calculus - What is$limlimits_{n rightarrow +infty} left(int_{a}^{b}e^{-nt^2}text{d}tright)^{1/n}$?

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

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 $$\begin{align}\tag{2} \int^b_a e^{-nt^2}\,dt = \frac{e^{-na^2}}{2na}+o\left(\frac{e^{-na^2}}{n}\right) \end{align}$$

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 $0The 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:
$$\begin{align} \lim_{n\to\infty} I_n = e^{-a^2} \end{align}$$