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Let's have integral $$ I(x) = \frac{1}{2\pi} \int \limits_{-\pi}^{\pi}e^{x\cos(\theta )}d \theta, \quad x \to +\infty . $$ How to use Laplace approximation for this integral and find first two summands of asymptotic expansion?

Edit:

$$ I(x) = 2\frac{1}{2\pi}\int \limits_{0}^{\pi}e^{xcos(\theta )}d \theta = \left|z^2 = 1 - cos(\theta ), \quad d\theta = \frac{2dz}{\sqrt{2 - z^2}}\right| = $$ $$ =4e^{x}\frac{1}{2\pi}\int \limits_{0}^{2} e^{-xz^2}\frac{dz}{\sqrt{2 - z^2}} = \left|\frac{1}{\sqrt{2 - z^2}} = \frac{1}{\sqrt{2}} + \frac{z^2}{4 \sqrt{2}} + O(z^4)\right| = $$ $$ =\sqrt{2}\frac{1}{2\pi}e^{x}\int \limits_{-\infty}^{\infty}e^{-xz^2}\left( 1 + \frac{z^2}{4} + O(z^{4})\right)dz = \sqrt{\frac{1}{2 \pi x}}e^{x}\left(1 + \frac{1}{8 x} + O\left(\frac{1}{x^2}\right)\right) . $$

glS
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John Taylor
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  • First, you're only interested in the maxima of $\cos \theta$ in the interval $[-\pi,\pi]$. There is only one, and it's not on the boundary of the interval. As for finding the first two terms of the asymptotic series, this thread outlines a few different methods for doing so. – Antonio Vargas Mar 09 '14 at 14:03
  • @AntonioVargas : so I can only integrate by parts my integral expression, after that expand $cos(\theta )$ near maximum and then use Laplace method for each summand? – John Taylor Mar 09 '14 at 14:17
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    You don't need to integrate by parts. Look at the answers to the question instead. – Antonio Vargas Mar 09 '14 at 14:20
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    @AntonioVargas : so the convenient way is to expand $cos(\theta )$ near zero, then to make substitution $z^2 = 1 - cos(\theta ) \Rightarrow \theta = arccos (1 - z^2)$, then represent $d \theta $ through $dz$, after that - to integrate each summand, and, finally, to leave only the summands with first two orders of $x$? – John Taylor Mar 09 '14 at 14:44
  • @AntonioVargas : can you check my solution (I had updated the question's body)? – John Taylor Mar 09 '14 at 15:07
  • The method is perfect. There's an issue with your second term at the end though. Instead of $1/4x$ you should have $1/8x$. Other than that it's right on. – Antonio Vargas Mar 10 '14 at 13:18
  • Also don't forget your factor of $1/2\pi$ in front (from the definition of $I(x)$ at the top). – Antonio Vargas Mar 10 '14 at 15:35
  • @AntonioVargas : thank you! – John Taylor Mar 10 '14 at 20:26

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