 Solution

Under local integrability and suitable bounds away from zero, Watson's lemma givesEach term follows by $u=\lambda t$ and the Gamma function integral; a finite local remainder is bounded in the same way, and the region away from zero is exponentially small.

For Laplace's method, assume an isolated nondegenerate global minimum $t_{\ast }$ of $p$, with $p^{\prime \prime }(t_{\ast })>0$ and suitable tail bounds. Expand $p,q$ near $t_{\ast }$ and set $t-t_{\ast }=v/\sqrt{z}$. The leading term is $e^{-zp(t_{\ast })}q(t_{\ast })\sqrt{2\pi /(z{p}^{\prime \prime }(t_{\ast }))}$; successive terms come from Gaussian moments of the Taylor expansions. Several equally low minima contribute additively, while degenerate minima need a different scale.

Here the integrand is entire in $t$, so deform the contour to the line at imaginary part $-\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the line at imaginary part $\pi$. The two horizontal tails sum to $2{\int }_0^{\infty }{e}^{-z\cosh x}dx=O(e^{-z}{z}^{-1/2})$. The vertical segment contributes $i{\int }_{-\pi }^{\pi }{e}^{z\cos \theta }d\theta$. Its maximum is at zero; putting $\theta =v/\sqrt{z}$ and using $\cos \theta =1-{\theta }^2/2+{\theta }^4/24+\cdots$ givesThe fourth Gaussian moment is three times the zeroth, soSince $\Gamma (1/2)=\sqrt{\pi }$, this has the required normalization.