An ordinary delta comb can be defined as
$$Ш(x) = \sum_{n=-\infty}^\infty \delta(x-n)$$
The (distributional) Fourier transform of this is equal to another delta comb with frequency inverse to the original. In the case of a unit-period delta comb, the Fourier transform is equal to itself.
Consider instead a delta comb which follows the pattern of a geometric series, with a weighting term:
$$\tilde{Ш}_r(x,a) = \sum_{n=-\infty}^\infty a^{nr}\delta(x-a^n)$$
Note we can also write that as
$$\tilde{Ш}_r(x,a) = \sum_{n=-\infty}^\infty x^{r}\delta(x-a^n)$$
This appears to be a tempered distribution for any $r>1$.
If it's simpler, I'd also be interested in an answer in which the lower bound on the summation is some arbitrary integer, such as 0 or a negative number...
$$\tilde{Ш}_r(x,a) = \sum_{n=i}^\infty x^{r}\delta(x-a^n)$$
What is the Fourier transform of $ \widetilde{Ш}_r(x,a)$? Does there exist a closed-form solution?