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Asymptotic behaviour of Legendre polynomial?

  1. Jun 29, 2008 #1
    Dear all,
    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!
  2. jcsd
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