$\int_{-1}^{1} x^{k+i} P_n(x)dx$, $P_n$ Legendre polynomial. I was wondering whether there is a way to say what $$\int_{-1}^{1} x^{k} P_n(x)dx$$ is, where $k,n$ are positive integers or zero and $P_n$ is the n-th Legendre polynomial? I am looking for an analytic result for this. Probably one could use this: Barne's integral: Barne's Integral But the set over which is integrated is different in this case.

We consider $$I(m,n)=\int_{-1}^{1}x^m P_n(x)dx\tag{1}$$

Since $x^{m}$ belongs to the linear span of $P_0(x),\ldots, P_{m}(x)$, the orthogonality implies that for $m
Now let us use the Rodrigues formula $$P_n(x)=\frac{1}{2^nn!}\left[\frac{d^n}{dx^n}(x^2-1)^n\right]$$ to integrate $n$ times by parts in (1). This yields $$I(m,n)=\frac{1}{2^n} {m \choose n} \int_{-1}^{1}x^{m-n}\left(1-x^2\right)^n dx.\tag{2}$$ Since $m-n$ is even by assumption, we can make the change of variables $t=x^2$, which reduces (2) to a beta function integral, with the final result $$I(m,n)=\frac{1}{2^{n}} {m \choose n}B\left(\frac{m-n+1}2,n+1\right)=2^{n+1} \frac{m!\left(\frac{m+n}{2}\right)!}{(m+n+1)!\left(\frac{m-n}{2}\right)!}$$