According to the so-called formula of Laplace (see, e.g., [1, Theorem 8.21.2]), Legendre polynomials admit the following asymptotic expansion: $$ P_n(\cos\theta) = \sqrt{2} (\pi n\sin\theta)^{-\frac12}\cos((n+\tfrac12)\theta - \tfrac{\pi}{4}) + O(n^{-\frac32}), \quad \text{as } \; n \to +\infty, $$ where the convergence is uniform for $\theta \in [\varepsilon,\pi-\varepsilon]$.

Question: I would like to understand the dependence of the error term on $\varepsilon$, but I haven't found anything useful in this direction in the literature so far. Do you know where could I find this kind of result (if available)?


[1] Szegö, G. Orthogonal polynomials, 1939.

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    $\begingroup$ I guess you are aware of the representation $P_n(\cos\theta)=\int_{\theta}^\pi \frac{\sin(n+\frac12)\phi}{(2\cos\theta-2\cos\phi)^{1/2}}d\phi$? Maybe it proves helpful for doing the asymptotic analysis... $\endgroup$ – Suvrit Jun 22 '17 at 22:57
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    $\begingroup$ If your question is about the behavior of $P_n$ in a neighborhood of 1, that is $\theta=0$, Hilb's formula [1, Thm 8.21.6] gives you the expansion of $P_n$ there in terms of the Bessel function $J_0$. If your question is really about how Laplace's formula diverges as $\epsilon$ tends to 0, then look at the Stieltjes generalization of Laplace's formula [1, Thm 8.21.5] which gives the next terms in the expansion of $P_n(\cos\theta)$ and an explicit bound for the error term. $\endgroup$ – user111 Jun 23 '17 at 7:22
  • $\begingroup$ Thanks @user111! That looks like a really good hint. I've definitely overlooked Stieltjes' generalization of Laplace's formula. $\endgroup$ – Paglia Jun 23 '17 at 15:55

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