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

Question

[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.

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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?

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Q3: Or can I do something like with the integer derivatives of the zeta at $s=0$ where I express it as the Dirichlet-series having the logarithms in the numerators?

[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...

Answer 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.

Answer 2

Here is possibly an answer. I followed the path at Q3 in my above question with a final procedure of divergent summation. The ansatz is to express the problem as a sum of a function evaluated at consecutive arguments, just in the sense of the Euler-MacLaurin-summation method, which requires a power series for the function, which gives the "general term" for the series.

While we don't have a Taylor series for the general term $ \sqrt{-\ln(1+x)} $ around $x=0$ we can rewrite this a bit and get a related series: $$ \sqrt{\ln(1+x)} = \sqrt{x \ln(1+x)/x} = \sqrt x \sqrt{ {\ln(1+x) \over x}} $$ and for the second root-expression we have a nice Taylor-series

$ \displaystyle \qquad \small f(x)=\sqrt{ {\ln(1+x) \over x}} = \sum_{k=0}^{\infty} a_k x^k \\ \\ \small \qquad \qquad = 1 - 1/4 x + 13/96 x^2 - 35/384 x^3 + 6271/92160 x^4 - 2211/40960 x^5 + O(x^6) $

We get then

$ \displaystyle \qquad \small \sqrt{\ln(1+x)} = \sum_{k=0}^{\infty} a_k x^{0.5+k } $

Now I assume, that it is possible to express formally the sum of that function for k over all consecutive integers by the double sum

$ \displaystyle \qquad \small \sum_{k=0}^\infty \sqrt{ - \ln(1+k)} = \sqrt{-1} \cdot \sum_{k=0}^{\infty} \left( a_k \sum_{x=0}^\infty x^{0.5+k } \right) = i \sum_{k=0}^{\infty} a_k \zeta(-0.5-k) $

and the goal to evaluate the fractional half-derivative of $\zeta$ at zero were then, using a procedure $\small \mathfrak N$ for strongly divergent series,

$ \displaystyle \qquad \small \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \qquad \sim - 0.200824660892 \ i $

Now conjectured: $$ \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \\ \zeta(0)^{(1/2)} \sim - 0.200824660892 \ i $$

The applicability and the correctness of the above sequence is not sure/not proven. Heuristics show, that the method applied to the fractional derivatives of the $\eta$ gives the expected results but for instance for the non-alternating zeta series we might need some integral as summand like in the Ramanujan-summation-method.

So far I can only hope that someone else does crosscheck the method and the recieved value.

Answer 3

Sorry I don't have a rep > 50 so this is just a comment but have you noticed that $\zeta(0)^{(1/2)}$ is eerily close to $\frac{\\i\zeta(0)^{(2)}}{10}$ in value?