Prove a recurrence formula of Legendre Polynomials In Legendre's Differential Equation,
How to prove that $$\int_{-1}^1x^2p_{n+1}(x)p_{n-1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$
I did many search also i did try by myself. But No success. Is there any way to solve this?
Thanks
 A: Consider the recurrence relation: $$\boxed {(2n+1)xP_n = (n+1)P_{n+1} + nP_{n-1}} \tag {1} $$
Replacing $n $ by $(n-1) $ and $(n+1) $ respectively, we get, $$xP_{n-1} = \frac {1}{2n-1}[nP_n + (n-1)P_{n-2}]  \tag {2} $$ $$xP_{n+1} = \frac {1}{2n+3}[(n+2)P_{n+2} + (n+1)P_n] \tag {3} $$
Multiplying $(2)$ and $(3)$ and integrating with respect to $x $ over $[-1,1] $, we get, $$\int_{-1}^{1} x^2P_{n-1}P_{n+1} \mathrm {d}x = \frac {1}{(2n-1)(2n+3)} n (n+1) \int_{-1}^{1} P_n^2 \mathrm{d}x $$ since all other integrals  on the RHS are zero by the orthogonal property of the Legendre polynomials. Thus, $$\int_{-1}^{1} x^2P_{n-1}P_{n+1} \mathrm {d}x = \frac {2n (n+1)}{(2n-1)(2n+1)(2n+3)} $$
Hope it helps. 
A: If you're given the recurrance relation, Rohan's answer is far cleaner, but consider this alternative proof outline:
Legendre polynomials are defined as solutions to Legendre's differential equation 
$$\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_n(x)]+n(n+1)P_n(x)=0$$
You can rearrange this to get a single $P_n(x)$ term on one side which could then be substituted into your integral. This rearrangement is
$$P_n(x)=\frac{-1}{n(n+1)}\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_n(x)]$$
$n$ is a dummy variable, so we can rewrite this in two forms for the purpose of playing nicely with your integral.
$$
\begin{align}
P_{n+1}(x)&=\frac{-1}{(n+1)(n+2)}\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{n+1}(x)] \\
P_{n-1}(x)&=\frac{-1}{(n-1)(n)}\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{n-1}(x)]
\end{align}
$$
You're going to have to apply the fundamental theorem of calculus (second part) to evaluate your integral. Note that all polynomials are Riemann integrable, so you won't run into any trouble applying the FTC. Before the FTC, you will probably need to use the product rule of derivatives and get a total algebraic mess. There might be some clever way to avoid this, but I couldn't find it.
While evaluating the integral with the FTC, note that $P_n(1) = 1$ and $P_n(-1)=-1\ \forall n>0$. You could either accept this on faith, or know that $P_1(x) = x$ and use induction with the recurrence relation.
