Integral expansions

Question

How to find the leadi­ng behaviour of $$\int_0^{\pi^2/2}\int_0^{\pi^2/2} e^{x \cos(\sqrt{q+s})}\,dq\,ds$$ as x ten­ds to infinity?If we ­consider the first in­tegral w.r.t dq and t­ry Laplace method, is­ it right to take q=0­ as the maximum? Plea­se help to proceed

Answer 1

We first perform the substitution $$y=s+t, z= s-t$$ in order to obtain $$I = \int_0^{\pi^2/2} \!dy\,y e^{x \cos\sqrt y} + \int_{\pi^2/2}^{\pi^2}\!dy\, (\pi^2-y)e^{x \cos\sqrt y}.$$

As $\cos \sqrt y$ is negative on the interval $y \in[\pi^2/2,\pi^2]$ the second integral is exponentially small for $x\to \infty$ and the dominant contribution comes from the first integral.

In the first integral, $\cos \sqrt y$ obtains its maximum at $y=0$. So employing Laplace's method, we obtain the leading order $$I \sim \int_0^\infty \!dy\,y e^{x (1-y/2)} =\frac{4 e^{x}}{x^2}$$