How $x^n$ is linearly represented by Legendre polynomials

Question

I recently come across a problem with respect to Legendre polynomial as follows.

Let $L^2[-1,1]$ be the Hilbert space of real valued square integrable functions on $[-1,1]$ equipped with the norm $\|f\|$= $\sqrt{\int_{{-1}}^1 {|f(x)|}^2 dx}$.
For any $n \in \mathbb{N}$, $P_n(x) := \frac{1}{2^n n!}\frac{{\rm d}^n (x^2-1)^n}{ {\rm d} x^n} $ is the classical Legendre polynomial of degree $ n $, orthogonal on $L^2[-1,1] $.

According to the results of Wikipedia, there are some explicit representations of $P_n(x)$ as follows:
\begin{equation}
\begin{aligned}
&P_{n}(x)=\frac{1}{2^{n}} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)^{2}(x-1)^{n-k}(x+1)^{k}, \\
&P_{n}(x)=\sum_{k=0}^{n}\left(\begin{array}{c}
n \\
k
\end{array}\right)\left(\begin{array}{c}
n+k \\
k
\end{array}\right)\left(\frac{x-1}{2}\right)^{k}, \\
&P_{n}(x)=\frac{1}{2^{n}} \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k}\left(\begin{array}{c}
n \\
k
\end{array}\right)\left(\begin{array}{c}
2 n-2 k \\
n
\end{array}\right) x^{n-2 k}.
\end{aligned}
\end{equation}

My question is that how $x^n$ is linearly represented by Legendre polynomials. In other words, $x^n$ can be linearly represented by $P_0(x),P_1(x), \cdots,P_n(x)$ as
\begin{equation}
x^n = \sum_{k=0}^n \alpha_k P_k(x).
\end{equation}
What are the explicit expressions for these coefficients $\alpha_k$?

I have obtained that $\alpha_n = \frac{2^n}{\binom{2n}{n}}$.I don't know if there is a similar simple and clear expression for $\alpha_k$ when $k$ is less than $n$.

Answer 1

\begin{align} \alpha_{n,k}&=\frac{2k+1}{2}\int_{-1}^1 x^n P_k(x)\,dx\\ &=\frac{2k+1}{2^{k+1}k!}\int_{-1}^1 x^n\frac{d^k}{dx^k}(x^2-1)^k\,dx\\ \color{gray}{[\text{IBP }k\text{ times}]} &=\frac{2k+1}{2^{k+1}k!}\int_{-1}^1(1-x^2)^k\frac{d^k x^n}{dx^k}\,dx\\ &=\frac{2k+1}{2^{k+1}}\binom{n}{k}\int_{-1}^1 x^{n-k}(1-x^2)^k\,dx. \end{align} This is a beta integral, evaluated in terms of factorials. The result is $$\alpha_{n,k}=\begin{cases}2^k\dfrac{(2k+1)n!}{(n+k+1)!}\dfrac{\big((n+k)/2\big)!}{\big((n-k)/2\big)!},&n-k\text{ is even}\\\hfill 0,\hfill&n-k\text{ is odd}\end{cases}$$