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.

This is a slightly modified Laplace's method. Write the integrand as
$$ \exp{(-2\log{x}-\lambda e^{-x})}. $$
The maximum occurs when the derivative of the exponent is zero, which is when
$$ -\frac{2}{x} +\lambda e^{-x} = 0. $$
Taking logs,
$$ x -\log{x} = \log{\lambda}-\log{2}, \tag{1} $$
and since $\log{x} \ll x$ for large $x$, a sensible first-order approximation for the maximum is $x=\log{\lambda}$, and you can feed that back into (1) to get a better approximation (which will lead to a series in $\log{\lambda}$, presumably). You then feed this into the usual Laplace's method, approximating the function using the Taylor series of the exponent around the maximum up to the second derivative.
