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Question Number 105232 by mathmax by abdo last updated on 27/Jul/20
calculate ∫_0 ^∞ ((cos(2x^2 ))/((4x^2 +9)^3 ))dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{2x}^{\mathrm{2}} \right)}{\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{9}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$
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...
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}^{\mathrm{2}} \right)}{\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{9}\right)^{\mathrm{3}} }\mathrm{dx}\:\Rightarrow\mathrm{2A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}^{\mathrm{2}} \right)}{\left(\mathrm{4x}^{\mathrm{2}} \:+\mathrm{9}\right)^{\mathrm{3}} }\mathrm{dx}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}^{\mathrm{2}} \right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{3}} }\mathrm{dx}\:=_{\mathrm{x}\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\frac{\mathrm{9}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \right)}{\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{3}} \left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\frac{\mathrm{3}}{\mathrm{2}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{162}}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\frac{\mathrm{9}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dt}\:\:=\frac{\mathrm{1}}{\mathrm{162}}\:\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{it}^{\mathrm{2}} } }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\:\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\mathrm{and} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \:\varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\:\frac{\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\:\frac{\mathrm{9iz}\:\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\frac{\left(\mathrm{9iz}\:\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{3}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\frac{\left(\mathrm{9iz}^{\mathrm{2}} −\mathrm{12}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\left(\mathrm{3iz}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\left(\mathrm{6iz}\:\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } +\mathrm{9iz}\left(\mathrm{3iz}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } \right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{3iz}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{8}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\left(\mathrm{6iz}−\mathrm{27z}^{\mathrm{3}} −\mathrm{36iz}\right)\left(\mathrm{z}+\mathrm{i}\right)\:\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } −\left(\mathrm{3iz}^{\mathrm{2}} −\mathrm{4}\right)\mathrm{e}^{\frac{\mathrm{9}}{\mathrm{2}}\mathrm{iz}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\left(−\mathrm{6}+\mathrm{27i}+\mathrm{36}\right)\left(\mathrm{2i}\right)\mathrm{e}^{−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{i}} −\left(−\mathrm{3i}−\mathrm{4}\right)\mathrm{e}^{−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{i}} }{\left(\mathrm{2i}\right)^{\mathrm{5}} }\:…\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$

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