calculate ∫_(−∞) ^(+∞) ((cos(2x)dx)/(x^4 +x^2 +1))


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Question Number 136034 by mathmax by abdo last updated on 18/Mar/21

$calculate \int_{- \infty}^{+ \infty} \frac{\cos (2 x) dx}{x^{4} + x^{2} + 1}$
	Answered by mathmax by abdo last updated on 19/Mar/21
	$Φ =∫_(−∞) ^(+∞) ((cos(2x))/(x^4 +x^2 +1))dx ⇒Φ=Re(∫_(−∞) ^(+∞) (e^(2ix) /(x^4 +x^2 +1))dx) let ϕ(z)=(e^(2iz) /(z^4 +z^2 +1)) poles of ϕ? z^4 +z^2 +1 =0 ⇒u^2 +u +1 =0 (z^2 =u) Δ=−3 ⇒u_1 =((−1+i(√3))/2) =e^((2iπ)/3) and u_2 =e^(−((2iπ)/3)) ⇒ϕ(z)=(e^(2iz) /((z^2 −e^((2iπ)/3) )(z^2 −e^(−((2iπ)/3)) ))) =(e^(2iz) /((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^((iπ)/3) ))) ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ (Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )) Res(ϕ,e^((iπ)/3) ) =(e^(2ie^((iπ)/3) ) /(2e^((iπ)/3) (2isin(((2π)/3))))) =(e^(2i((1/2)+((i(√3))/2))) /(4i.((√3)/2)))e^(−((iπ)/3)) =(1/(2i(√3))) e^(−(√3)) .e^(i−((iπ)/3)) =(1/(2i(√3)))e^(−(√3)) .e^((2iπ)/3) Res(ϕ,−e^(−((iπ)/3)) ) =(e^(−2ie^(−((iπ)/3)) ) /(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(e^(−2i((1/2)−((i(√3))/2))) /(4i.((√3)/2))) .e^((iπ)/3) =(1/(2i(√3))).e^(−(√3)) .e^(−i+((iπ)/3)) =(1/(2i(√3)))e^(−(√3)) .e^(−((2iπ)/3)) ⇒ ∫_R ϕ(z)dz =2iπ.(1/(2i(√3)))e^(−(√3)) {e^((2iπ)/3) +e^(−((2iπ)/3)) } =(π/( (√3)))e^(−(√3)) .(2cos(((2π)/3))) =−(π/( (√3)))e^(−(√(3 )) ) ⇒ ∫_(−∞) ^(+∞) ((cos(2x))/(x^4 +x^2 +1))dx =−(π/( (√3)))e^(−(√3))$
	$Φ = \int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{4} + x^{2} + 1} dx \Rightarrow Φ = Re (\int_{- \infty}^{+ \infty} \frac{e^{2 ix}}{x^{4} + x^{2} + 1} dx)$ $let φ (z) = \frac{e^{2 iz}}{z^{4} + z^{2} + 1} poles of φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow u^{2} + u + 1 = 0 (z^{2} = u)$ $Δ = - 3 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and u_{2} = e^{- \frac{2 i π}{3}}$ $\Rightarrow φ (z) = \frac{e^{2 iz}}{(z^{2} - e^{\frac{2 i π}{3}}) (z^{2} - e^{- \frac{2 i π}{3}})} = \frac{e^{2 iz}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{\frac{i π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π (Res (φ, e^{\frac{i π}{3}}) + Res (φ, - e^{- \frac{i π}{3}}))$ $Res (φ, e^{\frac{i π}{3}}) = \frac{e^{{2 ie}^{\frac{i π}{3}}}}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{2 i (\frac{1}{2} + \frac{i \sqrt{3}}{2})}}{4 i . \frac{\sqrt{3}}{2}} e^{- \frac{i π}{3}}$ $= \frac{1}{2 i \sqrt{3}} e^{- \sqrt{3}} . e^{i - \frac{i π}{3}} = \frac{1}{2 i \sqrt{3}} e^{- \sqrt{3}} . e^{\frac{2 i π}{3}}$ $Res (φ, - e^{- \frac{i π}{3}}) = \frac{e^{- {2 ie}^{- \frac{i π}{3}}}}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))}$ $= \frac{e^{- 2 i (\frac{1}{2} - \frac{i \sqrt{3}}{2})}}{4 i . \frac{\sqrt{3}}{2}} . e^{\frac{i π}{3}} = \frac{1}{2 i \sqrt{3}} . e^{- \sqrt{3}} . e^{- i + \frac{i π}{3}}$ $= \frac{1}{2 i \sqrt{3}} e^{- \sqrt{3}} . e^{- \frac{2 i π}{3}} \Rightarrow$ $\int_{R} φ (z) dz = 2 i π . \frac{1}{2 i \sqrt{3}} e^{- \sqrt{3}} {e^{\frac{2 i π}{3}} + e^{- \frac{2 i π}{3}}}$ $= \frac{π}{\sqrt{3}} e^{- \sqrt{3}} . (2 \cos (\frac{2 π}{3})) = - \frac{π}{\sqrt{3}} e^{- \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{4} + x^{2} + 1} dx = - \frac{π}{\sqrt{3}} e^{- \sqrt{3}}$
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