Let $\{p_n(x)\}_{n=0,1,..}$ be a set of monic orthogonal polynomial of order $n$ with respect to some weight function.
I have established through a series of excercises that they satistfy the recurrence relation \begin{align} p_n(x) = (x+B_n)p_{n-1}(x) -C_n p_{n-2}(x) \end{align} Now, I must show that the set of integrals \begin{equation} \{\int_{-\infty}^\infty \frac{p_n(x')}{x-x'}w(x')dx'\}_{n=0,1,..} \end{equation} satisfies the same recurrence relation. I have absolutely no idea of how to go about it.
EDIT: I manged to partially prove it, but I still need some input..
Define \begin{align} \{f_n(x)\}_{n=0,1,...} := \left\{ \int_{-\infty}^{\infty} \frac{p_n(x')}{x-x'}w_2(x')d x' \right\}. \end{align} From the recurrence relation for $p_n$, we know \begin{align} f_n(x)&=\int_{\infty}^{\infty} \frac{(x'+B_n)p_{n-1}(x')-C_np_{n-2}(x')}{x-x'}w_2(x')d x'\notag\\ &=\int_{\infty}^{\infty} \frac{x'p_{n-1}(x')}{x-x'}w_2(x')d x'+B_nf_{n-1}(x)-C_nf_{n-2}(x) \end{align}
Thus I will have proved the relation if I can show that \begin{align} xf_{n-1}(x)=\int_{\infty}^{\infty} \frac{x'p_{n-1}(x')}{x-x'}w_2(x')d x' \end{align}
I found the answer. For further reference:
\begin{align} \int_{-\infty}^{\infty} \frac{x'p_{n-1}(x')}{x-x'}w_2(x')d x' &= \int_{-\infty}^{\infty} \left[\frac{x'p_{n-1}(x')}{x-x'}-\frac{xp_{n-1}(x')}{x-x'}\right]w_2(x')d x' +xf_{n-1}(x)\notag\\ &= -\int_{-\infty}^{\infty} p_{n-1}(x')w_2(x')d x' +xf_{n-1}(x)\notag\\ &= -(p_{n-1},p_0)_2 +xf_{n-1}(x) \notag\\ &= xf_{n-1}(x) \end{align}
Explanation: We first add 0, then we use the fact that $p_n$ is monic and thus that $p_0=1$ to transform the integral into a inner product. With $n-1\neq 0$, this product equals zero and the sought identity follows. For $n=-1$ it breaks downs but so does the recurrence relation (it is defined for $n\geq1$).