TITLE: Evaluate $\lim\limits_{n \to \infty}\frac{(-1)^n}{n!}\int_0^n (x-1)(x-2)\cdots(x-n){\rm d}x$
QUESTION [15 upvotes]: Evaluate $$\lim\limits_{n \to \infty}\frac{(-1)^n}{n!}\int_0^n (x-1)(x-2)\cdots(x-n){\rm d}x\,.$$


\begin{align*} I_n:&=\frac{(-1)^n}{n!}\int_0^n\prod_{k=1}^{n}(x-k){\rm d}x=\frac{(-1)^n}{n!}\int_0^n\prod_{k=1}^{n}(n-k-x){\rm d}x\\&=\frac{(-1)^n}{n!}\int_0^n\prod_{k=0}^{n-1}(k-x){\rm d}x=\frac{1}{n!}\int_0^n\prod_{k=0}^{n-1}(x-k){\rm d}x\\ &=\int_0^n\binom{x}{n}{\rm d}x, \end{align*}
but it's hard to go forward.

REPLY [16 votes]: This supplements the answer by @WhatsUp with a detailed asymptotics of $I_n$ in terms of $\log n$.
$I_n=A_n+B_n$ where $A_n=\int_0^1 f_n(x)\,dx$, $B_n=\int_1^n f_n(x)\,dx$, and $f_n(x)=\prod_{k=1}^{n}(1-x/k)$. Since the answer shows that $B_n=\mathcal{O}(1/n)$ as $n\to\infty$, let's focus on $A_n$. We use (see [1] and [2]) $$\prod_{k=1}^\infty\left(1-\frac{x}{k}\right)e^{x/k}=\frac{e^{\gamma x}}{\Gamma(1-x)};\qquad H_n:=\sum_{k=1}^{n}\frac1k=\log n+\gamma+\mathcal{O}(1/n)$$ as $n\to\infty$ (we omit this remark in the sequel). Further, $\prod_{k=n+1}^\infty(1-x/k)e^{x/k}=1+\mathcal{O}(1/n)$ uniformly for $0\leqslant x\leqslant 1$ (this boils down to showing $\sum_{k=n+1}^\infty k^{-2}\asymp 1/n$), which implies $$f_n(x)=e^{-H_n x}\prod_{k=1}^{n}\left(1-\frac{x}{k}\right)e^{x/k}=\frac{e^{(\gamma-H_n)x}}{\Gamma(1-x)}\big(1+\mathcal{O}(1/n)\big)=\frac{n^{-x}}{\Gamma(1-x)}\big(1+\mathcal{O}(1/n)\big),$$ again uniformly for $0\leqslant x\leqslant 1$. Thus, $$A_n=g(\log n)+\mathcal{O}(1/n),\qquad g(\lambda)=\int_0^1\frac{e^{-\lambda x}\,dx}{\Gamma(1-x)}.$$ To get the asymptotics of $g(\lambda)$, Watson's lemma is applicable. For this, we need the power series of $1/\Gamma(1-x)$, which can be obtained (algorithmically at least) from the known series $$\log\Gamma(1-x)=\gamma x+\sum_{k=2}^\infty\frac{\zeta(k)}{k}x^k.$$ Thus, $$I_n\asymp\frac{1}{\log n}-\frac{\gamma}{(\log n)^2}+\frac{\gamma^2-\pi^2/6}{(\log n)^3}+\ldots\qquad(n\to\infty)$$ (as the asymptotics is in powers of $1/\log n$, all the $\mathcal{O}(1/n)$ terms can be simply neglected).