Nonstationary phase method for oscillatory integral

Question

I want to approximate an integral of the form $$\int_a^bf(t)e^{ig(t)}dt,$$where $f(t)$ is smooth, $g(t)$ is real-valued and smooth. The stationary phase method says that if $t_0\in [a,b]$ is such that $g'(t_0)=0$ then $$\int_a^bf(t)e^{ig(t)}dt\sim f(t_0)e^{ig(t_0)}\sqrt{\frac{2\pi}{i|g''(t_0)|}}.$$ My question is what happens if there is no such $t_0$? Suppose a stationary point of $g(t)$ lies outside $[a,b]$. How do we estimate this?

My intuition is that there being no stationary point means either there are way more oscillations, or way fewer oscillations. If there are more oscillations, it stands to reason that we expect a lot of cancellation and then we get a $O(1)$ estimate for the integral (hopefully). On the other hand if there isn't much oscillation this becomes harder.

Does anyone have any ideas about this?

Answer 1

For stationary phase, you usually consider the integral $$I(\lambda)=\int_a^b f(t) e^{i\lambda g(t)}\,dt$$ with $\lambda>0$ a large parameter. If there are no stationary points inside $[a,b]$, then you may integrate by parts using the fact that $f(t)e^{i\lambda g(t)}=\frac{f(t)}{i\lambda g'(t)} \frac{d}{dt}e^{i\lambda g(t)}$ as follows $$I(\lambda) = \frac{1}{\lambda}\frac{f(t)}{i g'(t)}e^{i\lambda g(t)}\bigg|_a^b - \frac{1}{i\lambda}\int_a^b (f(t)/g'(t))' e^{i\lambda g(t)}\,dt $$ Hence, your integral is $O(1/\lambda)$. Now, if $f(t)$ is compactly supported in $[a,b]$, then you can show that $I=O(\lambda^{-N})$ for any $N\ge 1$.