I'm a little confused about what you're doing but I understand your arguments about sexp as a continuum product.
We have A beautiful result I would like to show:
We start with the two identities that are already proven by others working on fractional calculus.
}{dt^{-s}} |_{t=0})
= \mathcal{J}f * \mathcal{J}g= \sum_{n=0}^{\infty} \frac{\Gamma(1-s)}{\Gamma(n + s+1) n!}(\mathcal{J} f)(n) (\mathcal{J} g)(-s-n))
And even more generally:
g(t)}{dt^{-s}} = \frac{d^{-s}f(t)}{dt^{-s}} * \frac{d^{-s}g(t)}{dt^{-s}})
whree the convolution is done over s, and the values at t are the same for both f and g.
Therefore: if
and 
 g(t) dt= \phi * \psi)
That means even more remarkably
 = \mathcal{Z} ((\mathcal{J}^{-1} f) \cdot (\mathcal{J}^{-1} g)) = (\mathcal{Z} \mathcal{J}^{-1} f) * (\mathcal{Z}\mathcal{J^{-1}}g))
That means = (Sf) * (Sg))
This has so much value for continuum sums. This is remarkable!
I have to properly justify this using the continuity of these operators over some hilbert space. That's the only way I can think of.
I would also like to standardize a notation that is very intuitive. If we take the continuum sum over the interval [a,b] we say:
 \, \sigma y = S f(b) - Sf (a))
This has all the linearity rules of the integral, and some own unique rules of its own.
We have A beautiful result I would like to show:
We start with the two identities that are already proven by others working on fractional calculus.
And even more generally:
whree the convolution is done over s, and the values at t are the same for both f and g.
Therefore: if
That means even more remarkably
That means
This has so much value for continuum sums. This is remarkable!
I have to properly justify this using the continuity of these operators over some hilbert space. That's the only way I can think of.
I would also like to standardize a notation that is very intuitive. If we take the continuum sum over the interval [a,b] we say:
This has all the linearity rules of the integral, and some own unique rules of its own.