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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 $x \rightarrow + \infty$?

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    I imagine you have had a look at (https://en.wikipedia.org/wiki/Laplace's_method) but maybe a more interessant one is (http://bolvan.ph.utexas.edu/~vadim/Classes/2011f/saddle.pdf) with the saddle point idea. – 2017-02-11
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    There are detailed proofs of this online; the point is basically that away from the maximum, $e^{x \phi(t)}$ is much smaller than $e^{x \phi(c)}$ because $x$ is large. Consider for example $f(t)=1,a=-1,b=1$ and $\phi(t)=-t^2$, to get some intuition. – 2017-02-11

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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 $e^{xM}$.