How to get the short time asymptotics of this integral? The integral is like this:
$$
\int_0^\infty \mathrm{d} x \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2}
$$
The short time asymptotics is like this (some constant maybe missing):
$$
\sim \frac{1}{\ln t}
$$
I don't know how to get this...
Edit: what's the meaning of short time?
The source I am reading says: 

At short times the integral rises or falls sharply as $\sim \frac{1}{\ln t}$

My guess that the meaning of short time is the time smaller than the first maximum or minimum of the integral.
 A: For $x$ real we have
$$
\left| \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} \right| \leq \frac{1}{1+x^2},
$$
we know that
$$
\frac{1}{1+x^2} \in L^1([0,\infty)),
$$
and for fixed $x \geq 0$ we calculate
$$
\lim_{t \to 0^+} \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \frac{1}{1+x^2}.
$$
Therefore we may conclude that
$$
\lim_{t \to 0^+} \int_0^\infty dx \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{2}.
$$
by the Dominated Convergence Theorem.
A: here is an idea:  i will look at the asymptotics of the simpler $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx  \text{ for small } t. $$
observe that the graph of $y = \frac{\cos(te^x)}{1+x^2}$ oscillates between the envelopes $y = \pm \frac 1{1+x^2}$ starts at $(0,1)$ and the first $x$ -intercept is $\ln(\pi/2) - \ln(t).$ there after the intercepts are spaced by $\ln 2, \ln 3. \cdots.$
therefore $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx=\int_0^{\ln(\pi/(2t)}\frac{\cos(te^x)}{1+x^2}\, dx +\cdots = \frac12 \times 1 \times \ln\left(\frac{\pi}{2t}\right) + \cdots$$ the estimate being the area of the triangle with base $ \ln\left(\frac{\pi}{2t}\right)$ and height $1.$
