Laplace method for $\int_0^1 e^{a(x-1)}\ln(-\ln(x)) \, dx$ The following integral
$$ \int_0^1 e^{a(x-1)}\ln(-\ln(x)) \, dx$$
looks like the one for which the Laplace method is applicable. The function under the integral looks like this:

($a$ increases from red to black), so the leading contribution to the integral should be at the points 1 and 0. I can split this into two integrals from functions having maximum at 0, however direct application of the Laplace method is problematic since the maxima are infinite. Can it be fixed with some variable change, or do I need something beyond the Laplace method (steepest descents or anything else)?

Let $I(a,\epsilon) = \int_\epsilon^{1/e} e^{-a(1-x)}\log\log\frac{1}{x}dx$ and $J(a,\epsilon) = \int_{1/e}^{1-\epsilon} e^{1a(1-x)}\log\log\frac{1}{x}dx$, i.e. we cut the integral at the unique root $1/e$
Then holds for all $\epsilon>0$:


*

*$0\le I(a,\epsilon) \le e^{-a(1-\frac{1}{e})} I(0,\epsilon)$

*$0\ge J(a,\epsilon) \ge e^{-a(1-\frac{1}{e})} J(0,\epsilon)$


Hence if $I(a) = I(a,0)$ exists (which im not sure it does), then $\lim_{a\to\infty}I(a) = 0$; same for $J(a)$.
