# Connection between Legendre polynomial and Bessel function

In Abramovitz and Stegun (Eq. 9.1.71) I found this curious relation $$\lim_{\nu\to\infty} \left[ \nu^\mu P_\nu^{-\mu}\left(\cos \frac{x}{\nu} \right) \right]= J_\mu(x) \qquad(1)$$ valid for $x>0$. In fact it can be used to obtain a rather good approximation $$P_\nu^{-\mu}(\cos\theta) \approx \frac{1}{\nu^\mu} J_\mu(\nu \theta)$$ of the Legendre polynomial in terms of a Bessel function for small $\theta$ (but $\nu\theta$ potentially large). This relation is a way to understand the eikonal approximation of wave scattering (which is the reason I noted it in the first place).

As I am looking into the eikonal approximation, I would appreciate if somebody could help me proving equation (1)?

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– Raymond Manzoni Apr 8 '12 at 0:16
@RaymondManzoni: thank you. They express the left side in terms of a confluent hypergeometric function and then show that it converges to the left hand side; in fact, they don't directly do it for $P_\nu^{-\mu}\left(\cos \frac{x}{\nu} \right)$ but rather for $P_\nu^{-\mu}\left(1- \frac{x^2}{2\nu^2} \right)$. This kind of proof is more complicated than I expected (I thought one could use some integral representation and then perform the limit). Does somebody know an more compact proof without referring to hypergeometric functions? – Fabian Apr 8 '12 at 5:07

Consider the differential equation for the associated Legendre polynomials, $$(1-z^2)w''(z) - 2z w(z) + \left(\nu(\nu+1) - \frac{\mu^2}{1-z^2}\right)w(z) = 0.$$ Change variables. Let $z = \cos \frac{x}{\nu}$. (Notice, for example, that $\frac{d}{dz} = -\frac{\nu}{\sin\frac{x}{\nu}} \frac{d}{dx}$.) In the limit $\nu\to\infty$ the DE takes the form $$x^2 w''(x) + x w'(x) + (x^2-\mu^2)w(x) = 0$$ which is, of course, Bessel's equation. Therefore,
$$\lim_{\nu\to\infty} P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \propto J_\mu(x).$$ Since it is getting late, I leave it as an exercise to find the constant.
Addendum: The argument above tells us that in the limit $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is some combination of solutions to Bessel's equation. The singular solution $Y_\mu(x)$ is ruled out since $P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)$ is not singular at $x=0$.
Using the integral representation for $-1<z<1$ and $\mathrm{Re}\,\mu > 0$, $$P_\nu^{-\mu}(z) = \frac{(1-z^2)^{-\mu/2}}{\Gamma(\mu)} \int_z^1 d t\, P_\nu(t)(t-z)^{\mu-1},$$ we find for $x\ll 1 \ll \nu$ that $$P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2\nu}\right)^\mu.$$ (Here we exploit the fact that for $x\ll 1$, $P_\nu\left(\cos \frac{x}{\nu}\right) = 1+O(x^2)$.) But for small $x$ we have $$J_\mu(x) \sim \frac{1}{\Gamma(\mu+1)} \left(\frac{x}{2}\right)^\mu$$ and so $$\lim_{\nu\to\infty} \left[\nu^\mu P^{-\mu}_\nu\left(\cos \frac{x}{\nu}\right)\right] = J_\mu(x).$$