# Asymptotic behaviour of Legendre polynomial?

1. Jun 29, 2008

### ismaili

Dear all,
I want to know $$P_\ell(z)\rightarrow?$$ as $$z\rightarrow\infty$$ where the $$\ell$$ is analytic continuated to complex plane.
I found my book writes
$$P_\ell(z) = \pi^{-1/2}\frac{\Gamma(\ell+\frac{1}{2})}{\Gamma(\ell + 1)}(2z)^\ell \,_2F_1(-\frac{1}{2}\ell,-\frac{1}{2}\ell+\frac{1}{2};-\ell+\frac{1}{2};z^{-2}) \quad+\quad \Big(\ell\rightarrow -\ell - 1\Big)\quad\cdots(eq1)$$
where the second term is the one with $$\ell$$ replaced by $$-\ell-1$$
Since, $$_2F_1(\cdots;z^{-2})\rightarrow 1$$ as $$z\rightarrow \infty$$, so we have
$$P_\ell(z)\rightarrow\pi^{-1/2}\frac{\Gamma(\ell+\frac{1}{2})}{\Gamma(\ell+1)}(2z)^\ell$$ with $$\text{Re}(\ell) > -\frac{1}{2}$$ and
$$P_\ell(z)\rightarrow\pi^{-1/2}\frac{\Gamma(-\ell-\frac{1}{2})}{\Gamma(-\ell)}(2z)^{-\ell-1}$$ with $$\text{Re}(\ell) < -\frac{1}{2}$$

My question is, how to derive eq(1), and how he decide the two condition $$\text{Re}(\ell) > -\frac{1}{2}$$ and $$\text{Re}(\ell) < -\frac{1}{2}$$?
These two conditions seem to make the arguments of Gamma functions in the numerator to have positive real parts, but I don't know why he want that.
Thanks for any instructions!