TITLE: Asymptotics for $\int\exp( x t / \log t)dt$
QUESTION [5 upvotes]: 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 $p$.

REPLY [4 votes]: 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
$$f(x) = -\frac{\gamma+\log x}{x} +
 \sum_{j=2}^\infty \frac{(-1)^j(j-1)}{j\,j!} e^j x^{j-1}.$$