[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?

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...

  • $\begingroup$ Just found another very nice (and much compact) treatize at mathpages.com/home/kmath616/kmath616.htm It focuses nicely at the exponential function... but still I'm not farther than in my updated Q3... $\endgroup$ Commented Apr 29, 2013 at 12:19
  • $\begingroup$ Here is how you derive this formula. $\endgroup$ Commented May 15, 2013 at 0:56
  • $\begingroup$ I'll link to a previous thread of yours, where I derived expressions that will pop up in the Maclaurin series for $\zeta(s)$. Semidifferentiate this series, replace $s$ with $0$, and you now have a series to start with. I might write something up late if I find the time. $\endgroup$ Commented Dec 12, 2015 at 8:14
  • $\begingroup$ @j.m.: yes, that would be very nice. I had no problem do semi-differentiate the terms of the power series, only the leadign term (containing seemingly a fractional power of zero) was the remaining problem when I tried the ansatz with the power series $\endgroup$ Commented Dec 12, 2015 at 9:59
  • $\begingroup$ A very silly question, but why isn't (-1)^0,5 not -i? $\endgroup$
    – Gerben
    Commented Nov 18, 2019 at 3:48

4 Answers 4


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.

  • $\begingroup$ Hmm, thanks! But I've plenty "sponsored links" at the top of my search-results (from google.de), and many others seem to provide only pdf's. Would you mind to provide an exact link? $\endgroup$ Commented Apr 28, 2013 at 23:49
  • $\begingroup$ fractional-calculus.com $\endgroup$
    – imranfat
    Commented Apr 28, 2013 at 23:54
  • $\begingroup$ Thank you again. Very nice, indeed! It is much general and I did not find an easy answer for my question so I delay the reading to tomorrow. Meanwhile, I think there should be some relatively simple solution - I were also happy to have some numerical approximation beforehand... $\endgroup$ Commented Apr 29, 2013 at 11:20


As is usually the case with Fractional Derivatives, there is also not "the Fractional Derivatives" for the Riemann Zeta Function $\zeta\left( \cdot \right)$. They differ depending on the method of derivation used and the type of Differential Operator. I'll use $\operatorname{D_{x}^{\alpha}}$ as a differential operator given by $\operatorname{D_{x}^{\alpha}} = \frac{\operatorname{d}^{\alpha}}{\operatorname{d}x^{\alpha}}$.

My Solutions And Their Derivations

Calculation Via Series Expansion

A series expansion appropriate for this question is the Laurent Series around $x = 1$ given by $$ \begin{align*} \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( 1 - x \right)^{k} \right] \tag{0.1}\\ \end{align*} $$ where $\gamma_{k}$ are the Stieltjes Constants and $\gamma_{0} = \gamma$ is the Euler-Mascheroni Constant. Rewrtiging this a bit gives us: $$ \begin{align*} \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( 1 - x \right)^{k} \right]\\ \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( --\left( 1 - x \right) \right)^{k} \right]\\ \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -\left( -1 + x \right) \right)^{k} \right]\\ \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -\left( x - 1 \right) \right)^{k} \right]\\ \zeta\left( x \right) &= \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( x - 1 \right)^{k} \right] \tag{0.2}\\ \end{align*} $$

With this we'll get: $$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{x - 1} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( x - 1 \right)^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{x - 1} \right] + \operatorname{D_{x}^{\alpha}}\left[ \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( x - 1 \right)^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{x - 1} \right] + \sum\limits_{k = 0}^{\infty}\left[ \operatorname{D_{x}^{\alpha}}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( x - 1 \right)^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{x - 1} \right] + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{-1} \right] + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{k} \right] \right] \tag{0.3}\\ \end{align*} $$

Now we only have the task of determining $\operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{k} \right]$. But we can make that even more general ($\operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right]$) without making it that much harder.

