We begin with a density given by $$ \tag 1 K(\xi)=\sum_{k=1}^K p_k\delta\left(\xi -\xi_k\right) $$
The question is how to prove the following $$ \tag 2 \int_0^{max(\xi)}K(\xi)\frac{(z\xi)^{1-K}}{1-z\xi}d\xi=\sum_{k=1-K}^{K}z^kM_k $$
where the moment equation $$ \tag 3 \sum_{k=1}^Kp_k(\xi_k)^m=M_m, $$ holds. I don't understand how the summation limits in (2) are derived and why the term $(z\xi)^{1-K}$ is needed. I also don't understand how this relates to the formula for the Stiltjes transformation given here
Motivation: The reasoning behind this is to find the values of $\xi_k$. This is done using the theorem, which states that if we would define $$ f(z)=\int_0^\infty\frac{d\phi(u)}{1-zu} $$ for a real, non-decreasing and bounded function $\phi(x)$ for $x\in[0,\infty)$, for $\forall z \in \mathbb{C}\setminus \left ( \mathbb{R}^+ \right ) $ exists the series expansion in terms of the moments $f_j$ $$ f(z)=\sum\limits_{j=0}^\infty f_j z^j $$
By using the $[N+1][N]$ Pade expansion it can be shown that
$$
f(z)=\frac{a_0+a_1z+a_2z^2+\dots +a_{N+1}z^{N+1}}{b_0+b_1z+b_2z^2+\dots +a_{N}z^{N}}=\sum\limits_{i=1}^N\frac{w_i}{z-z_i}.
$$
In that case the poles $z_j$ are simple and real. The proof of the above statement can be found in this paper. (pdf link)
