The asymptotic $\prod_{k=1}^n\int_{(k-1)/n}^{k/n}f(t)\mathrm{d}t\sim A/(Bn)^n$

Question

Let $P_f(n)=\prod_{k=1}^n\int_{(k-1)/n}^{k/n}f(t)\mathrm{d}t$ be the product of integral parts of a function $f$. For simplicity, assume $f$ is smooth and positive-valued on $(0,1)$. I noticed some limits in this question and this question can be rephrased as asymptotics:

• $\displaystyle P_f(n)\sim \frac{2\cos(\frac{\pi}{2\sqrt{3}})}{(en)^n}$ for $f(t)=\sqrt{t(1-t)}$
• $\displaystyle P_f(n)\sim \frac{4\cosh^2(\frac{\pi}{2\sqrt{3}})}{(2n)^n}$ for $f(t)=1-\cos(2\pi t)$
• $\displaystyle P_f(n)\sim\frac{2}{(e^2n)^n}$ for $f(t)=t(1-t)$
• $\displaystyle P_f(n)\sim\frac{2\cosh(\frac{\pi}{2\sqrt{3}})}{(e^2n)^n}$ for $f(t)=t^2$

All of the asymptotics are of the form $A/(Bn)^n$ for some constants $A$ and $B$. Taking logs, dividing by $n$, and approximating $n\int_{(k-1)/n}^{k/n}f(t)\mathrm{d}t\approx f(\frac{2k-1}{2n})$, we can say $-\ln B$ is approximated by a midpoint Riemann sum yielding, potentially,

$$ B=\exp\Big(-\int_0^1\ln f(t)\mathrm{d}t\Big)=\prod_0^1 f(t)^{-\mathrm{d}t}, $$

a product integral. This agrees with the four examples above. But how do we find $A$? With the ol' adding-and-subtracting trick, I get $\ln A=X+Y$ where $X$ and $Y$ are given by

$$ X=\lim_{n\to\infty} n\Big[-\int_0^1\ln f(t)\mathrm{d}t+\sum_{k=1}^n\frac{1}{n}\ln f\big(\frac{2k-1}{2n}\big)\Big] $$

$$ Y=\lim_{n\to\infty} \sum_{k=1}^n \ln\Big(\frac{\int_{(k-1)/n}^{k/n}f(t)\mathrm{d}t}{\frac{1}{n}f(\frac{2k-1}{2n})}\Big) $$

The former seems to be in want of an Euler-Maclaurin formula for product integrals. The latter I assume needs to use Newton-Mercator for $\ln(1+u)$, but if $f\to 0$ as $t\to0$ or $t\to1$ then I'm not sure $u\to0$ uniformly as $n\to\infty$. Also possible is expanding $f(t)$ as a Taylor series around $\frac{2k-1}{2n}$ in each integral but I think this will hit the same wall.

Ideally it may be possible to find a complete asymptotic series for $\ln P_f(n)$ as $n\to\infty$ with coefficients expressed in terms of $f$, or at least an algorithm to generate such a series.

Answer 1

1. Main Result

In this post, we will prove the next technical result. If you are only interested in how this answers OP's question, you can just check the definition $\text{(2)}$ of the constant $C_{\alpha}$ and jump directly to the corollary below.

Theorem. Suppose

1. $K = K_n$ is a sequence of positive integers such that $K/n \to \ell$ for some $\ell \in (0, \infty)$ as $n \to \infty$.
2. $f(t) = t^{\alpha} g(t)$, where $\alpha \geq 0$ and $g(t)$ is a function which is strictly positive and $C^2$ on $[0, \overline{\ell}]$, where $\overline{\ell} = \sup_n K_n / n$.

