Problems regarding integrals involving Legendre polynomials: $\int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$

Question

I am finding difficulty doing this integral involving Legendre polynomials. $$\int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$ I have two strategies in my mind both of them have failed to produce results. One is that I could somehow use the orthogonality of Legendre polynomials after using Bonnet's recursion formula to get Legendre polynomials, to simplify the integrals, or I could use the Rodrigues formula by doing integration by parts. The first approach fails because of the $x^2$ in the integral. The second approach is not giving me integrated parts, where limits can be applied easily.How do I do this?

Answer 1

By Bonnet's formula, we have \[ (2n-1)xP_{n-1} = nP_n + (n-1)P_{n-2} \] and \[ (2n+3)xP_{n+1} = (n+2)P_{n+2} + (n+1)P_n \] so \begin{align*} \int_{-1}^1 x^2 P_{n-1}P_{n+1}\; dx &= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 \bigl( nP_n + (n-1)P_{n-2}\bigr)\bigl((n+2)P_{n+2} + (n+1)P_n\bigr)\; dx\\ &= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 n(n+1)P_n^2\; dx\\ &= \frac {2n(n+1)}{(2n-1)(2n+1)(2n+3)} \end{align*}

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I have another denominator, I saw. But I will show also by example mine is right ;-): We have ($n=1$) that $P_0 = 1$, $P_2(x) = \frac 12(3x^2 - 1)$. It holds \begin{align*} \frac 12 \int_{-1}^1 (3x^4 - x^2) \; dx &= \frac 22\left(\frac 35 - \frac 13\right)\\ &= \frac 4{15}\\ &= \frac{2\cdot 1 \cdot (1+1)}{(2\cdot 1 - 1)(2 \cdot 1 \mathbin{{\color{red}+}} 1)(2\cdot 1 + 3)} \end{align*}