What is the half-derivative of zeta at $s=0$ (and how to compute it)?

[Update 3:] I gave a new partial answer following the ansatz in question Q3. I leave the other parts of the question untouched, they are also partially answered in specialized other questions in MSE.

I'm trying to understand the concept of fractional derivatives and am fiddling with the examples at wikipedia. The a'th derivative of a monomial in x, where a can be fractional is accordingly $$ {d^a \over dx^a} x^m = { \Gamma(1+m) \over \Gamma (1+m-a) } x^{m-a} $$.

Q1: But what happens for some function $f(x)$ if I want to evaluate the half-derivative at zero?

Originally I'm interested in the fractional derivatives of the zeta at zero. Because I thought that the monomial-halfiterative is the most easy one to understand I tried first the power-series expression of the zeta-function $$ \zeta(x) = - {1 \over 1-x} + \sum_{k=0}^\infty w_k x^k $$ where $w_k$ are some coefficients beginning with $w_0=0.5, w_1=0.081... , w_2=-0.0031... , \cdots $
But if I want to find the (1/2)'th derivative at $x=0$ I need definitions how I should handle the fractional powers of zero.

Q2: How can I evaluate the fractional derivative of the leading fraction ${1 \over 1-x}$? Can I do better than to express the fraction by its power series and do the derivations termwise at the monomials?

[update]: concerning Q3, I've now used the alternating-zeta version and assumed, that $$ {d^{1/2} \over dx^{1/2}} {1 \over k^x}={d^{1/2} \over dx^{1/2}} \exp(x(-\log(k))) =(-\log(k))^{1/2} {1 \over k^x} = i \cdot (\log(k))^{1/2} {1 \over k^x} $$ and then I set $x=0$ and evaluate the alternating series $$ \eta(0)^{(1/2)} \underset{\mathfrak E}{=} \sum_{k=0}^\infty \left((-1)^k log(1+k)^{1/2}\cdot i \right)\sim - 0.347006596200 \cdot i $$ where $\mathfrak E $ means Euler-summation of the divergent series. However, even if that result is meaningful this does not yet help much because I've now no further idea how I could use the Euler's zeta/eta-conversion-formula here. (I have just developed the conversion scheme for the integer derivatives, but that transforms to an infinite series if this is at all generalizable to fractional indexes)

[update2]: I tried the Riemann-Liouville-formula for the half-derivative as given by @J.M. in MSE, but the result is inconclusive. First, I need to handle zeros in the denominators, and second, if I replace them by limiting expressions with $\epsilon \to 0$ the numerical integration seems to diverge either to $-\infty$ or $- i \infty$ depending on whether I approach zero from positive or from negative values. So I need some help even for this...

Solution 1:

Here is a tip: Google "super calculus" I just did it, it's the second hit. It deals with fractional calculus and it enables you for example to calculate sums of rediculous series. Not for the everyday mathematician though.