"Euler Method" For $\operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{\pm c} \right]$ ($c \geq 0$)

Leonhard Euler came up with a nice method for some Fractional Derivatives (see Euler's Approach). He has sorted the derivatives of the function by order and is trying to find a pattern in it. He then tried to transfer this pattern from integer orders to fractional ones, which I always affectionately call the "Euler method": $$ \begin{align*} \operatorname{D_{x}^{0}}\left[ \left( a \cdot x + b \right)^{c} \right] &= \left( a \cdot x + b \right)^{c}\\ \operatorname{D_{x}^{1}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a \cdot c \cdot \left( a \cdot x + b \right)^{c - 1}\\ \operatorname{D_{x}^{2}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{2} \cdot c \cdot \left( c - 1 \right) \cdot \left( a \cdot x + b \right)^{c - 2}\\ \operatorname{D_{x}^{3}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{3} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( a \cdot x + b \right)^{c - 3}\\ \operatorname{D_{x}^{4}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{4} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( c - 3 \right) \cdot \left( a \cdot x + b \right)^{c - 4}\\ \operatorname{D_{x}^{5}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{5} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdot \left( c - 3 \right) \cdot \left( c - 4 \right) \cdot \left( a \cdot x + b \right)^{c - 5}\\ &\cdots\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{n} \cdot c \cdot \left( c - 1 \right) \cdot \left( c - 2 \right) \cdots \left( c - n + 1 \right) \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{n} \cdot \frac{c!}{\left( c - n \right)!} \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{\alpha} \cdot \frac{\Gamma\left( c + 1 \right)}{\Gamma\left( c - \alpha + 1 \right)} \cdot \left( a \cdot x + b \right)^{c - \alpha},\, &\text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $$

$$\fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{\alpha} \cdot \frac{\Gamma\left( c + 1 \right)}{\Gamma\left( c - \alpha + 1 \right)} \cdot \left( a \cdot x + b \right)^{c - \alpha},\, \text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $} \tag{0.4}$$ wich can be simplefied via using the Falling Factorial $\left( x \right)_{y}$.


$$ \begin{align*} \operatorname{D_{x}^{0}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( a \cdot x + b \right)^{-c}\\ \operatorname{D_{x}^{1}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a \cdot c \cdot \left( a \cdot x + b \right)^{-c - 1}\\ \operatorname{D_{x}^{2}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{2} \cdot c \cdot \left( c + 1 \right) \cdot \left( a \cdot x + b \right)^{-c - 2}\\ \operatorname{D_{x}^{3}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a^{3} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( a \cdot x + b \right)^{-c - 3}\\ \operatorname{D_{x}^{4}}\left[ \left( a \cdot x + b \right)^{c} \right] &= a^{4} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( c - 3 \right) \cdot \left( a \cdot x + b \right)^{-c - 4}\\ \operatorname{D_{x}^{5}}\left[ \left( a \cdot x + b \right)^{c} \right] &= -a^{5} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdot \left( c + 3 \right) \cdot \left( c + 4 \right) \cdot \left( a \cdot x + b \right)^{-c - 5}\\ &\cdots\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{c} \right] &= \left( -a \right)^{n} \cdot c \cdot \left( c + 1 \right) \cdot \left( c + 2 \right) \cdots \left( c + n - 1 \right) \cdot \left( a \cdot x + b \right)^{c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{n}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{n} \cdot \frac{\left( c + n -1 \right)!}{\left( c - 1 \right)!} \cdot \left( a \cdot x + b \right)^{-c - n},\, &\text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{\alpha} \cdot \frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} \cdot \left( a \cdot x + b \right)^{-c - \alpha},\, &\text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $$

$$\fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \left( a \cdot x + b \right)^{-c} \right] &= \left( -a \right)^{\alpha} \cdot \frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} \cdot \left( a \cdot x + b \right)^{-c - \alpha},\, \text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $} \tag{0.5}$$

You also can write this in terms of the Rising Factorial $c^{\left( \alpha \right)}$ ($\frac{\Gamma\left( c + \alpha \right)}{\Gamma\left( c \right)} = c^{\left( \alpha \right)}$).

We can work with that. We obtain: $$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{-1} \right] + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{k} \right] \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \left( -1 \right)^{\alpha} \cdot \frac{\Gamma\left( 1 + \alpha \right)}{\Gamma\left( 1 \right)} \cdot \left( 1 \cdot x - 1 \right)^{-1 - \alpha} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( -1 \right)^{\alpha} \cdot \frac{\Gamma\left( k + \alpha \right)}{\Gamma\left( k \right)} \cdot \left( 1 \cdot x - 1 \right)^{-k - \alpha} \right]\\ \operatorname{D_{x}^{\alpha}}\left[ \zeta\left( x \right) \right] &= \left( -1 \right)^{\alpha} \cdot \Gamma\left( 1 + \alpha \right) \cdot \frac{1}{\left( x - 1 \right)^{1 + \alpha}} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( -1 \right)^{\alpha} \cdot \frac{k \cdot \Gamma\left( k + \alpha - 1 \right)}{\Gamma\left( k + 1 \right)} \cdot \frac{1}{\left( x - 1 \right)^{k + \alpha}} \right]\\ \end{align*} $$

