What is the asymptotic behaviour of the function $e !n-n!$ , where $!n = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$ is the subfactorial of $n$. I tried Wolfram Alpha but the series for n=$\infty$ is pretty complicated. There should be a simplier function doing the job.
The function arises from the integrals
$\int_{0}^{1}e^xx^ndx$ = $(-1)^n(e !n-n!)$ for every positive integer $n$.
It's best to work directly with the integral
$$ I(n) = \int_0^1 e^x x^n\, dx. $$
Make the substitution $x = e^{-t}$ to put it into the form
$$ I(n) = \int_0^\infty e^{e^{-t}-t} e^{-nt} \,dt. $$
Watson's lemma tells us that we can expand the integrand as a power series about $t=0$
$$ e^{e^{-t}-t} = e-2 e t+\frac{5 e t^2}{2} + \cdots $$
and integrate term-by-term to obtain an asymptotic expansion for the integral. Thus
$$ \begin{align} I(n) &\approx e \int_0^\infty e^{-nt}\,dt - 2e \int_0^\infty te^{-nt}\,dt + \frac{5e}{2} \int_0^\infty t^2 e^{-nt}\,dt + \cdots \\ &= \frac{e}{n} - \frac{2 e}{n^2} + \frac{5 e}{n^3} + \cdots. \end{align} $$