Integral with Legendre polynomial: $\int_{-1}^{1}x^{n+2k}P_{n}(x)dx$

Question

How to compute the following integral?

$$\int_{-1}^{1} x^{n+2k}P_{n}(x) dx $$

where $P_n(x)$ is the Legendre function, and $n, k = 1,2, \cdots.$

Answer 1

Hint. By using Rodrigues' formula, $$ P_n(x)=\frac{1}{n!\space2^n}\frac{d^n}{dx^n}(x^2-1)^n, $$ one may integrate by parts $n$ times obtaining $$ \begin{align} \int_{-1}^{1} x^{n+2k}P_{n}(x) dx&=\frac{(-1)^n}{n!\space2^n}\frac{(n+2k)!}{(2k+1)!}\int_{-1}^{1} x^{2k}(x^2-1)^n dx \\\\&=\frac{(n+2k)!}{n!\space2^n\:(2k+1)!}\int_0^{1} u^{k-1/2}(1-u)^n du \end{align} $$ that is, for $n=1,2,3,\ldots$ and $k=1,2,3,\ldots$, using the Euler beta function we have

$$ \int_{-1}^{1} x^{n+2k}P_{n}(x) \:dx=\frac{2^n(n+2k)!\:(n+k)!}{(2n+2k+1)!\:k!}. $$