Asymptotic answer of Fourier transform radially symmetric function with a ring of minimum I am looking for the two-dimensional Fourier transform of the following function
$
f(\vec{r}) = \exp\left[-\left(r-r_0\right)^2\right/ 2]\quad r\in\left(0,\infty\right)
$
in the limit of large $r_0$. The Fourier transform is defined as
$
\tilde{f}\left(k\right) = \int_0^{\infty}rdr\int_0^{2\pi}d\phi\exp\left[\dot{\iota}\vec{r}\cdot\vec{k}\right]f\left(r\right)
$
This can be simplified to be
$
\tilde{f}\left(k\right) = 2\pi\int_0^{\infty}rdrJ_0\left(kr\right)f\left(r\right),
$
where $J_{0}\left(x\right)$ is the zeroth order Bessel function. I am looking for the asymptotic expression for the the above integral in the limit of $r_0$ being large. Let us do a change of variable $t\equiv r-r_{0}$ , then
$
\tilde{f}\left(k\right) = 2\pi\int_{-r_{0}}^{\infty}tdtJ_{0}\left(k\left(t+r_{0}\right)\right)e^{-\frac{t^{2}}{2}}+2\pi\int_{-r_{0}}^{\infty}r_{0}dtJ_{0}\left(k\left(t+r_{0}\right)\right)e^{-\frac{t^{2}}{2}}
$
Is there any way to evaluate find the Fourier transform/evaluate the integral in the limit of large $r_0$?
 A: We want to approximate
$$\tag{1}
I(a)=2 \pi\int\limits_{-a}^\infty dt \ e^{-t^2/2}(a+t)J_0(k(a+t))
$$
For large $a$ and fixed $k$. This is the integral in the last equation of the question, with $r_0$ replaced by $a$. For large $z$, we may approximate $J_0(z)$ with $\sqrt{\frac{2}{\pi z}}\cos(z-\pi/4)$. As $a\to\infty$, the integrand only contributes significantly near $t=0$, so we expand the $\sqrt{t-a}$ around $t=0$. We then extend the integration range by replacing $-a$ with $-\infty$ in the lower limit. Let $A(a)$ be our approximation, then
$$\tag{2}
A(a)=2\pi \int\limits_{-\infty}^\infty dt \ e^{-t^2/2} \cos(k(a+t)-\pi/4) \left[ \sqrt{\frac{2a}{\pi k}}+\frac{t}{\sqrt{2 \pi a k}}\right]
$$
The resulting integrals yield
$$\tag{3}
A(a)=\frac{2\pi}{\sqrt{ak}} e^{-k^2/2}\left[2a \sin(ak+\pi/4)+k \cos(ak+\pi/4) \right]
$$
Here is a plot of the ratio $A/I$, where $I$ was evaluated numerically from (1) and $k=2.1$.

There are some points where the approximation seems bad: this is because the zeros of $A$ and $I$ are not exactly coincident, and why we do not have asymptotic equality between $A$ and $I$.
PS: the type of integral you have for $\tilde{f}$ is a Hankel transform. Unfortunately, searching for Hankel transform of shifted Gaussian did not yield anything fruitful.
