Evaluate Ω if Ω = ∫_0 ^( ∞) ((cos(mx))/(x^4 +x^2 +1))dx


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Question Number 126383 by Lordose last updated on 20/Dec/20

$Evaluate Ω if$ $Ω = \int_{0}^{\infty} \frac{\cos (mx)}{x^{4} + x^{2} + 1} dx$
	Answered by mathmax by abdo last updated on 20/Dec/20
	$I =∫_0 ^∞ ((cos(mx))/(x^4 +x^2 +1))dx ⇒2I =∫_(−∞) ^(+∞) ((cos(mx))/(x^4 +x^2 +1))dx=Re(∫_(−∞) ^(+∞) (e^(imx) /(x^4 +x^2 +1))dx) let ϕ(z)=(e^(imz) /(z^4 +z^2 +1)) ]poles of ϕ? z^4 +z^2 +1=0 ⇒u^2 +u+1=0 (u=z^2 ) Δ=−3 ⇒u_1 =((−1+i(√3))/2)=e^((i2π)/3) and u_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ ϕ(z)=(e^(imz) /((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) =(e^(imz) /((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/3) )+Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =(e^(ime^((iπ)/3) ) /(2e^((iπ)/3) (2isin((π/3))))) =(1/(4i×((√3)/2))) e^(−((iπ)/3)) e^(im((1/2)+((i(√3))/2))) =(1/(2i(√3))) e^(i((m/2)−(π/3))) e^(−((√3)/2)m) Res(ϕ,−e^(−((iπ)/3)) ) =(e^(ime^(−((iπ)/3)) ) /(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(1/(2i(√3))) e^((iπ)/3) e^(im((1/2)−((i(√3))/2))) =(1/(2i(√3))) e^(i((m/2)+(π/3))) e^(((√3)/2)m) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(1/(2i(√3))){ e^(−((√3)/2)m) e^(i((m/2)−(π/3))) +e^(((√3)/2)m) e^(i((m/2)+(π/3))) } =(π/( (√3))){e^(−((m(√3))/2)) (cos((m/2)−(π/3))+i sin((m/2)−(π/3))) +e^((m(√3))/2) ( cos((m/2)+(π/3))+isin((m/2)+(π/3)))} ⇒ I =(1/2)Re(∫...)=(π/( 2(√3))){ e^(−((m(√3))/2)) cos((m/2)−(π/3))+e^((m(√3))/(2 )) cos((m/2)+(π/3))}$
	$I = \int_{0}^{\infty} \frac{\cos (mx)}{x^{4} + x^{2} + 1} dx \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{\cos (mx)}{x^{4} + x^{2} + 1} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{imx}}{x^{4} + x^{2} + 1} dx)$ $let φ (z) = \frac{e^{imz}}{z^{4} + z^{2} + 1}] poles of φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow u^{2} + u + 1 = 0 (u = z^{2})$ $Δ = - 3 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and u_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow$ $φ (z) = \frac{e^{imz}}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})} = \frac{e^{imz}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{3}}) + Res (φ, - e^{- \frac{i π}{3}})}$ $Res (φ, e^{\frac{i π}{3}}) = \frac{e^{{ime}^{\frac{i π}{3}}}}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{π}{3}))} = \frac{1}{4 i \times \frac{\sqrt{3}}{2}} e^{- \frac{i π}{3}} e^{im (\frac{1}{2} + \frac{i \sqrt{3}}{2})}$ $= \frac{1}{2 i \sqrt{3}} e^{i (\frac{m}{2} - \frac{π}{3})} e^{- \frac{\sqrt{3}}{2} m}$ $Res (φ, - e^{- \frac{i π}{3}}) = \frac{e^{{ime}^{- \frac{i π}{3}}}}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = \frac{1}{2 i \sqrt{3}} e^{\frac{i π}{3}} e^{im (\frac{1}{2} - \frac{i \sqrt{3}}{2})}$ $= \frac{1}{2 i \sqrt{3}} e^{i (\frac{m}{2} + \frac{π}{3})} e^{\frac{\sqrt{3}}{2} m} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{1}{2 i \sqrt{3}} {e^{- \frac{\sqrt{3}}{2} m} e^{i (\frac{m}{2} - \frac{π}{3})} + e^{\frac{\sqrt{3}}{2} m} e^{i (\frac{m}{2} + \frac{π}{3})}}$ $= \frac{π}{\sqrt{3}} {e^{- \frac{m \sqrt{3}}{2}} (\cos (\frac{m}{2} - \frac{π}{3}) + i \sin (\frac{m}{2} - \frac{π}{3}))$ $+ e^{\frac{m \sqrt{3}}{2}} (\cos (\frac{m}{2} + \frac{π}{3}) + isin (\frac{m}{2} + \frac{π}{3}))} \Rightarrow$ $I = \frac{1}{2} Re (\int . . .) = \frac{π}{2 \sqrt{3}} {e^{- \frac{m \sqrt{3}}{2}} \cos (\frac{m}{2} - \frac{π}{3}) + e^{\frac{m \sqrt{3}}{2}} \cos (\frac{m}{2} + \frac{π}{3})}$
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