Problem: find the asymptotic expansion of the following integral:

$$I_n=\int_0^1\exp(x^n)dx$$

as $n \to \infty$. This expansion, while at first pedestrian-seeming, turns out to have a very interesting set of coefficients.

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

$$K_m = \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}$$

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.

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