Hey, so I read your post trying to find the half-derivative of zeta. It's very important to remember in fractional calculus that the differintegral always depends on a base point.

The standard form of any differintegral is written as,

\(

I_a^{-z} f(x) = \frac{1}{\Gamma(z)}\int_a^x f(t)(x-t)^{z-1}\,dt

\)

Now each, \( I_a \) provides different solutions for each value \( a \). Particularly, the Gamma function derivative of monomials works as,

\(

I_a^{-z} (x-a)^n = \frac{\Gamma(n+1)}{\Gamma(n+z + 1)}(x-a)^{n+z}\\

\)

But, for \( b \neq a \),

\(

I_a^{-z} (x-b)^n \neq \frac{\Gamma(n+1)}{\Gamma(n+z + 1)}(x-b)^{n+z}\\

\)

Additionally, we get a lot of problems when considering transcendental functions for \( a \) finite. Which is epitomized by,

\(

I_a^{-z} e^{x} \neq e^x\\

\)

To get the above identity, we use Riemann-Liouville's differintegral, which is when we set \( a = -\infty \) (or more accurately, we set \( a = \infty \) the point on the Riemann Sphere). This differintegral will not converge on monomials without A LOT of massaging and discussion of distribution theory (something I think I wasted way too much time on). But, if we write,

\(

\frac{d^z}{dw^z} \vartheta(w) = \frac{1}{\Gamma(-z)}( \sum_{n=0}^\infty \vartheta^{(n)}(w)\frac{(-1)^n}{n!(n-z)} + \int_1^\infty \vartheta(w-x)x^{-z-1}\,dx)\\

\)

This is probably the best analytic continuation of the Riemann Liouville I came up with. And if you note,

\(

\frac{d^z}{dw^z} e^{\lambda w} = \lambda^z e^{\lambda w}\\

\)

Which means it's the differintegral which preserves the exponential. Even if \( \lambda \) is complex, we can choose different branches of \( \lambda^z \) which corresponds to different branches of the Riemann-Liouville integral (which is sort of like integrating in different directions and rotating the direction collecting powers of \( e^{2\pi i k z} \)).

Now, as to the zeta function, you have two choices. Expand the zeta function into a Taylor series, and differintegrate term wise. Or consider the zeta function as a transcendental function (which is a much better choice in my opinion).

The trouble is, the zeta function has decay as \( \Re(s) \to \infty \) rather than \( \Re(s) \to -\infty \). This isn't much of a problem really, as we just rotate the integration and add a exponential factor.

\(

\frac{d^{-z}}{dw^{-z}}\zeta(w) = \frac{e^{(2k+1)\pi i z}}{\Gamma(z)}\int_0^\infty \zeta(w+x)x^{z-1}\,dx\\

\)

For some \( k \in \mathbb{Z} \). This will almost equal the expected result if it weren't for the pesky first term of the zeta function, the constant \( 1 \). Let's ignore it briefly and write,

\(

\frac{d^{z}}{dw^{z}}\zeta(w) -1= \sum_{n=2}^\infty \frac{e^{(2k+1)\pi i z}\log(n)^z}{n^{w}}\\

\)

To uncover the value at \( 1 \) we have to use the above analytic continuation, this is,

\(

\frac{d^z}{dw^z} 1 = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty 1^{(n)} \frac{(-1)^n}{n!(n-z)} + \int_1^\infty x^{-z-1}\,dx)\\

= \frac{1}{\Gamma(-z)}( \frac{-1}{z} + \frac{1}{z}) = 0\,\,\text{for}\,\,z\neq 0\\

\)

Which is where we have to use pesky distribution theory. If \( \delta(0) = 1 \) and \( \delta(z) = 0 \) for all other \( z \in \mathbb{C} \), then we can conclude,

\(

\frac{d^z}{dw^z} \zeta(w) = \delta + \sum_{n=2}^\infty \frac{e^{(2k+1)\pi i z}\log(n)^z}{n^{w}}\\

\)

Which tells us that \( \frac{d^z}{dw^z} \zeta(w) \) is holomorphic for \( \Re(w) > 1 \) and \( \Re(z) > 0 \) (which avoids all that pesky \( \delta \) nonsense).

