In addition to a former question. I wish to show that:
$$\int_{-1}^1 x^n P_n (x) dx = \frac{n}{2n+1}\int_{-1}^1 x^{n-1} P_{n-1} (x) dx$$
Assuming the result of:
$$P'_{n+1}-P'_{n-1} = (2n+1)P_{n}$$
Multiplying both sides with $x^n$. We obtain:
$$x^nP'_{n+1}-x^nP'_{n-1} = x^n(2n+1)P_{n}$$
Now integrating both sides:
$$\int_{-1}^1x^nP'_{n+1}-x^nP'_{n-1}dx = \int_{-1}^1 x^n(2n+1)P_{n}dx$$
Now have to integrate by parts on the lefthand side. This becomes:
$$\int_{-1}^1 x^n(P'_{n+1}-P'_{n-1})dx= x^n[P_{n+1}-P_{n-1}]^{1}_{-1} - \int_{-1}^1 nx^{n-1}(P_{n+1}-P_{n-1})dx$$But here I am stuck. Any further help appreciated again.
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 $$ \begin{align} \int_{-1}^{1} x^{n}P_{n}(x) dx&=\frac{(-1)^n}{n!\space2^n}\cdot n!\int_{-1}^{1} (x^2-1)^n dx \\\\&=\frac{1}{2^n}\int_0^{1} u^{-1/2}(1-u)^n du \end{align} $$ that is, using the Euler beta function we have