I'm struggling to find the Legendre expansion for $x^n$ (exercise 15.1.17 from Mathematical Methods for Physicists).
I'm trying to evaluate the following integral:
$$a_m = \frac{2m+1}{2} \int_{-1}^{1} x^n P_m(x) dx$$
I know the answer depends whether $n$ is odd or even, as well as if $m$ is odd or even. I also know that for $m>n$ the coefficients are zero.
I tried using the Rodrigues formula and integrated by parts $m$-times. Assuming $m$ and $n$ are both even and that $m\leq n$, I find that:
$$a_m = \frac{2m+1}{2^{m+1}} (-1)^m \begin{pmatrix} n \\ m \end{pmatrix} \int_{-1}^{1} x^{n-m} (x^2-1)^m dx$$
since everytime I integrate by parts the first term has an odd function evaluated in a symmetric interval, thus remaining only the integral parts and the derivative disappearing after integrating by parts $m$ times.
Using the Binomial Theorem, I have arrived in
$$a_m = \frac{2m+1}{2^m} \begin{pmatrix} n \\ m \end{pmatrix} \sum_{k=0}^{m} (-1)^k \begin{pmatrix} m \\ k \end{pmatrix} \frac{1}{(n-m-2k+1)}$$
for $m$ and $n$ even, which doesn't seem to match the answer sheet:
$$x^{2m} = \sum_{n=0}^{m} \frac{2^{2n}(4n+1)(2m)!(m+n)!}{(2m+2n+1)!(m-n)!} P_{2n}(x)$$
Thanks in advance!