Asymptotic expansion of exp of exp

Question

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.

Answer 1

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.

Answer 2

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

$$ \int_1^\infty x^{-2} \exp(-\lambda e^{-x})\,dx = (\log \lambda)^{-1} - \gamma(\log \lambda)^{-2} + \left(\gamma^2 + \tfrac{\pi^2}{6}\right)(\log \lambda)^{-3} + \cdots. $$

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