I am looking for a nontrivial example of a singular distribution that when convolved with a Gaussian distribution has a pdf of a 'simple' form.
I let 'simple' be something that you interpret yourself.
Singular distributions are an import class of distributions that are often 'swept under the carpet.' I would like to see a nice illustrative example of how to work with such distributions.
One way to do this is to give a characteristic function $\phi(t)$ that when multiplied by $e^{-t^2/2}$ has a simple Fourier inverse.
However, I don't have a good choice of the characteristic function $\phi(t)$ that would lead to a meaningful result.
For example, for the Cantor distribution, the characteristic function is given by \begin{align} \phi(t)=e^\frac{it}{2} \prod_{i=1}^\infty \cos \left( \frac{t}{3^k} \right). \end{align} However, it and is not easy to work with this characteristic function \begin{align} \phi(t) e^{-\frac{t^2}{2}}. \end{align} In particular, it is difficult to fuind its Fourier inverse.
Edit: By singular distributions I mean: A singular distribution is a probability distribution concentrated on a set of Lebesgue measure zero, where the probability of each point in that set is zero.
Edit 2: Another approach we can take is to look at the convolution directly. That is look at the $U=X+V$ where $V$ is has a singular distribution and $X$ is Gaussian, in this case, the pdf of $U$ is given by \begin{align} f_U(u)=E\left[ \frac{1}{\sqrt{2 \pi}} e^{-\frac{(u-V)^2}{2}} \right]. \end{align}
I was wondering if we can come up with a sequence of random variables $V_n$ that converges in distribution to some $V$ with a singular distribution, for which we can compute the limit \begin{align} \lim_{ n \to \infty}E\left[ \frac{1}{\sqrt{2 \pi}} e^{-\frac{(u-V_n)^2}{2}} \right]. \end{align}