No problem. I know it doesn't converge for e^s because its exponential derivative doesn't approach zero fast enough as it goes to negative infinity. It stays constant. So what we say is e^s is non-integral analytic.
Sorry should have said that. We can also solve for:
\( \mathcal{E} e^{e^s} = g(s) \)
or any function \( f \) that grows fast enough.
and find:
\( e^{e^s} = \int_{-\infty}^{\infty} g(t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
My guess is that it is integral analytic; (i.e; converges for some s in this expression); but I haven't proved that.
Basically this manipulation works on functions that grow a certain rate. I'm going to try and prove what that rate is regarding that I have to take the integral transformations into consideration.
I actually prefer this only being on this website until I write the paper. I tend to get over anxious and post things that I forget to check. Does this make more sense at what I was trying to get at? I'm more interested in the fact that I have an integral expression for the inverse of the iterated Riemann-Liouville differintegral.
I just didn't want to say that out-loud
The point of having expressions for \( e^s, \sin(s),... \) is that I can express multiplication and addition.
Sorry should have said that. We can also solve for:
\( \mathcal{E} e^{e^s} = g(s) \)
or any function \( f \) that grows fast enough.
and find:
\( e^{e^s} = \int_{-\infty}^{\infty} g(t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
My guess is that it is integral analytic; (i.e; converges for some s in this expression); but I haven't proved that.
Basically this manipulation works on functions that grow a certain rate. I'm going to try and prove what that rate is regarding that I have to take the integral transformations into consideration.
I actually prefer this only being on this website until I write the paper. I tend to get over anxious and post things that I forget to check. Does this make more sense at what I was trying to get at? I'm more interested in the fact that I have an integral expression for the inverse of the iterated Riemann-Liouville differintegral.
I just didn't want to say that out-loud
The point of having expressions for \( e^s, \sin(s),... \) is that I can express multiplication and addition.