2))) −((√(2−(√2)))/2)×(1/( (√2))) and Φ =(π/(2(√(√2))))(2+(√2))cos(((3π)/8))$
	$Φ = \int_{0}^{\infty} \frac{x^{2} + 2}{x^{4} + {2 x}^{2} + 2} dx \Rightarrow Φ = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{x^{2} + 2}{x^{4} + {2 x}^{2} + 2} dx let$ $φ (z) = \frac{z^{2} + 2}{z^{4} + {2 z}^{2} + 2} poles of φ ?$ $z^{4} + {2 z}^{2} + 2 = 0 \Rightarrow t^{2} + 2 t + 2 = 0 (t = x^{2})$ $Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow t_{1} = - 1 + i and t_{2} = - 1 - i$ $t_{1} = \sqrt{2} (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \sqrt{2} e^{\frac{i 3 π}{4}} and t_{2} = \sqrt{2} e^{- \frac{i 3 π}{4}}$ $\Rightarrow φ (z) = \frac{z^{2} + 2}{(z^{2} - \sqrt{2} e^{\frac{i 3 π}{4}}) (z^{2} - \sqrt{2} e^{- \frac{i 3 π}{4}})} let α = \sqrt{\sqrt{2}} \Rightarrow$ $φ (z) = \frac{z^{2} + 2}{(z - α e^{\frac{i 3 π}{8}}) (z + α e^{\frac{i 3 π}{8}}) (z - α e^{- \frac{i 3 π}{8}}) (z + α e^{- i \frac{3 π}{8}})}$ $\Rightarrow \int_{R} φ (z) dz = 2 i π {Res (φ, α e^{\frac{i 3 π}{8}}) + Res (φ, - α e^{- \frac{i 3 π}{8}})}$ $Res (φ, α e^{\frac{i 3 π}{8}}) = \frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{2 α e^{\frac{i 3 π}{8}} \sqrt{2} (2 isin (\frac{3 π}{4}))}$ $= \frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{4 i α} e^{- \frac{i 3 π}{8}}$ $Res (φ, - α e^{- \frac{i 3 π}{8}}) = \frac{α^{2} e^{- \frac{3 π i}{4}} + 2}{- 2 α e^{- \frac{i 3 π}{8}} (- \sqrt{2}) (2 isin (\frac{3 π}{4})}$ $= \frac{α^{2} e^{- i \frac{3 π}{4}} + 2}{4 i α} e^{\frac{i 3 π}{8}} \Rightarrow$ $\int_{R} φ (z) dz = 2 i π {\frac{α^{2} e^{\frac{i 3 π}{4}} + 2}{4 i α} e^{- \frac{i 3 π}{8}} + \frac{α^{2} e^{- \frac{i 3 π}{4}} + 2}{4 i α} e^{\frac{i 3 π}{8}}}$ $= \frac{π}{2 α} {α^{2} e^{\frac{i 3 π}{8}} + {2 e}^{- \frac{i 3 π}{8}} + α^{2} e^{- \frac{i 3 π}{8}} + {2 e}^{\frac{i 3 π}{8}}}$ $= \frac{π}{2 α} {2 α^{2} \cos (\frac{3 π}{8}) + 4 \cos (\frac{3 π}{8})}$ $= \frac{π}{2 α} (2 α^{2} + 4) \cos (\frac{3 π}{8}) = \frac{π}{α} (α^{2} + 2) \cos (\frac{3 π}{8})$ $= \frac{π}{\sqrt{\sqrt{2}}} (\sqrt{2} + 2) \cos (\frac{3 π}{8}) = 2 Φ$ $\cos (\frac{3 π}{8}) = \cos (\frac{π}{8} + \frac{π}{4}) = \cos (\frac{π}{8}) \cos (\frac{π}{4}) - \sin (\frac{π}{8}) \sin (\frac{π}{4})$ $= \frac{\sqrt{2 + \sqrt{2}}}{2} \times \frac{1}{\sqrt{2}} - \frac{\sqrt{2 - \sqrt{2}}}{2} \times \frac{1}{\sqrt{2}}$ $and Φ = \frac{π}{2 \sqrt{\sqrt{2}}} (2 + \sqrt{2}) \cos (\frac{3 π}{8})$
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