calculate ∫_0 ^∞ ((cos(2x^2 ))/((4x^2 +9)^3 ))dx


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Question Number 105232 by mathmax by abdo last updated on 27/Jul/20

$calculate \int_{0}^{\infty} \frac{\cos ({2 x}^{2})}{{({4 x}^{2} + 9)}^{3}} dx$
	Answered by mathmax by abdo last updated on 27/Jul/20
	$A =∫_0 ^∞ ((cos(2x^2 ))/((4x^2 +9)^3 ))dx ⇒2A =∫_(−∞) ^(+∞) ((cos(2x^2 ))/((4x^2 +9)^3 ))dx =(1/4^3 ) ∫_(−∞) ^(+∞) ((cos(2x^2 ))/((x^2 +(9/4))^3 ))dx =_(x =(3/2)t) (1/4^3 ) ∫_(−∞) ^(+∞) ((cos((9/2)t^2 ))/(((9/4))^3 (t^2 +1)^3 ))(3/2)dt =(1/(162)) ∫_(−∞) ^(+∞) ((cos((9/2)t^2 ))/((t^2 +1)^3 )) dt =(1/(162)) Re(∫_(−∞) ^(+∞) (e^((9/2)it^2 ) /((t^2 +1)^3 ))dt) let ϕ(z) =(e^((9/2)iz^2 ) /((z^2 +1)^3 )) poles of ϕ? ϕ(z) =(e^((9/2)iz^2 ) /((z−i)^3 (z+i)^3 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) and Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) { (e^((9/2)iz^2 ) /((z+i)^3 ))}^((2)) =(1/2)lim_(z→i) { ((9iz e^((9/2)iz^2 ) (z+i)^3 −3(z+i)^2 e^((9/2)iz^2 ) )/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) {(((9iz (z+i)−3)e^((9/2)iz^2 ) )/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) {(((9iz^2 −12)e^((9/2)iz^2 ) )/((z+i)^4 ))}^((1)) =(3/2)lim_(z→i) {(((3iz^2 −4)e^((9/2)iz^2 ) )/((z+i)^4 ))}^((1)) =(3/2)lim_(z→i) (((6iz e^((9/2)iz^2 ) +9iz(3iz^2 −4)e^((9/2)iz^2 ) )(z+i)^4 −4(z+i)^3 (3iz^2 −4)e^((9/2)iz^2 ) )/((z+i)^8 )) =(3/2)lim_(z→i) (((6iz−27z^3 −36iz)(z+i) e^((9/2)iz^2 ) −(3iz^2 −4)e^((9/2)iz^2 ) )/((z+i)^5 )) =(3/2)×(((−6+27i+36)(2i)e^(−(9/2)i) −(−3i−4)e^(−(9/2)i) )/((2i)^5 )) ...be continued...$
	$A = \int_{0}^{\infty} \frac{\cos ({2 x}^{2})}{{({4 x}^{2} + 9)}^{3}} dx \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{\cos ({2 x}^{2})}{{({4 x}^{2} + 9)}^{3}} dx$ $= \frac{1}{4^{3}} \int_{- \infty}^{+ \infty} \frac{\cos ({2 x}^{2})}{{(x^{2} + \frac{9}{4})}^{3}} dx =_{x = \frac{3}{2} t} \frac{1}{4^{3}} \int_{- \infty}^{+ \infty} \frac{\cos (\frac{9}{2} t^{2})}{{(\frac{9}{4})}^{3} {(t^{2} + 1)}^{3}} \frac{3}{2} dt$ $= \frac{1}{162} \int_{- \infty}^{+ \infty} \frac{\cos (\frac{9}{2} t^{2})}{{(t^{2} + 1)}^{3}} dt = \frac{1}{162} Re (\int_{- \infty}^{+ \infty} \frac{e^{\frac{9}{2} {it}^{2}}}{{(t^{2} + 1)}^{3}} dt) let$ $φ (z) = \frac{e^{\frac{9}{2} {iz}^{2}}}{{(z^{2} + 1)}^{3}} poles of φ ?$ $φ (z) = \frac{e^{\frac{9}{2} {iz}^{2}}}{{(z - i)}^{3} {(z + i)}^{3}} residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) and$ $Res (φ, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{3}}}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{9 iz e^{\frac{9}{2} {iz}^{2}} {(z + i)}^{3} - 3 {(z + i)}^{2} e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{6}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{(9 iz (z + i) - 3) e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{4}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{({9 iz}^{2} - 12) e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{4}}}^{(1)}$ $= \frac{3}{2} \lim_{z \to i} {\frac{({3 iz}^{2} - 4) e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{4}}}^{(1)}$ $= \frac{3}{2} \lim_{z \to i} \frac{(6 iz e^{\frac{9}{2} {iz}^{2}} + 9 iz ({3 iz}^{2} - 4) e^{\frac{9}{2} {iz}^{2}}) {(z + i)}^{4} - 4 {(z + i)}^{3} ({3 iz}^{2} - 4) e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{8}}$ $= \frac{3}{2} \lim_{z \to i} \frac{(6 iz - {27 z}^{3} - 36 iz) (z + i) e^{\frac{9}{2} {iz}^{2}} - ({3 iz}^{2} - 4) e^{\frac{9}{2} {iz}^{2}}}{{(z + i)}^{5}}$ $= \frac{3}{2} \times \frac{(- 6 + 27 i + 36) (2 i) e^{- \frac{9}{2} i} - (- 3 i - 4) e^{- \frac{9}{2} i}}{{(2 i)}^{5}} . . . be continued . . .$
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