Menu Close

let-u-n-0-sin-nx-2-x-2-6-dx-1-calculate-u-n-and-lim-u-n-n-2-find-nature-of-u-n-and-calaculate-it-3-find-nature-of-u-n-2-

Tinku Tara 0 Comments
Pin




Question Number 54777 by maxmathsup by imad last updated on 10/Feb/19
let u_n =∫_0 ^∞ ((sin(nx^2 ))/(x^2 +6))dx 1) calculate u_n and lim u_n (n→+∞) 2) find nature of Σ u_n and calaculate it. 3) find nature of Σ u_n ^2
$${let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({nx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{6}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{u}_{{n}} \:\:\:{and}\:{lim}\:{u}_{{n}} \left({n}\rightarrow+\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:\:{and}\:{calaculate}\:{it}. \\ $$$$\left.\mathrm{3}\right)\:{find}\:{nature}\:\:{of}\:\Sigma\:{u}_{{n}} ^{\mathrm{2}} \\ $$
Commented by Abdo msup. last updated on 11/Feb/19
1) we have u_n =_(x=(√6)t) ∫_0 ^∞ ((sin(6nt^2 ))/(6(1+t^2 ))) (√6)dt =(1/( (√6)))∫_0 ^∞ ((sin(6nt^2 ))/(1+t^2 )) dt ⇒2u_n =(1/( (√6))) ∫_(−∞) ^(+∞) ((sin(6nt^2 ))/(1+t^2 ))dt ⇒2(√6)u_n =Im( ∫_(−∞) ^(+∞) (e^(i6nt^2 ) /(1+t^2 ))dt) let ϕ(z) =((e^(6inz^2 ) )/(z^2 +1)) the poles of ϕ are i and −i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Rez(ϕ,i) Rez(ϕ,i) =lim_(z→i) (z−i)ϕ(z) =(e^(−6ni) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−6ni) /(2i)) =π{cos(6n)−i sin(6n)}⇒ 2(√6)u_n =−π sin(6n) ⇒u_n =−(π/(2(√6))) sin(6n) 2) the sequence (u_n ) diverges ⇒Σ u_n diverges also u_n ^2 =(π^2 /(24))sin^2 (6n) diverges ⇒Σ u_n ^2 diverges
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{u}_{{n}} =_{{x}=\sqrt{\mathrm{6}}{t}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{6}{nt}^{\mathrm{2}} \right)}{\mathrm{6}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\sqrt{\mathrm{6}}{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{6}{nt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:\Rightarrow\mathrm{2}{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\:\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\mathrm{6}{nt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{6}}{u}_{{n}} ={Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\mathrm{6}{nt}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right)\:{let}\: \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{\mathrm{6}{inz}^{\mathrm{2}} } \:\:}{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Rez}\left(\varphi,{i}\right) \\ $$$${Rez}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)\:=\frac{{e}^{−\mathrm{6}{ni}} }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\mathrm{6}{ni}} }{\mathrm{2}{i}}\:=\pi\left\{{cos}\left(\mathrm{6}{n}\right)−{i}\:{sin}\left(\mathrm{6}{n}\right)\right\}\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{6}}{u}_{{n}} =−\pi\:{sin}\left(\mathrm{6}{n}\right)\:\Rightarrow{u}_{{n}} =−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}\:{sin}\left(\mathrm{6}{n}\right) \\ $$$$\left.\mathrm{2}\right)\:{the}\:{sequence}\:\left({u}_{{n}} \right)\:{diverges}\:\Rightarrow\Sigma\:{u}_{{n}} \:{diverges} \\ $$$${also}\:\:{u}_{{n}} ^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}{sin}^{\mathrm{2}} \left(\mathrm{6}{n}\right)\:{diverges}\:\Rightarrow\Sigma\:{u}_{{n}} ^{\mathrm{2}} \:{diverges} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *