What is the asymptotic growth rate of $$f(x) = \int\_e^\infty e^{ - x t / \log t} dt$$ as $x \to 0$?


As an example of what is meant by "growth rate" consider $$g(x) = \int\_e^\infty e^{-x t} dt = \frac {e^{-ex}} x \approx x^{-1}.$$


By comparing to $g$, it is easy to see that $f(x) \approx x^{-1 - o(1) }$, but we would like to know what the lower order correction is.


We are guessing that $f(x) \approx \log^p(1/x)/x$ for some $p >0$, but unsure how to show this or find $p$.



Asymptotics for $\int\exp( -x t / \log t)dt$
Denote $t/\log t=y$. Then $y$ increases from $e$ to $\infty$ when $x$ goes from $e$ to $\infty$, and $dt(1/\log t-1/\log^2 t)=dy$, thus $dt\sim \log t\cdot dy\sim \log y\cdot dy$ for large $t$ (since $\log y=\log t-\log\log t\sim \log t$). Over any finite interval, the integral is uniformly bounded, thus we get $$f(x)\sim \int\_e^\infty e^{-xy}\log y dy.$$
Here denote $xy=\tau$ to get
$$
\int\_e^\infty e^{-xy}\log y dy=x^{-1}\int\_{ex}^\infty e^{-\tau}\log (\tau/x) d\tau=x^{-1}\left(\int\_{ex}^\infty e^{-\tau}\log \tau d\tau-\log x \int\_{ex}^\infty e^{-\tau}d\tau\right)\\
=-\frac{\log x+O(1)}x.
$$


Therefore, $f(x)\sim -\frac{\log x}x$.


