Question Number 60263 by maxmathsup by imad last updated on 19/May/19
![let U_n =∫_0 ^∞ (e^(−n[x^2 ]) /(x^2 +3)) dx 1) calculate U_n interms of n 2) find lim_(n→+∞) n U_n 3)determine nature of the serie Σ U_n](https://www.tinkutara.com/question/Q60263.png)
$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{n}\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{n}\:{U}_{{n}} \\ $$$$\left.\mathrm{3}\right){determine}\:{nature}\:{of}\:{the}\:{serie}\:\:\Sigma\:{U}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 20/May/19
![1) we have 2U_n =∫_(−∞) ^(+∞) (e^(−n[x^2 ]) /(x^2 +3)) dx let ϕ(z) =(e^(−n[z^2 ]) /(z^2 +3)) we have ϕ(z) = (e^(−n[z^2 ]) /((z−i(√3))(z+i(√3)))) so the poles of ϕ are +^− i(√3) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i(√3)3 Res(ϕ,i) =lim_(z→i(√3)) (z−i(√3))ϕ(z) =lim_(z→i) (e^(−n[z^2 ]) /(z+i(√3))) =(e^(−n[−3]) /(2i(√3))) =(e^(3n) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(3n) /(2i(√3))) =(π/( (√3))) e^(3n) ⇒ U_n =(π/(2(√3))) e^(3n) . 2) lim_(n→+∞) nU_n =+∞ 3) its clear that Σ U_n diverges ..](https://www.tinkutara.com/question/Q60345.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{U}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{e}^{−{n}\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} \:+\mathrm{3}}\:{dx}\:\:\:{let}\:\varphi\left({z}\right)\:=\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{\mathrm{3}}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\mathrm{3}\right. \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \left({z}−{i}\sqrt{\mathrm{3}}\right)\varphi\left({z}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{{z}+{i}\sqrt{\mathrm{3}}}\:=\frac{{e}^{−{n}\left[−\mathrm{3}\right]} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{{e}^{\mathrm{3}{n}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{\mathrm{3}{n}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:{e}^{\mathrm{3}{n}} \:\Rightarrow\:{U}_{{n}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:{e}^{\mathrm{3}{n}} \:. \\ $$$$\left.\mathrm{2}\right)\:\:{lim}_{{n}\rightarrow+\infty} \:{nU}_{{n}} =+\infty \\ $$$$\left.\mathrm{3}\right)\:{its}\:{clear}\:{that}\:\Sigma\:{U}_{{n}} \:{diverges}\:.. \\ $$