Integrals of rapidly oscillating phase functions. Why is it the case that if the integrand of a given integral is rapidly oscillating, then all the contributions to the integral cancel out apart from where the function is stationary?
For example, consider an integral of the form $$I=\int_{-\infty}^{+\infty}F(x)e^{i\phi(x)}dx$$ where $\phi(x)$ is a rapidly oscillating function. As I've read, the only non-trivial contributions to this integral will be about the point where $\phi(x)$ is stationary, say at $x=x_{s}$, such that we can approximate the integral as $$I\simeq F(x_{s})e^{i\phi(x_{s})}\int_{-\infty}^{+\infty}e^{i\phi''(x_{s})(x-x_{s})^{2}/2}=\sqrt{\frac{2\pi}{i\phi''(x_{s})}}F(x_{s})e^{i\phi(x_{s})}$$ I believe this is known as the "Method of Stationary Phase".
Is the intuition that, because the integrand is rapidly oscillating, the neighbouring contributions to the integral will cancel one another out (since the integrand will be oscillating sinusoidally, if it is doing so rapidly, then neighbouring phases $\phi(x)$ will vary rapidly in value, and so neighbouring values of the integrand $F(x)e^{i\phi(x)}$ will cancel one another out on average), however, at the stationary point of the phase $\phi(x)$ (within the integration region), i.e. the point at which $\phi'(x)=0$, the phase won't be changing very much, and so contributions near this region will be "in phase", thus adding up instead of cancelling one another?!
Furthermore, if the integrand is oscillating, but not rapidly, then this cancellation will not occur, since neighbouring values will not necessarily negate one another (since the oscillations will be sinusoidal, if it is oscillating slowly, then neighbouring phases $\phi(x)$ will not necessarily differ much in value and so won't necessarily cancel out on average)?!
 A: It's better to use the more general steepest descent method, as in general there may not be such stationary purely imaginary points on the real axis. The general method is to deform the contour so that it picks up points in the complex plane where you do have such stationary phases. For each such point you rewrite the integral by performing a conformal transform such that the exponential becomes exactly $\exp(-w^2)$ this then gets multiplied by the Jacobian $\frac{dz}{dw}$, expanding this factor is series then yields an asymptotic series. So, each saddle point then yields an asymptotic series that all contribute to the integral.
The expansion parameter is then $\epsilon$ when you replace $\phi(x)$ by $\frac{\phi(x)}{\epsilon}$. So, it's wise to put in this $\epsilon$ and then consider the convergence for $\epsilon = 1$. In general, asymptotic series will start to diverge after a number of terms that decreases with $\epsilon$. The best approximation is obtained by truncating the series after the smallest term. This is called the superasymptotic approximation, the remainder term is then of the form $\exp\left(-\frac{a}{\epsilon^b}\right)$, which is beyond all orders in $\epsilon$. There exists specialized methods to go beyond the asymptotically method, which involves resumming the divergent tail.
