How fast does the function $f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \frac{e^{xt}}{t^t} \, dt$ grow? Let $x$ be a positive real number and $f(x):=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $. How fast does this function grow ? In other words can we find a good asymptote for $f(x)$ as $x$ goes to $+\infty$ ?
Can we show one of these two limits converges to a constant : 
A) $\lim_{x\to+\infty} \dfrac{\ln(f(x))}{P(x)} $
B) $\lim_{x\to+\infty} \dfrac{f(x)}{P(x)} $
For some polynomial $P(x)$ ?
I know $f(z)$ is an entire function , so I tried using Taylor series.
However the derivatives of $f$ are similar looking and Hence I do not know their growth rate either !?
$$\frac{d f(x)}{d x^k} = \lim_{\epsilon\to0}\int_\epsilon^\infty \frac{e^{xt}}{t^{t-k}}\,dt.$$
Since by Taylor's theorem I need the derivatives of $f(x)$, so I am stuck on how to prove any growth rate or limit.
I considered replacing the integral with an infinite sum but that did not work for me.
I assume one way is to use contour integrals but I'm not sure how that would work.
 A: First we can rewrite the integral as
$$
f(x) = \int_0^\infty \exp\left\{x t - t\log t\right\}\,dt.
$$
The function $g(t) = xt - t\log t$ has a maximum when $t = e^{x-1}$.  This suggests the change of variables
$$
t = e^{x-1}(1+s),
$$
which transforms our integral into
$$
f(x) = \exp\left\{e^{x-1} + x - 1\right\} \int_{-1}^{\infty} \exp\left\{e^{x-1} \Bigl[s - s\log(1+s) - \log(1+s)\Bigr]\right\}\,ds.
$$
(I've found this trick to be pretty useful.  I learned it while reading about the asymptotics for the Gamma function and I also used it in this answer.)  Near $s=0$ we have
$$
s - s\log(1+s) - \log(1+s) \sim -\frac{1}{2}s^2,
$$
so the Laplace method yields immediately
$$
\begin{align}
&\int_{-1}^{\infty} \exp\left\{e^{x-1} \Bigl[s - s\log(1+s) - \log(1+s)\Bigr]\right\}\,ds \\
&\qquad \sim \int_{-\infty}^{\infty} \exp\left\{-\frac{1}{2}e^{x-1}s^2\right\}\,ds \\
&\qquad = \sqrt{2\pi} e^{-(x-1)/2}
\end{align}
$$
as $x \to \infty$.  Thus
$$
f(x) \sim \sqrt{2\pi} \exp\left\{e^{x-1} + \frac{x-1}{2}\right\}
$$
as $x \to \infty$.  If one desired they could obtain more terms of the asymptotic expansion using the method in this answer.  For example, by including the next term we get
$$
f(x) = \sqrt{2\pi} \exp\left\{e^{x-1} + \frac{x-1}{2}\right\} \left[1 + \frac{5}{24}e^{1-x} + O\left(e^{-2x}\right)\right]
$$
as $x \to \infty$.
