 Solution

For the Laplace method, under the usual regularity ($p$ continuous at zero and $q$ sufficiently smooth near its isolated minimum), write $q(t)=q(0)+q^{\prime \prime }(0)t^2/2+o(t^2)$. On any interval bounded away from zero, continuity and uniqueness of the minimum give a positive gap in $q$, making that contribution exponentially smaller. Set $t=s/\sqrt{x}$ near zero. The leading integral isA local quadratic lower bound on $q-q(0)$ justifies dominated convergence after this rescaling. Thus, when $p(0)\ne 0$,If $p(0)=0$, the displayed coefficient vanishes and further local terms of $p$ determine the first nonzero asymptotic term; one must not interpret a zero coefficient as an asymptotic equivalence.

For the first Bessel integral, take $q(\theta )=-\cos \theta$, $p(\theta )=\cos (\nu \theta )$. The endpoint minimum is $q(0)=-1$, $q^{\prime \prime }(0)=1$, $p(0)=1$, so after dividing by $\pi$ its leading term is $e^x/\sqrt{2\pi x}$.

For the second integral, $\cosh t\ge 1+t^2/2$ gives, for fixed real $\nu$,Completing the square proves the last bound for large $x$. It is exponentially smaller than the first integral; the fixed prefactor $\sin (\nu \pi )$ does not change this. Therefore