Assume $f,\phi$ are smooth on $[a,b]$ and that $\phi$ achieves its maximum at $c\in (a,b)$. Consider the Laplace integral $$I(x) = \int_a^b f(t) e^{x \phi(t)} \text{d} t$$
The first step of Laplace's method says that if we replace this integral with $I(x,\epsilon) = \int_{c-\epsilon}^{c+\epsilon} f(t) e^{x \phi(t)} \text{d} t$ then we make exponentially small error. Why is that? How can we prove that $I(x) - I(x,\epsilon)$ is exponentially small compared to $I(x)$ as $x \rightarrow + \infty$?
Basically for $t \in [a,c-\epsilon] \cup [c+\epsilon,b]$ there is a $M < \phi(c)$ such that $\varphi(t) < M$. Then, for $x > 0$,
$$ \left| \int_{[a,b]\setminus (c-\epsilon,c+\epsilon)} f(t) e^{x\phi(t)}\,dt \right| < e^{xM} \int_{[a,b]\setminus (c-\epsilon,c+\epsilon)} |f(t)| dt. $$
The leading order term for the part of the integral over $(c-\epsilon,c+\epsilon)$ is within a polynomial factor of $e^{x\phi(c)}$, which is exponentially larger than $e^{xM}$.