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


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

$calculate \int_{- \infty}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{2} {({2 x}^{2} + 5)}^{2}}$
	Answered by mathmax by abdo last updated on 18/Jul/20
	$A =∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^2 (2x^2 +5)^2 )) let ϕ(z) =(1/((z^2 +z+1)^2 (2z^2 +5)^2 )) poles of ϕ? z^2 +z +1 =0→Δ =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3) and z_2 =e^(−((i2π)/3)) 2z^2 +5 =0 ⇒z^2 +(5/2) =0 ⇒z^2 =−(5/2) ⇒z =+^− i(√(5/2)) ⇒ϕ(z) =(1/(4(z−e^((i2π)/3) )^2 (z+e^((i2π)/3) )^2 (z−i(√(5/2)))^2 (z+i(√(5/2)))^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((i2π)/3) )+Res(ϕ,i(√(5/2)))} Res(ϕ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){ (z−e^((i2π)/3) )^2 ϕ(z)}^((1)) =lim_(z→e((i2π)/3)) {(1/(4(z+e^((i2π)/3) )^2 (z^2 +(5/2))^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) −(1/4)×((2(z+e^((i2π)/3) )(z^2 +(5/2))^2 +4z (z^2 +(5/2))(z+e^((i2π)/3) )^2 )/((z+e^((i2π)/3) )^4 (z^2 +(5/2))^4 )) =−(1/4)lim_(z→e^((i2π)/3) ) {(2/((z+e^((i2π)/3) )^3 (z^2 +(5/2))^2 )) +((4z)/((z+e^((i2π)/3) )^2 (z^2 +(5/2))^3 ))} ...be continued...$
	$A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + x + 1)}^{2} {({2 x}^{2} + 5)}^{2}} let φ (z) = \frac{1}{{(z^{2} + z + 1)}^{2} {({2 z}^{2} + 5)}^{2}} poles of φ ?$ $z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and z_{2} = e^{- \frac{i 2 π}{3}}$ ${2 z}^{2} + 5 = 0 \Rightarrow z^{2} + \frac{5}{2} = 0 \Rightarrow z^{2} = - \frac{5}{2} \Rightarrow z = \bar{+} i \sqrt{\frac{5}{2}}$ $\Rightarrow φ (z) = \frac{1}{4 {(z - e^{\frac{i 2 π}{3}})}^{2} {(z + e^{\frac{i 2 π}{3}})}^{2} {(z - i \sqrt{\frac{5}{2}})}^{2} {(z + i \sqrt{\frac{5}{2}})}^{2}}$ $residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i 2 π}{3}}) + Res (φ, i \sqrt{\frac{5}{2}})}$ $Res (φ, e^{\frac{i 2 π}{3}}) = \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i 2 π}{3}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to e \frac{i 2 π}{3}} {\frac{1}{4 {(z + e^{\frac{i 2 π}{3}})}^{2} {(z^{2} + \frac{5}{2})}^{2}}}^{(1)}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} - \frac{1}{4} \times \frac{2 (z + e^{\frac{i 2 π}{3}}) {(z^{2} + \frac{5}{2})}^{2} + 4 z (z^{2} + \frac{5}{2}) {(z + e^{\frac{i 2 π}{3}})}^{2}}{{(z + e^{\frac{i 2 π}{3}})}^{4} {(z^{2} + \frac{5}{2})}^{4}}$ $= - \frac{1}{4} \lim_{z \to e^{\frac{i 2 π}{3}}} {\frac{2}{{(z + e^{\frac{i 2 π}{3}})}^{3} {(z^{2} + \frac{5}{2})}^{2}} + \frac{4 z}{{(z + e^{\frac{i 2 π}{3}})}^{2} {(z^{2} + \frac{5}{2})}^{3}}}$ $. . . be continued . . .$
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