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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.

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  • $\begingroup$ @Ty. I think you didn't understand that he did a change of variable $x \mapsto n - x$. $\endgroup$
    – WhatsUp
    Jul 4, 2020 at 14:57
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    $\begingroup$ The function peaks close to the origin where it behaves as something like $\sim e^{-\log(n) x}$ and will be close to zero otherwise so I would expect the integral to behave as $\mathcal{O}(1/\log(n))$. Suspect if you multiply $I_n$ by $\log(n)$ then you might get a finite non-zero limit. $\endgroup$
    – Winther
    Jul 4, 2020 at 15:14

2 Answers 2

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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).

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  • $\begingroup$ Just a remark: The $A_n$'s are related to the Bernoulli number of the second kind (or Gregory coefficients): en.wikipedia.org/wiki/Gregory_coefficients $\endgroup$
    – Gary
    Jul 4, 2020 at 17:14
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    $\begingroup$ @Gary: The Gregory coefficients are $\int_0^1\binom{x}{n}\,dx$, but $A_n=\int_{n-1}^n\binom{x}{n}\,dx$. $\endgroup$
    – metamorphy
    Jul 4, 2020 at 17:17
  • $\begingroup$ Ok, I see it now. $\endgroup$
    – Gary
    Jul 4, 2020 at 17:23
  • $\begingroup$ +1 and I learnt a new thing called Watson's lemma. $\endgroup$
    – Paramanand Singh
    Jul 5, 2020 at 1:51
  • $\begingroup$ If $n$ is odd, then the substitution $x - (n + 1)/2 = u$ converts the integral from 1 to $n$ to $\int_{(1-n)/2}^{-(1-n)/2} u(u^2 - 1)\cdots\left(u^2 - ((n - 1)/2)^2\right)\,du = 0$ since $\int_{-a}^a f(x)dx = 0 $ when $f$ is an odd function. $\endgroup$
    – PolyaPal
    Jul 7, 2020 at 18:10
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Let $I_{n, d}$ be the integral $\int_{d - 1}^d\binom x n dx$. Then your $I_n$ is just $\sum_{d = 1}^n I_{n, d}$.

We are going to treat the terms $I_{n, d}$ in several cases.

As a first observation, note that on the interval $[d - 1, d)$, we have: \begin{eqnarray} n!\left|\binom x n\right| &=& x \cdot (x - 1)\cdots (x - (d - 1)) \cdot (d - x) \cdot (d + 1 - x) \cdots (n - 1 - x)\\ &\leq& d \cdot (d - 1) \cdots 1 \cdot 1\cdot 2 \cdots (n - d)\\ &=& d! (n - d)! \end{eqnarray} which gives $\left|\binom x n\right| \leq \binom n d ^{-1}$.

Now for our $I_{n, d}$, we have the following cases.


Case 1: $d = 1$ or $d = n - 1$.

For these two $d$, we have $\left|\binom x n\right| \leq n^{-1}$, hence $I_{d, n} \leq n^{-1}$. Taking limit yields $\lim_{n\rightarrow \infty} I_{n, 1} + I_{n, n - 1} = 0$.


Case 2: $2\leq d \leq n - 2$.

In this case, we have $\left|\binom x n\right| \leq \frac 2{n(n - 1)}$ for sufficiently large $n$ (e.g. $n \geq 4$). Therefore we can bound the sum: $$\sum_{d = 2}^{n - 2} \left|I_{n, d}\right|\leq (n - 3)\cdot \frac 2{n(n - 1)}.$$ Taking limit again gives $\lim_{n \rightarrow\infty} \sum_{d = 2}^{n - 2} I_{n, d} = 0$.


Case 3: $d = n$.

This is the only remaining case. We write $J_n$ for $I_{n, n}$ and rewrite the integral as $$ J_n = \int_{n - 1}^n\binom x n dx = \int_0^1 \frac{x \cdot (x + 1) \cdots (x + n - 1)}{n!}dx.$$

Write $g_n(x) = \frac{x \cdot (x + 1) \cdots (x + n - 1)}{n!}$. Obviously we have $0\leq g_n(x) \leq 1$ for all $x\in[0, 1]$.

Furthermore, it is a simple exercise, based on the divergence of the harmonic series, that $\lim_{n\rightarrow \infty} g_n(x) = 0$ for any $x\in[0, 1)$.

Now the magic happens by applying the dominated convergence theorem. It tells us that the limit $\lim_{n\rightarrow\infty} J_n$ is $0$.


Combining all three cases, we get $\lim_{n\rightarrow\infty} I_n = 0$.

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    $\begingroup$ @Downvoters Please consider leaving a comment mentioning the reason for the downvote. $\endgroup$
    – WhatsUp
    Jul 4, 2020 at 16:18
  • $\begingroup$ +1 Sometimes they downvote by msitake $\endgroup$ Jul 4, 2020 at 19:24

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