(I asked this in MSE before but there was only a general reference which did not help for my specific question)

I think I understood the concept of fractional derivatives applied to monomials to consider some general cases and also applied to exponential terms as they occur in the formal expression of the Riemann zeta as Dirichlet series and their integer or fractionally indexed derivatives.

For integer indexed derivatives of the zeta at zero I can numerically evaluate the alternating zeta (Dirichlet's eta) instead using some common version of divergent summation, say Euler-summation ($\mathfrak E$) and then formally refer to $$ {d^m \over dx^m } \eta (0) =\eta^{(m)}(0) \underset{\mathfrak E}= \sum _{k=0}^\infty (-1)^k (-\log(1+k))^m$$
and then express the m'th derivatives of the zeta at zero recursively using a binomial formula (and writing $ \zeta_m =\zeta^{(m)}(0)$ and $ \eta_m =\eta^{(m)}(0) $ and $u=\log(1/2)$ for convenience) $$ \zeta_m = - \left(\eta_m + 2 \cdot \sum_{k=0}^{m-1} \left[ \binom{m}{k}u^{m-k} \zeta_k\right]\right) $$

However, I cannot generalize this to the according fractional derivatives, in this question the half-derivative, because the finite binomial sum would become an infinite series, likely also divergent and where I also would not know, in which direction I should re-interpret the binomial sum which is symmetric in its indexes for the case of integer exponents.

For the half-derivative of the $\eta$ I get by the Euler-summation and using the setting $\sqrt{-\log(1+k)} = i\sqrt{\log(1+k)} $ the approximation $$\eta_{1/2} \sim -0.347006596200 i $$

Q1: How could I express the half-derivative $\zeta_{0.5} = \zeta^{(0.5)}(0) $ formally
Q2: and what is a meaningful value?


1 Answer 1



Indeed, the results can be checked numerically for $k\in\mathbb N^*$. So a meaningful value for $\zeta^{(1/2)}(0)$

could be something around $-\dfrac{\sqrt\pi}2$ . But, then again, $\zeta^{(0)}(0)=\zeta(0)=-\dfrac12$ is not a very good

approximation of $-0!=-1$, so, for values of k in between $0$ and $1$, the approximation may lack

relevance. Nevertheless, I am quite confident that a value somewhere in between $-\dfrac{\sqrt\pi}2$ and

$-\dfrac{\sqrt\pi}4$ is most likely. Hope this helps.

  • $\begingroup$ Very nice - at least an approach! Thank you. I'll look at it tomorrow. $\endgroup$ Jun 6, 2014 at 15:29

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