# Evaluate $lim _{nto infty }int _{0}^{1}nx^ne^{x^2}dx$

Mathematics Asked by Unknown x on November 4, 2020

Evaluate $$lim _{nto infty }int _{0}^{1}nx^ne^{x^2}dx.$$

I applied the mean value thorem of integral to $$int _{0}^{1}nx^ne^{x^2}dx.$$ We get $$cin (0,1):$$

$$int _{0}^{1}nx^ne^{x^2}dx=(1-0)nc^ne^{c^2}.$$ Taking limit ($$lim_{nto infty}$$)on the both side,

We get, $$lim_{nto infty}int _{0}^{1}nx^ne^{x^2}dx=lim_{nto infty} nc^ne^{c^2}=0.$$

My answer in the examination was wrong. I don’t know the correct answer. Where is my mistake?

As the Taylor series of $$e^{x^2}$$ converges uniformly in $$[0,1]$$, $$int_0^1n x^nsum_{k=0}^inftyfrac{x^{2k}}{k!}dx=int_0^1 nsum_{k=0}^infty frac{x^{2k+n}}{k!}dx=sum_{k=0}^inftyfrac n{(2k+n+1)k!}\=sum_{k=0}^inftyfrac 1{k!}-sum_{k=0}^inftyfrac{2k+1}{(2k+n+1)k!}.$$

The second term vanishes because $$sum_{k=0}^inftyfrac{2k+1}{(2k+n+1)k!}

Correct answer by Yves Daoust on November 4, 2020

$$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$$ begin{align} lim _{n to infty}int_{0}^{1}nx^{n}expo{x^{2}}dd x & = lim _{n to infty}bracks{nint_{0}^{1} exppars{nlnpars{1 - x} + pars{1 - x}^{2}}dd x} \[5mm] & = lim _{n to infty}bracks{nint_{0}^{infty} expo{1 -pars{n + 2}x}dd x} \[5mm] & = expo{}lim _{n to infty}{n over n + 2} = bbx{largeexpo{}} \ & end{align}

See Laplace's Method.

Answered by Felix Marin on November 4, 2020

A bit late answer but maybe worth noting it.

Partial integration gives

$$I_n :=int_0^1 underbrace{nx^{n-1}}_{u'}cdotunderbrace{xe^{x^2}}_{v}dx= left.x^{n+1}e^{x^2}right|_0^1- underbrace{int_0^1 x^n(1+2x^2)e^{x^2}dx}_{J_n=}=e-J_n$$

Now, $$J_n$$ can be easily estimated as follows $$0leq J_n leq 3eint_0^1x^ndx=frac{3e}{n+1}stackrel{nto infty}{longrightarrow}0$$

Hence, $$I_n stackrel{nto infty}{longrightarrow} e$$.

Answered by trancelocation on November 4, 2020

Alternatively we could compute this limit directly via the substitution $$u = x^{n+1}$$

$$lim_{ntoinfty} frac{n}{n+1}int_0^1e^{u^{frac{2}{n+1}}}:du to int_0^1e:du = e$$

by dominated convergence.

Answered by Ninad Munshi on November 4, 2020

Your proof is incorrect. The issue is that the $$c$$ which you choose may depend on $$n$$.

It turns out that the correct answer is in fact $$e$$. This is because for any continuous $$f : [0, 1] to mathbb{R}$$, we have

$$limlimits_{n to infty} intlimits_0^1 n x^n f(x) dx = f(1)$$

This follows from the Stone-Weierstrass theorem as follows:

First, we prove that $$limlimits_{n to infty} intlimits_0^1 n x^n f(x) dx = f(1)$$ for every $$f$$ of the form $$f(x) = x^m$$. We then extend this to all $$f$$ polynomial quite easily.

Suppose now that we have continuous $$g : [0, 1] to mathbb{R}$$. Given arbitrary $$epsilon > 0$$, let $$w = frac{epsilon}{3}$$. Take polynomial $$f$$ s.t. $$|f - g| < w$$ (uniform norm) which is possible by Stone-Weierstrass, and take $$N$$ s.t. for all $$n geq N$$, $$left|intlimits_0^1 n x^n f(x) dx - f(1)right| < w$$. Then we have

$$begin{equation} begin{split} left| intlimits_0^1 n x^n g(x) dx - g(1) right| &leq left|intlimits_0^1 n x^n g(x) dx - intlimits_0^1 n x^n f(x) dxright| + left|intlimits_0^1 n x^n f(x) dx - f(1)right| + left|g(1) - f(1)right| \[10pt] &= left|intlimits_0^1 n x^n (g(x) - f(x)) dx right| + |g(1) - f(1)| + left|intlimits_0^1 n x^n f(x) dx - f(1)right| \[10pt] &< left|intlimits_0^1 n x^n (g(x) - f(x)) dx right| + w + w \[6pt] &leq intlimits_0^1 n x^n left|g(x) - f(x)right| dx + 2w \ &leq w intlimits_0^1 n x^n dx + 2w \ &= w frac{n}{n + 1} + 2w \[6pt] &<3w \[6pt] &= epsilon end{split} end{equation}$$

And therefore $$limlimits_{n to infty} intlimits_0^1 n x^n g(x) dx = g(1)$$

Answered by Doctor Who on November 4, 2020

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