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Does anyone know where to find (or how to obtain) expressions for the Legendre functions for large degree, to second order? For example, to first order the expressions are $$ P_n(\cosh(x)) ~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \frac{1}{\sqrt{\pi n}}\frac{e^{(n+1/2)x}}{\sqrt{2\sinh(x)}} \\ Q_n(\cosh(x))~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \sqrt{\frac{\pi}{n}}\frac{e^{-(n+1/2)x}}{\sqrt{2\sinh(x)}}, $$ for integer $n$. These are apparently given on page 305-306 of "The theory of spherical and ellipsoidal harmonics" by E. W. Hobson, although I can't seem to access this book.

I am looking for the second order of these expressions - corrections $\mathcal{O}(1/n)$ relative to the above expressions.

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A series with precise error estimates is derived in Error bounds for a uniform asymptotic expansion of the Legendre function:

$$P_n(\cosh x)=\left(\frac{x}{\sinh x}\right)^{1/2}\sum_{\nu=0}^{\infty}c_\nu(x)\frac{I_\nu[(n+1/2)x]}{(n+1/2)^\nu},$$ $$c_0=1,\;\;c_1=\frac{1}{8}(\coth x-1/x),$$ $$c_2=-\frac{1}{16}(1+3/x^2-(3/x)\coth x)+\frac{9}{128}x^{-2}(1-x\coth x)^2.$$

(There is a recursion relation for higher order $c_\nu$'s.)
An alternative representation involving only the Bessel function $I_0$ is given in equation 5.1 of that paper.

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In addition to the Carlo Beenakker's answer. The following asymptotic expansion was proved in https://www.sciencedirect.com/science/article/pii/0041555365901345?via%3Dihub (Asymptotic formulae for legendre functions, by N.K.Chukhrukidze. Russion version can be downloaded from here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=zvmmf&paperid=7560&option_lang=eng): $$P_n(\cosh{x})=A_0(x)\left[ I_0[(n+1/2)x]\left(1+\frac{A_1(x)}{(n+1/2)^2} -\frac{4A_3(x)-xA_4(x)}{(n+1/2)^4x}\\+\frac{24A_4(x)}{(n+1/2)^6x^2}+\ldots\right)+I_1[(n+1/2)x]\left(\frac{A_1(x)}{n+1/2}-\frac{2A_2(x)-xA_3(x)}{(n+1/2)^3x}+\\\frac{8[A_3(x)-xA_4(x)]}{(n+1/2)^5x^2}-\frac{48A_4(x)}{(n+1/2)^7x^3}+\ldots \right)\right],$$ where $$A_0(x)=\sqrt{\frac{x}{\sinh{x}}},\;\;A_1(x)=\frac{x\coth{x}-1}{8x},\\A_2(x)=\frac{9x^2\coth^2{x}+6x\coth{x}-8x^2-15}{128x^2},\\A_3(x)=\frac{5\left(15x^3\coth^3{x}+27x^2\coth^2{x}-11x^3\coth{x}+21x\coth{x}-24x^2-63\right)}{1024x^3},A_4(x)=\frac{7}{32768x^4}\left(525x^4\coth^4{x}+1500x^3\coth^3{x}-720x^4\coth^2{x}\\+2430x^2\coth^2{x}-1600x^3\coth{x}+1980x\coth{x}+192x^4-2160x^2-6435\right).$$

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Actually I got hold of Hobson's book and it has the expressions to second order anyway. They are: $$ P_n(\cosh x) ~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \frac{1}{\sqrt{\pi n}}\frac{e^{(n+1/2)x}}{\sqrt{2\sinh x}}\bigg(1-\frac{2-\coth x}{8n}\bigg)\\ Q_n(\cosh x)~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \sqrt{\frac{\pi}{n}}\frac{e^{-(n+1/2)x}}{\sqrt{2\sinh x}}\bigg(1-\frac{2+\coth x}{8n}\bigg). $$ The expression for $P_n$ agrees with Carlo's answer with the asymptotic form of the Bessel function to second order and summing up to $\nu=1$.

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