What is$\lim\limits_{n \rightarrow +\infty} \left(\int_{a}^{b}e^{-nt^2}\text{d}t\right)^{1/n}$?

Question

For $\left(a,b\right)) \in \left(\mathbb{R}^{*+}\right)^2$. Let $\left(I_n\right)_{n \in \mathbb{N}}$ be the sequence of improper integrals defined by $$ \left(\int_{a}^{b}e^{-nt^2}\text{d}t\right)^{1/n} $$ I'm asked to calculate the limit of $I_n$ when $ \ n \rightarrow +\infty$.

I've shown that $$ \int_{x}^{+\infty}e^{-t^2}\text{d}t \underset{(+\infty)}{\sim}\frac{e^{-x^2}}{2x} $$ However, how can I use it ? I wrote that $$ \int_{a}^{b}e^{-nt^2}\text{d}t=\frac{1}{\sqrt{n}}\int_{\sqrt{n}a}^{\sqrt{n}b}e^{-t^2}\text{d}t $$ Hence I wanted to split it in two integrals to use two times the equivalent but i cannot sum them so ... Any idea ?

Answer 1

With $f(x)=e^{-x^{2}}$, by letting $M:=f(x_{0})=\max_{x\in[a,b]}f(x)>0$, we have \begin{align*} \int_{a}^{b}f(x)^{n}dx\leq(b-a)M^{n}, \end{align*} so \begin{align*} \left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq(b-a)^{1/n}M, \end{align*} so \begin{align*} \limsup_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq M. \end{align*} On the other hand, let $\epsilon>0$ be such that $\epsilon<M$, then there is some open interval $I$ of $x_{0}$ such that $f(x)>M-\epsilon$ for all $x\in I\cap[a,b]$, then \begin{align*} \int_{a}^{b}f(x)^{n}dx\geq\int_{I\cap[a,b]}f(x)^{n}\geq\int_{I\cap[a,b]}(M-\epsilon)^{n}dx=|I\cap[a,b]|(M-\epsilon)^{n}, \end{align*} so \begin{align*} \left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq|I\cap[a,b]|^{1/n}(M-\epsilon), \end{align*} so \begin{align*} \liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq M-\epsilon, \end{align*} now taking $\epsilon\downarrow 0$, then \begin{align*} \liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\geq M. \end{align*} In short, \begin{align*} M\leq\liminf_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq\limsup_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}\leq M, \end{align*} so \begin{align*} \lim_{n\rightarrow\infty}\left(\int_{a}^{b}f(x)^{n}dx\right)^{1/n}=M. \end{align*}

Answer 2

Indeed, this case it is easy since by continuity, for any $\varepsilon $ such that $0<\varepsilon< e^{-a^2}$ there exists $d>0$ such that for $t\in (a, a+d)\subset(a,b)$ we have

$$|e^{-t^2}-e^{-a^2}|\le\varepsilon \implies e^{-t^2} \ge e^{-a^2}-\varepsilon>0$$ therefore for all $n$, we get

$$ (e^{-a^2}-\varepsilon)d^{1/n}\le \left(\int_a^{a+d}e^{-nt^2}dt\right)^{1/n}\le \left(\int_a^be^{-nt^2}dt\right)^{1/n}\le e^{-a^2}$$ you obtain the result by squeezing and letting $\varepsilon \to0$ afthermath. That is $$ e^{-a^2} =\lim_{n\to\infty}\left(\int_a^b e^{-nt^2}dt\right)^{1/n}$$

Generally, It is well known that for continuous function
$$ \max_{[a,b]}|f| =\lim_{n\to\infty}\left(\int_a^b|f(t)|^ndt\right)^{1/n}$$

Answer 3

First answer. This has some problems but now it is fixed.

So you have the result: \begin{align}\tag{1} \int^{\infty}_x e^{-t^2}\,dt = \frac{e^{-x^2}}{2x}+o\left(\frac{e^{-x^2}}{x}\right) \ \ \ \text{as} \ \ x\to\infty \end{align} In your last step, you had a mistake. It would be: \begin{align} \int^b_a e^{-nt^2}\,dt &= \frac{1}{\sqrt[]{n}}\int^{\sqrt[]{n}b}_{\sqrt[]{n}a}e^{-t^2}\,dt\\ & = \frac{1}{\sqrt[]{n}}\left(\int^{\infty}_{\sqrt[]{n}a}e^{-t^2}\,dt - \int^\infty_{\sqrt[]{n}b}e^{-t^2}\,dt \right)\\ \end{align} Assume $0<a<b$ $(\star)$. First note that $$\frac{e^{-nb^2}}{n}=o\left(\frac{e^{-na^2}}{n}\right)$$ as $n\to\infty$ (we will avoid writing this from now on). So use $(1)$ to get: \begin{align}\tag{2} \int^b_a e^{-nt^2}\,dt = \frac{e^{-na^2}}{2na}+o\left(\frac{e^{-na^2}}{n}\right) \end{align} For $n$ large enough we can take $n$-th root on both sides of $(2)$ to get: \begin{align} \left(\int^b_a e^{-nt^2}\,dt\right)^{1/n}&=\left[\frac{e^{-na^2}}{2na}+o\left(\frac{e^{-na^2}}{n}\right)\right]^{1/n}\\ &=e^{-a^2}\frac{1}{n^{1/n}(2a)^{1/n}}\left[1+o\left(1\right)\right]^{1/n}\\ &\to e^{-a^2} \end{align} Where we have used $c_n^{1/n}\to 1$ for $c_n$ strictly positive and bounded away from $0$ and the fact that $\sqrt[n]{n}\to 1$.

$(\star)$: If you allow $a=0$, then something similar can be done which is even easier.

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Edit One can also come up with the asymptotics of the integral: \begin{align} I_n^n=\int^b_a e^{-nt^2}\,dt \end{align} Assume $0<a<b$. Note that $t^2$ is monotonically increasing and using The Laplace Method, we get: \begin{align} I_n^n\sim \frac{e^{-na^2}}{2an} \end{align} Taking $n$-th root we obtain the result: \begin{align} \lim_{n\to\infty} I_n = e^{-a^2} \end{align}