Question Number 103974 by abdomsup last updated on 18/Jul/20

calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +3)^2 ))

Answered by OlafThorendsen last updated on 18/Jul/20

R(x) = (1/((x^2 −x+1)(x^2 +3)^2 )) R(x) = ((x−2)/(7(x^2 +3)^2 ))+((5x−3)/(49(x^2 +3)))−((5x−8)/(x^2 −x+1)) ∫_(−∞) ^(+∞) R(x)dx = −(2/7)∫_(−∞) ^(+∞) (dx/((x^2 +3)^2 )) (I) −(3/(49))∫_(−∞) ^(+∞) (dx/(x^2 +3)) (J) −∫_(−∞) ^(+∞) ((5x−8)/(x^2 −x+1)) (K) (we don′t keep odd functions) I = −(2/7)[(x/(6(x^2 +3)))+(1/(6(√3)))arctan(x/(√3))]_(−∞) ^(+∞) I = −(π/(21(√3))) J = −(3/(49))[(1/(√3))arctan(x/(√3))]_(−∞) ^(+∞) J = −((3π)/(49(√3))) K = −[((11)/(√3))arctan((1/(√3))−((2x)/(√3)))+(5/2)ln(x^2 −x+1)]_(−∞) ^(+∞) K = ((11π)/(√3)) ∫_(−∞) ^(+∞) R(x)dx = I+J+K = −(π/(21(√3)))−((3π)/(49(√3)))+((11π)/(√3)) = ((1601π)/(147(√3))) sorry, i′m not sure of my result.

Commented bymathmax by abdo last updated on 19/Jul/20

nevermind you do a work thanks sir.

Answered by mathmax by abdo last updated on 19/Jul/20

I =∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +3)^2 )) let ϕ(z) =(1/((z^2 −z +1)(z^2 +3)^2 )) poles 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) =(1/((z−e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z−i(√3))^2 (z+i(√3))^2 )) rwsidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ ,e^((iπ)/3) ) +Res(ϕ,i(√3))} Res(ϕ ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^(i(π/3)) )ϕ(z) = (1/(2isin((π/3))(e^((2iπ)/3) +3)^2 )) =(1/(2i((√3)/2)(e^((2iπ)/3) +3)^2 )) Res(ϕ,i(√3)) =lim_(z→i(√3)) (1/((2−1)!)){(z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) {(1/((z^2 −z+1)(z+i(√3))^2 ))}^((1)) =lim_(z→i(√3)) −(((2z−1)(z+i(√3))^2 +2(z+i(√3))(z^2 −z+1))/((z^2 −z+1)^2 (z+i(√3))^4 )) =lim_(z→i(√3)) −(((2z−1)/((z^2 −z+1)^2 (z+i(√3))^2 )) +(2/((z^2 −z+1)(z+i(√3))^3 ))) =−{((2i(√3)−1)/((−3−i(√3)+1)^2 (2i(√3))^2 )) +(2/((−3−i(√3)+1)(2i(√3))^3 ))} =−{((2i(√3)−1)/((2+i(√3))^2 (−12))) +(2/((−2−i(√3))(−24i)))} ....be continued...