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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.

ZaWarudo
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    Yes, it is known that $ 1 - t < e^{ - t} < 1 - t + \frac{{t^2 }}{2} $ for any $t>0$. – Gary Jan 03 '22 at 05:05
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    This works. You can use the Lemma here once you know the value without the exponent is $1-1/(3n)+o(1/n).$ https://math.stackexchange.com/a/1451245/7933 – Thomas Andrews Jan 03 '22 at 05:19
  • If $c_n\xrightarrow{n\rightarrow\infty}c$, then $\lim_n\Big(1+\frac{c_n}{n}\Big)^n=e^c$. – Mittens May 26 '22 at 13:39

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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 $n>3$).

Claude Leibovici
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