I need to find the following asymptotic expansion as $t\rightarrow \infty$ :
$\int_{0}^{e^{-1}}e^{-t\sqrt{-y\ln y}}{\rm d}y. $
Introducing the new variable (related to the left branch of the Lambert function) : $u=-e^{\ln y}\ln y\Longleftrightarrow y=\exp\left(W_{-1}\left(-u\right)\right)$ and ${\rm d}y=-\frac{{\rm d}u}{1+W_{-1}\left(-u\right)}$, we have :
$\int_{0}^{e^{-1}}e^{-t\sqrt{-y\ln y}}{\rm d}y=-\int_{0}^{e^{-1/2}}\frac{e^{-t\sqrt{u}}}{1+W_{-1}\left(-u\right)}{\rm d}u$. Unfortunately, from there I can not say much..
Numerically it seems that the integral is pretty close to $1 / (t^2\ln t)$ (c.f Mathematica)
Let's be very candid: let $u = -\ln y$ so that $\mathrm dy = - e^{-u}\, \mathrm du$, and write your equation as $$ I(t) = \int_{1}^{\infty} e^{-u -t \sqrt{u e^{-u}}} \, \mathrm du $$ Then invoke dominated convergence to obtain $$ \lim_{t \to \infty} I(t) = \int_{1}^\infty \lim_{t_\to \infty}e^{-u -t \sqrt{u e^{-u}}} \, \mathrm du = \int_{1}^\infty e^{-u}\,\mathrm du = e^{-1} $$