Asymptotic expansion of the integral $\int_0^1 e^{x^n} dx$ for $n \to \infty$ 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!
 A: Let $n \geq 1$. A first crude estimate can be obtained as follows. Substitute $x \leftarrow x^{1/n}$ to get
$$ I_n = \frac{1}{n}\int_0^1 x^{\frac{1}{n} - 1} e^x dx.
$$
Now we can estimate $e^x$ on the interval $[0,1]$ to get
$$
\frac{1}{n}\int_0^1 x^{\frac{1}{n}-1}(1+x)\, dx < I_n < \frac{1}{n}\int_0^1 x^{\frac{1}{n} - 1} (1 + (e-1)x)\, dx
$$
or
$$
1 + \frac{1}{n+1} < I_n < 1 + \frac{e-1}{n+1}.
$$
A: 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 $1$
$$ I = \rm e( 1 - 1 + \frac{1}{2} - \frac{1}{3!}+\frac{1}{4!} - \dots)=\rm e \rm e^{-1}=1. $$
A: You can expand the exponential in a Taylor series quite accurately:
$$\exp{\left ( x^n \right )} = 1 + x^n + \frac12 x^{2 n} + \ldots$$
Because $x \in [0,1]$, this series converges rapidly as $n \to \infty$.
Then the integral is
$$1 + \frac{1}{n+1} + \frac12 \frac{1}{2 n+1} + \ldots = \sum_{k=0}^{\infty}\frac{1}{k!} \frac{1}{k n+1}$$
We can rewrite this as
$$\begin{align}I_n&=1+\frac{1}{n} \sum_{k=1}^{\infty} \frac{1}{k \cdot k!} \left ( 1+\frac{1}{k n} \right )^{-1}\\ &= 1+\frac{1}{n} \sum_{m=0}^{\infty} \frac{(-1)^m}{n^m} \: \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}\\ &= 1+\sum_{m=1}^{\infty} (-1)^{m+1}\frac{K_m}{n^m} \end{align}$$
where 
$$K_m = \sum_{k=1}^{\infty} \frac{1}{k^{m} k!}$$
To first order in $n$:
$$I_n \sim 1+\frac{K_1}{n} \quad (n \to \infty)$$
where
$$K_1 = \sum_{k=1}^{\infty} \frac{1}{k\, k!} = \text{Ei}(1) - \gamma \approx 1.3179$$
This checks out numerically in Mathematica.
BONUS
As a further check, I computed the following asymptotic approximation:
$$g(n) = 1+\frac{K_1}{n} -\frac{K_2}{n^2} $$
where 
$$K_2 = \sum_{k=1}^{\infty} \frac{1}{k^2 k!} \approx 1.1465$$
I computed 
$$\log_2{\left[\frac{\left|g\left(2^m\right)-I_{2^m}\right|}{I_{2^m}}\right]}$$
for $m \in \{1,2,\ldots,9\}$  The results are as follows
$$\left(
\begin{array}{cc}
 1 & -4.01731 \\
 2 & -6.56064 \\
 3 & -9.26741 \\
 4 & -12.0963 \\
 5 & -15.0028 \\
 6 & -17.9538 \\
 7 & -20.9287 \\
 8 & -23.916 \\
 9 & -26.9096 \\
\end{array}
\right)$$
Note that the difference between successive elements is about $-3$; because this is a log-log table, that means that this error is $O(1/n^3)$ and that the approximation is correct.
