I am trying to determine the asymptotic expansion of the Legendre function as described in a Bender and Orszag text, but have been unable to - some solutions are online, but they approach it in different forms and I'd like to solve it using the below integral definition of the Legendre polynomial.

Specifically, defining:

$$P_n(x)=\frac{1}{\pi}\int_0^\infty(x+cos(\theta)\sqrt{x^2-1})^n d\theta $$

Show that for large n:

$$P_n(x) \sim \frac{1}{\sqrt{2\pi n}}\frac{(x+(x^2-1)^{1/2})^{n+1/2}}{(x^2-1)^{1/4}}$$

Any help is appreciated!


1 Answer 1


This is an application of the famous saddle-point approximation. Write $$ P_n(x)=\int e^{nf(\theta)}d\theta$$ with $f(\theta)=\log(x+\cos(\theta)\sqrt{x^2-1})$.

Saddle-point gives $$P_n(x)\approx\sqrt{\frac{2\pi}{nf''(\theta_0)}}e^{nf(\theta_0)},$$ where $\theta_0$ is the solution to $f'(\theta_0)=0$.

In your case $\theta_0=0$ and this formula gives what you are looking for.

  • $\begingroup$ How to get this if we have the asymptotic mentioned if we have the legendre polynomial not the integral? @marcel $\endgroup$
    – user652838
    Dec 15, 2020 at 9:17
  • $\begingroup$ @user726608 sorry, I didn't understand your question $\endgroup$
    – Marcel
    Dec 15, 2020 at 13:11
  • $\begingroup$ see mathoverflow.net/q/379065/11260 $\endgroup$ Dec 17, 2020 at 9:53
  • $\begingroup$ @CarloBeenakker I know you are not the author of this answer, but nevertheless (since Marcel seems to not be active) isn't $P_n(x)$ defined for $x\in [-1,1]$ making the asymptotic form given a complex number? Or am I missing something and one can prove that the asymptotic form is indeed real for $x$ in the domain of the Legendre polynomials? $\endgroup$
    – user7896
    Apr 21 at 13:13
  • $\begingroup$ @user7896 -- does this help? mathoverflow.net/a/379091/11260 $\endgroup$ Apr 21 at 13:47

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