calculate ∫_0 ^∞ ((cos(3x))/((x^2 +3)^2 ))dx


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Question Number 97619 by mathmax by abdo last updated on 08/Jun/20

$calculate \int_{0}^{\infty} \frac{\cos (3 x)}{{(x^{2} + 3)}^{2}} dx$
	Answered by mathmax by abdo last updated on 11/Jun/20
	$I =∫_0 ^∞ ((cos(3x))/((x^2 +3)^2 ))dx changement x =(√3)t give I =∫_0 ^∞ ((cos(3(√3)t))/(9(t^2 +1)^2 ))(√3)dt =((√3)/9) ∫_0 ^∞ ((cos(3(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18)) ∫_(−∞) ^(+∞) ((cos(3(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18)) Re(∫_(−∞) ^(+∞) (e^(3(√3)it) /((t^2 +1)^2 ))) let considere the complex function ϕ(z) =(e^(3i(√3)z) /((z^2 +1)^2 )) ⇒ ϕ(z) =(e^(3i(√3)z) /((z−i)^2 (z+i)^2 )) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(3i(√3)z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((3i(√3)e^(3i(√3)z) (z+i)^2 −2(z+i)e^(3i(√3)z) )/((z+i)^4 )) =lim_(z→i) (((3i(√3)(z+i)−2)e^(3i(√3)z) )/((z+i)^3 )) =(((−6(√3)−2) e^(−3(√3)) )/((2i)^3 )) =(((−6(√3)−2)e^(−3(√3)) )/(−8i)) =(((3(√3)+1)e^(−3(√3)) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((3(√3)+1)e^(−3(√3)) )/(4i)) =(π/2)(3(√3)+1)e^(−3(√3)) ⇒ I =((√3)/(18))×(π/2)(3(√3)+1)e^(−3(√3)) =((π(√3))/(36))(3(√3)+1)e^(−3(√3))$
	$I = \int_{0}^{\infty} \frac{\cos (3 x)}{{(x^{2} + 3)}^{2}} dx changement x = \sqrt{3} t give I = \int_{0}^{\infty} \frac{\cos (3 \sqrt{3} t)}{9 {(t^{2} + 1)}^{2}} \sqrt{3} dt$ $= \frac{\sqrt{3}}{9} \int_{0}^{\infty} \frac{\cos (3 \sqrt{3} t)}{{(t^{2} + 1)}^{2}} dt = \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{\cos (3 \sqrt{3} t)}{{(t^{2} + 1)}^{2}} dt = \frac{\sqrt{3}}{18} Re (\int_{- \infty}^{+ \infty} \frac{e^{3 \sqrt{3} it}}{{(t^{2} + 1)}^{2}})$ $let considere the complex function φ (z) = \frac{e^{3 i \sqrt{3} z}}{{(z^{2} + 1)}^{2}} \Rightarrow$ $φ (z) = \frac{e^{3 i \sqrt{3} z}}{{(z - i)}^{2} {(z + i)}^{2}} residus tbeorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i)$ $Res (φ, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= \lim_{z \to i} {\frac{e^{3 i \sqrt{3} z}}{{(z + i)}^{2}}}^{(1)} = \lim_{z \to i} \frac{3 i \sqrt{3} e^{3 i \sqrt{3} z} {(z + i)}^{2} - 2 (z + i) e^{3 i \sqrt{3} z}}{{(z + i)}^{4}}$ $= \lim_{z \to i} \frac{(3 i \sqrt{3} (z + i) - 2) e^{3 i \sqrt{3} z}}{{(z + i)}^{3}} = \frac{(- 6 \sqrt{3} - 2) e^{- 3 \sqrt{3}}}{{(2 i)}^{3}} = \frac{(- 6 \sqrt{3} - 2) e^{- 3 \sqrt{3}}}{- 8 i}$ $= \frac{(3 \sqrt{3} + 1) e^{- 3 \sqrt{3}}}{4 i}$ $\Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{(3 \sqrt{3} + 1) e^{- 3 \sqrt{3}}}{4 i} = \frac{π}{2} (3 \sqrt{3} + 1) e^{- 3 \sqrt{3}}$ $\Rightarrow I = \frac{\sqrt{3}}{18} \times \frac{π}{2} (3 \sqrt{3} + 1) e^{- 3 \sqrt{3}} = \frac{π \sqrt{3}}{36} (3 \sqrt{3} + 1) e^{- 3 \sqrt{3}}$
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