Asymptotic expansion of exp of exp I am having difficulties trying to find the asymptotic expansion of $I(\lambda)=\int^{\infty}_{1}\frac{1}{x^{2}}\exp(-\lambda\exp(-x))\mathrm{d}x$ as $\lambda\rightarrow\infty$ up to terms of order $O((\ln\lambda)^{-2})$. How does $(\ln\lambda)^{-1}$ appear as a small parameter? Please help. Thank you.

If we substitute $e^{-x} = y$ and integrate by parts the integral becomes
$$
\begin{align}
\int_1^\infty x^{-2} \exp(-\lambda e^{-x})\,dx &= \int_0^{1/e} y^{-1} (\log y)^{-2} e^{-\lambda y}\,dy \\
&= e^{-\lambda/e} - \lambda \int_0^{1/e} (\log y)^{-1} e^{-\lambda y}\,dy. \tag{1}
\end{align}
$$
It was proved by Erdélyi in [1] that for $a$ real, $b>0$, and $0 < c < 1$,
$$
\int_0^c (-\log t)^a t^{b-1} e^{-\lambda t}\,dt \sim \lambda^{-b} \sum_{n=0}^{\infty} (-1)^n \binom{a}{n} \Gamma^{(n)}(b) (\log \lambda)^{a-n}
$$
as $\lambda \to \infty$.  So, setting $a=-1$ and $b=1$ we get
$$
\int_0^{1/e} (-\log y)^{-1} e^{-\lambda y}\,dy \sim \lambda^{-1} \sum_{n=0}^{\infty} \Gamma^{(n)}(1) (\log \lambda)^{-n-1}.
$$
Since each term in this asymptotic series is larger than $e^{-\lambda/e}$, we conclude upon substituting this into $(1)$ that

$$
\int_1^\infty x^{-2} \exp(-\lambda e^{-x})\,dx \sim \sum_{n=0}^{\infty} \Gamma^{(n)}(1) (\log \lambda)^{-n-1}
$$
  as $\lambda \to \infty$.

The first few terms of this expansion are
$$
\int_1^\infty x^{-2} \exp(-\lambda e^{-x})\,dx = (\log \lambda)^{-1} - \gamma(\log \lambda)^{-2} + \left(\gamma^2 + \tfrac{\pi^2}{6}\right)(\log \lambda)^{-3} + \cdots.
$$

[1] A. Erdélyi, General asymptotic expansions of Laplace integrals, Archive for Rational Mechanics and Analysis, 7 (1961), No. 1, pp. 1-20.
[Article page on SpringerLink]
