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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)

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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 $$ \lim_{t \to \infty} I(t) = \int_{1}^\infty \lim_{t_\to \infty}e^{-u -t \sqrt{u e^{-u}}} \, \mathrm du = \int_{1}^\infty e^{-u}\,\mathrm du = e^{-1} $$

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  • $\begingroup$ 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 $\endgroup$ – flo3299 Oct 29 '18 at 15:28

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