Integral asymptotic expansion of $\int_{0}^{\infty} \frac{e^{-x \cosh t}}{\sqrt{\sinh t}}dt$ for $x \to \infty$

Question

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

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?

Answer 1

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

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

or

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

as $x \rightarrow \infty$.

Answer 2

Changing the variable of integration from $t$ to $s$ using $t = \cosh^{-1}(1 + s)$ transforms the integral into the form $$ {\rm e}^{ - x} \int_0^{ + \infty } {{\rm e}^{ - xs} s^{ - 3/4} (2 + s)^{ - 3/4} {\rm d}s} . $$ Applying Watson's lemma directly leads to the asymptotic expansion $$ {\rm e}^{ - x} \int_0^{ + \infty } {{\rm e}^{ - xs} s^{ - 3/4} (2 + s)^{ - 3/4} {\rm d}s} \sim \frac{\Gamma \big( {\tfrac{1}{4}} \big)}{\sqrt 2}\frac{{{\rm e}^{ - x} }}{{ (2x)^{1/4} }}\sum\limits_{n = 0}^\infty \frac{{\left( {\frac{1}{4}} \right)_n \left( {\frac{3}{4}} \right)_n }}{{n!}}\frac{{( - 1)^n }}{{(2x)^n }} , $$ as $x\to+\infty$. Here, $(a)_n = \Gamma(a+n)/\Gamma(a)$ represents the Pochhammer symbol.