Now, you're safest bet to uncover the fractional derivative at zero using the Riemann Liouville differintegral is to analytically continue this Dirichlet series (which isn't hard at all considering the complexity of the dirichlet coefficients). This will give you a function,

\(

\frac{d^z}{dw^z}\zeta(w) \,\,\text{is holomorphic for}\,\,w \neq 1\,\Re(z) > 0\\

\)

Remembering there is a choice of which branch you are using, and there is a different differintegral for each \( k \in \mathbb{Z} \).

Now the other way, is fairly simpler, but I absolutely hate it for power series. Largely because it will not behave well on the prototypical power series \( e^w \). This is much plainer and it is just differintegrating termwise (which is boring).

Anywho, that's my take on the fractional derivative of \( \zeta \).

Regards, James

EDIT: It's important to remember that the Riemann-Liouville integral is largely meant for when our function has decay to zero in some direction, as it goes to infinity. If the function decays to a constant, like how \( \zeta \to 1 \) then the differintegration can't be holomorphic at \( z=0 \), we will have a discontinuity here. This also happens if we have \( \vartheta(w) = \mathcal{O}(w^k) \) for monomials, which would mean the differintegration is holomorphic for \( \Re(z) > k \). I thought it may be helpful to write out the formula for the differintegrations of monomials using Riemann-Liouville.

\(

\Gamma(-z)\frac{d^z}{dw^z}w^k = \sum_{n=0}^\infty \frac{k!}{(k-n)!} w^{k-n} \frac{(-1)^n}{n!(n-z)} + \int_1^\infty (w-x)^{k}x^{-z-1}\,dx\\

= \sum_{n=0}^k \binom{k}{n} (-1)^n \frac{w^{k-n}}{n-z} + \sum_{n=0}^k \binom{k}{n} (-1)^{n} \frac{w^{k-n}}{z-n}\\

\)

And again we have to use a type of distribution where \( \frac{d^z}{dw^z}w^k = \frac{k!}{(k-n)!}w^{k-n} \) for \( z = n\,\,0 \le n \le k \) and it equals zero everywhere else. Which is a perfectly natural interpolation of \( \frac{k!}{(n-k)!}w^{k-n} \) but looks a lot uglier than the typical formula for a monomial. But, when you enter the distribution field of fractional calculus this formula can be very helpful, especially when correcting errors, and interpreting things like \( w^k + e^w \)--it essentially gives us justification for throwing away \( w^k \), especially when we let \( \Re(z) > k \). It preserves the semi-group property and the relationship \( \frac{d}{dw}\frac{d^z}{dw^z} = \frac{d^{z+1}}{dw^{z+1}} \); however it really only does this trivially. This is rather unconventional--and is because we are analyticially continuing the operator as a semi-group and sort of ignoring the differential relationship--I haven't investigated it much but it's nice. So it disagrees, somewhat, with what we expect a differintegral to do--especially for negative real values (when we fractionally iterate the integral); because, well, it doesn't integrate at all--it just nulls it.

The alternative way to do this, is to no longer insist that \( \frac{d^z}{dw^z} \) interpolates differentiation necessarily, but is instead just a semigroup. And part of the kernel of this semi-group is the space of all polynomials,

\(

\frac{d^z}{dw^z} : \mathbb{C}[w] \to \{0\}\\

\)

I like this interpretation because it essentially restricts the Riemann-Liouville differintegral to non-polynomials; and only works with functions expressible by power-series with an infinite number of terms. Both interpretations are equally valid though; largely because in the distribution sense we are only worried about a measure zero set of values (so we can throw it away almost everywhere); they're equivalent almost everywhere.

Non-integer monomials behave nice though, thought I'd add this too.

\(

\Gamma(-z)\frac{d^z}{dw^z}\sqrt{w} = \sum_{n=0}^\infty \frac{\Gamma(\frac{1}{2}+1)}{\Gamma(\frac{1}{2}+1-n)} w^{\frac{1}{2}-n} \frac{(-1)^n}{n!(n-z)} + \int_1^\infty \sqrt{w-x}x^{-z-1}\,dx\\

= \sum_{n=0}^\infty \binom{1/2}{n} w^{\frac{1}{2}-n}\frac{(-1)^n}{n-z} + \sum_{n=0}^\infty \binom{1/2}{n} w^{\frac{1}{2}-n}\frac{(-1)^n}{z-n}\\

=0\\

\)

Which means even more functions are in the kernel. This produces some trouble when talking about the logarithm, but only if we think about it dramatically. Remembering these are semigroups, we still have,

