I want to do the following integral:
$$ I=\int_{-1}^{1} x^n P_{n}(x) \rm{d}x $$
WITHOUT using Rodrigues' formula. I'm required to use
$$ P_{n}(x) = \sum_{r=0}^{[n/2]} \frac{(-1)^r (2n-2r)!}{2^n r! (n-r)! (n-2 r)!} x^{n-2 r}. $$
Substituing $ P_{n}(x) $ into I, I got
$$ I = 2 \sum_{r=0}^{[n/2]} \frac{ (-1)^{r} (2n-2r)!}{2^n r! (n-r)!(n-2r)!(2n-2r+1)} $$
, but I don't know how I can proceed.
Making use of the fact that the set of Legendre polynomials $P_n$ form an orthogonal base for all polynomials of degree less or equal to n, we can state that $$x^n = \sum_{j=0}^n b_j P_j(x)$$ The coefficients $b_j$ can be derived by using the internal product defined on functions on $[-1, 1]$: $$b_j = \int_{-1}^1 x^n P_j(x)dx$$ $P_n$ is a polynomial of degree $n$, so we can write $$P_n(x) = \sum_{k=0}^n a_k x^k$$
The highest order term has coefficient $b_n$ and that is the one you are looking for initially. If we decompose $x^n$ into a linear combination of $P_n, P_{n-1}, ..., P_1$, starting with the highest order term thus trying to find the factor that $P_n$ has to be multiplied with, it is clear that this is equal to $1/a_n$.
This value can be derived to be equal to $$b_n = \frac{1}{a_n} = \frac{2^nn!^2}{(2n)!} =\frac{n!}{1\cdot3\cdot\cdot\cdot(2n-3)(2n-1)}$$ which you can also derive from the formula you gave, by looking at the term with $r=0$.
See for instance (Choice of the First Term in Legendre Polynomials). Note that there the highest order term of $P_n(x)$ has been given a coefficient named $a_0$, but that is our $a_n$ here.
