Sum of terms involving Beta function Let $n, m\in\mathbb{N}^*$, and $N=\{1,\ldots, n\}$.
$\forall x\in \{0,1,\ldots, m\}^n$, we denote: $m(x)=\vert \{i\in N\vert x_i=m\} \vert, s(x)=\vert \{i\in N\vert x_i>0\}\vert$.
Example: For $n=6, m=4, x=(0, 1, 2, 4, 4, 4)$, we have $m(x)=3$ and $s(x)=5$. 
It's possible to calculate : 
$$F(x) = \sum_{y\geq x} B(n-s(y)+1, m(y)+1), \forall x\in \{0,1,\ldots, m\}^n.$$
$B$ denotes the beta function.
$\forall x, y \in \{0,1,\ldots, m\}^n, x\geq y \Longrightarrow x_i ≥ y_i, \forall i, j\in N$. Example: $(3,2,4)\geq (2,1,4)$.
Thanks for your help !!!
 A: We want to compute
$$ F(x)=\int_{0}^{1}\sum_{y\geq x} z^{n-s(y)} (1-z)^{m(y)}\,dz = \int_{0}^{1}\sum_{s=s(x)}^n\sum_{m=m(x)}^{n} C_{s,m} z^{n-s}(1-z)^m\,dz$$
where $C_{s,m}$ stands for the number of $n$-tuples $\geq x$ having specific values of $s$ (number of positive components) and $m$ (number of maximized components). Obviously $s\geq m$, so we may state that $o(x)$ are the number components of $x$ equal to $0$, $p(x)$ are the number of positive components of $x$ that are $<m$ and $m(x)$ is the same as before. By this way $o+p+m=n$ and we want to compute
$$ \int_{0}^{1}\sum_{p=p(x)}^{n}\sum_{m=m(x)}^{n} C_{p,m} z^{n-p-m}(1-z)^m\,dz.\tag{1}$$
Let now $d_0(z)=1, d_1(z)=d_2(z)=\ldots=d_{m-1}(z)=\frac{1}{z}$, $d_m(z)=\frac{1-z}{z}$ and
$$ f_z(w) = \prod_{h=1}^{n}\sum_{k=x_h}^{m} d_k(z) w^k. \tag{2}$$
It is pretty clear how $(1)$ and $(2)$ are related:
$$ (1) = \int_{0}^{1} z^n f_z(1)\,dz\tag{3}$$
hence $(1)$ is given by the integral over $(0,1)$ of a product of $n$ terms, of the form $ z+(m-1)+(1-z)=m$ or $t+(1-z)$, with $t\in[0,m-1]$, according to the value of $x_h$.
