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For fixed $x>1$, write $a=\sqrt{x^2-1}$ and $A=x+a$. The maximum of the integrand occurs at $\theta =0$, andHere is the required endpoint Laplace method derivation. Outside a fixed neighborhood of zero, the integrand is exponentially smaller than $A^n$. Inside that neighborhood, use $\theta =(A/(na){)}^{1/2}u$. The quadratic term gives $e^{-u^2/2}$ and the higher terms vanish in the scaled limit. The Gaussian integral on the half-line is $\sqrt{\pi /2}$, soThis is the Hyperbolic Debye asymptotic for Legendre polynomials. At the endpoint the integrand is identically one, giving $P_n(1)=1$ exactly.

The localization width in the Laplace method is $O((na{)}^{-1/2})$. It becomes order one when $n\sqrt{x-1}=O(1)$, so the Legendre boundary layer at x equals one has $q=2$. With $x=1+\nu /n^2$ and fixed $\nu \ge 0$,uniformly in $\theta$. The boundary-layer answer is thereforeThis real integral representation of the modified Bessel function I0 gives one at $\nu =0$, agreeing with the endpoint value. Applying the same endpoint Laplace method to the integral for $I_0(z)$ gives $I_0(z)\sim e^z/\sqrt{2\pi z}$. Thus, as $\nu \to \infty$, the inner result is $e^{\sqrt{2\nu }}/\sqrt{2\pi \sqrt{2\nu }}$. The outer result has precisely this limit in the overlap $1\ll \nu \ll n$. This proves both matches rather than extending the fixed-$x$ formula directly to $x=1$.

For the derivatives, the power series $I_0(\sqrt{2\nu })=1+\nu /2+{\nu }^2/16+\cdots$ suggests their leading orders. To justify differentiation, expand the original integral itself near $x=1$. Its first three even cosine moments giveOdd cosine moments vanish, and higher even moments start at cubic order. Expanding at $x=1$ yieldsFor large $n$, the graph starts at $(1,1)$ with a very large positive slope and positive curvature. It rises through the $n^{-2}$ boundary layer and then follows the rapidly increasing outer asymptotic expansion. The sketch below shows both scales.