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Question Number 56189 by maxmathsup by imad last updated on 11/Mar/19

let u_n =∫_(−∞) ^∞ ((sin(nx^2 ))/(x^2 +x +n)) dx 1) calculate u_n 2) find lim_(n→+∞) u_n 3) study the serie Σ u_n

$${let}\:{u}_{{n}} =\int_{−\infty} ^{\infty} \:\:\:\frac{{sin}\left({nx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{x}\:+{n}}\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{u}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{serie}\:\Sigma\:{u}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 12/Mar/19

we have u_n =Im(∫_(−∞) ^(+∞) (e^(inx^2 ) /(x^2 +x +n))dx) let ϕ(z) =(e^(inz^2 ) /(z^2 +z+n)) poles of ϕ? z^2 +z +n =0 →Δ =1−4n =(i(√(4n−1)))^2 ⇒the roots are z_1 =((−1+i(√(4n−1)))/2) and z_2 =((−1−i(√(4n−1)))/2) ⇒ ϕ(z) =(e^(inz^2 ) /((z−z_1 )(z−z_2 ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) =(e^(inz_1 ^2 ) /(z_1 −z_2 )) =(e^(inz_1 ^2 ) /(i(√(4n−1)))) but z_1 ^2 =(1/4){1−2i(√(4n−1))−4n+1} =(1/4){2−4n −2i(√(4n−1))} =(1/2){1−2n −i(√(4n−1))} ⇒ inz^2 =((in)/2)(1−2n −i(√(4n−1))} =(n/2)(√(4n−1)) +((n(1−2n))/2) i ⇒ e^(inz^2 ) =e^((n/2)(√(4n−1))) {cos(((n(1−2n))/2))+i sin(((n(1−2n))/2))} ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(i(√(4n−1)))) e^((n/2)(√(4n−1))) { cos(((n(1−2n))/2))+i sin(((n(1−2n))/2))} ⇒ u_n =−((2π)/(√(4n−1))) e^((n/2)(√(4n−1))) sin(((n(2n−1))/2)) with n integr and n≥1 .

$${we}\:{have}\:{u}_{{n}} ={Im}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{inx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+{x}\:+{n}}{dx}\right)\:\:{let}\:\varphi\left({z}\right)\:=\frac{{e}^{{inz}^{\mathrm{2}} } }{{z}^{\mathrm{2}} \:+{z}+{n}}\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} \:+{z}\:+{n}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}{n}\:=\left({i}\sqrt{\mathrm{4}{n}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{the}\:{roots}\:{are}\: \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}}{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}}{\mathrm{2}}\:\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{inz}^{\mathrm{2}} } }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\:\:{residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\frac{{e}^{{inz}_{\mathrm{1}} ^{\mathrm{2}} } }{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\frac{{e}^{{inz}_{\mathrm{1}} ^{\mathrm{2}} } }{{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}}\:\:{but}\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{1}−\mathrm{2}{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}−\mathrm{4}{n}+\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{2}−\mathrm{4}{n}\:−\mathrm{2}{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−\mathrm{2}{n}\:−{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}\right\}\:\Rightarrow \\ $$$${inz}^{\mathrm{2}} \:=\frac{{in}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{n}\:−{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}\right\}\:=\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{n}−\mathrm{1}}\:\:+\frac{{n}\left(\mathrm{1}−\mathrm{2}{n}\right)}{\mathrm{2}}\:{i}\:\Rightarrow \\ $$$${e}^{{inz}^{\mathrm{2}} } \:={e}^{\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{n}−\mathrm{1}}} \left\{{cos}\left(\frac{{n}\left(\mathrm{1}−\mathrm{2}{n}\right)}{\mathrm{2}}\right)+{i}\:{sin}\left(\frac{{n}\left(\mathrm{1}−\mathrm{2}{n}\right)}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{4}{n}−\mathrm{1}}}\:{e}^{\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{n}−\mathrm{1}}} \left\{\:{cos}\left(\frac{{n}\left(\mathrm{1}−\mathrm{2}{n}\right)}{\mathrm{2}}\right)+{i}\:{sin}\left(\frac{{n}\left(\mathrm{1}−\mathrm{2}{n}\right)}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$${u}_{{n}} =−\frac{\mathrm{2}\pi}{\sqrt{\mathrm{4}{n}−\mathrm{1}}}\:{e}^{\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{n}−\mathrm{1}}} {sin}\left(\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}}\right)\:\:{with}\:{n}\:\:{integr}\:{and}\:{n}\geqslant\mathrm{1}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 12/Mar/19

2) no limit for (u_n ) but ∣u_n ∣ →+∞ 3) Σ u_n is a divergent serie .

$$\left.\mathrm{2}\right)\:{no}\:{limit}\:{for}\:\left({u}_{{n}} \right)\:{but}\:\mid{u}_{{n}} \mid\:\rightarrow+\infty \\ $$$$\left.\mathrm{3}\right)\:\Sigma\:{u}_{{n}} \:{is}\:{a}\:{divergent}\:{serie}\:. \\ $$

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