The integrand seems extremely easy:


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!

  • $\begingroup$ The existing answers are better, but note that $e^{x^n}$ is concave and so lies above its tangent line (and obviously exceeds $1$ as well). Note that the slope of the tangent line to $y=e^{x^n}$ at $x=1$ equals $ne$. Therefore $I_n$ is larger than the integral of the piecewise linear function going through the three points $(0,1)$, $(1-1/ne,1)$, and $(1,e)$, which is $1+1/2en$. A similar argument involving secant lines would probably yield an upper bound of the same order of magnitude. $\endgroup$ Apr 29, 2013 at 6:59
  • $\begingroup$ The integral seems easy, and the behavior of the expansion is expected, but the coefficients are totally not trivial. This turned out to be a very cool problem - thanks for posting! $\endgroup$
    – Ron Gordon
    Apr 29, 2013 at 13:42

3 Answers 3


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

$$\begin{align}I_n&=1+\frac{1}{n} \sum_{k=1}^{\infty} \frac{1}{k \cdot k!} \left ( 1+\frac{1}{k n} \right )^{-1}\\ &= 1+\frac{1}{n} \sum_{m=0}^{\infty} \frac{(-1)^m}{n^m} \: \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}\\ &= 1+\sum_{m=1}^{\infty} (-1)^{m+1}\frac{K_m}{n^m} \end{align}$$


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

To first order in $n$:

$$I_n \sim 1+\frac{K_1}{n} \quad (n \to \infty)$$


$$K_1 = \sum_{k=1}^{\infty} \frac{1}{k\, k!} = \text{Ei}(1) - \gamma \approx 1.3179$$

This checks out numerically in Mathematica.


As a further check, I computed the following asymptotic approximation:

$$g(n) = 1+\frac{K_1}{n} -\frac{K_2}{n^2} $$


$$K_2 = \sum_{k=1}^{\infty} \frac{1}{k^2 k!} \approx 1.1465$$

I computed


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.

  • $\begingroup$ Very good, thanks! $\endgroup$
    – Yai0Phah
    Apr 29, 2013 at 6:59
  • $\begingroup$ Since $\exp(x^n)=\sum_0^\infty x^{nk}/k!$ converges absolutely on $[0,1]$, the interchange of summation and integration is justified. The interchange of summation is valid for the absolute convergence of the double sum. Am I right? $\endgroup$
    – Yai0Phah
    Apr 29, 2013 at 7:18
  • $\begingroup$ @FrankScience: absolutely correct. $\endgroup$
    – Ron Gordon
    Apr 29, 2013 at 7:20
  • $\begingroup$ @Ron Gordon The first few terms are numerically $\left\{\frac{1.3179}{n},-\frac{1.1465}{n^2},\frac{1.0694}{n^3},-\frac{1.03348}{n^4},\frac{1.01635}{n^5}\right\}$ Neglecting the difference from unity in the numerator and summing up the asmptotic series the asymptotic behaviour of the integral can be written in this approximation as $\frac{n+2}{n+1}$ $\endgroup$ Nov 7, 2019 at 10:05

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 $$ 1 + \frac{1}{n+1} < I_n < 1 + \frac{e-1}{n+1}. $$

  • $\begingroup$ -- very nice! -- $\endgroup$ Apr 29, 2013 at 7:00

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 $1$

$$ I = \rm e( 1 - 1 + \frac{1}{2} - \frac{1}{3!}+\frac{1}{4!} - \dots)=\rm e \rm e^{-1}=1. $$


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