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I want to know [tex] P_\ell(z)\rightarrow? [/tex] as [tex] z\rightarrow\infty [/tex] where the [tex]\ell[/tex] is analytic continuated to complex plane.

I found my book writes

[tex]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)[/tex]

where the second term is the one with [tex]\ell[/tex] replaced by [tex]-\ell-1[/tex]

Since, [tex]_2F_1(\cdots;z^{-2})\rightarrow 1[/tex] as [tex]z\rightarrow \infty[/tex], so we have

[tex]P_\ell(z)\rightarrow\pi^{-1/2}\frac{\Gamma(\ell+\frac{1}{2})}{\Gamma(\ell+1)}(2z)^\ell[/tex] with [tex]\text{Re}(\ell) > -\frac{1}{2}[/tex] and

[tex]P_\ell(z)\rightarrow\pi^{-1/2}\frac{\Gamma(-\ell-\frac{1}{2})}{\Gamma(-\ell)}(2z)^{-\ell-1}[/tex] with [tex]\text{Re}(\ell) < -\frac{1}{2}[/tex]

My question is, how to derive eq(1), and how he decide the two condition [tex]\text{Re}(\ell) > -\frac{1}{2}[/tex] and [tex]\text{Re}(\ell) < -\frac{1}{2}[/tex]?

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!

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# Asymptotic behaviour of Legendre polynomial?

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