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

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

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

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

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

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    $\begingroup$ Your $c \in (0, 1)$ depends on $n$. $\endgroup$ – sera Aug 12 at 6:37
  • $\begingroup$ @sera but $f(x)=nx^ne^{x^2}$ is a continuous function on $[0,1]$ $\endgroup$ – Unknown x Aug 12 at 6:42
  • $\begingroup$ @Unknownx The point $c$ is different for all $n$. You are applying the mean value theorem for integrals for different continuous functions. $\endgroup$ – Keen-ameteur Aug 12 at 6:45
  • $\begingroup$ By the way, are you also asking what the correct approach to showing this would be? $\endgroup$ – Keen-ameteur Aug 12 at 6:46
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    $\begingroup$ Does this answer your question? Find $\displaystyle\lim_{n\to \infty} \int_0^1 nx^n e^{x^2} dx$ $\endgroup$ – sera Aug 12 at 6:57

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

The second term vanishes because $$\sum_{k=0}^\infty\frac{2k+1}{(2k+n+1)k!}<\frac 1n\sum_{k=0}^\infty\frac{2k+1}{k!}.$$

Hence $$e.$$

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  • $\begingroup$ Just out of curiosity, how did you come up eith the last equality? $\endgroup$ – tommy1996q Aug 12 at 8:01
  • $\begingroup$ @tommy1996q: which one ? Notice the minor typo, just fixed. $\endgroup$ – Yves Daoust Aug 12 at 8:03
  • $\begingroup$ Where you equate $\frac{n}{(2k+n+1)k!}$ with those other 2 addendums, the last = you’ve written $\endgroup$ – tommy1996q Aug 12 at 14:40
  • $\begingroup$ @tommy1996q: reduce to common denominator. $\endgroup$ – Yves Daoust Aug 12 at 14:41
  • $\begingroup$ Oh, I’d misread it and thought it was some obscure identity. Not gonna lie, feel really dumb right now $\endgroup$ – tommy1996q Aug 12 at 20:32

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

$\lim\limits_{n \to \infty} \int\limits_0^1 n x^n f(x) dx = f(1)$

This follows from the Stone-Weierstrass theorem as follows:

First, we prove that $\lim\limits_{n \to \infty} \int\limits_0^1 n x^n f(x) dx = f(1)$ 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 $n \geq N$, $\left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| < w$. Then we have

\begin{equation} \begin{split} \left| \int\limits_0^1 n x^n g(x) dx - g(1) \right| &\leq \left|\int\limits_0^1 n x^n g(x) dx - \int\limits_0^1 n x^n f(x) dx\right| + \left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| + \left|g(1) - f(1)\right| \\[10pt] &= \left|\int\limits_0^1 n x^n (g(x) - f(x)) dx \right| + |g(1) - f(1)| + \left|\int\limits_0^1 n x^n f(x) dx - f(1)\right| \\[10pt] &< \left|\int\limits_0^1 n x^n (g(x) - f(x)) dx \right| + w + w \\[6pt] &\leq \int\limits_0^1 n x^n \left|g(x) - f(x)\right| dx + 2w \\ &\leq w \int\limits_0^1 n x^n dx + 2w \\ &= w \frac{n}{n + 1} + 2w \\[6pt] &<3w \\[6pt] &= \epsilon \end{split} \end{equation}

And therefore $\lim\limits_{n \to \infty} \int\limits_0^1 n x^n g(x) dx = g(1)$

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

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

by dominated convergence.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim _{n \to \infty}\int_{0}^{1}nx^{n}\expo{x^{2}}\dd x & = \lim _{n \to \infty}\bracks{n\int_{0}^{1} \exp\pars{n\ln\pars{1 - x} + \pars{1 - x}^{2}}\dd x} \\[5mm] & = \lim _{n \to \infty}\bracks{n\int_{0}^{\infty} \expo{1 -\pars{n + 2}x}\dd x} \\[5mm] & = \expo{}\lim _{n \to \infty}{n \over n + 2} = \bbx{\large\expo{}} \\ & \end{align}

See Laplace's Method.

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A bit late answer but maybe worth noting it.

Partial integration gives

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

Now, $J_n$ can be easily estimated as follows $$0\leq J_n \leq 3e\int_0^1x^ndx=\frac{3e}{n+1}\stackrel{n\to \infty}{\longrightarrow}0$$

Hence, $I_n \stackrel{n\to \infty}{\longrightarrow} e$.

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