TITLE: Estimate $\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt$ accurately.
QUESTION [10 upvotes]: How can I obtain good asymptotics for $$\gamma_n=\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt\text{ ? }$$
[This has been already done] In particular, I would like to obtain asymptotics that show $$\sum_{n\geqslant 0}\gamma_nz^n$$
converges for every $z\in\Bbb C$. 
N.B.: The above are the coefficients when expanding $$\Gamma \left( z \right) = \sum\limits_{n \geqslant 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\frac{1}{{n + z}}}  + \sum\limits_{n \geqslant 0} {{\gamma _n}{z^n}} $$
ADD Write $${c_n} = \int\limits_0^\infty  {{t^n}{e^{ - {e^t}}}dt}  = \int\limits_0^\infty  {{e^{n\log t - {e^t}}}dt} $$
We can use something similar to Laplace's method with the expansion $${p_n}\left( x \right) = g\left( {{\rm W}\left( n \right)} \right) + g''\left( {{\rm W}\left( n \right)} \right)\frac{{{{\left( {x - {\rm W}\left( n \right)} \right)}^2}}}{2}$$
where $g(t)=n\log t-e^t$. That is, let $$\begin{cases} w_n={\rm W}(n)\\
   {\alpha _n} = n\log {w_n} - {e^{{w_n}}}  \\
   {\beta _n} = \frac{n}{{w_n^2}} + {e^{{w_n}}} \end{cases} $$
Then we're looking at something asymptotically equal to $${C_n} = \exp {\alpha _n}\int\limits_0^\infty  {\exp \left( { - {\beta _n}\frac{{{{\left( {t - {w_n}} \right)}^2}}}{2}} \right)dt} $$

REPLY [9 votes]: To get an asymptotic estimate we can use a trick similar to one used to find the asymptotics of $n!$.
You wrote
$$
c_n = \int_0^\infty t^n e^{-e^t}\,dt = \int_0^\infty \exp\left(n\log t - e^t\right)\,dt = \int_0^\infty \exp f_n(t)\,dt
$$
and saw that $f_n(t)$ has a maximum at $t = W(n)$.  We thus scale the integration variable by
$$
t = W(n)(1+s),
$$
which yields
$$
\begin{align}
c_n &= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n\left[\log(1+s) - W(n)^{-1} e^{W(n)s}\right]\right\}\,ds \\
&= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n g_n(s)\right\}\,ds
\end{align}
$$
The largest contribution to the integral now comes from a neighborhood of $s=0$, and there we have
$$
g_n(s) = -\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2 + \cdots.
$$
The details of the Laplace method can be worked out as usual to arrive at the conclusion that
$$
\begin{align}
c_n &\sim W(n)^{n+1} \int_{-\infty}^{\infty} \exp\left\{n\left[-\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2\right]\right\}\,ds \\
&= W(n)^{n+1} e^{-n/W(n)} \sqrt{\frac{2\pi}{n(1+W(n))}}.
\end{align}
$$
Below is a plot of $c_n$ in blue and this asymptotic in purple for $1 < n < 10$.  Below that is a plot of $W(n)^{-n-1} e^{n/W(n)} c_n$ in blue and $\sqrt{\frac{2\pi}{n(1+W(n))}}$ in purple for $1 < n < 50$.


We can then use the estimate
$$
\log n - \log\log n < W(n) < \log n
$$
from this answer, true for $n > e$, to find that
$$
c_n < (\log n)^{n+1} e^{-n/\log n} \sqrt{\frac{2\pi}{n(1+\log n - \log\log n)}}
$$
for $n$ large enough.  Since $n! \geq (n/e)^n \sqrt{2\pi n}$ we obtain the asymptotic bound
$$
\begin{align}
\gamma_n &= \frac{c_n}{n!} \\
&< \left(\frac{e \log n}{n}\right)^{n+1} e^{-1-n/\log n} (1+\log n-\log\log n)^{-1/2}
\end{align}
$$
which clearly decreases faster than exponentially.  From this it follows that $\sum \gamma_n z^n$ converges for all $z$.