Integrating Legendre's Polynomials

Question

I wanted to find the value of: $$ \def\LegP{\operatorname{LegendreP}} \int_{-1}^{1} \LegP[n, x] \frac{d}{dx} (\LegP[n+1,x]) \, dx. $$

Answer 1

$\int_{-1}^1 P_n P_{n+1}^{'}dx = \int_{-1}^1 P_n ((2n+1)P_n + P_{n-1}^{'})dx = (2n+1)\int_{-1}^1 P_n^2dx + \int_{-1}^1 P_n P_{n-1}^{'}dx $ = 2

Here I used the following:

$$(2n+1)P_n = P_{n+1}^{'} - P_{n-1}^{'}$$ $$\int_{-1}^1 P_n^2dx = \frac{2}{2n+1}$$ (see Legendre Polynomials: proofs)

And the fact that $P_{n-1}^{'}$ is a polynomial of degree $n-2$ and thus can be expanded into a linear combination of $P_{n-2}$ to $P_0$, and $\int_{-1}^1 P_n P_m dx = 0$ if $m \ne n$