Tinkutara Equation Editor: Math Forum Question #: 102165


Question Number 102165 by mathmax by abdo last updated on 07/Jul/20

	Answered by MAB last updated on 07/Jul/20

	Answered by mathmax by abdo last updated on 07/Jul/20
	$I =∫_(−∞) ^(+∞) ((x^2 dx)/((x^2 −x+1)^3 )) let ϕ(z) =(z^2 /((z^2 −z+1)^3 )) pole of ϕ z^2 −z+1 =0 →Δ =−3 ⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) and z_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ϕ(z) =(z^2 /((z−z_1 )^3 (z−z_2 )^3 )) the poles of ϕ are e^((iπ)/3) and e^(−((iπ)/3) ) (triples) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/3) ) and Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((3−1)!)){(z−e^((iπ)/3) )^3 ϕ(z)}^((2)) =(1/2)lim_(z→e^((iπ)/3) ) {(z^2 /((z−e^(−((iπ)/3)) )^3 ))}^((2)) =lim_(z→e^((iπ)/3) ) {((2z(z−e^(−((iπ)/3)) )^3 −3(z−e^(−((iπ)/3)) )^2 z^2 )/((z−e^(−((iπ)/3)) )^6 ))}^((1)) =lim_(z→e^((iπ)/3) ) { ((2z(z−e^(−((iπ)/3)) )−3z^2 )/((z−e^(−((iπ)/3)) )^4 ))}^((1)) =lim_(z→e^((iπ)/3) ) {((−z^2 −2z e^(−((iπ)/3)) )/((z−e^(−((iπ)/3)) )^4 ))}^((1)) −lim_(z→e^((iπ)/3) ) (((2z+2e^(−((iπ)/3)) )(z−e^(−((iπ)/3)) )^4 −4(z−e^(−((iπ)/3)) )^3 (z^2 +2z e^(−((iπ)/3)) ))/((z−e^(−((iπ)/3)) )^8 )) =−lim_(z→e^((iπ)/3) ) (((2z+2e^(−((iπ)/3)) )(z−e^(−((iπ)/3)) )−4(z^2 +2z e^(−((iπ)/3)) ))/((z−e^(−((iπ)/3)) )^5 )) =− ((4cos((π/3))2isin((π/3))−4{e^((2iπ)/3) +2})/((2isin((π/3)))^5 )) =−((8×(1/2)×((√3)/2)i−8 −4e^((i2π)/3) )/(32i(((√3)/2))^5 )) =((2(√3)i −8 −4(−(1/2)+((i(√3))/2)))/(32i(((√3)/2))^5 )) =(6/(32i((√3))^5 ))×(2^5 /(√3)) =(6/(i((√3))^6 )) =(6/(3^3 i)) =(2/(9i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(2/(9i)) =((4π)/9) ⇒ I =((4π)/9)$
	Commented bymathmax by abdo last updated on 07/Jul/20

	Commented byMAB last updated on 07/Jul/20