Is there no analytical Fourier transform of the Laplacian of a 2D Gaussian? If not, is there at least a good analytical approximation?

Question

Convolution of an image with Laplacian of Gaussian kernel can enhance features of a certain size while removing high spatial frequencies and also very slowly varying "background".

If we take the Fourier transform of a Laplacian of a Gaussian in two dimensions, we get a positive, circular "donut" that goes smoothly to zero approaching zero and infinity radius.

Instead of direct convolution of the signal with a kernel, we can multiply both of their Fourier transforms then inverse transform the product.

I wanted to try several other parameterized "donut-like" filters in Fourier space to compare.

In Wolfram Alpha the functional form of the Laplacian of a 2D Gaussian is easily found, but its Fourier is not. I get only

no result found in terms of standard functions or distributions

1. Is it correct that there is none?
2. Are there good approximations or rapidly converging series for this, or do I have to generate this convolution kernel numerically?

note: there is always a difference of Gaussians kernel which is obviously amenable to Fourier transform.

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Answer 1

Maybe I'm missing something, but in arbitrary dimension we should have $$ f(\mathbf{r}) = \int e^{-i\mathbf{k} \cdot \mathbf{r}} \tilde{f}(\mathbf{k}) \, d^n \mathbf{k}, $$ where $\tilde{f}(\mathbf{k}) = \mathcal{F}[f(\mathbf{r})]$, by the properties of the inverse Fourier transform. This then implies that $$ \nabla^2 f(\mathbf{r}) =\int \nabla^2 (e^{-i\mathbf{k} \cdot \mathbf{r}}) \tilde{f}(\mathbf{k}) \, d^n \mathbf{k} = -\int e^{-i\mathbf{k} \cdot \mathbf{r}} \mathbf{k}^2 \tilde{f}(\mathbf{k}) \, d^n \mathbf{k}. $$ Applying the Fourier Transform to both sides, we then obtain \begin{align*} \mathcal{F} [\nabla^2 f(\mathbf{r})] &= \int e^{i \mathbf{k} \cdot \mathbf{r}} [\nabla^2 f(\mathbf{r})] \, d^n \mathbf{r}\\ &= - \iint e^{i (\mathbf{k} - \mathbf{k}')\cdot \mathbf{r}} \mathbf{k'}^2 \tilde{f}(\mathbf{k}') \, d^n \mathbf{k}' \, d^n \mathbf{r} \\ &= - \int \delta(\mathbf{k} - \mathbf{k'}) \mathbf{k'}^2 \tilde{f}(\mathbf{k'}) \, d^n \mathbf{k'} \\ &= - \mathbf{k}^2 \tilde{f}(\mathbf{k}) \end{align*} So the Fourier transform of the Laplacian of a function is simply $-\mathbf{k}^2$ times the Fourier transform of the function itself. This explains the "donut shape" of the transformed Gaussian you mentioned, since it automatically goes to zero when $\mathbf{k} = 0$.

Why Wolfram Alpha couldn't figure this out, I cannot say.