Proving a result related to Legendre polynomials

Question

Prove $$J= \int _{-1}^1 xP_nP_{n-1}dx=\frac {2n}{4n^2-1}$$ where $P_n $ is the Legendre polynomial of degree n . $$\text {**My attempt**} $$ Using by parts assuming $u=x,v=P_nP_{n-1} $ the first part vanishes and we get $$\int _{-1}^1 (\int (P_nP_{n-1} ) $$ now I dont know how to handle this indefinite integral even if I substitute Rodrigue's formula. Another thought was using generating function but that too didn't work.

Answer 1

Start with the recursion formula $$(n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x)$$ Rearrange and multiply with $P_{n-1}(x)$ $$(2n+1) x P_n(x)P_{n-1}(x) = (n+1) P_{n+1}(x)P_{n-1}(x) + n P_{n-1}(x)P_{n-1}(x)$$ Then integrate $$(2n+1)\int_{-1}^{1} x P_n(x)P_{n-1}(x)dx = \int_{-1}^{1}(n+1) P_{n+1}(x)P_{n-1}(x)dx + \int_{-1}^{1}n P_{n-1}(x)P_{n-1}(x)dx$$ and use the orthogonal property on the RHS $$(2n+1)\int_{-1}^{1} x P_n(x)P_{n-1}(x) dx = 0 + n \frac{2}{2(n-1)+1}$$ $$\int_{-1}^{1} x P_n(x)P_{n-1}(x) dx = n \frac{2}{(2n-1)(2n+1)}$$ and finally get the result

$$\int_{-1}^{1} x P_n(x)P_{n-1}(x)dx = \frac{2n}{4n^2-1}$$