The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$

Question

The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$ is _____

I tried by taking the odd $n$ values as in that case, the integral I suppose was easier to calculate. So, denote $I_n=n\int_{0}^{1}e^{x^2}x^ndx$, then we have : $I_1=\frac{e-1}{2},\\I_3=\frac{3}{2},\\I_5=5(\frac{e}{2}-1),\\I_7=7(3-e)$

Then I tried calculating (random) values using calculator as integration was getting cumbersome. $I_{31}=2.488$, $I_{51}=2.57$.

I do not see any kind of recurrence so that I can find a general term for odd $n$. I also tried the method given here. The idea was that since integral doesn't depend on $n$ considering $I(n)=\int_{0}^{1}e^{x^2}x^ndx$, then $I'(n)=\int_{0}^{1}x^n\ln(x)e^{x^2}dx$, but this also doesn't lead me to any conclusion as well.

I believe I couldn't get to a proper approach to tackle this question. Can someone please help with the idea that involves in solving these type of questions?

This is related to the concept of approximation to the identity. In general, if $f$ is continuous on $[0,1]$, then $$\lim_{n\to\infty}\int_{0}^{1}nx^nf(x),\mathrm{d}x=f(1).$$ For this reason, we may simply say that $nx^n,\mathrm{d}x$ converges to the unit point mass $\delta_1(\mathrm{d}x)$ on $[0,1]$. For the proof, check this posting. — Sangchul Lee, Jul 10 '20 at 05:24

score 6 · Answer 1 · answered Jul 10 '20 at 05:22

Clearly $$\int_0^1 nx^n\,dx\le I_n\le e\int_0^1 nx^n\,dx.$$ Thus $$\frac{n}{n+1}\le I_n\le e\frac{n}{n+1}.$$ So, if the limit exists, it's between $1$ and $e$ (and certainly is non-zero).

Most of the weight of the integral $I_n$ comes from near $1$, so for $\newcommand{\ep}{\epsilon}\ep>0$ we have $$I_n\ge\int_{1-\ep}^1nx^n e^{x^2}\,dx\ge e^{(1-\ep)^2}\int_{1-\ep}^1nx^n\,dx =e^{(1-\ep)^2}\frac n{n+1}(1-(1-\ep)^{n+1}).$$ Then $\liminf I_n\ge e^{(1-\ep)^2}$, and this is enough to deduce $\lim I_n=e$.

Graviton · Answer 2 · 2020-07-10T05:27:19.513

Wolfralm Alpha calculates

$$\int_0^1nx^ne^{x^2}dx=-\frac{1}{2}i^{1-n}n\Bigg(\Gamma\bigg(\frac{n+1}{2}\bigg)-\Gamma\bigg(\frac{n+1}{2},-1\bigg)\Bigg)$$

No clue if this monstrosity can be simplified, but again using Wolfram alpha to find the lmit, it says

(no result found in terms of standard mathematical functions)

Yet it gives us a series expansion at $n=\infty$

$$\Bigg(e-\frac{3e}{n}+\frac{13e}{n^2}-\frac{71e}{n^3}+\frac{547e}{n^4}+O\Big(\big(\frac{1}{n}\big)^5\Big)\Bigg)+2^{-n/2}e^{-n(1+\pi i)/2}n^{n/2}O\Big(\big(\frac{1}{n}\big)^{11/2}\Big)$$

So perhaps the limit is just $e$ or $e-\epsilon$ for some $\epsilon>0$

score 1 · Answer 3 · answered Jul 10 '20 at 05:30

Roughly speaking, the function $n x^n$, for $n$ large, concentrates its mass $n/(n+1)$ near $x=1$ (you can think it like a Dirac delta concentrated in $x=1$), so that the sequence converges to $e$.

To prove rigorously this statement, let $0<a<1$ and split the integral into $$ I_n = \int_0^a f_n(x)\, dx, \quad J_n = \int_0^a f_n(x)\, dx, \qquad f_n(x) := n x^n e^{x^2}. $$ Since, in $[0,a]$, the sequence $(f_n)$ converges uniformly to $0$, we have that $I_n \to 0$.

On the other hand, $$ n\, x^n\, e^{a^2} \leq f_n(x) \leq n\, x^n \qquad \forall x\in [a,1], $$ hence $$ \frac{n}{n+1}\, e^{a^2} \leq J_n \leq \frac{n}{n+1}\, e, \qquad \forall n\in\mathbb{N}. $$ Since these inequalities hold for every $a\in (0,1)$, we can finally conclude that the initial limit is $e$.

score 1 · Answer 4 · answered Feb 08 '21 at 15:54

A bit late answer because it was a duplicate somewhere else.

Simple partial integration gives:

\begin{eqnarray*}I_n =\int_0^1nx^ne^{x^2}dx & = & \frac n{n-1}\int_0^1(n-1)x^ne^{x^2}dx \\ & = & \frac n{n-1}\left(\left.x^ne^{x^2}\right|_0^1-2\int_0^1x^{n+1}e^{x^2}dx\right)\\ & = & \frac n{n-1}\left(e-J_n\right) \end{eqnarray*}

with

$$0\leq J_n \leq 2e\int_0^1x^{n+1}dx =\frac{2e}{n+1}\stackrel{n\to\infty}{\longrightarrow}0$$

Hence, $\boxed{\lim_{n\to\infty}I_n = e}$.

The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$

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