Sum of all derivatives of Dirac Delta function

Question

I wonder whether this expression has any meaning?

$$\pi \sum _{n=0}^{\infty } i^n \delta ^{(n)}(x)$$

I encountered it while trying to make it into a Maclaurin series but unfortunately it has no factorial in the denominator.

Answer 1

$$T_k = \sum_{n=0}^k i^n \delta^{(n)}$$ Is a distribution but $\lim_{k \to \infty} T_k$ doesn't converge in the sense of distributions,

It converges only in the sense of analytic functionals acting on set of analytic functions satisfying $$\lim_{k \to \infty} (1+\epsilon)^k f^{(k)}(0)=0$$

This means in particular the (bilateral) Laplace transform of $T$ isn't well-defined away from $|s|< 1$.

Most of the interesting analytic functionals appear from the inverse Laplace transform.

In contrary to distributions there is no easy way to check if two sequences of distributions converge to the same analytic functional, for example $\delta(.+1)= \sum_{n=0}^\infty \frac{\delta^{(n)}}{n!}$.