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Question Number 67342 by mathmax by abdo last updated on 26/Aug/19
find the value of ∫_(−∞) ^∞ ((sin(2x^2 ))/((x^2 −x +3)^3 ))dx
$${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{\infty} \:\:\:\frac{{sin}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} −{x}\:+\mathrm{3}\right)^{\mathrm{3}} }{dx} \\ $$
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...
$${let}\:\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)^{\mathrm{3}} }{dx}\:\Rightarrow\:{I}\:={Im}\:\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} −{z}\:+\mathrm{3}\right)^{\mathrm{3}} }{dz}\right) \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} −{z}\:+\mathrm{3}\right)^{\mathrm{3}} }\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{2}} −{z}\:+\mathrm{3}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{12}\:=−\mathrm{11}\:=\left({i}\sqrt{\mathrm{11}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\Rightarrow{W}\left({z}\right)\:=\frac{{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{3}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{let}\:{apply}\:{residus}\:{theorem} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{z}_{\mathrm{1}} \right)\:\:\:\left({z}_{\mathrm{1}} {is}\:{a}\:{triple}\:{pole}\right) \\ $$$$\left.{Res}\left({W},{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{3}} {W}\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \right\} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:\:{we}\:{have} \\ $$$$\left\{\frac{{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:=\left\{\frac{\mathrm{4}{iz}\:{e}^{\mathrm{2}{iz}^{\mathrm{2}} } \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{3}\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} {e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\left\{\frac{\mathrm{4}{iz}\:{e}^{\mathrm{2}{iz}^{\mathrm{2}} } \left({z}−{z}_{\mathrm{2}} \right)−\mathrm{3}{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \:=\left\{\frac{\left(\mathrm{4}{iz}^{\mathrm{2}} −\mathrm{4}{iz}_{\mathrm{2}} {z}\right){e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\left\{\left(\mathrm{8}{iz}\:−\mathrm{4}{iz}_{\mathrm{2}} \right){e}^{\mathrm{2}{iz}^{\mathrm{2}} } \:\:\:+\mathrm{4}{iz}\:{e}^{\mathrm{2}{iz}^{\mathrm{2}} } \left(\mathrm{4}{iz}^{\mathrm{2}} −\mathrm{4}{iz}_{\mathrm{2}} {z}\right)\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{4}\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} \left\{\mathrm{4}{iz}^{\mathrm{2}} −\mathrm{4}{iz}_{\mathrm{2}} {z}\right\}{e}^{\mathrm{2}{iz}^{\mathrm{2}} } \right.}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{8}} } \\ $$$$=\frac{\left\{\left(\mathrm{8}{iz}−\mathrm{4}{iz}_{\mathrm{2}} −\mathrm{16}{z}^{\mathrm{3}} \:+\mathrm{16}{z}_{\mathrm{2}} {z}^{\mathrm{2}} \:\right)\left({z}−{z}_{\mathrm{2}} \right)−\mathrm{16}{iz}^{\mathrm{2}} \:+\mathrm{16}{iz}_{\mathrm{2}} {z}\right\}{e}^{\mathrm{2}{iz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{5}} } \\ $$$$\mathrm{2}{Res}\left({W},{z}_{\mathrm{1}} \right)\:=\frac{\left\{\left(\mathrm{8}{iz}_{\mathrm{1}} −\mathrm{4}{iz}_{\mathrm{2}} −\mathrm{16}{z}_{\mathrm{1}} ^{\mathrm{3}} \:+\mathrm{16}{z}_{\mathrm{2}} {z}_{\mathrm{1}} ^{\mathrm{2}} \right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)−\mathrm{16}{iz}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{16}{iz}_{\mathrm{1}} {z}_{\mathrm{2}} \right\}{e}^{\mathrm{2}{iz}_{\mathrm{1}} ^{\mathrm{2}} } }{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{5}} } \\ $$$${with}\:{z}_{\mathrm{1}} −{z}_{\mathrm{2}} ={i}\sqrt{\mathrm{11}}\:\:\:\:\:{rest}\:{to}\:{finich}\:{the}\:{calculus}… \\ $$

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