Theorem: The Legendre polynomials $P_m(x)$ and $P_n(x)$ satisfy:

$\int_{-1}^1 P_m(x)P_n(x)dx = \frac{2}{2n+1}\delta_{mn}$

Proof:

Step 1: Legendre polynomials satisfy Legendre's equation:

$\frac{d}{dx}\left[(1-x^2)\frac{dP_n}{dx}\right] + n(n+1)P_n = 0$

This is a Sturm-Liouville problem with $p(x) = 1-x^2$, $q(x) = 0$, $w(x) = 1$, and eigenvalues $\lambda_n = n(n+1)$.

Step 2: Write the equation for $P_m$ and $P_n$:

$\frac{d}{dx}\left[(1-x^2)\frac{dP_m}{dx}\right] + m(m+1)P_m = 0$

$\frac{d}{dx}\left[(1-x^2)\frac{dP_n}{dx}\right] + n(n+1)P_n = 0$

Step 3: Multiply the first equation by $P_n$ and the second by $P_m$:

$P_n\frac{d}{dx}\left[(1-x^2)\frac{dP_m}{dx}\right] + m(m+1)P_m P_n = 0$

$P_m\frac{d}{dx}\left[(1-x^2)\frac{dP_n}{dx}\right] + n(n+1)P_m P_n = 0$

Step 4: Subtract the second from the first:

$P_n\frac{d}{dx}\left[(1-x^2)\frac{dP_m}{dx}\right] - P_m\frac{d}{dx}\left[(1-x^2)\frac{dP_n}{dx}\right] + [m(m+1) - n(n+1)]P_m P_n = 0$

Step 5: The first two terms combine as:

$\frac{d}{dx}\left[(1-x^2)\left(P_n\frac{dP_m}{dx} - P_m\frac{dP_n}{dx}\right)\right]$

Thus:

$\frac{d}{dx}\left[(1-x^2)\left(P_n\frac{dP_m}{dx} - P_m\frac{dP_n}{dx}\right)\right] = [n(n+1) - m(m+1)]P_m P_n$

Step 6: Integrate from $-1$ to $1$:

$\left[(1-x^2)\left(P_n\frac{dP_m}{dx} - P_m\frac{dP_n}{dx}\right)\right]_{-1}^1 = [n(n+1) - m(m+1)]\int_{-1}^1 P_m P_n dx$

The left side vanishes because $(1-x^2) = 0$ at both $x = \pm 1$.

Step 7: For $m \neq n$, we have $n(n+1) - m(m+1) \neq 0$, so:

$\int_{-1}^1 P_m(x)P_n(x)dx = 0 \quad (m \neq n)$

Step 8: For normalization ($m = n$), use Rodrigues' formula:

$P_n(x) = \frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$

Compute:

$\int_{-1}^1 P_n^2(x)dx = \frac{1}{(2^n n!)^2}\int_{-1}^1 \frac{d^n}{dx^n}(x^2-1)^n \cdot \frac{d^n}{dx^n}(x^2-1)^n dx$

Integrate by parts $n$ times (boundary terms vanish):

$= \frac{(-1)^n}{(2^n n!)^2}\int_{-1}^1 (x^2-1)^n \cdot \frac{d^{2n}}{dx^{2n}}(x^2-1)^n dx$

Since $(x^2-1)^n$ is a polynomial of degree $2n$, its $2n$-th derivative is $(2n)!$:

$= \frac{(2n)!}{(2^n n!)^2}\int_{-1}^1 (x^2-1)^n dx$

Evaluating this integral (using the beta function):

$\int_{-1}^1 (1-x^2)^n dx = 2\int_0^1 (1-x^2)^n dx = \frac{2^{n+1}(n!)^2}{(2n+1)!}$

Therefore:

$\int_{-1}^1 P_n^2(x)dx = \frac{(2n)!}{(2^n n!)^2} \cdot \frac{2^{n+1}(n!)^2}{(2n+1)!} = \frac{2}{2n+1}$

Theorem: Any function $f(x) \in L^2([-1,1])$ can be expanded as:

$f(x) = \sum_{n=0}^{\infty}c_n P_n(x), \quad c_n = \frac{2n+1}{2}\int_{-1}^1 f(x)P_n(x)dx$

Proof Sketch:

The Legendre polynomials form a complete orthogonal basis because:

1. They are eigenfunctions of a self-adjoint operator
2. The Weierstrass approximation theorem shows polynomials are dense in $C([1,1])$
3. Gram-Schmidt applied to monomials $\{1, x, x^2, \ldots\}$ produces the Legendre polynomials
4. Since finite linear combinations of $P_n$ can approximate any continuous function, the $P_n$ span the space

Completeness guarantees:

$\sum_{n=0}^{\infty}|c_n|^2\frac{2}{2n+1} = \int_{-1}^1 |f(x)|^2dx$

(Parseval's theorem for Legendre expansion).