We know that an oscillatory integral in $n$ dimensions is an integral of the form \begin{equation*} I(\lambda)=\int_{\mathbb{R}^n}e^{i\lambda\phi(x)}f(x)dx \end{equation*} where $\phi\in C^{\infty}$ is the phase function and $f\in C^{\infty}_{c}$ is the amplitude.
The behaviour of $I(\lambda)$ is determined by the points $x_0$ where the gradient of the phase is zero (stationary phase), ie, when $\phi'(x)=0$. However, when $\phi'(x)\neq 0$, the decay to zero is ideal. My question is, why is this the case? What is the intuition behind this? What happens to the integral if $\phi'(x)\neq 0$ and we have non-stationary phase?
If anyone is aware of any good references or understands why, please let me know.
Thanks in advance for the help.
If $\nabla \phi \ne 0$, then defining $$ L = \frac{1}{i \lambda} \frac{\nabla \phi}{\lvert \nabla \phi \rvert^2} \nabla $$ we have $L e^{i \lambda \phi} = e^{i \lambda \phi}$. Now integration by parts gives the adjoint $$ L^\ast = \frac{1}{\lambda} \nabla \left( \frac{\nabla \phi}{\lvert \nabla \phi \rvert^2} \right). $$ Importantly, we have $(L^\ast)^N f = O(\lambda^{-N})$, so that we get arbitrary polynomial decay on $I(\lambda)$ in $\lambda$.
The decay of $I$ in the case $\nabla \phi = 0$ at one point and the Hessian of $\phi$ nondegenerate here is $O(\lambda^{-n/2})$.
The following comments represent my (possibly incorrect, certainly deficient) intuition. I would recommend caution in absorbing any of this.
We see that there is a correspondence between movement of $\phi$ and decay of $I$. Very roughly, the integral $I$ is an oscillating function integrated against a smooth function, and approaches a local averaging process of $f$ as $\lambda \to \infty$, so long as $\phi$ "moves". Since $f$ has smoothness, it is locally close to its average value, and so we get decay.
Note that $\nabla \phi \ne 0$ on $\text{supp}(f)$ and so we have $\lvert \nabla \phi \rvert \gtrsim 1$, so that $\phi$ is locally linear (in a uniform sense, by compactness of $\text{supp}(f)$). Thus we can transfer intuition from the case $\phi(\xi) = \xi$ for large $\lambda$, where this is just the fact that the Fourier transform of a Schwartz function is Schwartz.
When $\nabla \phi(0) = 0$, the Hessian of $\phi$ is nondegenerate at $0$, and $\nabla \phi \ne 0$ otherwise, we get $\lvert \nabla \phi(\xi) \rvert \gtrsim \lvert \xi \rvert$ from Taylor expansion. Then one can localize the contribution to $I$ at $\xi = 0$, getting the weaker $-n/2$ decay rate.
The reference I used to write this answer is section 4.2.2 of the first volume of Classical and Multilinear Harmonic Analysis by C. Muscalu and W. Schlag.