Legendre integral transform

Remarks

The integrals converge because $\abs{P_n(x)} \le 1$ on $[-1, 1]$. Let $f \in L^2([-1, 1])$, the $L^2$ space. Then the inverse transform is the Fourier–Legendre series $$f(x) = \sum_{n=0}^{\infty} \frac{2n + 1}{2}\, \tilde f(n)\, P_n(x),$$ with convergence in $L^2([-1, 1])$; this follows from $\int_{-1}^{1} P_m P_n\, dx = \frac{2}{2n+1} \delta_{mn}$ and the completeness of the Legendre polynomials. If $f$ is twice continuously differentiable on $[-1, 1]$, integration by parts gives $\int_{-1}^{1} P_n(x)\, \frac{d}{dx}\big[(1 - x^2) f'(x)\big]\, dx = -n(n+1)\, \tilde f(n)$, the rule that makes the transform useful for Legendre's equation.

Example: $f(x) = x^2$. Since $x^2 = \frac{1}{3} P_0(x) + \frac{2}{3} P_2(x)$, orthogonality gives $\tilde f(0) = \frac{1}{3} \cdot 2 = \frac{2}{3}$, $\tilde f(2) = \frac{2}{3} \cdot \frac{2}{5} = \frac{4}{15}$, and $\tilde f(n) = 0$ otherwise. Directly, $\tilde f(0) = \int_{-1}^{1} x^2\, dx = \frac{2}{3}$.

Non-example: $f(x) = \frac{1}{1 - x}$ is not integrable on $[-1, 1]$, and $\tilde f(0) = \int_{-1}^{1} \frac{dx}{1 - x}$ diverges.