Question Number 103593 by mathmax by abdo last updated on 16/Jul/20

calculate  ∫_(−∞) ^∞    (dx/((x^2  +x +1)^2 (2x^2  +5)^2 ))

Answered by mathmax by abdo last updated on 18/Jul/20

A =∫_(−∞) ^(+∞)  (dx/((x^2  +x+1)^2 (2x^2  +5)^2 ))  let ϕ(z) =(1/((z^2  +z+1)^2 (2z^2  +5)^2 )) poles of ϕ?  z^2  +z +1 =0→Δ =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3)  and z_2 =e^(−((i2π)/3))    2z^2  +5 =0 ⇒z^2 +(5/2) =0 ⇒z^2  =−(5/2) ⇒z =+^− i(√(5/2))  ⇒ϕ(z) =(1/(4(z−e^((i2π)/3) )^2 (z+e^((i2π)/3) )^2 (z−i(√(5/2)))^2 (z+i(√(5/2)))^2 ))  residus theorem give  ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((i2π)/3) )+Res(ϕ,i(√(5/2)))}  Res(ϕ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) )    (1/((2−1)!)){ (z−e^((i2π)/3) )^2  ϕ(z)}^((1))   =lim_(z→e((i2π)/3))   {(1/(4(z+e^((i2π)/3) )^2 (z^2  +(5/2))^2 ))}^((1))   =lim_(z→e^((i2π)/3) )      −(1/4)×((2(z+e^((i2π)/3) )(z^2  +(5/2))^2  +4z (z^2  +(5/2))(z+e^((i2π)/3) )^2 )/((z+e^((i2π)/3) )^4 (z^2 +(5/2))^4 ))  =−(1/4)lim_(z→e^((i2π)/3) )      {(2/((z+e^((i2π)/3) )^3 (z^2  +(5/2))^2 )) +((4z)/((z+e^((i2π)/3) )^2 (z^2  +(5/2))^3 ))}  ...be continued...