Question Number 205534 by hardmath last updated on 23/Mar/24
$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{n}\:\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:\mathrm{e}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:? \\ $$
Answered by Mathspace last updated on 24/Mar/24
$${x}\in\left[\mathrm{0},\mathrm{1}\right]\:\Rightarrow\:{e}^{{x}^{\mathrm{2}} } \in\left[\mathrm{1},{e}\right]\:\Rightarrow \\ $$$${nx}^{{n}} {e}^{{x}^{\mathrm{2}} } \leqslant{nx}^{{n}} \:\Rightarrow{pour}\:{a}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$$\int_{\mathrm{0}} ^{{a}} {nx}^{{n}} {e}^{{x}^{\mathrm{2}} } {dx}\leqslant{n}\int_{\mathrm{0}} ^{{a}} \:{x}^{{n}} {dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{a}} {nx}^{{n}} {e}^{{x}^{\mathrm{2}} } \leqslant\frac{{n}}{{n}+\mathrm{1}}{a}^{{n}+\mathrm{1}} \:\:\:\Rightarrow \\ $$$${lim}_{{a}\rightarrow\mathrm{1}} \int_{\mathrm{0}} ^{{a}} {nx}^{{n}} {e}^{{x}^{\mathrm{2}} } {dx}\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \left({lim}_{{a}\rightarrow\mathrm{1}} \int_{\mathrm{0}} ^{{a}} {nx}^{{n}} {e}^{{x}^{\mathrm{2}} } {dx}\right)=\mathrm{0} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {e}^{{x}^{\mathrm{2}} } {dx}=\mathrm{0} \\ $$
Commented by Berbere last updated on 24/Mar/24
$${nx}^{{n}} \leqslant{nx}^{{n}} {e}^{{x}^{\mathrm{2}} } \leqslant{nx}^{{n}} {e} \\ $$
Answered by Berbere last updated on 24/Mar/24
$${I}_{\mathrm{0}} =\mathrm{0} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$$$\frac{{I}_{{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {e}^{{x}^{\mathrm{2}} } \leqslant{e}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} =\frac{{e}}{{n}+\mathrm{1}}\rightarrow\mathrm{0}….\left({E}\right) \\ $$$${I}_{{n}+\mathrm{1}} \overset{{IBP}} {=}\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}=\left({n}+\mathrm{1}\right)\left[_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}}{x}^{{n}} \right]−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$“{x}^{{n}+\mathrm{1}} {e}^{{x}^{\mathrm{2}} } =\left({x}^{{n}} \right)_{{u}} .\left({xe}^{{x}^{\mathrm{2}} } \right)_{{v}'} '' \\ $$$$=\left({n}+\mathrm{1}\right)\frac{{e}}{\mathrm{2}}−\frac{\left({n}+\mathrm{1}\right){n}}{\mathrm{2}\left({n}−\mathrm{1}\right)}{I}_{{n}−\mathrm{1}} \\ $$$$\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\frac{{e}}{\mathrm{2}}−\frac{{n}}{\mathrm{2}}.\frac{{I}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}};{applie}\:\left({E}\right),\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\overset{\infty} {\rightarrow}\mathrm{0} \\ $$$$\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\rightarrow\mathrm{0}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{I}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}.\frac{{n}}{{e}}=\mathrm{1}\Rightarrow{I}_{{n}−\mathrm{1}} \sim\frac{{e}\left({n}−\mathrm{1}\right)}{{n}}\overset{\infty} {\rightarrow}{e} \\ $$
Commented by hardmath last updated on 24/Mar/24
$$\mathrm{Yes}\:\mathrm{answer}\:=\:\mathrm{e},\:\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{thankyou} \\ $$
Commented by Berbere last updated on 24/Mar/24
$${withe}\:{Pleasur} \\ $$