TITLE: Evaluate $\lim_{n \to \infty} \left(\int_0^1 e^{-x^2/n} dx\right)^n$
QUESTION [8 upvotes]: Evaluate $\lim_{n \to \infty} \left(\int_0^1 e^{-x^2/n} dx\right)^n$.

I've tried this: from Taylor's expansion at $t_0=0$ of the function $e^t$, I get that for any $t \in \mathbb{R}$ it is $e^t \ge 1+t$. Moreover, for the Lagrange's remainder of the Taylor's expansion there exists $c(t)$ on the segment of endpoints $0$ and $t$ such that
$$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}e^{c(t)}$$
When $t \le 0$, it is $\frac{t^3}{6}e^{c(t)}\le0$; hence for $t \le 0$ it is $e^t\le1+t+\frac{t^2}{2}$.
Since $-\frac{x^2}{n} \le 0$ for any $x\in[0,1]$ and for any $n\in\mathbb{N}$, it is
$$1-\frac{x^2}{n}\le e^{-x^2/n} \le 1-\frac{x^2}{n}+\frac{x^4}{2n^2}$$
For monotonicity of integral, integrating in the inequality in the interval $[0,1]$ it is
$$1-\frac{1}{3n} \le \int_0^1 e^{-x^2/n} dx\le1-\frac{1}{3n}+\frac{1}{10n^2}$$
Since $1-\frac{1}{3n} \ge 0$ for any $n\in\mathbb{N}$, for the monotonicity of the $n$-th power it is
$$\left(1-\frac{1}{3n}\right)^n \le \left(\int_0^1 e^{-x^2/n} dx\right)^n \le \left(1-\frac{1}{3n}+\frac{1}{10n^2}\right)^n$$
Since the limit preserves non strict inequalities, it is
$$\lim_{n\to\infty} \left(1-\frac{1}{3n}\right)^n\le \lim_{n \to \infty} \left(\int_0^1 e^{-x^2/n} dx\right)^n \le \lim_{n\to\infty} \left(1-\frac{1}{3n}+\frac{1}{10n^2}\right)^n$$
So, since for $n\to\infty$ it is $\left(1-\frac{1}{3n}\right)^n \to e^{-1/3}$ and $\left(1-\frac{1}{3n}+\frac{1}{10n^2}\right)^n \to e^{-1/3}$, for the squeeze theorem it follows that
$$\lim_{n \to \infty} \left(\int_0^1 e^{-x^2/n} dx\right)^n=e^{-1/3}$$
Could this work? I am unsure about the use of the Taylor formula for $t \le 0$ and the various monotonicity arguments I made.

REPLY [3 votes]: We could look at the asymptotics
$$I_n=\int_0^1 e^{-\frac{x^2}{n}}\,dx=\sqrt{n} \int_0^{\frac{1}{\sqrt{n}} } e^{-t^2}\,dt=\frac{1}{2} \sqrt{n\pi } \,\, \text{erf}\left(\frac{1}{\sqrt{n}}\right)$$
$$J_n=I_n^n=\Bigg[\frac{1}{2} \sqrt{n\pi } \,\, \text{erf}\left(\frac{1}{\sqrt{n}}\right) \Bigg]^n$$
Taking logarithm and using the series expansion of the error function for small arguments
$$\log(J_n)=n\,\log\Bigg[1-\frac{1}{3 n}+\frac{1}{10 n^2}-\frac{1}{42
   n^3}+O\left(\frac{1}{n^4}\right) \Bigg]=-\frac{1}{3}+\frac{2}{45 n}-\frac{8}{2835   n^2}+O\left(\frac{1}{n^3}\right)$$
$$J_n=e^{\log(J_n)}=\frac{1}{\sqrt[3]{e}}\Bigg[1+\frac{2}{45 n}-\frac{26}{14175 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$ which, for sure, shows the limit, how it is approached and gives a shortcut method for the evaluation of $J_n$ (the relative error being smaller than $0.001$% as soon as $n>3$).