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$$\int_{0}^{\infty} \frac{e^{-x \cosh t}}{\sqrt{\sinh t}}dt$$

I'm trying to use Laplace's method to find the leading asymptotic behavior as $x$ goes to positive infinity, but I'm having some trouble. Could someone help me?

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Here is a related problem. – Mhenni Benghorbal Jan 24 '13 at 21:36
Please do not delete the contents of your questions. If you're looking to get answers and then erase the evidence, this may not be the site for you. – Rahul Jan 25 '13 at 6:40

In Laplace's method, we recognize that the main contribution to the integral is in the neighborhood of the minimum of the exponent, here at $t=0$. It appears that the slowly varying part of the integrand has a singularity there, but this singularity is integrable. Thus, we may approximate as follows, noting that $\sinh{t} \sim t$ and $\cosh{t} \sim 1+\frac{t^2}{2}$ as $t \rightarrow 0$:

$$I(x) = \int_{0}^{\infty} dt \: \frac{e^{-x \cosh t}}{\sqrt{(\sinh t)}} \sim e^{-x} \int_{0}^{\infty} dt \: \frac{e^{-x t^2/2}}{\sqrt{t}} $$

This latter integral may be evaluated any number of ways; here, we substitute $v = x t^2$, $dv = 2 \sqrt{x v} dt$, and get

$$I(x) \sim \frac{1}{2} e^{-x} x^{-1/4} \int_{0}^{\infty} dv \: v^{-3/4} e^{-v/2} $$


$$I(x) \sim 2^{-3/4} \Gamma \left ( \frac{1}{4} \right ) e^{-x} x^{-1/4} $$

as $x \rightarrow \infty$.

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