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So, I just need a hint for proving $$\lim_{n\to \infty} \int_0^1 e^{-nx^2}\, dx = 0$$

I think maybe the easiest way is to pass the limit inside, because $e^{-nx^2}$ is uniformly convergent on $[0,1]$, but I'm new to that theorem, and have very limited experience with uniform convergence. Furthermore, I don't want to integrate the Taylor expansion, because I'm not familiar with that. So, I want to prove it in a way I'm more familiar with, if possible. So far I've tried:

  1. Show that $e^{-nx^2}$ is a monontone decreasing sequence with limit $0$. Then use the monotone property of integrals but I think this argument would just end circularly with passing the limit out of the integration operator.

  2. Bound $e^{-nx^2}$ by 0 and some other $f(x)$ like $\cos^n x$ or $(1-\frac{x^2}{2})^n$ and then use the squeeze theorem. But the integrals of those functions seem to be a little bit out of my math range to analyze.

But I have a feeling that there's something much simpler here that I'm missing.

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  • $\begingroup$ Try letting $x= y/\sqrt n.$ $\endgroup$ – zhw. Nov 12 '17 at 1:16
  • $\begingroup$ How did you pick your right hand side? Looks like for $\int_a^b f(x) dx$ you chose $\epsilon f(a) + (1- \epsilon)f(\epsilon)$ which looks like the definition of the convex function, but I'm missing how the integral is related. $\endgroup$ – Zduff Nov 12 '17 at 2:06
  • $\begingroup$ @Sangchul Lee: You should make this an answer. imho, this is the best and most elementary answer. $\endgroup$ – marty cohen Nov 12 '17 at 4:19
  • $\begingroup$ @martycohen, Thank you for the suggestion, I moved it to an answer. $\endgroup$ – Sangchul Lee Nov 12 '17 at 5:01
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$$0\leq\int_{0}^{1}e^{-nx^2}\,dx \leq \int_{0}^{1}\frac{dx}{1+n x^2} = \frac{\arctan\sqrt{n}}{\sqrt{n}}\leq \frac{\pi}{2\sqrt{n}}.$$

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There are many ways to prove the limit, and better inputs would lead to a better quantitative bounds.

  1. Let $\epsilon \in (0, 1)$. Since $x \mapsto e^{-nx^2}$ is decreasing on $[0, 1]$, we have

    $$ 0 \leq \int_{0}^{1}e^{-nx^2} \,dx = \int_{0}^{\epsilon}e^{-nx^2} \,dx + \int_{\epsilon}^{1}e^{-nx^2} \,dx \leq \epsilon + (1-\epsilon)e^{-n\epsilon^2}. $$

    So taking $n\to\infty$, we obtain

    $$ 0 \leq \liminf_{n\to\infty} \int_{0}^{1}e^{-nx^2} \,dx \leq \limsup_{n\to\infty} \int_{0}^{1}e^{-nx^2} \,dx \leq \epsilon. $$

    Since this holds for all $\epsilon > 0$, taking $\epsilon \downarrow 0$ proves the desired limit.

  2. Notice that $I = \int_{0}^{\infty}e^{-x^2} \, dx < \infty$. This can be proved in a various way, but one easy trick is to observe that for $R > 1$ we have the following uniform bound

    $$ \int_{0}^{R} e^{-x^2} \, dx \leq \int_{0}^{1}e^{-x^2} \, dx + \underbrace{\int_{1}^{R} e^{-x} \, dx}_{=e^{-1} - e^{-R}} \leq \int_{0}^{1} e^{-x^2} \,dx + e^{-1}$$

    So it follows that

    $$ \int_{0}^{1} e^{-nx^2} \, dx \stackrel{\sqrt{n}x = y}{=} \frac{1}{\sqrt{n}} \int_{0}^{\sqrt{n}} e^{-y^2} \, dy \leq \frac{I}{\sqrt{n}} \xrightarrow{n\to\infty} 0. $$

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$$ \int_0^1 e^{-nx^2} dx < \int_0^{\infty} e^{-nx^2} dx = \sqrt{\frac{\pi}{4n}} \to 0 $$

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    $\begingroup$ Correct answer but a bit hasty. Maybe you can think about keeping the $\int_{0}^1$ and find a better bound for the integrand. The point is to find an function that can be integrated or antiderivative can be found. The use of Improper integral is really not in the best interest of OP. $\endgroup$ – DeepSea Nov 12 '17 at 1:17
  • $\begingroup$ Ok, it's the first thing that came to mind. OP didn't clarify how much they know, and it's a pretty common identity. $\endgroup$ – Dylan Nov 12 '17 at 1:24
  • $\begingroup$ Sorry, yeah, I haven't undergone a rigorous study of improper integrals yet. Or that identiity. Should have said something about that. $\endgroup$ – Zduff Nov 12 '17 at 1:26
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Letting $y=x/\sqrt n$ shows the integral equals

$$\frac{1}{\sqrt n} \int_0^{\sqrt n} e^{-y^2}\, dy <\frac{1}{\sqrt n} \int_0^{\infty} e^{-y^2}\, dy .$$

Since the last integral converges, the desired limit is $0.$

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