orthogonality of Lagrange basis with legend nodes

Question

I came across the following statement and I don't know how to justify it.

If $L_i$ is a Lagrange basis, and $x$ is a zero of Legendre polynomial, then

$$ \int_{-1}^{1} L_i(x)L_j(x) dx = \delta_{ij}w_j$$

where $w_j$ are the Gauss–Legendre quadrature weights. How to justify this statement?

Answer 1

This statement is false. The integral $$ \int_{-1}^1 L_i(x)L_j(x)dx = \delta_{ij}c_j = \delta_{ij}\frac{2}{2j+1} $$ (see https://en.wikipedia.org/wiki/Legendre_polynomials) whereas the weights for Gauss-Legendre are (for $n^\textrm{th}$ order GL quadrature) $$ w_i = \frac{2}{(1-x_i)^2\left[L^{'}_n(x_i)\right]^2} $$ where $x_i$ are the zeros of the $n^\textrm{th}$ order Legendre polynomial (see https://en.wikipedia.org/wiki/Gauss%E2%80%93Legendre_quadrature). These two are clearly different.