Question Number 148565 by mathmax by abdo last updated on 29/Jul/21

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

Answered by Ar Brandon last updated on 29/Jul/21

Υ=∫_(−∞) ^(+∞) ((x^2 dx)/((x^2 −x+1)(x^2 +x+1)))=2∫_0 ^∞ ((x^2 dx)/((x^2 −x+1)(x^2 +x+1))) =2∫_0 ^1 ((x^2 dx)/((x^2 −x+1)(x^2 +x+1)))+2∫_1 ^∞ ((x^2 dx)/((x^2 −x+1)(x^2 +x+1))) =2∫_0 ^1 ((x^2 dx)/((x^2 −x+1)(x^2 +x+1)))+2∫_0 ^1 (dx/((x^2 −x+1)(x^2 +x+1))) =2∫_0 ^1 ((x^2 +1)/((x^2 −x+1)(x^2 +x+1)))dx=2∫_0 ^1 ((x^2 +1)/(x^4 +x^2 +1))dx =2∫_0 ^1 ((1+(1/x^2 ))/(x^2 +1+(1/x^2 )))dx=2∫_0 ^1 ((1+(1/x^2 ))/((x−(1/x))^2 +3))dx =2∫_0 ^1 ((d(x−(1/x)))/((x−(1/x))^2 +3))=(2/( (√3)))[arctan(((x^2 −1)/( (√3)x)))]_0 ^1 =(π/( (√3)))

Answered by mathmax by abdo last updated on 29/Jul/21

Ψ=∫_(−∞) ^(+∞) (x^2 /((x^2 −x+1)(x^2 +x+1)))dx let Λ(z)=(z^2 /((z^2 −z+1)(z^2 +z+1))) poles of Λ? z^2 −z+1=0→Δ=−3⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) and z_2 =((1−i(√3))/2)=e^(−((iπ)/3)) z^2 +z+1=0→Δ=−3 ⇒x_1 =((−1+i(√3))/2)=e^((2iπ)/3) ⇒x_2 =e^(−((2iπ)/3)) ⇒ Λ(z)=(z^2 /((z−e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z−e^((2iπ)/3) )(z−e^(−((2iπ)/3)) ))) ∫_(−∞) ^(+∞) Λ(z)dz=2iπ{Res(Λ,e^((iπ)/3) )+Res(Λ,e^((2iπ)/3) )} Res(Λ,e^((iπ)/3) ) =(e^((2iπ)/3) /(2isin((π/3))×))(((e^((iπ)/3) −1))/(e^(iπ) −1)) =(1/(−4i×((√3)/2)))(−1−e^((2iπ)/3) ) =((1+e^((2iπ)/3) )/(2i(√3))) Res(Λ,e^((2iπ)/3) ) =(e^((4iπ)/3) /(2isin(((2π)/3))))×((e^((2iπ)/3) +1)/2) =((1+e^((4iπ)/3) )/(4i×((√3)/2)))=((1+e^((4iπ)/3) )/(2i(√3))) ⇒ ∫_(−∞) ^(+∞) Λ(z)dz=((2iπ)/(2i(√3))){1+e^((2iπ)/3) +1−e^((iπ)/3) } =(π/( (√3))){2 −(1/2)+((i(√3))/2) −(1/2)−i((√3)/2)} =(π/( (√3))) ⇒Ψ=(π/( (√3))) .