find the value of ∫_(−∞) ^∞ ((sin(2x^2 ))/((x^2 −x +3)^3 ))dx


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Question Number 67342 by mathmax by abdo last updated on 26/Aug/19

$f i n d t h e v a l u e o f \int_{- \infty}^{\infty} \frac{s i n (2 x^{2})}{{(x^{2} - x + 3)}^{3}} d x$
Commented by mathmax by abdo last updated on 26/Aug/19
$let I = ∫_(−∞) ^(+∞) ((sin(2x^2 ))/((x^2 −x+3)^3 ))dx ⇒ I =Im ( ∫_(−∞) ^(+∞) (e^(2iz^2 ) /((z^2 −z +3)^3 ))dz) let W(z) =(e^(2iz^2 ) /((z^2 −z +3)^3 )) poles of W? z^2 −z +3 =0 →Δ =1−12 =−11 =(i(√(11)))^2 ⇒z_1 =((1+i(√(11)))/2) z_2 =((1−i(√(11)))/2) ⇒W(z) =(e^(2iz^2 ) /((z−z_1 )^3 (z−z_2 )^3 )) let apply residus theorem ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,z_1 ) (z_1 is a triple pole) Res(W,z_1 ) =lim_(z→z_1 ) (1/((3−1)!)){(z−z_1 )^3 W(z)}^((2)) } =lim_(z→z_1 ) (1/2){ (e^(2iz^2 ) /((z−z_2 )^3 ))}^((2)) we have {(e^(2iz^2 ) /((z−z_2 )^3 ))}^((2)) ={((4iz e^(2iz^2 ) (z−z_2 )^3 −3(z−z_2 )^2 e^(2iz^2 ) )/((z−z_2 )^6 ))}^((1)) ={((4iz e^(2iz^2 ) (z−z_2 )−3e^(2iz^2 ) )/((z−z_2 )^4 ))}^((1)) ={(((4iz^2 −4iz_2 z)e^(2iz^2 ) )/((z−z_2 )^4 ))}^((1)) =(({(8iz −4iz_2 )e^(2iz^2 ) +4iz e^(2iz^2 ) (4iz^2 −4iz_2 z)(z−z_2 )^4 −4(z−z_2 )^3 {4iz^2 −4iz_2 z}e^(2iz^2 ) )/((z−z_2 )^8 )) =(({(8iz−4iz_2 −16z^3 +16z_2 z^2 )(z−z_2 )−16iz^2 +16iz_2 z}e^(2iz^2 ) )/((z−z_2 )^5 )) 2Res(W,z_1 ) =(({(8iz_1 −4iz_2 −16z_1 ^3 +16z_2 z_1 ^2 )(z_1 −z_2 )−16iz_1 ^2 +16iz_1 z_2 }e^(2iz_1 ^2 ) )/((z_1 −z_2 )^5 )) with z_1 −z_2 =i(√(11)) rest to finich the calculus...$
$l e t I = \int_{- \infty}^{+ \infty} \frac{s i n (2 x^{2})}{{(x^{2} - x + 3)}^{3}} d x \Rightarrow I = I m (\int_{- \infty}^{+ \infty} \frac{e^{2 {i z}^{2}}}{{(z^{2} - z + 3)}^{3}} d z)$ $l e t W (z) = \frac{e^{2 {i z}^{2}}}{{(z^{2} - z + 3)}^{3}} p o l e s o f W ?$ $z^{2} - z + 3 = 0 \to Δ = 1 - 12 = - 11 = {(i \sqrt{11})}^{2} \Rightarrow z_{1} = \frac{1 + i \sqrt{11}}{2}$ $z_{2} = \frac{1 - i \sqrt{11}}{2} \Rightarrow W (z) = \frac{e^{2 {i z}^{2}}}{{(z - z_{1})}^{3} {(z - z_{2})}^{3}} l e t a p p l y r e s i d u s t h e o r e m$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, z_{1}) (z_{1} i s a t r i p l e p o l e)$ $R e s (W, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(3 - 1)!} {{(z - z_{1})}^{3} W (z)}^{(2)}}$ $= {l i m}_{z \to z_{1}} \frac{1}{2} {\frac{e^{2 {i z}^{2}}}{{(z - z_{2})}^{3}}}^{(2)} w e h a v e$ ${\frac{e^{2 {i z}^{2}}}{{(z - z_{2})}^{3}}}^{(2)} = {\frac{4 i z e^{2 {i z}^{2}} {(z - z_{2})}^{3} - 3 {(z - z_{2})}^{2} e^{2 {i z}^{2}}}{{(z - z_{2})}^{6}}}^{(1)}$ $= {\frac{4 i z e^{2 {i z}^{2}} (z - z_{2}) - 3 e^{2 {i z}^{2}}}{{(z - z_{2})}^{4}}}^{(1)} = {\frac{(4 {i z}^{2} - 4 {i z}_{2} z) e^{2 {i z}^{2}}}{{(z - z_{2})}^{4}}}^{(1)}$ $= \frac{{(8 i z - 4 {i z}_{2}) e^{2 {i z}^{2}} + 4 i z e^{2 {i z}^{2}} (4 {i z}^{2} - 4 {i z}_{2} z) {(z - z_{2})}^{4} - 4 {(z - z_{2})}^{3} {4 {i z}^{2} - 4 {i z}_{2} z} e^{2 {i z}^{2}}}{{(z - z_{2})}^{8}}$ $= \frac{{(8 i z - 4 {i z}_{2} - 16 z^{3} + 16 z_{2} z^{2}) (z - z_{2}) - 16 {i z}^{2} + 16 {i z}_{2} z} e^{2 {i z}^{2}}}{{(z - z_{2})}^{5}}$ $2 R e s (W, z_{1}) = \frac{{(8 {i z}_{1} - 4 {i z}_{2} - 16 z_{1}^{3} + 16 z_{2} z_{1}^{2}) (z_{1} - z_{2}) - 16 {i z}_{1}^{2} + 16 {i z}_{1} z_{2}} e^{2 {i z}_{1}^{2}}}{{(z_{1} - z_{2})}^{5}}$ $w i t h z_{1} - z_{2} = i \sqrt{11} r e s t t o f i n i c h t h e c a l c u l u s . . .$
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