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Question Number 136034 by mathmax by abdo last updated on 18/Mar/21
calculate ∫_(−∞) ^(+∞)  ((cos(2x)dx)/(x^4  +x^2  +1))
calculate+cos(2x)dxx4+x2+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=+cos(2x)x4+x2+1dxΦ=Re(+e2ixx4+x2+1dx)letφ(z)=e2izz4+z2+1polesofφ?z4+z2+1=0u2+u+1=0(z2=u)Δ=3u1=1+i32=e2iπ3andu2=e2iπ3φ(z)=e2iz(z2e2iπ3)(z2e2iπ3)=e2iz(zeiπ3)(z+eiπ3)(zeiπ3)(z+eiπ3)+φ(z)dz=2iπ(Res(φ,eiπ3)+Res(φ,eiπ3))Res(φ,eiπ3)=e2ieiπ32eiπ3(2isin(2π3))=e2i(12+i32)4i.32eiπ3=12i3e3.eiiπ3=12i3e3.e2iπ3Res(φ,eiπ3)=e2ieiπ32eiπ3(2isin(2π3))=e2i(12i32)4i.32.eiπ3=12i3.e3.ei+iπ3=12i3e3.e2iπ3Rφ(z)dz=2iπ.12i3e3{e2iπ3+e2iπ3}=π3e3.(2cos(2π3))=π3e3+cos(2x)x4+x2+1dx=π3e3

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