<s>[INST] 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.
 [/INST] 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.
</s>