Lower Bound on Oscillatory Integral

Let $p,y\in\mathbb{R}^d\setminus\{0\},\beta>0$ be given and fixed and define for all $\alpha>0$, $$I(\alpha) := \int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-\alpha\beta \|x-y\|^2)f(x)\mathrm{d}x$$ where $f:\mathbb{R}^d\to[0,1]$ is some bump function (smooth, non-negative, of compact support).

I am interested in estimating (or obtaining a lower bound) on $|I(\alpha)|$ for very large values of $\alpha$. In particular, to see that $|I(\alpha)|\geq g(\alpha)$ as $\alpha\to\infty$, for some simple $g$ which vanishes at infinity (say, Gaussian).

$I(\alpha)$ is essentially the Fourier transform of a bump function and a Gaussian evaluated very far away from the origin. So I tried to use the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian, but it's still not clear to me how this helps, because I don't know quantitative estimates on the Fourier transform of a bump function.

Then I came across: this website which claims that "When controlling an oscillatory integral, bump functions and bounded phase corrections are not very important". So I replaced $f$ with another Gaussian:

If we take as a model for $f$ the function $$ f(x) = \chi_{[-1,1]}(\|x\|)\exp\left(1-\frac{1}{1-\|x\|^2}\right)$$

then we replace it with some Gaussian $\tilde{f}(x) := \exp\left(-\delta \|x\|^2\right)$ where $\delta>0$ is some parameter to be adjusted later (say, $\delta=2$).

We then obtain \begin{align}I(\alpha)&=\int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-\alpha\beta \|x-y\|^2-\delta \|x\|^2)\mathrm{d}x\\&=\exp(-\frac{\alpha\beta\delta}{\alpha\beta+\delta}\|y\|^2)\int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-(\alpha\beta+\delta) \|x-\frac{\alpha\beta}{\alpha\beta+\delta}y\|^2)\mathrm{d}x\\&=\exp(-\frac{\alpha\beta\delta}{\alpha\beta+\delta}\|y\|^2)(\frac{\pi}{\alpha\beta+\delta})^{\frac{d}{2}}\exp\left(-\frac{\alpha^2}{4(\alpha\beta+\delta)}\|p\|^2+\mathrm{i}\alpha p \cdot y\right)\,.\end{align}

However, how do you estimate the error of replacing $f$ by a Gaussian?

We can do a similar exercise replacing $f$ by a Taylor approximation to its second degree, e.g..

Most of the texts I read about estimating oscillatory integrals deal with the case that the phase is rather complicated. However here it is just the Fourier transform, whose gradient never vanishes.

Here's an example, not using bump functions admittedly, where $I(\alpha) \equiv 0$.

Taking $\beta = 1$ for simplicity, a slight modification of the Gaussian example included in the question shows that if we define $f_{z}(x) = e^{-\| x - z \|^{2}}$ for $z \in \mathbb{R}^{d}$, then \begin{align} I_{z}(\alpha) &= \int_{\mathbb{R}^{d}} \exp(i \alpha p \cdot x - \alpha \| x - y \|^{2}) f_{z}(x) \, dx \\ &= \left( \frac{\pi}{\alpha + 1} \right)^{d/2} \exp \left( -\frac{\alpha^{2}}{4(\alpha + 1)} \| p \|^{2} -\frac{\alpha}{\alpha + 1} \| y - z \|^{2} \right) e^{i \alpha p \cdot y}. \end{align}

The key property that I want to focus on here is that if $z_1$ and $z_2$ are such that $\| y - z_1 \| = \| y - z_2 \|$, then $I_{z_1} \equiv I_{z_2}$.

So if we define $f(x) = f_{z_1}(x) - f_{z_2}(x)$ for such $z_1$ and $z_2$, the resulting oscillatory integral is identically zero.

This isn't a representative example, of course, but I would suspect that bounding $I(\alpha)$ away from $0$ as $\alpha \to \infty$ would be pretty hard in general.

The form of the integral is very similar to that of the FBI (Fourier–Bros–Iagolnitzer) transform, so the behavior of $I(\alpha)$ might depend on local properties of $f$ at $y$, and in the direction of $p$.