How to calculate $\int_{-1}^1 \frac{P_n(x)}{x-x_k} \, dx$

Question

I'm told that

$$\int_{-1}^1 \frac{P_n(x)}{x-x_k} \, dx=\frac{2}{n P_{n-1}\left(x_k\right)}$$

,where $P_n(x)$ is Legendre polynomial, and $x_k$ is one of its root. I'm just wondering how to prove it.

Answer 1

Legendre Polynomials satisfy the recurrence relation $$(k+1)P_{k+1}(x)+kP_{k-1}(x)=(2k+1)xP_k(x)$$ Multiplying both sides by $P_k(y)$ we get $$(k+1)P_{k+1}(x)P_{k}(y)+kP_{k-1}(x)P_{k}(y)=(2k+1)xP_k(x)P_{k}(y)\tag{1}$$ Interchanging the roles of $x$and $y$ we get $$(k+1)P_{k+1}(y)P_{k}(x)+kP_{k-1}(y)P_{k}(x)=(2k+1)yP_k(x)P_{k}(y)\tag{2}$$ Subtracting $(1)$ from $(2)$ we get $$(k+1)\Delta_{k}(x,y)-k\Delta_{k-1}(x,y)=(2k+1)(y-x)P_k(x)P_{k}(y), \text{with $ \Delta_{k}(x,y)=P_{k+1}(y)P_k(x)-P_{k+1}(x)P_k(y)$}\tag{3}$$ Adding $(3)$ for $k=1,2,\ldots,n-1$ we get $$n\Delta_{n-1}(x,y)-\Delta_{0}(x,y)=(y-x)\sum_{k=1}^{n-1}(2k+1)P_k(x)P_{k}(y)$$ Thus $$n\frac{P_{n}(y)P_{n-1}(x)-P_{n}(x)P_{n-1}(y)}{y-x}=\sum_{k=0}^{n-1}(2k+1)P_k(x)P_{k}(y)\tag4$$ Now, taking $y=x_\ell$ a root of $ P_n$, we get $$n\frac{P_{n}(x)P_{n-1}(x_\ell)}{x-x_\ell}=\sum_{k=0}^{n-1}(2k+1)P_k(x)P_{k}(x_\ell)\tag5$$ Integrating on $[-1,1]$ and using the orthogonality of Legendre polynomials we get $$nP_{n-1}(x_\ell)\int_{-1}^1\frac{P_{n}(x)}{x-x_\ell}dx=2$$ Hence $$\int_{-1}^1\frac{P_{n}(x)}{x-x_\ell}dx=\frac{2}{nP_{n-1}(x_\ell)}$$ as desired.