\(

\Gamma(-z)\frac{d^z}{dw^z} \log(w) = \sum_{n=0}^\infty \log^{(n)}(w) \frac{(-1)^n}{n!(n-z)} + \int_1^\infty \log(w)w^{-z-1}\,dw\\

\)

Which is holomorphic for \( \Re(z) > 0 \) and is a semi-group. However we lose right side associativity, \( \frac{d^z}{dw^z} \frac{d}{dw} = \frac{d^{z+1}}{dw^{z+1}} \). This isn't a problem, we just have to remember to be very careful once we analytically continue the Riemann-Liouville integral. The correct manner of interpreting this, is that MOST OF THE TIME \( \frac{d^z}{dw^z} = \frac{d^n}{dw^n} \) when \( z=n \) but NOT ALWAYS. Especially once the differintegral is extended to a larger space of functions. However, the semi-group property is still conserved. And the reason it works for the logarithm, is because we are NOT performing/allowing the differintegral identity \( \frac{d^{z-1}}{dw^{z-1}} \frac{1}{w} = \frac{d^z}{dw^z}\log(w) \) (recall that \( \frac{d^{-1}}{dw^{-1}} \) is integration with initial point at infinity, which is infinite for \( \frac{1}{w} \), this is meaningless no matter how we slice it).

So try to remember that \( \frac{d^z}{dw^z} \) is a semi-group acting on a space of integrable functions at infinity (with certain decay conditions)--and in this space interpolates the derivative. When we extend the operator (analytically continue the operator), it is still a semi-group but it no longer necessarily interpolates the derivative. For all of our purposes this is great news. We don't really care that it interpolates the derivative as much as we care that the semi-group property holds.

This whole edit (which I must've done ten or twelve times so far, lol) is an argument that The Riemann Liouville integral can be analytically continued and has a very robust kernel. Also, it's meant as an argument, that this function is infinitely more important/effective for transcendental/infinite power series functions. As a nice end to end, not all algebraic functions suffer this problem.

\(

\Gamma(-z) \frac{d^z}{dw^z} \frac{1}{1-w} = \int_0^\infty \frac{x^{-z-1}}{1-w+x}\,dx\\

\)

Equals a transcendental function and is not \( 0 \) everywhere. It equals something with the \( \sin \) function I can't remember now, but,

\(

\int_0^\infty \frac{x^{-z-1}}{1+x}\,dx =\frac{-\pi}{\sin\pi z}\\

\)

And we can use a power series expansion to derive the general formula.

\(

\frac{d^z}{dw^z} \frac{1}{1-w} = \sum_{n=0}^\infty \lim_{q\to n} (\frac{-\pi}{\Gamma(-z-q)\sin\pi (z+q)}) \frac{w^n}{n!}\\

\)

This has a closed form but I'm too lazy to go through it right now, lol. But it does equal \( \frac{n!}{(1-w)^{n+1}} \) for \( z = n \).

PS:

I'm sorry to hear you're having problems doing math like you used to. That must be very hard, and I can't even imagine. I appreciate you reading what I've written. I'd say just keep at the math, because even if it's hard, it's the type of thing that can only help your brain

.

Ironically enough, I'm rather horrible with matrices and linear algebra (at best, I have a working rudimentary knowledge), analysis and integrals and the sort we're always my strong suit. It's always lovely to see matrix people coming to similar results as analysis people. I like to think of the history of Schrodinger's mathematical formalism and Heisenberg's mathematical formalism; one with complex analysis, the other with matrices; and the eventual equivalence proven by Neumann. When I read what you wrote in the initial post I found it strikingly similar to the uniqueness condition imposed by Ramanujan's master theorem; especially when you pulled out the incomplete Gamma function. It took me a very long time to make sense of the equations, and the exact language needed. But if,

\(

f(z) = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty f(n) \frac{(-1)^n}{n!(n-z)} + h(s))\\

\)

For a holomorphic function \( h \). There is only ONE function \( h \) which makes our function \( f \) belong to the space we want. Which, if we change \( h \) to \( \tilde{h} \) a holomorphic function, to get a function \( \tilde{f} \); we still get \( \tilde{f}(n) = f(n) \), but \( \tilde{f} \neq f \). This implies that \( \tilde{f} \) is not in our nice space. And that there is only one function \( h \) which gives us our uniqueness condition. And this function happens to be the incomplete Mellin transform/(incomplete Gamma function). And then, the added benefit is that in this space it must preserve the semi-group property! And the semi-group property of the differintegral is easily translatable into functional equations!