# Explicit representation of weights for Newton-Cotes

For the closed Newton-Cotes quadrature over $[x_1, x_n]$, the coefficients $H_{n,i}$ for $$\int_{x_1}^{x_n} f(x)\:\text{d}x = h \sum_{i=1}^n H_{n,i} \; f(x_i)$$ are given explicitly by $$H_{n,r+1} =\frac{(-1)^{n-r}}{r!(n-r)!}\int_0^n \frac{\prod_{k=0}^n (t-k)}{t-r}\:\text{d}t;$$ see http://mathworld.wolfram.com/Newton-CotesFormulas.html.

Is there a similar formula for the weights of the open Newton-Cotes scheme?

The general expression is $$H_{n,k} = \frac{1}{\Psi'(x_k)} \int_a^b \frac{\Psi(x)}{x-x_k}\;\text{d}x$$ with $$\Psi(x) = \prod_{k\in S_n} (x-x_k)$$ with $S_n$ in the set of quadrature points; see, e.g., http://www.doiserbia.nb.rs/img/doi/0354-5180/2013/0354-51801304649M.pdf (2).
Choosing $a=0, b=n$ and $$\text{closed:}\quad S_n = \{k: 0\le k \le n\}\\ \text{open:}\quad S_n = \{k: 1\le k \le n-1\}$$ makes clear that this is an entirely rational expression. For open Newton-Cotes, it is $$H_{n,r}=\frac{(−1)^{n−r+1}}{(r-1)!(n−r-1)!}\int_{0}^n \frac{\prod_1^n(t-k)}{t-r}\;\text{d}t.$$