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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

$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{n}\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{n}\:{U}_{{n}} \\ $$$$\left.\mathrm{3}\right){determine}\:{nature}\:{of}\:{the}\:{serie}\:\:\Sigma\:{U}_{{n}} \\ $$

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 ..

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{U}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{e}^{−{n}\left[{x}^{\mathrm{2}} \right]} }{{x}^{\mathrm{2}} \:+\mathrm{3}}\:{dx}\:\:\:{let}\:\varphi\left({z}\right)\:=\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{\mathrm{3}}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\mathrm{3}\right. \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \left({z}−{i}\sqrt{\mathrm{3}}\right)\varphi\left({z}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{{e}^{−{n}\left[{z}^{\mathrm{2}} \right]} }{{z}+{i}\sqrt{\mathrm{3}}}\:=\frac{{e}^{−{n}\left[−\mathrm{3}\right]} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{{e}^{\mathrm{3}{n}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{\mathrm{3}{n}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\sqrt{\mathrm{3}}}\:{e}^{\mathrm{3}{n}} \:\Rightarrow\:{U}_{{n}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:{e}^{\mathrm{3}{n}} \:. \\ $$$$\left.\mathrm{2}\right)\:\:{lim}_{{n}\rightarrow+\infty} \:{nU}_{{n}} =+\infty \\ $$$$\left.\mathrm{3}\right)\:{its}\:{clear}\:{that}\:\Sigma\:{U}_{{n}} \:{diverges}\:.. \\ $$

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