## A Proof of the Orthogonality of the Legendre Polynomials

This post was prompted by this question at Mathematics Stack Exchange. This proof of the orthogonality of the Legendre polynomials is from Special Functions and Their Applications by N. N. Lebedev, a book that I highly recommend.

We begin with Legendre’s differential equation

[(1-x^{2})P^{\prime}_{n}(x)]^{\prime} +n(n+1)P_{n}(x) = 0,\quad n \in \mathbb{Z}_{0}^{+}
\label{eq:lp1}
\tag{1}

The first step is to multiple equation \eqref{eq:lp1} by $$P_{m}(x)$$ and subtract it from equation \eqref{eq:lp1} written for $$m$$ and multiplied by $$P_{n}(x)$$.

[(1-x^{2})P^{\prime}_{m}(x)]^{\prime}P_{n}(x) \,-\, [(1-x^{2})P^{\prime}_{n}(x)]^{\prime}P_{m}(x) + [m(m+1)-n(n+1)]P_{m}(x)P_{n}(x) = 0

Rearrangement yields

\{(1-x^{2})[P^{\prime}_{m}(x)P_{n}(x)-P^{\prime}_{n}(x)P_{m}(x)]\}^{\prime} + (m-n)(m+n+1)P_{m}(x)P_{n}(x) = 0
\label{eq:lp2}
\tag{2}

Integrating equation \eqref{eq:lp2} from -1 to 1, the first term goes to 0 and we are left with

(m-n)(m+n+1) \int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0

or

\int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0, \quad m \ne n