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:

or any function that grows fast enough.

and find:

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 is that I can express multiplication and addition.

Sorry should have said that. We can also solve for:

or any function that grows fast enough.

and find:

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 is that I can express multiplication and addition.