 Solution

PutOn $[0,2]$, its global maximum is the endpoint $t=2$, whereand the amplitude $\log t$ has value $\log 2$. The endpoint form of Laplace's method therefore givesIndeed, with $u=2-t$, Taylor expansion gives $\varphi (2-u)=2-5u+O(u^2)$ and $\log (2-u)=\log 2+O(u)$; the scale contributing to the integral is $u=O(x^{-1})$. Outside any fixed neighbourhood of $2$, $\varphi$ is bounded above by $2-\delta$, so that part is exponentially smaller. This also controls the integrable logarithmic singularity at zero.