Legendre Polynomials: proofs $\int_{-1}^1P_n^2(x)dx=\frac{2}{(2n+1)}$ Does any one know, how to compute any of those two things?


*

*The relationship between Legendre polynomials and Shifted Legendre Polynomials.


*$\displaystyle\int_{-1}^1P_n^2(x)dx=\dfrac{2}{(2n+1)}$ for $n\geq0$.

I tried to use Bonnet's equation:
$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$
but I couldn't move. Thanks :)
Edit: The second problem refers to regular Legendre Polynomials.
 A: I will answer your question about determining the value of $\int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
$$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcommand{\bracket}[1]{\left(#1\right)}$
\begin{equation}
\begin{split}
I&=\int_{-1}^1 P_n(x)^2dx
\\ &=\frac{1}{2^{2n}n!^2}\int_{-1}^1 [(x^2-1)^n]^{(n)} \cdot [(x^2-1)^n]^{(n)}dx
\\ &=\frac{1}{2^{2n}n!^2} \bracket{\partial{[(x^2-1)^n]^{(n)}\cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx}
\\ &=\frac{1}{2^{2n}n!^2} \bracket{0-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx}
\\ &=...
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \int_{-1}^1 [(x^2-1)^n]^{(2n)}\cdot [(x^2-1)^n]^{(0)}dx
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot \int_{-1}^1 (x^2-1)^ndx
\end{split}
\end{equation}
Now
\begin{equation}
\begin{split}
\int_{-1}^1 (x^2-1)^ndx&=\int_{-1}^1 (x+1)^n(x-1)^ndx
\\ &=\bracket{\partial{\frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx}
\\ &=\bracket{0 - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx}
\\ &=...
\\ &=(-1)^n\int_{-1}^1 \frac{n!\cdot (x+1)^{2n}}{(2n)!}\cdot n!(x-1)^0dx
\\ &=(-1)^n\frac{n!^2}{(2n)!}\int_{-1}^1 (x+1)^{2n}dx
\\ &=(-1)^n\frac{n!^2}{(2n)!}\partial{\frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
\\ &=(-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1}
\end{split}
\end{equation}
So finally we get our desired value:
\begin{equation}
\begin{split}
I&=\int_{-1}^1 P_n(x)^2dx
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot (-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1}
\\ &=\frac{2}{2n+1}
\end{split}
\end{equation}
Q.E.D.
A: Hints:
Part 1:
See Shifted Legendre Polynomials.
I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.
Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?
Part 2:
Try evaluating the integral using Rodrigues' Formula.
