\begin{align} I_n=\int_0^1 x^ne^xdx = a_ne+b_n \end{align} where $a_n,b_n \in \mathbb{Z}$ and $n\geq0$ is a integer.
How to find a bound for $a_n$ in the form $a_n<f(n) \: ?$
I'm not sure what are the most appropriate tags for this question.
Using Taylor series for $e^x$, $\int_0^1 x^ne^xdx = \sum_{k=0}^{\infty} \frac{1}{k!(n+k+1)}$. Maybe evaluating this series would help?
To give a example of what I'm asking:
\begin{align} &I_0=e-1\\ &I_1= 1\\ &I_2= e-2\\ &I_3=6-2e\\ &I_4=9e-24\\ \end{align}
Above, the number $a_n$ that multiplies $e$ is of the form $|a_n|<n^2$, if $n\leq4$. I'm looking for a estimation that is valid for all $n\geq0$.
$$ \int_{0}^{1} f(x) e^{x}\,dx = \left[\left(f(x)-f'(x)+f''(x)-f'''(x)+\ldots\right)e^x\right]_{0}^{1}$$ gives that, up to the sign, $b_n$ is the number of elements of $S_n$ (i.e. n!) and $a_n$ is the number of permutations without fixed points (derangements) in $S_n$. Of course the signs of $a_n$ and $b_n$ are always opposite since
$$ \int_{0}^{1} x^n e^{x}\,dx \sim \int_{0}^{1} x^n(1+(e-1)x)\,dx \sim \frac{e}{n} $$ as $n\to +\infty$, so $|a_n|\sim\frac{n!}{e}$. A marvelous strategy for finding accurate rational approximations of $e$ (actually the convergents of its continued fraction) is to consider the similar (Beuker-like) integrals
$$\int_{0}^{1}x^n(1-x)^n e^{x}\,dx < \frac{e}{4^n}.$$