The differential equation for the Legendre polynomials $P_n(x)$ is given by: $(1 - x^2) \frac{d^2P_n}{dx^2} - 2x \frac{dP_n}{dx} + n(n + 1)P_n = 0$. Now I want to prove that $\begin{equation} \int_{-1}^{1} P_l(x) P_m (x) = \dfrac{2}{2l+1} \delta_{lm} \end{equation} $.
It's possible to prove the orthogonal relation for $l \neq m$ case just by Legendre's differential equation.
But for $l =m $ case, we gonna use Rodrigue's formula: $P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left[(x^2 - 1)^n\right] $ or, Generating function: $ \frac{1}{\sqrt{1 - 2tx + t^2}} = \sum_{n=0}^{\infty} P_n(x) t^n $ or, Recurrance relation: $(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x),$ to archive the orthogonal integral form.
But is there any way to prove this orthogonal relation without using them solely from Legendre's differential equation?