[Math] The L2 Norm of Legendre Polynomials

Question

I need help in proving:$$\int_{-1}^{1}P_n^2(x)dx=\frac{2}{2n+1}$$ using the following formula:$$xP'_n-P'_{n-1}=nP_n,\ n=1,2,…$$
(where $P_n$ are the legendre polynomials).
Thanks!

Answer 1

The intent of the problem seems to be to apply integration by parts. Let $u(x)=P_n^2(x)$, $du=2P_n(x)P_n^{\prime}(x)dx$, $dv=dx$, $v(x)=x$, $\int u\,dv=uv-\int v\,du$, so $$\begin{align}\int_{-1}^1P_n^2(x)dx&=\left.xP_n^2(x)\right|_{-1}^1-2\int_{-1}^1xP_n(x)P_n^{\prime}(x)dx\\ &=(1)P_n^2(1)-(-1)P_n^2(-1)-2\int_{-1}^1P_n(x)\left(P_{n-1}^{\prime}(x)+nP_n(x)\right)dx\\ &=P_n^2(1)+P_n^2(-1)-2\int_{-1}^1P_n(x)P_{n-1}^{\prime}(x)dx-2n\int_{-1}^1P_n^2(x)dx\\ &=\frac{1+1-0}{2n+1}=\frac2{2n+1}\end{align}$$ We hope you already know the normalization of the Legendre polynomials: $P_n(1)=1$ and the parity: $P_n(-x)=(-1)^nP_n(x)$ and the orthogonality: $$\int_{-1}^1P_n(x)P_m(x)dx=0$$ for $m<n$ which implies that $$\int_{-1}^1P_n(x)q(x)dx = 0$$ for any $q(x)\in\mathcal{P}_{n-1}$ in particular for $q(x)=P_{n-1}^{\prime}(x)$.