Well about a year ago I posted my paper on the bounded analytic hyper-operators. The result was straight forward and involved expressing functions using the Mellin transform (or the exponential differintegral as I expressed it). One question that arose is the following,
If we define)
to be the n'th super root
 = x)
is the function
 = \sum_{n=0}^\infty \Psi(n+1,x)\frac{u^n}{n!})
differintegrable? And if differintegrated, does it equal the super root for arbitrary
?
Well I recently thought more about this question, and using an invaluable Lemma I proved I've found that in fact, for x>1 and
we do have
|u^{-\sigma}\,du < \infty)
Therein defining the function for > 0)
 = \frac{1}{\Gamma(1-z)}(\sum_{n=0}^\infty \Psi(n+1,x)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty g(-u,x)u^{-z}\,du))
Now here's the really beautiful part, and sadly where I am stuck.
If I can show that \neq 0)
then we can invert this function across 
, to get 
the function ) = x)
.
Taking this function observe the following
 = \Psi(z,x)^{F(z,\Psi(z,x))} = \Psi(z,x)^x)
is factorizable in for x close enough to 1.
Just as well
 = F(z+1,\Psi(z,x)))
is factorizable in for x close enough to 1.
Factorizable implies we can use the natural identity theorem. This means if
then 
Well...
 = \Psi(n,x)^x)
and
 = F(n+1,\Psi(n,x)) = (^{n+1}\Psi(n,x)) = \Psi(n,x)^{^n\Psi(n,x)} = \Psi(n,x)^x)
and therefore
 = a(z))
What this means, unraveling the cryptic writing, is if)
then
} = F(z+1,y))
TETRATION!
Therefore the requirement boils down into showing the rather painful fact that
for > M)
for some M > 0. This will successfully construct a tetration function. I'm at a loss frankly on how to show this. The rest fell into place rather easily, nothing too exhaustive was required.
It is obviously true for most x>1, but not for all. There may be exceptional values. Therefore we do have a tetration solution using the differintegral given some conditions... Numerical evidence seems to support the result too but I've only tried a little bit.
If we define
is the function
differintegrable? And if differintegrated, does it equal the super root for arbitrary
Well I recently thought more about this question, and using an invaluable Lemma I proved I've found that in fact, for x>1 and
Therein defining the function for
Now here's the really beautiful part, and sadly where I am stuck.
If I can show that
Taking this function observe the following
Just as well
Factorizable implies we can use the natural identity theorem. This means if
Well...
and
and therefore
What this means, unraveling the cryptic writing, is if
TETRATION!
Therefore the requirement boils down into showing the rather painful fact that
for
It is obviously true for most x>1, but not for all. There may be exceptional values. Therefore we do have a tetration solution using the differintegral given some conditions... Numerical evidence seems to support the result too but I've only tried a little bit.