Problems regarding integrals involving Legendre polynomials 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? 
 A: 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*}

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*}
A: Recall the more general form of the orthogonality condition:
$$\int_{-1}^1 x^j P_k(x)\,\mathrm dx=0\qquad\text{if }j\neq k$$
Use this to simplify your integral to
$$2^{-n+1}\binom{2n-2}{n-1}\int_{-1}^1 x^{n+1} P_{n+1}(x)\,\mathrm dx$$
Now, there is the identity
$$\int_0^1 x^{n+2\rho}P_n(x)\,\mathrm dx=\frac{\left(2\rho+1\right)_n}{2^{n+1}\left(\rho+\tfrac12\right)_{n+1}}$$
Using this gives the result
$$\begin{align*}
\int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)\mathrm dx&=2^{-n+1}\binom{2n-2}{n-1}\frac{(n+1)!}{2^{n+1}\left(\tfrac12\right)_{n+2}}\\
&=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}
\end{align*}$$
