Associated Legendre polynomials of fractional order My question concerns the associated or generalized Legendre polynomials. They are labeled by two numbers $m$ and $l$; i.e., $P_l^m(x)$, for $x \in [-1,1]$. Usually one assumes that $m$ and $l$ are both integers, but I suspect that it is possible for them to be half integers. If this is correct, what is the meaning of formulas like 
$$P_l^{-m}(x)=(-1)^m \frac{(l-m)!}{(l+m)!}P_l^m(x)$$
or Rodrigues's formula? 
Thanks
 A: It is perfectly alright to have (associated) Legendre functions of half-integer order (or in fact, any arbitrary complex order). The key is to define them in terms of Gaussian hypergeometric functions, e.g.
$$P_\ell^m(z)=\frac{(1+z)^{m/2}}{(1-z)^{m/2}}\frac1{\Gamma(1-m)}{}_2 F_1\left({{-\ell\quad\ell+1}\atop{1-m}} \mid \frac{1-z}{2}\right)$$
to use one particular normalization.
For $\ell$ a half integer, one could use the formulae
$$\begin{align*}
P_{-\frac12}^m(z)&=\frac2{\pi}K\left(\frac{1-z}{2}\right)\\
P_\frac12^m(z)&=\frac2{\pi}\left(2E\left(\frac{1-z}{2}\right)-K\left(\frac{1-z}{2}\right)\right)
\end{align*}$$
where $K(m)$ and $E(m)$ are complete elliptic integrals with parameter $m$, along with the usual recursion relations over $\ell$ and $m$.
For half-integer $m$ we can start the recursion relations with
$$\begin{align*}
P_0^{-\frac12}(z)&=\frac2{\sqrt\pi}\sqrt[4]{\frac{1-z}{1+z}}\\
P_0^\frac12(z)&=\frac1{\sqrt\pi}\sqrt[4]{\frac{1+z}{1-z}}
\end{align*}$$
and for both $\ell$ and $m$ half-integer, the recursion relations take the initial values
$$\begin{align*}
P_{-\frac12}^{-\frac12}(z)&=2\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\arcsin\left(\sqrt{\frac{1-z}{2}}\right)\\
P_{-\frac12}^\frac12(z)&=\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\\
P_\frac12^{-\frac12}(z)&=\sqrt{\frac2{\pi}}\sqrt[4]{1-z^2}\\
P_\frac12^\frac12(z)&=\sqrt{\frac2{\pi}}\frac{z}{\sqrt[4]{1-z^2}}
\end{align*}$$
Note that for arbitrary order, the Rodrigues formula can be suitably interpreted as a Riemann-Liouville differintegral (i.e., differentiation/integration to arbitrary complex order). I might edit this answer for a demonstration later.
