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


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

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - x + 1) {(x^{2} + 3)}^{2}}$
	Answered by OlafThorendsen last updated on 18/Jul/20

	$R (x) = \frac{1}{(x^{2} - x + 1) {(x^{2} + 3)}^{2}}$ $R (x) = \frac{x - 2}{7 {(x^{2} + 3)}^{2}} + \frac{5 x - 3}{49 (x^{2} + 3)} - \frac{5 x - 8}{x^{2} - x + 1}$ $\int_{- \infty}^{+ \infty} R (x) d x =$ $- \frac{2}{7} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 3)}^{2}} (I)$ $- \frac{3}{49} \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 3} (J)$ $- \int_{- \infty}^{+ \infty} \frac{5 x - 8}{x^{2} - x + 1} (K)$ $(we {don}^{'} t keep odd functions)$ $I = - \frac{2}{7} {[\frac{x}{6 (x^{2} + 3)} + \frac{1}{6 \sqrt{3}} \arctan \frac{x}{\sqrt{3}}]}_{- \infty}^{+ \infty}$ $I = - \frac{π}{21 \sqrt{3}}$ $J = - \frac{3}{49} {[\frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}}]}_{- \infty}^{+ \infty}$ $J = - \frac{3 π}{49 \sqrt{3}}$ $K = - {[\frac{11}{\sqrt{3}} \arctan (\frac{1}{\sqrt{3}} - \frac{2 x}{\sqrt{3}}) + \frac{5}{2} \ln (x^{2} - x + 1)]}_{- \infty}^{+ \infty}$ $K = \frac{11 π}{\sqrt{3}}$ $\int_{- \infty}^{+ \infty} R (x) d x = I + J + K$ $= - \frac{π}{21 \sqrt{3}} - \frac{3 π}{49 \sqrt{3}} + \frac{11 π}{\sqrt{3}}$ $= \frac{1601 π}{147 \sqrt{3}}$ $sorry, i^{'} m not sure of my result .$
	Commented by mathmax 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
	$I =∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +3)^2 )) let ϕ(z) =(1/((z^2 −z +1)(z^2 +3)^2 )) poles of ϕ z^2 −z+1 =0→Δ =−3 ⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) and z_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ϕ(z) =(1/((z−e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z−i(√3))^2 (z+i(√3))^2 )) rwsidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ ,e^((iπ)/3) ) +Res(ϕ,i(√3))} Res(ϕ ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^(i(π/3)) )ϕ(z) = (1/(2isin((π/3))(e^((2iπ)/3) +3)^2 )) =(1/(2i((√3)/2)(e^((2iπ)/3) +3)^2 )) Res(ϕ,i(√3)) =lim_(z→i(√3)) (1/((2−1)!)){(z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) {(1/((z^2 −z+1)(z+i(√3))^2 ))}^((1)) =lim_(z→i(√3)) −(((2z−1)(z+i(√3))^2 +2(z+i(√3))(z^2 −z+1))/((z^2 −z+1)^2 (z+i(√3))^4 )) =lim_(z→i(√3)) −(((2z−1)/((z^2 −z+1)^2 (z+i(√3))^2 )) +(2/((z^2 −z+1)(z+i(√3))^3 ))) =−{((2i(√3)−1)/((−3−i(√3)+1)^2 (2i(√3))^2 )) +(2/((−3−i(√3)+1)(2i(√3))^3 ))} =−{((2i(√3)−1)/((2+i(√3))^2 (−12))) +(2/((−2−i(√3))(−24i)))} ....be continued...$
	$I = \int_{- \infty}^{+ \infty} \frac{dx}{(x^{2} - x + 1) {(x^{2} + 3)}^{2}} let φ (z) = \frac{1}{(z^{2} - z + 1) {(z^{2} + 3)}^{2}} poles of φ$ $z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} and z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}$ $\Rightarrow φ (z) = \frac{1}{(z - e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) {(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2}} rwsidus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{3}}) + Res (φ, i \sqrt{3})}$ $Res (φ, e^{\frac{i π}{3}}) = \lim_{z \to e^{\frac{i π}{3}}} (z - e^{i \frac{π}{3}}) φ (z) = \frac{1}{2 isin (\frac{π}{3}) {(e^{\frac{2 i π}{3}} + 3)}^{2}} = \frac{1}{2 i \frac{\sqrt{3}}{2} {(e^{\frac{2 i π}{3}} + 3)}^{2}}$ $Res (φ, i \sqrt{3}) = \lim_{z \to i \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i \sqrt{3})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to i \sqrt{3}} {\frac{1}{(z^{2} - z + 1) {(z + i \sqrt{3})}^{2}}}^{(1)}$ $= \lim_{z \to i \sqrt{3}} - \frac{(2 z - 1) {(z + i \sqrt{3})}^{2} + 2 (z + i \sqrt{3}) (z^{2} - z + 1)}{{(z^{2} - z + 1)}^{2} {(z + i \sqrt{3})}^{4}}$ $= \lim_{z \to i \sqrt{3}} - (\frac{2 z - 1}{{(z^{2} - z + 1)}^{2} {(z + i \sqrt{3})}^{2}} + \frac{2}{(z^{2} - z + 1) {(z + i \sqrt{3})}^{3}})$ $= - {\frac{2 i \sqrt{3} - 1}{{(- 3 - i \sqrt{3} + 1)}^{2} {(2 i \sqrt{3})}^{2}} + \frac{2}{(- 3 - i \sqrt{3} + 1) {(2 i \sqrt{3})}^{3}}}$ $= - {\frac{2 i \sqrt{3} - 1}{{(2 + i \sqrt{3})}^{2} (- 12)} + \frac{2}{(- 2 - i \sqrt{3}) (- 24 i)}} . . . . be continued . . .$
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