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I am trying to find the first term in the asymptotic expansion of

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{1}{s^2}e^{t(s-m\sqrt{s^2-1})} ds $$

where $0<m<1$, $c<1$, as $t$ approaches $\infty$ with $m$ fixed.

I think I am supposed to used the method of steepest descents to deform the path from a line parallel to the imaginary axis to one where $Im(s-m\sqrt{s^2-1})$ is a constant, but am unsure how to actually proceed.

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    $\begingroup$ You mention that your contour should be parallel to the imaginary axis, so did you intend the bounds for the integral to be $c \pm i\infty$ instead of $c\pm\infty$? $\endgroup$ – Antonio Vargas Dec 1 '15 at 9:35
  • $\begingroup$ Yeah I did, that was my bad - thanks for catching that! $\endgroup$ – meanderingthroughmath Dec 2 '15 at 10:29
  • $\begingroup$ The main idea is to deform your contour so that it passes through the saddle point at the relevant solution to $\frac{d}{ds} (s-m\sqrt{s^2-1}) = 0$. $\endgroup$ – Antonio Vargas Dec 2 '15 at 10:37

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