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Given a real number $a > 0$, I need to find $$ \lim_{n \to \infty} \left( \int_0^a e^{x^2/n} dx\right)^n $$

I really have no idea where to start, should I write down the function below the integral and somehow use the dominant convergence theorem?

gt6989b
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popmaria
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    Just like the feedback you received on your last problem, please explain the background behind this problem or any other details that would improve the quality of this question. Remember question quality is about making this problem as interesting to would be solvers as it is providing better context for where such techniques or ideas belong. – Ninad Munshi Jun 16 '21 at 19:29
  • Play around with this in Wolfram Alpha and see if you notice any patterns. – Cameron Williams Jun 16 '21 at 19:35
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    Letting $x=u\sqrt n$ the integral is $$\sqrt{n}\int_0^{a/\sqrt n} e^{u^2}du$$ Not sure if that helps. – Thomas Andrews Jun 16 '21 at 19:35
  • No idea where to start? How about $\lim_{n \rightarrow \infty} e^{\frac{x^2}{n}}$ – Phil H Jun 16 '21 at 19:38
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    You are going to need $a<1,$ because $e^{x^2/n}\geq 1$ so the value is at least $a^n.$ – Thomas Andrews Jun 16 '21 at 19:41
  • @ThomasAndrews over here ruining my kind of crappy hint – Cameron Williams Jun 16 '21 at 19:44
  • For $f(n)^n$ to converge, with $f$ differentiable, you either need it to converge to zero, or you need $$f(n)=1+\frac bn+o\left(\frac 1n\right)$$ – Thomas Andrews Jun 16 '21 at 19:54
  • How can you multiply $n^n$ to the value and not affect the limit? @OliverDiaz – Thomas Andrews Jun 16 '21 at 19:56
  • @ThomasAndrews: the only difficulty is when $a=1$. The cases $0<a<1$ and $a>1$ are simple. – Mittens Jun 16 '21 at 20:09
  • @popmaria Do you know Taylor's formula? Writing $\mathrm{e}^{\frac{x^2}{n}}=1+\frac{x^2}{n}+\mathcal{O}{n\to+\infty}\left(\frac{1}{n^2}\right)$ then integrating, you will eventually compute $\lim{n\to+\infty}\exp\left[n\left(\ln(a)+\ln\left(1+\frac{a^2}{3n}+\mathcal{O}_{n\to+\infty}\left(\frac{1}{n^2}\right)\right)\right)\right]$ whose value depends on $a$. – Nicolas Jun 16 '21 at 20:13
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    You may rewrite your expression as $$\Big(\int^a_0 e^{x^2/n},dx \Big)^n=\left(a\tfrac{\sqrt{n}}{a}\int^{\tfrac{a}{\sqrt{n}}}0 e^{u^2},du\right)^n$$ Notice that $\lim{h\rightarrow0}\frac{1}{h}\int^h_0 e^{u^2},du=1$. When $a=1$ one has $$\Big( \sqrt{n}\int^{1/\sqrt{n}}_0e^{u^2}-1+1,du\Big)^n=\left(\frac{ n^{3/2}\int^{n^{-1/2}}_0(e^{u^2}-1),du}{n} +1\right)^n=$$ Notice that $$\frac{1}{h^3}\int^h_0(e^{u^2}-1),du\sim\frac{e^{h^2}-1}{3h^2}\xrightarrow{h\rightarrow0}\frac13$$ Other cases $0<a<1$ and $a>1$ are trivial. – Mittens Jun 16 '21 at 20:37
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    @OliverDiaz directing that to me is pointless. That is the site policy now apparently. I'm only doing my best to try to make sure everyone goes with what the majority wants. – Ninad Munshi Jun 16 '21 at 20:46

2 Answers2

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$$I_n=\int_{0}^a e^{x^2/n}dx =\sum_{k=0}^\infty \frac{a^{2k+1}}{(2k+1)n^k k!}$$

This converges to $a$ as $n\to\infty.$ In particular, if $0<a<1$ $I_n^n\to 0,$ and if $a>1,$ then $I_n^n\to\infty.$

So the only interesting case is $a=1.$

Then $n(I_n-1)\to \frac{1}{3}.$

So, by the Lemma here, $I^n_n\to e^{1/3}.$

Paresseux Nguyen
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Thomas Andrews
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If $a<1,$ then

$$\left (\int_0^a e^{x^2/n}\, dx \right)^n < (a\cdot e^{1/n})^n = a^n\cdot e \to 0.$$

If $a>1,$ then as Thomas Andrews pointed out, we have

$$\left(\int_0^a e^{x^2/n}\,dx\right)^n > \left(\int_0^a 1\,dx\right)^n = a^n \to \infty.$$

For $a=1,$ let's use $e^u= 1 +u + o(u),$ which follows from the definition of the derivative of $e^u$ at $0.$ We then get

$$\int_0^1 e^{x^2/n}\, dx = \int_0^1 (1+x^2/n + o(1/n))\,dx$$ $$ = 1+1/(3n)+ o(1/n).$$

Raising to the $n$th power, we get $(1+1/(3n)+ o(1/n))^n \to e^{1/3}.$

zhw.
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