Then the asymptotic formula

$$ \prod_{k=1}^{K} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(t) \, \mathrm{d}t \sim \frac{C_{\alpha}}{n^K} \biggl[ \prod_{0}^{K/n} f(t)^{\mathrm{d}t} \biggr]^n \tag{1} $$

holds as $n \to \infty$, where

\begin{align*} C_{\alpha} &:= \lim_{n\to\infty} (2\pi n)^{\alpha/2} \prod_{k=1}^{n} \frac{k^{\alpha+1} - (k-1)^{\alpha+1}}{(\alpha + 1)k^{\alpha}}. \tag{2} \end{align*}

We will postpone the proof to the end. As a corollary, we also obtain:

Corollary. Suppose $f$ is of the form $f(t) = t^{\alpha}(1-t)^{\beta}g(t)$ for some $\alpha, \beta \geq 0$ and a strictly positive $C^2$ function $g(t)$ on $[0, 1]$. Then

$$ P_n(f) = \prod_{k=1}^{n} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(t) \, \mathrm{d}t \sim \frac{C_{\alpha} C_{\beta}}{n^n} \biggl[ \prod_{0}^{1} f(t)^{\mathrm{d}t} \biggr]^n \tag{3} $$

Proof. Write

\begin{align*} \prod_{k=1}^{n} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(t) \, \mathrm{d}t &= \Biggl[ \prod_{k=1}^{\lfloor n/2\rfloor} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(t) \, \mathrm{d}t \Biggr] \Biggl[ \prod_{k=1}^{n-\lfloor n/2\rfloor} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(1-t) \, \mathrm{d}t \Biggr] \end{align*}

and apply the theorem to each of the products in the right-hand side. $\square$

---

2. Examples

In this section, we provide some examples for sanity check. In doing so, we will utilize the identity

$$ C_0 = 1, \qquad C_1 = \sqrt{2}, \qquad C_2 = 2 \cosh \left(\frac{\pi}{2 \sqrt{3}}\right). $$

Example 1. Consider $f(t) = t(1-t)$. Then $ \prod_{0}^{1} f(t)^{\mathrm{d}t} = \frac{1}{e^2} $, hence by $\text{(3)}$,

$$ P_n(f) \sim \frac{C_1^2}{(e^2 n)^n} = \frac{2}{(e^2 n)^n}. $$

Example 2. Let $f(t) = 1 - \cos(2\pi t)$ and note that $g(t) = \frac{1 - \cos(2\pi t)}{t^2(1-t)^2}$ is a smooth function which is positive on $[0, 1]$. Also, we can check that $ \prod_{0}^{1} f(t)^{\mathrm{d}t} = \frac{1}{2} $. So by the corollary,

$$ P_n(f) \sim \frac{C_2^2}{(2n)^n} = \frac{4 \cosh^2 \left(\frac{\pi}{2 \sqrt{3}}\right)}{(2n)^n}. $$

---

3. Proof of Theorem

Step 1. Write $x_k = \frac{k}{n}$ for simplicity. Invoking the definition $\text{(2)}$, we get

\begin{align*} \prod_{k=1}^{K} \int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t = \prod_{k=1}^{K} \frac{k^{\alpha+1} - (k-1)^{\alpha+1}}{(\alpha+1)n^{\alpha+1}} \sim \frac{C_{\alpha}}{(2\pi K)^{\alpha/2}} \prod_{k=1}^{K} \frac{k^{\alpha}}{n^{\alpha+1}}. \end{align*}

Then by invoking the Stirling's approximation, this further reduces to

\begin{align*} \prod_{k=1}^{K} \int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t \sim \frac{C_{\alpha}}{n^K} \left( \frac{K}{ne} \right)^{K\alpha} = \frac{C_{\alpha}}{n^K} \biggl[ \prod_{0}^{K/n} (t^{\alpha})^{\mathrm{d}t} \biggr]^n, \end{align*}

proving the desired asymptotic formula when $g \equiv 1$.

