Hi everyone. I've been doing some exhaustive research into the Weyl differintegral and I've found a very beautiful approach to solving hyperoperators.
I'll be very frank. so I'll just state the following results:
If |_{w=0} = \phi(z) = \frac{1}{\Gamma(z)}\int_0^\infty f(-w)w^{z-1}\,dw)
Is the Weyl differintegral at zero, then I've shown some equivalencies. If
is kosher then it is defined for at least the left half plane ending at the line
and as well
for all
and
for
and for some 
Then:
 = \sum_{n=0}^\infty \phi(-n-z)\frac{w^n}{n!})
As Well we obtain a new expression for
which is what I'm going for:
\Gamma(z) = \sum_{n=0}^\infty \phi(-n)\frac{(-1)^n}{n!(z+n)} + \int_1^\infty f(-w)w^{z-1}\,dw)
I am currently writing a paper on this. The results are solid. I've checked them very well. I can explain how I got to them, but they take a bit to write up.
So now we say, by the following.
Define the function:
}{x[z]y} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\vartheta_x(y,-w)w^{-z-1}\,dw)
where
is an entire function of order zero defined by  = \sum_{n=0}^\infty \frac{w^n}{(x[n]y)n!})
Let us assume there exists an analytic continuation of natural hyperoperators (natural z) for natural values of x (in the first argument), and complex arguments in the second argument (in y). And If
and
both are kosher
then 1/x[z+1]y is kosher
and
} {x[z+1](y+1)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n+1]y+1)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty \frac{(-w)^k}{(x[k+1](y+1))k!})w^{-z-1}\,dw)
But!!! Since}{x [z] (x[z+1]y)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n](x[n+1]y)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty\frac{(-w)^k}{(x[k](x[k+1]y))k!} w^{-z-1}\,dw)
We see we are only talking about the naturals and the results are EQUIVALENT!!!!!!!
Okay, I know. This is a big if. Essentially you need to prove that the following integral on the right converges for
, that
for
when
and that
for some
. Where
is the following function
\omega(z,x,y) = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\sum_{k=0}^\infty \frac{(-w)^k}{(x[k]y)k!}w^{-z-1}\,dw)
If this happens and
is holomorphic in
for all natural
. THEN WE HAVE AN ANALYTIC CONTINUATION OF HYPER OPERATORS!!!!!!!!!!!!!
QUESTIONS COMMENTS!
I have the evaluation that:
I'll be very frank. so I'll just state the following results:
If
Is the Weyl differintegral at zero, then I've shown some equivalencies. If
Then:
As Well we obtain a new expression for
I am currently writing a paper on this. The results are solid. I've checked them very well. I can explain how I got to them, but they take a bit to write up.
So now we say, by the following.
Define the function:
where
Let us assume there exists an analytic continuation of natural hyperoperators (natural z) for natural values of x (in the first argument), and complex arguments in the second argument (in y). And If
then 1/x[z+1]y is kosher
and
But!!! Since
We see we are only talking about the naturals and the results are EQUIVALENT!!!!!!!
Okay, I know. This is a big if. Essentially you need to prove that the following integral on the right converges for
If this happens and
QUESTIONS COMMENTS!
I have the evaluation that: