TITLE: Asymptotic expansion of the integral $\int_0^1 e^{x^n} dx$ for $n \to \infty$
QUESTION [10 upvotes]: The integrand seems extremely easy:
$$I_n=\int_0^1\exp(x^n)dx$$
I want to determine the asymptotic behavior of $I_n$ as $n\to\infty$. It's not hard to show that $\lim_{n\to\infty}I_n=1$ follows from Lebesgue's monotone convergence theorem. However, when I try to obtain more precise results, I confront difficulties. Since it's not of the canonical form of Laplace's method, I have no powerful tools to estimate $I_n$.
Is there any good approach for that? I will be pleased if we can obtain the asymptotic expansion of $I_n$. Thanks!

REPLY [2 votes]: A related technique. Here is a start. Integrating by parts gives,
$$I= {{\rm e}}-{\frac {{{\rm e}}\,n}{1+n}}+{\frac {{{\rm e}}\,{n}^{
2}}{2\,{n}^{2}+3\,n+1}}-{\frac {{{\rm e}}\,{n}^{3}}{6\,{n}^{3}+11\,{
n}^{2}+6\,n+1}}$$
$$+{\frac{{{\rm e}}\,{n}^{4}}{24\,{n}^{4}+50\,{n}^{3}+
35\,{n}^{2}+10\,n+1}}-\int _{0}^{1}\!{\frac {{n}^{5}{{\rm e}^{{
x}^{n}}}{x}^{5\,n+1}}{x \left( 4\,n+1 \right)  \left( 3\,n+1 \right)\left( 1+2\,n \right)  \left( 1+n \right) }}{dx}.$$
From the above, we can see that as $n\to \infty$, the integral approaches $1$
$$ I = \rm e( 1 - 1 + \frac{1}{2} - \frac{1}{3!}+\frac{1}{4!} - \dots)=\rm e \rm e^{-1}=1. $$