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let-A-n-0-e-nx-2-sin-x-n-dx-with-n-integr-not-0-1-calculate-A-n-2-find-lim-n-A-n-

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Question Number 37285 by math khazana by abdo last updated on 11/Jun/18
let A_n = ∫_0 ^∞ e^(−nx^2 ) sin((x/n))dx with n integr not 0 1) calculate A_n 2) find lim_(n→+∞) A_n
$${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}^{\mathrm{2}} } {sin}\left(\frac{{x}}{{n}}\right){dx}\:\:{with}\:{n}\:{integr}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 14/Jun/18
A_n =∫_0 ^(+∞) e^(−nx^2 ) cos( (x/n))dx .
$${A}_{{n}} =\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{nx}^{\mathrm{2}} } \:{cos}\left(\:\frac{{x}}{{n}}\right){dx}\:. \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
changement (x/n)=t give A_n = ∫_0 ^∞ e^(−nn^2 t^2 ) cos(t) ndt =n ∫_0 ^∞ e^(−n^3 t^2 ) cost dt =(n/2) ∫_(−∞) ^(+∞) e^(−n^3 t^2 +it) dt (2/n) A_n = ∫_(−∞) ^(+∞) e^(−{(n(√n)t)^2 −it}) dt =∫_(−∞) ^(+∞) e^(−{ (n(√n)t)^2 −2 (i/(n(√n)))(n(√n)t) + ((i/(n(√n))))^2 −((i/(n(√n))))^2 }) dt = ∫_(−∞) ^(+∞) e^(−(n(√n) t −(i/(n(√n))))^2 −(1/n^3 )) dt =_(n(√n)t −(i/(n(√n))) = u) e^(−(1/n^3 )) ∫_(−∞) ^(+∞) e^(−u^2 ) du = (√π) e^(−(1/(n^3 ))) ⇒ A_n =((n(√π))/2) e^(−(1/(n^3 ))) .
$${changement}\:\frac{{x}}{{n}}={t}\:{give} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nn}^{\mathrm{2}} {t}^{\mathrm{2}} } \:{cos}\left({t}\right)\:{ndt} \\ $$$$={n}\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{n}^{\mathrm{3}} {t}^{\mathrm{2}} } \:{cost}\:{dt} \\ $$$$=\frac{{n}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{n}^{\mathrm{3}} {t}^{\mathrm{2}} \:+{it}} {dt} \\ $$$$\frac{\mathrm{2}}{{n}}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left\{\left({n}\sqrt{{n}}{t}\right)^{\mathrm{2}} −{it}\right\}} {dt} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left\{\:\left({n}\sqrt{{n}}{t}\right)^{\mathrm{2}} \:−\mathrm{2}\:\frac{{i}}{{n}\sqrt{{n}}}\left({n}\sqrt{{n}}{t}\right)\:+\:\left(\frac{{i}}{{n}\sqrt{{n}}}\right)^{\mathrm{2}} \:−\left(\frac{{i}}{{n}\sqrt{{n}}}\right)^{\mathrm{2}} \right\}} {dt} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left({n}\sqrt{{n}}\:{t}\:−\frac{{i}}{{n}\sqrt{{n}}}\right)^{\mathrm{2}} \:\:−\frac{\mathrm{1}}{{n}^{\mathrm{3}} }} \:{dt} \\ $$$$=_{{n}\sqrt{{n}}{t}\:−\frac{{i}}{{n}\sqrt{{n}}}\:=\:{u}} {e}^{−\frac{\mathrm{1}}{{n}^{\mathrm{3}} }} \:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:{du} \\ $$$$=\:\sqrt{\pi}\:\:{e}^{−\frac{\mathrm{1}}{{n}^{\mathrm{3}} \:\:}} \:\:\Rightarrow \\ $$$${A}_{{n}} =\frac{{n}\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\frac{\mathrm{1}}{{n}^{\mathrm{3}} \:\:}} \:\:\:. \\ $$
Commented by math khazana by abdo last updated on 17/Jun/18
2) its clear that lim_(n→+∞) A_n =+∞ .
$$\left.\mathrm{2}\right)\:{its}\:{clear}\:{that}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \:=+\infty\:. \\ $$

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