Identity relating hypergeometric function and Legendre polynomial

Question

In my notes I have written down the following relation:

$_2F_1(a,a+\frac{1}{2};c;z)=2^{c-1}z^{(1-c)/2}(1-z)^{-a+(c-1)/2}L_{2a-c}^{1-c}\big(\frac{1}{\sqrt{1-z}}\big)\ ,$

where $_2F_1(a,b;c;z)$ is the hypergeometric function and $L_{\alpha}^{\beta}(z)$ is the associated Legendre polynomial. I then use this for the special case $c=1$, reducing to the normal Legendre polynomial.

However now I can't confirm the above identity. I've looked through Wikipedia, Mathworld, Wolfram functions and dlfm, to no avail.

Am I missing it in those sites, or did I get it somewhere else? Did I make it up? Does it refer to another function than Legendre?

Answer 1

Let $m=1-c$, $n=2a-c$, $x=\frac{1}{\sqrt{1-z}}$. Then on the left hand side one has $$ _2F_1(a,a+\frac{1}{2};c;z)=F\left(\frac{n-m+1}{2},\frac{n-m}{2}+1;1-m;1-\frac{1}{x^2}\right),\tag{1} $$ and on the right hand side: $$ 2^{-m}x^{n-m+1}(x^2-1)^{m/2}{L}^{m}_{n}(x).\tag{2} $$

Using eq. (15.8.17) one can write eq. $(1)$ as $$ x^{n-m+1}F\left(n-m+1,-n-m;1-m,\frac{1-x}{2}\right), $$ and after using Euler's transformation (15.1.8) as $$ x^{n-m+1}\left(\frac{1+x}{2}\right)^{-m}F\left(-n,n+1;1-m,\frac{1-x}{2}\right)=\\ \Gamma(1-m)2^mx^{n-m+1}\frac{(x-1)^{\frac{m}{2}}}{(x+1)^{\frac{3m}{2}}}\cdot \frac{1}{\Gamma(1-m)}\left(\frac{x+1}{x-1}\right)^{\frac{m}{2}}F\left(-n,n+1;1-m,\frac{1-x}{2}\right).\quad \ (3) $$

According to Whittaker and Watson the associated Legendre function is given by $$ P_n^m(x)=\frac{1}{\Gamma(1-m)}\left(\frac{x+1}{x-1}\right)^{m/2}F\left(-n,n+1;1-m;\frac{1-x}{2}\right). $$

Comparing eqs. $(2)$ and $(3)$ one obtains $$ {L}^{m}_{n}(x)=\frac{\Gamma(1-m)4^m}{(x+1)^{2m}}P_n^m(x). $$ When $m=0$ $(c=1)$ one has $L_n^0(x)=P_n(x)$.