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.
 A: 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 $p$.
