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I have come across an integral that I would like to asymptotically evaluate (to leading order at least) which I have seen no mention of in standard textbooks.

I want to evaluate an integral of the form

$$ \int_{-1}^1 h(x) \frac{e^{kf(x)}}{e^{ikg(x)}-\lambda} d x $$

where $h$ is a smooth nicely behaved function, $f$ is a complex function, $g$ is a real function and $|\lambda|>1$. $f$ and $g$ depend on a parameter as well.

Ideas so far are to expand the integral using binomial theorem which then gives an infinite sum to evaluate, or try and write it in terms of $sin$, $cos$ and $sec$.

I would also like to think about the case where $|\lambda|<1$ and how this changes things.

OmniJames
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  • Interesting question. Since the denominator is bounded and bounded away from zero I would suspect that its oscillatory contributions affect only the higher-order asymptotics, so you might have something like $$\int_{-1}^1 h(x) \frac{e^{kf(x)}}{e^{ikg(x)}-\lambda} ,dx \asymp \int_{-1}^{1} h(x) e^{kf(x)},dx,$$ which you could then approach with the saddle point method. I'd like to play around with some specific examples of $f(x)$, $g(x)$, and $h(x)$ if you have them. – Antonio Vargas Dec 09 '13 at 17:06
  • The functions are something like $$f(x)=i\sqrt{1-x^2}a+ibx$$ and $$g(x)=\sqrt{1-x^2}c$$ and $a$, $b$ and $c$ are fixed parameters that I would like to get an answer in terms of. I don't think $h(x)$ matters too much, but it has no singularities in the interval $[-1,1]$. – OmniJames Dec 10 '13 at 09:31

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