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 Berbere last updated on 24/Mar/24
![I_0 =0 I_1 =(1/2)(e−1) (I_n /n)=∫_0 ^1 x^n e^x^2 ≤e∫_0 ^1 x^n =(e/(n+1))→0....(E) I_(n+1) =^(IBP) (n+1)∫_0 ^1 x^(n+1) e^x^2 dx=(n+1)[_0 ^1 (e^x^2 /2)x^n ]−((n(n+1))/2)∫_0 ^1 x^(n−1) e^x^2 dx “x^(n+1) e^x^2 =(x^n )_u .(xe^x^2 )_(v′) ” =(n+1)(e/2)−(((n+1)n)/(2(n−1)))I_(n−1) (I_(n+1) /(n+1))=(e/2)−(n/2).(I_(n−1) /(n−1));applie (E),(I_(n+1) /(n+1))→^∞ 0 (I_(n+1) /(n+1))→0⇒lim_(n→∞) (I_(n−1) /(n−1)).(n/e)=1⇒I_(n−1) ∼((e(n−1))/n)→^∞ e](Q205539.png)
$${I}_{\mathrm{0}} =\mathrm{0} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$$$\frac{{I}_{{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {e}^{{x}^{\mathrm{2}} } \leqslant{e}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} =\frac{{e}}{{n}+\mathrm{1}}\rightarrow\mathrm{0}....\left({E}\right) \\ $$$${I}_{{n}+\mathrm{1}} \overset{{IBP}} {=}\left({n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}=\left({n}+\mathrm{1}\right)\left[_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}}{x}^{{n}} \right]−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$``{x}^{{n}+\mathrm{1}} {e}^{{x}^{\mathrm{2}} } =\left({x}^{{n}} \right)_{{u}} .\left({xe}^{{x}^{\mathrm{2}} } \right)_{{v}'} '' \\ $$$$=\left({n}+\mathrm{1}\right)\frac{{e}}{\mathrm{2}}−\frac{\left({n}+\mathrm{1}\right){n}}{\mathrm{2}\left({n}−\mathrm{1}\right)}{I}_{{n}−\mathrm{1}} \\ $$$$\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\frac{{e}}{\mathrm{2}}−\frac{{n}}{\mathrm{2}}.\frac{{I}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}};{applie}\:\left({E}\right),\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\overset{\infty} {\rightarrow}\mathrm{0} \\ $$$$\frac{{I}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\rightarrow\mathrm{0}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{I}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}.\frac{{n}}{{e}}=\mathrm{1}\Rightarrow{I}_{{n}−\mathrm{1}} \sim\frac{{e}\left({n}−\mathrm{1}\right)}{{n}}\overset{\infty} {\rightarrow}{e} \\ $$
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
$${withe}\:{Pleasur} \\ $$