Gegenbauer transform

Remarks

The polynomials $C_n^\lambda$ are orthogonal for $w$, and $F \in L^2(w)$ is recovered as $F = \sum_{n \ge 0} \frac{f(n)}{h_n} C_n^{\lambda}$ with $h_n = \int_{-1}^1 w\, (C_n^\lambda)^2\, dx$. If $F$ is smooth enough and the boundary terms vanish, the transform of $w^{-1} \frac{d}{dx}\left( (1 - x^2)^{\lambda + 1/2} F' \right)$ is $-n(n + 2\lambda) f(n)$.

Example: $\lambda = 1/2$ gives $w = 1$ and $C_n^{1/2} = P_n$, the Legendre polynomials. For $F(x) = x$ the transform is $f(1) = \int_{-1}^1 x^2\, dx = 2/3$ and $f(n) = 0$ for $n \ne 1$, since $x = P_1(x)$.

Non-example: for $\lambda = -1/2$ the weight $(1 - x^2)^{-1}$ is not integrable near $\pm 1$, so $F = 1$ has no transform.

The value $\lambda = 0$ is excluded because $C_n^0 = 0$ for $n \ge 1$ in the usual normalization; the Chebyshev polynomials of the first kind take its place.