Let $f^{<n>}(x)$ denote the n-fold convolution of $f$ with itself (also called convolution power). Is it possible to write the following series $\sum_{n=0}^{\infty}\frac{i^n}{n!}f^{<n>}(x)$ in closed form?
Progress so far:
So far I've seen several articles in which methods to evaluate $f^{<n>}(x)$ efficiently are presented. However, I am interested in the evaluation of this series and not of $f^{<n>}$ for particular values.
In the convolution power wikipedia article a convolutional exponential of the form
$$\exp^*(x) = \delta(0) + \sum_{n=1}^{\infty}\frac{i^n}{n!}f^{<n>}(x)$$
can be defined, but no closed form is given. I also tried looking around for this so called convolutional exponential, but I was unable to find anything.
I have the feeling that the closed form of this series is somehow related to the exponential $e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}$ but I have not been able to find a solution so far.
An additional solution is possible by using the Fourier transform ($\mathcal{F}$). By using the Fourier transform, one can arrive to a solution to the series of the form $\mathcal{F}^{-1}(e^{i\mathcal{F}f})$ by using the exponential series property. However, I would like to avoid going to the Fourier domain and back.
Thank you all in advance! :)