Prove: $\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$ I'm trying to prove:
$$\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$$
My attempt consisted in applying the Rodrigues formula:
$$\int_{-1}^1 x^n P_n(x)dx = \dfrac{1}{2^n n!} \int_{-1}^1 x^n \dfrac{d^n}{dx^n}(x^2-1)^ndx$$
and integrating by parts n times:
$$\int_{-1}^1 x^nP_n(x)dx=\dfrac{1}{2^n n!}\left(x^n \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \bigg|_{-1}^1 + (-1)^{n} n!\int_{-1}^1 (x^2-1)^ndx\right)$$
The first term vanishes at the extremes. And the remaining integral I solved through substitution:
$$t = x^{2} \rightarrow dx = \dfrac{dt}{2\sqrt{t}}$$
$$\dfrac{1}{2}\int^{1}_{0}(t - 1)^{n}\dfrac{dt}{2\sqrt{t}} = \dfrac{1}{2}\int^{1}_{0}(t - 1)^{n} t^{-1/2} dt$$
Using the beta function:
$$(-1)^{n}\dfrac{1}{2}\int^{1}_{0}(1 - t)^{n} t^{-1/2} dt =\dfrac{(-1)^{n}}{2}\beta(1/2, n + 1) =\dfrac{(-1)^{n}}{2}\dfrac{\Gamma(1/2)\Gamma(n + 1)}{\Gamma(n + 3/2)}$$
I rewrote the gamma functions as
$$\Gamma(1/2) = \sqrt{\pi}$$
$$\Gamma(n + 1) = n!$$
and $$\Gamma(n + 3/2) = \dfrac{(n+1)(2n)!\sqrt{\pi}}{2^{2n}n!}$$
Finally I reached:
$$(-1)^{n} n!\int_{-1}^1 x^{n-1} \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \dfrac{(-1)^{n}n!2}{2^{n}n!}\dfrac{(-1)^{n}}{2}\dfrac{\sqrt{\pi}2^{2n}n!n!}{(n+1)(2n)!\sqrt{\pi}} = \dfrac{2^{n}n!n!}{(n+1)(2n)!}$$
I tried simplifying furter, but I didn't get anywhere
 A: A method based upon the series for $P_{n}(x)$ is seen by the following.
Using
$$ P_{n}(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, (2 x)^{n-2 k}}{k! \, (n-2 k)!} $$
then
\begin{align}
I &= \int_{-1}^{1} x^n \, P_{n}(x) \, dx \\
&= \sum_{k} a_{k}^{n} \, 2^{n-2 k} \, \int_{-1}^{1} x^{2 (n-k)} \, dx \\
&= \sum_{k} a_{k}^{n} \, 2^{n-2k} \frac{1}{2n -2 k + 1} \, \left( 1^{2n -2k +1} - (-1)^{2n -2k +1} \right) \\
&= \sum_{k} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, 2^{n -2k +1}}{k! \, (n -2k)! \, (2n -2k +1) } \\
&= \frac{2^{n+1} \, \left(\frac{1}{2}\right)_{n}}{n! \, (2n+1)} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\
&= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\
&= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, \frac{n!}{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}} \\
&= \frac{n!}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \\
&= \frac{2^{n+1}}{\binom{2n}{n} \, (2n+1)}.
\end{align}
A: We'll use the following:

*

*$$\int_{-1}^1 P_n(x) \cdot P_n(x) = \frac{2}{2n+1}$$


*The leading coefficient of $P_n(x)$ ( coefficient of $x^n$) equals $\frac{\binom{2n}{n}}{2^n}$


*$$\int_{-1}^1 x^k P_n(x) d x = 0$$ for all $0 \le k \le n-1$.
From here we conclude
$$\int_{-1}^1 x^n P_n(x)\, dx = \int_{-1}^1 \frac{2^n}{\binom{2n}{n}} P_n(x) \cdot P_n(x)\, d x = \frac{2^n}{\binom{2n}{n}} \cdot \frac{2}{2n+1}$$
