I want to find the closed form of the following $$\int\limits_{t<t_p<\cdots<t_1<1} t_1^{n_1-1}\cdots t_p^{n_p-1}dt_1\cdots dt_p,$$ where $p,n_j\ (j=1,2,\ldots,p)$ are positive integers.
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Let $F(\{n_1,\cdots,n_p\},t)$ denote the integral. Then we have the following recursive formula:
$$ F(\{n_1,\cdots,n_p\},t) = \int_{t}^{1} \tau^{n_p-1} F(\{n_1,\cdots,n_{p-1}\}, \tau) \, \mathrm{d}\tau, \qquad F(\varnothing, t) = 1. $$
If we write $s_k = n_1 + \cdots + n_k$ and $s_0 = 0$, then
$$ F(\{n_1, \cdots, n_p\}, t) = \sum_{k=0}^{p} \frac{t^{s_p - s_k}}{\prod_{j\neq k} (s_k - s_j)}. \tag{*} $$
Indeed, $\text{(*)}$ is trivial if $p = 0$, and assuming $\text{(*)}$ for $p$,
\begin{align*} F(\{n_1,\cdots,n_{p+1}\},t) &= \int_{t}^{1} \tau^{n_{p+1}-1} F(\{n_1,\cdots,n_p\}, \tau) \, \mathrm{d}\tau \\ &= \int_{t}^{1} \left( \sum_{k=0}^{p} \frac{\tau^{s_{p+1} - s_k - 1}}{\prod_{j\neq k} (s_k - s_j)} \right) \, \mathrm{d}\tau \\ &= \sum_{k=0}^{p} \frac{1 - t^{s_{p+1} - s_k}}{(s_{p+1} - s_k) \prod_{j\neq k} (s_k - s_j)} \\ &= \sum_{k=0}^{p} \frac{1}{(s_{p+1} - s_k) \prod_{j\neq k} (s_k - s_j)} + \left( \sum_{k=0}^{p} \frac{t^{s_{p+1} - s_k}}{\prod_{j\neq k} (s_k - s_j)} \right) \end{align*}
The sum in the constant term can be shown to be $1/\prod_{j=1}^{p} (s_{p+1} - s_j)$, and so, $\text{(*)}$ holds for $p+1$. Now the desired claim follows the induction on $p$.
Sangchul Lee
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