Using Rodrigues' formula and integrating by parts $n$ times, prove that $$\int_{-1}^1P_n^2(x)dx=\frac{2}{2n+1}$$
where $P_n(x)$ is a Legendre polynomial.
I tried this way
Let $$f(x)=(x^2-1)^s$$ $$\int_{-1}^1f^{(s)}(x)f^{(s)}(x)dx=-\int_{-1}^1f^{(s+1)}(x)f^{(s-1)}(x)dx$$ (the identity that I have not been able to demonstrate).
Then apply this formula repeatedly and obtain
$$2(2s)!\int_{0}^1(1-x^2)^sdx$$
I know what to use a replacement $x=\cos\theta$ and transform this integral en
$\int_{0}^{\pi/2} \sin^{2s+1} t \;dt$
I do not know how to continue.
Best Answer
Recall the Rodrigues' formula $$ P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left(x^2-1\right)^n, $$ then $$ \mathcal{I}(n)=\int_{-1}^1 P_n^2(x)\ dx=\frac1{4^n(n!)^2}\int_{-1}^1\frac{d^n}{dx^n}\left(x^2-1\right)^n\frac{d^n}{dx^n}\left(x^2-1\right)^n\ dx. $$ Using IBP yields \begin{align} \mathcal{I}(n)&=\frac1{4^n(n!)^2}\left[\left.\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\right|_{-1}^1-\int_{-1}^1\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\frac{d^{n+1}}{dx^{n+1}}\left(x^2-1\right)^n\ dx\right]. \end{align} Hence IBP $n$ times yields \begin{align} \mathcal{I}(n)&=\frac{(-1)^n}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2-1\right)^n\ dx\\ &=\frac{(-1)^n(2n)!}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\ dx\\ \end{align} Setting $x=2t-1$ yields \begin{align} \mathcal{I}(n) &=\frac{(2n)!}{(n!)^2}\int_{0}^1t^n\left(1-t\right)^n\ dt\\ &=\frac{(2n)!}{(n!)^2}\cdot\text{B}(n+1,n+1)\\ &=\frac{(2n)!}{(n!)^2}\cdot\frac{(n!)^2}{(2n+1)!}\\ &=(2n)!\cdot\frac{1}{(2n+1)(2n)!}\\ &=\large\color{blue}{\frac1{2n+1}},\tag{Q.E.D.} \end{align} where $\text{B}(x,y)$ is a Beta function.