find ∫_0 ^∞ ((xsin(2x))/((x^2 +x+1)^2 ))dx


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Question Number 130530 by mathmax by abdo last updated on 26/Jan/21

$find \int_{0}^{\infty} \frac{xsin (2 x)}{{(x^{2} + x + 1)}^{2}} dx$
Commented by mathmax by abdo last updated on 27/Jan/21

$the question is find \int_{- \infty}^{+ \infty} \frac{xsin (2 x)}{{(x^{2} + x + 1)}^{2}} dx$
Commented by mathmax by abdo last updated on 28/Jan/21
$A=∫_(−∞) ^(+∞) ((xsin(2x))/((x^2 +x+1)^2 ))dx ⇒A =Im(∫_(−∞) ^(+∞) ((xe^(2ix) )/((x^2 +x+1)^2 ))dx)let ϕ(z)=((ze^(2iz) )/((z^2 +z+1)^2 )) poles of ϕ? z^2 +z+1 =0→Δ=−3 ⇒z_1 =((−1+i(√3))/2)=e^((i2π)/3) and z_2 =e^(−((i2π)/3)) ⇒ ϕ(z)=((ze^(2iz) )/((z−e^((i2π)/3) )^2 (z−e^(−((i2π)/3)) )^2 )) so the poles of ϕ are z_i (doubles) and ∫_R ϕ(z)dz =2iπ Res(ϕ,e^((i2π)/3) ) Res(ϕ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){(z−e^((i2π)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((i2π)/3) ) {((ze^(2iz) )/((z−e^(−((i2π)/3)) )^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) (((e^(2iz) +2iz e^(2iz) )(z−e^(−((i2π)/3)) )^2 −2ze^(2iz) (z−e^(−((i2π)/3)) ))/((z−e^(−((i2π)/3)) )^4 )) =lim_(z→e^((i2π)/3) ) ((e^(2iz) (1+2iz)(z−e^(−((i2π)/3)) )−2ze^(2iz) )/((z−e^(−((2iπ)/3)) )^3 )) =((e^(2ie^((i2π)/3) ) {(1+2ie^((i2π)/3) )(2isin(((2π)/3))−2e^((i2π)/3) })/((2isin(((2π)/3)))^3 )) =((e^(2i(−(1/2)+((i(√3))/2))) {(1+2i(−(1/2)+((i(√3))/2))(i(√3))−2(−(1/2)+((i(√3))/2))})/((i(√3))^3 )) =((e^(−(√3)) e^(−i) {(1−i−(√3))i(√3)+1−i(√3)))/(−3(√3)i)) =((e^(−(√3)) e^(−i) {i(√3)+(√3)−3i+1−i(√3)})/(−3(√3)i)) =((e^(−(√3)) (cos1−isin1)(1+(√3)−3i))/(−3(√3)i)) =((e^(−(√3)) {(1+(√3))cos(1)−3icos(1)−i(1+(√3))sin(1)−3sin(1))/(−3(√3)i)) ⇒ ∫_R ϕ(z)dz =2iπ×(...) =−(2/(3(√3))){....} ⇒A =−(2/(3(√3)))(−3cos(1)−(1+(√3))sin(1)} =(2/(3(√3)))(3cos(1)+(1+(√3))sin(1))$
$A = \int_{- \infty}^{+ \infty} \frac{xsin (2 x)}{{(x^{2} + x + 1)}^{2}} dx \Rightarrow A = Im (\int_{- \infty}^{+ \infty} \frac{{xe}^{2 ix}}{{(x^{2} + x + 1)}^{2}} dx) let$ $φ (z) = \frac{{ze}^{2 iz}}{{(z^{2} + z + 1)}^{2}} poles of φ ?$ $z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and z_{2} = e^{- \frac{i 2 π}{3}} \Rightarrow$ $φ (z) = \frac{{ze}^{2 iz}}{{(z - e^{\frac{i 2 π}{3}})}^{2} {(z - e^{- \frac{i 2 π}{3}})}^{2}} so the poles of φ are z_{i} (doubles)$ $and \int_{R} φ (z) dz = 2 i π Res (φ, e^{\frac{i 2 π}{3}})$ $Res (φ, e^{\frac{i 2 π}{3}}) = \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i 2 π}{3}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} {\frac{{ze}^{2 iz}}{{(z - e^{- \frac{i 2 π}{3}})}^{2}}}^{(1)}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{(e^{2 iz} + 2 iz e^{2 iz}) {(z - e^{- \frac{i 2 π}{3}})}^{2} - {2 ze}^{2 iz} (z - e^{- \frac{i 2 π}{3}})}{{(z - e^{- \frac{i 2 π}{3}})}^{4}}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{e^{2 iz} (1 + 2 iz) (z - e^{- \frac{i 2 π}{3}}) - {2 ze}^{2 iz}}{{(z - e^{- \frac{2 i π}{3}})}^{3}}$ $= \frac{e^{{2 ie}^{\frac{i 2 π}{3}}} {(1 + {2 ie}^{\frac{i 2 π}{3}}) (2 isin (\frac{2 π}{3}) - {2 e}^{\frac{i 2 π}{3}}}}{{(2 isin (\frac{2 π}{3}))}^{3}}$ $= \frac{e^{2 i (- \frac{1}{2} + \frac{i \sqrt{3}}{2})} {(1 + 2 i (- \frac{1}{2} + \frac{i \sqrt{3}}{2}) (i \sqrt{3}) - 2 (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}}{{(i \sqrt{3})}^{3}}$ $= \frac{e^{- \sqrt{3}} e^{- i} {(1 - i - \sqrt{3}) i \sqrt{3} + 1 - i \sqrt{3})}{- 3 \sqrt{3} i}$ $= \frac{e^{- \sqrt{3}} e^{- i} {i \sqrt{3} + \sqrt{3} - 3 i + 1 - i \sqrt{3}}}{- 3 \sqrt{3} i}$ $= \frac{e^{- \sqrt{3}} (\cos 1 - isin 1) (1 + \sqrt{3} - 3 i)}{- 3 \sqrt{3} i}$ $= \frac{e^{- \sqrt{3}} {(1 + \sqrt{3}) \cos (1) - 3 icos (1) - i (1 + \sqrt{3}) \sin (1) - 3 \sin (1)}{- 3 \sqrt{3} i} \Rightarrow$ $\int_{R} φ (z) dz = 2 i π \times (. . .)$ $= - \frac{2}{3 \sqrt{3}} {. . . .} \Rightarrow A = - \frac{2}{3 \sqrt{3}} (- 3 \cos (1) - (1 + \sqrt{3}) \sin (1)}$ $= \frac{2}{3 \sqrt{3}} (3 \cos (1) + (1 + \sqrt{3}) \sin (1))$
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