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
- $K = K_n$ is a sequence of positive integers such that $K/n \to \ell$ for some $\ell \in (0, \infty)$ as $n \to \infty$.
- $f(t) = t^{\alpha} g(t)$, 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. $\square$
