Hey everyone. I thought I'd post this theorem, perhaps someone has some uses for it.
Theorem:
A.) If
is holomorphic for  > -a-\epsilon>-1)
for some 
and  < C e^{\alpha |\Im(z)| + \rho|\Re(z)|})
for 
and 
.
B.) for some
and 
we have ) = F(1+n) - 1)
Then, for > 1-a)
we have ) = F(1+z) - 1)
Proof:
Well this is rather easy:
F(1-\xi)x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(1+n)\frac{(-x)^n}{n!})
Which follows by cauchy's residue formula and the bounds of F (the gamma function along with x small enough pulls the arc next to our line integral to zero at infinity). For those who don't see,
 = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+z)} + \int_1^\infty e^{-t}t^{z-1}\,dt)
where the right term is entire in z and only contribute asymptotics, observe stirlings asymptotic formula
 \sim \sqrt{2\pi} z^{z-1/2}e^{-z})
Therefore this holds.
Now observe that by a similar argument:
F(\log_b(1-\xi))x^{-\xi}\,d\xi = \sum_{n=0}^\infty F(\log_b(1+n))\frac{(-x)^n}{n!}= \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x))
And of course, by another similar argument:
(F(1-\xi)-1)x^{-\xi}\,d\xi = \sum_{n=0}^\infty (F(1+n)-1)\frac{(-x)^n}{n!} = g(x))
Therefore since the kernel of this integral transform is zero (its a modified fourier transform). On the line
we have  -1 = F(\log_b(1-\xi)))
. Therefore since both functions are analytic we get the desired. 
I'm wondering, does anyone see any uses for this?
I know with some formal manipulation we can say that, if = 1+n)
and  = \log_b(n+1))
and 
is holo and is invertible which satisfies the bounds above. Then } = G(z+1))
Theorem:
A.) If
B.) for some
Then, for
Proof:
Well this is rather easy:
Which follows by cauchy's residue formula and the bounds of F (the gamma function along with x small enough pulls the arc next to our line integral to zero at infinity). For those who don't see,
where the right term is entire in z and only contribute asymptotics, observe stirlings asymptotic formula
Therefore this holds.
Now observe that by a similar argument:
And of course, by another similar argument:
Therefore since the kernel of this integral transform is zero (its a modified fourier transform). On the line
I'm wondering, does anyone see any uses for this?
I know with some formal manipulation we can say that, if