Expansion of the associated Legendre polynomials $P^m_l(\cos\vartheta)$ for $\vartheta \rightarrow 0$

Question

A simple test with Mathematica indicates that for the associated Legendre polynomials $P^m_l$ the following relation should hold: $$ \lim_{\vartheta\rightarrow 0} P^m_l(\cos\vartheta) =a_{lm} \vartheta^{|m|}+{\cal O}(\vartheta^{|m|+1}).$$ However, I cannot find an asymptotic expansion like this anywhere. Can the value of the first expansion coefficient $a_{lm}$ be determined explicitly as a function of $l$ and $m$?

Here are the first few values of $a_{lm}$

\begin{array}{r | c | c | c | c |c } m\text{\\}l & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 1 &1 &1 &1 & 1 \\ 1 & &-1 &-3 &-6 & -10\\ 2 & & & 3&15 &45 \\ 3 & & & & -15 & -105\\ 4 & & & & & 105 \end{array}

The closest formula I could find, was the trigonometric expanion formula in Abramowitz, Stegun (8.7.1 p. 335). But it is explicitly only valid for $0<\vartheta<\pi$ and it does not yield the values shown inside the table.

Answer 1

In view the Rodrigues formula for the associated Legendre polynomials, one finds $$a_{nm}=\lim_{\vartheta\rightarrow 0} \vartheta^{-m}P^m_n(\cos\vartheta)= (-1)^m\frac{1}{2^n n!} \lim_{x\rightarrow 1}\frac{\partial ^{m+n}}{\partial x^{m+n}}\left(x^2-1\right)^n.$$ Then use that $$\lim_{x\rightarrow 1}\frac{\partial ^{m+n}}{\partial x^{m+n}}\left(x^2-1\right)^n=\sum _{j=0}^n \binom{n}{j} \frac{(-1)^j (2 n-2 j)! }{(n-2 j-m)!},$$ as follows from the binomial expansion of $(x^2-1)^n$. The sum over $j$ can be evaluated further, I arrive at $$a_{nm}=\frac{(-1)^m (n+m)!}{2^m m! (n-m)!}.$$

Answer 2

For completeness, here are the slightly modified expressions for $m<0$: $$a_{lm}={1\over 2^{-m} (-m)!},$$

as well as the corresponding coefficients for the limit of $x\rightarrow -1$ or $\vartheta \rightarrow \pi$ respectively.

$m\ge 0$: $$a_{lm}= {(-1)^l (l+m)!\over 2^m m! (l-m)!}$$

and $m<0$: $$a_ {lm} = {(-1)^{l+m}\over 2^{-m} (-m)!}$$