that $$\label{quotient}
\frac{\widetilde{P}_n(x)}{P_{n}(x)}=\sum_{j=1}^{n} \frac{\lambda_{j}}{x-x_{j}}, \quad \lambda_{j}>0, \quad j=1,\ldots,n.$$ Observe that $$\frac{\displaystyle \prod_{i=1}^{\ell}\left(x-t_i\right)}{P_{n}(x)}=\frac{1}{\displaystyle  \prod_{i=1}^{\ell}\left(x-t_i\right)^{2d_i-1}q(x)}\left[p_{n-1}(x)\sum_{j=1}^{n} \frac{\lambda_{j}}{x-x_{j}}+p_{n}(x)\right].$$ This means that the above function satisfies that $$\frac{1}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right]=\mathcal{O}\left(\frac{1}{z^{n-\ell}}\right) \quad \mbox{as} \quad z \to \infty,$$ which is a holomorphic functions on $\displaystyle \overline{\mathbb{C}}\setminus \left(\left\{x_1,\ldots,x_{n}\right\}\setminus S\right).$ For each $\nu=0,\ldots, n-\ell-2$ we have then $$\frac{z^{\nu}}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right]=\mathcal{O}\left(\frac{1}{z^{2}}\right) \quad \mbox{as} \quad z\to \infty,$$ also holomorphic functions on $\displaystyle \overline{\mathbb{C}}\setminus \left(\left\{x_1,\ldots,x_{n}\right\}\setminus S\right).$ Set the elements $\displaystyle y_j \in \left\{x_1,\ldots,x_{n}\right\}\setminus S$, $j=1, \ldots, n-\ell$ with $y_1 < y_2 < \cdots < y_{n-\ell}$, and $\widetilde{\lambda}_j$, $j=1, \ldots, n-\ell$ the coefficients $\lambda$’s defined in (\[quotient\]) corresponding to points $y_j$. Also let $\lambda_j^{\prime}$ denote the $\lambda$’s of $t_j$, $j=1, \ldots, \ell$. Call $F$ the set of the roots of the polynomial $q$ defined in (\[positivepol\]). Consider a closed integration path $\Gamma$ with winding number $1$ for all its interior points. Denote $\mbox{Ext} (\Gamma)$ and $\mbox{Int}(\Gamma)$ the unbounded and bounded connected components respectively of the complement of $\Gamma$. Take $\Gamma$ so that $I\subset \mbox{Int}(\Gamma)$ and $F\subset \mbox{Ext}(\Gamma)$. From Cauchy’s Theorem and the above two conditions, it follows that $$0=\frac{1}{2 \pi i} \int_{\Gamma} \frac{z^{\nu}}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right] \, \mbox{d} z$$ $$=\frac{1}{2 \pi i} \int_{\Gamma} \frac{\displaystyle z^{\nu} p_{n-1}(z)\sum_{j=1}^{n-\ell} \frac{\widetilde{\lambda}_{j}}{z-y_{j}}\,  \mbox{d} z}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)} + \frac{1}{2 \pi i} 