Asymptotics for $\int\exp( -x t / \log t)dt$ 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 $p$.
 A: THIS ANSWER IS INCORRECT, SEE THE COMMENTS.
According to Maple,
$$ f(x) = -\frac{\gamma+\log x}{x}
+\tfrac14 e^2 x - \tfrac19 e^3x^2
+ \tfrac{1}{32}e^4x^3 - \tfrac{1}{150}e^5x^4
+ O(x^5).$$
A convergent series is available. The exact value of the integral is $\frac{1}{x}(e^{-ex} + \mathrm{Ei}(1,ex))$ where Ei is the exponential integral. Using the known Taylor series of the exponential integral, we get
$$f(x) = -\frac{\gamma+\log x}{x} +
 \sum_{j=2}^\infty \frac{(-1)^j(j-1)}{j\,j!} e^j x^{j-1}.$$
A: 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, $f(x)\sim -\frac{\log x}x$.
A: Take any real $a>e$. Then
\begin{equation*}
    f(x)=f_{1,a}(x)+f_{2,a}(x), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
    f_{1,a}(x):=\int_e^a dt\,e^{-xg(t)},\quad 
    f_{2,a}(x):=\int_a^\infty dt\,e^{-xg(t)}, \tag{2}\label{2}
\end{equation*}
\begin{equation*}
    g(t):=\frac t{\ln t}. \tag{3}\label{3}
\end{equation*}
Note that
\begin{equation*}
    0\le f_{1,a}(x)\le a. \tag{4}\label{4}
\end{equation*}
The function $g\colon[e,\infty)\to[e,\infty)$ is continuous and increasing from $g(e)=e$ to $g(\infty-)=\infty$, with
\begin{equation*}
    g'(t)=\frac{\ln t-1}{\ln^2 t}>0
\end{equation*}
for $t\in(e,\infty)$. So, $g$ has the inverse $h:=g^{-1}$, which is continuous and increasing from $h(e)=e$ to $h(\infty-)=\infty$, and also $h$ is differentiable on $(e,\infty)$. So, making the substitution $s=g(t)\iff t=h(s)$, we have
\begin{equation*}
    f_{2,a}(x)=\int_{g(a)}^\infty ds\,h'(s)e^{-xs}=\int_{g(a)}^\infty ds\,\frac{\ln^2 h(s)}{\ln h(s)-1}e^{-xs}. \tag{5}\label{5}
\end{equation*}
Note that $h(s)\sim s\ln s$ as $s\to\infty$. So, there exists a function $A$ from a right-neighborhood of $0$ to $(e,\infty)$ such that
\begin{equation*}
    A(x)\to\infty,\quad A(x)=o\Big(\frac1x\ln\frac1x\Big),\quad
    xg(A(x))\to0, \tag{6}\label{6}
\end{equation*}
and
\begin{equation*}
    \text{$h(s)\sim s\ln s$, and hence $\ln h(s)\sim\ln s$, } \\
    \text{uniformly in $s\in[g(A(x)),\infty)$;}\tag{7}\label{7}
\end{equation*}
all limits here and in what follows are taken for $x\downarrow0$.
It follows by \eqref{5}, \eqref{6}, \eqref{7} that
\begin{equation*}
\begin{aligned}
    f_{2,A(x)}(x)&\sim\int_{g(A(x))}^\infty ds\,\ln s\,e^{-xs} \\  
    &=\int_{xg(A(x))}^\infty \frac{du}x\,\Big(\ln u+\ln\frac1x\Big)\,e^{-u} \\  
    &=\frac1x\,\int_{xg(A(x))}^\infty du\,\ln u\,e^{-u}
    +\frac1x\,\ln\frac1x\,\int_{xg(A(x))}^\infty du\,e^{-u} \\
        &=\frac{1+o(1)}x\,\int_0^\infty du\,\ln u\,e^{-u}
    +\frac{1+o(1)}x\,\ln\frac1x\,\int_0^\infty du\,e^{-u} \\  
    &\sim\frac1x\,\ln\frac1x. 
\end{aligned}
\tag{8}\label{8}
\end{equation*}
It also follows from \eqref{4} and \eqref{6} that
\begin{equation}
    f_{1,A(x)}(x)=o\Big(\frac1x\,\ln\frac1x\Big). \tag{9}\label{9}
\end{equation}
Finally, by \eqref{1}, \eqref{8}, \eqref{9},
\begin{equation}
    f(x)\sim\frac1x\,\ln\frac1x. 
\end{equation}
