# Laplace's method with Lambert function

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 $$\lim_{t \to \infty} I(t) = \int_{1}^\infty \lim_{t_\to \infty}e^{-u -t \sqrt{u e^{-u}}} \, \mathrm du = \int_{1}^\infty e^{-u}\,\mathrm du = e^{-1}$$
• Thanks but I am pretty sure your answer is wrong as for u close to infinity the term $ue^{-u}$ will be very close to 0 – flo3299 Oct 29 '18 at 15:28