Step 2. Now, the general case follows once we prove that

\begin{align*} \prod_{k=1}^{K} \frac{\int_{x_{k-1}}^{x_{k}} g(t)t^{\alpha} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} \sim \biggl[ \prod_{0}^{K/n} g(t)^{\mathrm{d}t} \biggr]^n. \tag{4} \end{align*}

To this end, define $\xi_k$'s by

$$ \xi_k = \frac{\int_{x_{k-1}}^{x_{k}} t^{\alpha+1} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} \in [x_{k-1}, x_k] $$

and then note that

\begin{align*} \frac{\int_{x_{k-1}}^{x_{k}} g(t)t^{\alpha} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} &= \frac{\int_{x_{k-1}}^{x_{k}} [g(\xi_k) + g'(\xi_k)(t - \xi_k) + \mathcal{O}(n^{-2})] t^{\alpha} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} \\ &= g(\xi_k) + \mathcal{O}(n^{-2}), \end{align*}

where the implicit bound does not depend on $k$ since $g$ is $C^2$. Also, since $g$ is strictly positive on $[0, \overline{\ell}]$, its reciprocal $1/g$ is bounded. Hence we get

\begin{align*} \log \biggl[ \frac{\int_{x_{k-1}}^{x_{k}} g(t)t^{\alpha} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} \biggr] &= \log g(\xi_k) + \mathcal{O}(n^{-2}). \end{align*}

Arguing similarly, but now with $\eta_k = \frac{x_{k-1} + x_k}{2}$, we get

$$ n \int_{x_{k-1}}^{x_k} \log g(t) \, \mathrm{d}t = \log g(\eta_k) + \mathcal{O}(n^{-2}). $$

The difference between these two quantities are roughly proportional to $|\xi_k - \eta_k|$, so we need to estimate it. First, it is clear that $|\xi_k - \eta_k| \leq \frac{1}{n}$. Next, if we fix $\epsilon \in (0, 1)$ and consider the range $k > n^{1-\epsilon}$, then

$$ \xi_k = \eta_k \cdot \frac{\alpha+1}{\alpha+2} \frac{ (1 + \frac{1}{2n\eta_k})^{\alpha+2} - (1 - \frac{1}{2n\eta_k})^{\alpha+2} }{ (1 + \frac{1}{2n\eta_k})^{\alpha+1} - (1 - \frac{1}{2n\eta_k})^{\alpha+1} } = \eta_k + \mathcal{O}(n^{-1-\epsilon}). $$

Hence, the logarithmic difference between the left-hand side and the right-hand side of $\text{(4)}$ can be estimated as

\begin{align*} &\Biggl| \log \prod_{k=1}^{K} \frac{\int_{x_{k-1}}^{x_{k}} g(t)t^{\alpha} \, \mathrm{d}t}{\int_{x_{k-1}}^{x_{k}} t^{\alpha} \, \mathrm{d}t} - \log \biggl[ \prod_{0}^{K/n} g(t)^{\mathrm{d}t} \biggr]^n \Biggr| \\ &\leq \sum_{k=1}^{K} \left| \log g(\xi_k) - \log g(\eta_k) \right| + \mathcal{O}(Kn^{-2}) \\ &\leq M \sum_{k=1}^{K} |\xi_k - \eta_k| + \mathcal{O}(Kn^{-2}), \qquad M = \sup_{[0, \overline{\ell}]} |g'/g|, \\ &\leq M \biggl( \sum_{k < n^{1-\varepsilon}} |\xi_k - \eta_k| +\sum^{K}_{k \geq n^{1-\varepsilon}} |\xi_k - \eta_k| \biggr) + \mathcal{O}(Kn^{-2}), \qquad M = \sup_{[0, \overline{\ell}]} |g'/g|, \\ &\leq M n^{-\epsilon} + \mathcal{O}(K n^{-1-\epsilon}) + \mathcal{O}(Kn^{-2}), \end{align*}

which converges to $0$ as $n \to \infty$. This completes the proof of $\text{(4)}$, concluding the main theorem as required. $\square$