Results with Legendre Polynomials I am studying Legendre's differential equation and trying to use it to establish various results but struggling as I am not really sure how to manipulate the Legendre polynomials.  I begin with:
$$\frac{d}{dz}\bigg((1-z^{2})\frac{dP_{n}}{dz}\bigg) + n(n-1)P_{n}=0.$$
I have managed to show that
$$n(n+1) \int_{-1}^{1}P_{m} (z) P_{n}(z) \, dz =  \int_{-1}^{1}(1-z^2)P_{m}'(z)P_{n}'(z) \, dz$$
I then used this to deduce the orthogonality condition and now need to use it to deduce:
$$\int_{-1}^{1}z^{k}P_{n}(z) \, dz=0.$$
for $k=0,1,...,n-1$. I have also shown that:
$$P_{n+1}'-P_{n-1}' = (2n+1)P_{n}.$$
I have to use this result to deduce that:
$$\int_{-1}^{1}z^{n}P_{n} (z) \, dz=\bigg( \frac{n}{2n+1} \bigg) \int_{-1}^{1}z^{n-1}P_{n-1} (z) \, dz.$$
From this I need to deduce that:
$$\int_{-1}^{1}z^{n}P_{n} (z) \, dz= \frac{2^{n+1}(n!)^2}{(2n+1)!}.$$
In manipulating the polynomials I am able to use the fact that $P_{n}(1)=1$, $P_{n}(-1)=(-1)^n$, and $P_{0}(z)=1$.
 A: I will use Rodrigues formula for the Legendre polynomials $P_n (x)$, namely
$$P_n (x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2 - 1)^n].$$
We start by considering the integral
$$I = \int_{-1}^1 f(x) P_n (x) \, dx.$$
Substituting Rodrigues formula for $P_n (x)$ into the above integral followed by integrating by parts $n$ times, recoginising the term $(x^2 - 1)$ repeatedly cancels at either of the end-points, one will be left with
$$I = \frac{(-1)^n}{2^n n!} \int_{-1}^1 f^{(n)}(x) (x^2 - 1)^n \, dx.$$
Now if $f(x) = x^k$ where $k = 0,1,2,\ldots, n - 1$, then
$$f^{(n)} (x) = \frac{d^n}{dx^n} [x^k] = 0.$$
Thus
$$\int_{-1}^1 x^k P_n (x) \, dx = 0, \qquad k = 0,1,2,\ldots,n - 1.$$
Next, if $f(x) = x^n$ then
$$f^{(n)} (x) = \frac{d^n}{dx^n} [x^n] = n!.$$
Thus
\begin{align*}
\int_{-1}^1 x^n P_n (x) \, dx &= \frac{(-1)^n}{2^n} \int_{-1}^1 (x^2 - 1) \, dx\\
&= \frac{1}{2^n} \int_{-1}^1 (1 - x^2) \, dx\\
&= \frac{1}{2^{n - 1}} \int_0^1 (1 - x^2)^n \, dx,
\end{align*}
since the integral is even.
This last integral can be readily found in terms of the beta function. Enforcing a substution of $x \mapsto \sqrt{x}$ the integral becomes
$$\int_0^1 (1 - x^2)^n \, dx = \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx = \frac{1}{2} \text{B} \left (\frac{1}{2}, n + 1 \right ).$$
As
\begin{align*}
\text{B} \left (\frac{1}{2}, n + 1 \right ) &= \frac{\Gamma (1/2) \Gamma (n + 1)}{\Gamma (n + 3/2)} = \frac{\sqrt{\pi} n!}{(n + 1/2) \Gamma (n + 1/2)}.
\end{align*}
Now as
$$\Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$$
we can write
$$\text{B} \left (\frac{1}{2}, n + 1 \right ) = \frac{2^{2n + 1} (n!)^2}{(2n + 1)!}.$$
Thus
$$\int_{-1}^1 (1 - x^2)^n \, dx = \frac{2^{2n} (n!)^2}{(2n + 1)!},$$ 
giving
$$\int_{-1}^1 x^n P_n (x) \, dx = \frac{2^{n + 1} (n!)^2}{(2n + 1)!}.$$