Choosing $\alpha = \frac{1}{2}$, the main branch of this function and $x = 0$ gives: $$ \begin{align*} \operatorname{D_{x}^{\frac{1}{2}}}\left[ \zeta\left( 0 \right) \right] &= \left( -1 \right)^{0.5} \cdot \Gamma\left( 1 + 0.5 \right) \cdot \frac{1}{\left( 0 - 1 \right)^{1 + 0.5}} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \left( -1 \right)^{k} \cdot \left( -1 \right)^{0.5} \cdot \frac{k \cdot \Gamma\left( k + 0.5 - 1 \right)}{\Gamma\left( k + 1 \right)} \cdot \frac{1}{\left( 0 - 1 \right)^{k + 0.5}} \right]\\ \operatorname{D_{x}^{\frac{1}{2}}}\left[ \zeta\left( 0 \right) \right] &= -\Gamma\left( 1 + 0.5 \right) + \sum\limits_{k = 0}^{\infty}\left[ \frac{\gamma_{k}}{k!} \cdot \frac{k \cdot \Gamma\left( k - 0.5 \right)}{\Gamma\left( k + 1 \right)} \right]\\ \end{align*} $$

Calculation Via Using Special Differential Operators

Riemann–Liouville Operator

The Riemann–Liouville Fractional Operator, on the other hand, uses the formulas for simplifying mutible integrals into single integral:: $$ \begin{align*} \operatorname{_{a}D^{-v}_{x}}\left[ f\left( x \right) \right] \equiv \frac{1}{\Gamma\left( v \right)} \cdot \int\limits_{a}^{x} f\left( u \right) \cdot \left( x - u \right)^{v - 1} \, \operatorname{d}u \tag{1}\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ ist the Complte Gamma Function.

This operator has a so-called Convolutional Formula, wich is just spezial case with $a = 0$, but it also takes advantage of the fact that integrating and deriving cancel each other out: $$ \begin{align*}\operatorname{_{0}D^{v}_{x}}\left[ f\left( x \right) \right] \equiv \frac{1}{\Gamma\left( 1 - v \right)} \cdot \operatorname{D^{1}_{x}}\left[ \int\limits_{0}^{x} f\left( u \right) \cdot \left( x - u \right)^{-v}\, \operatorname{d}u \right] \tag{2}\\ \end{align*} $$

This would give us: $$ \begin{align*} \operatorname{_{0}D^{0.5}_{x}}\left[ \zeta\left( x \right) \right] \equiv \frac{1}{\Gamma\left( 1 - 0.5 \right)} \cdot \operatorname{D^{1}_{x}}\left[ \int\limits_{0}^{x} \zeta\left( u \right) \cdot \left( x - u \right)^{-0.5}\, \operatorname{d}u \right] \tag{2}\\ \end{align*} $$

But this does not converge at $x = 0$. It diverges to $\hat{\infty}$ (Complex Infinity) according to Wolfram|Alpha, which is probably also the reason why the calculation with other operators is also extremely difficult.

That was a lot to write... owo"


This mostly depends on the differenti aloperator tha you use. Principles give you a value or set of values ​​that describe that location ($x = 0$). For sure. Your series expansion might be usefull for $x > 1$, but it would not be a particularly spectacular solution. Your definition of $0^{0}$ might change it a bit but you would not get a classical solution of Fractional Calculus. Depending on the application context, it might not be quite what you are looking for. It could even lead to contradictions in the context of an FDE.

You should use a series expansion for this, where not all summants add up to $0$ for a $x$. However, this series development should also be easy to derive fractionally. I looked through many series expansions of the Riemann Zeta Function and found only one series expansion that fulfills all of this: Laurent Series around $x = 1$


There are many different ways to achieve this. The easiest way would probably be to use Euler's method: For this you can look at my derivation from the equation complex $\left( 0.5 \right)$.

  • 1
    $\begingroup$ Whoa, that's a lot of work - thank you very much! I'll need a certain time to catch it all up, but it all looks much coherent, so I think I'll get it soon and get a deeper understanding of this fractional-derivative thingie. The last given formula in the Euler-derivation-part seems to be usable; I'll try to evaluate it using Pari/GP and a nontrivial subset of the Stieltjes-numbers (needs some setup of my old software-enviroment), to see what the half derivative of zeta at zero actually is (and whether my guessed value has been at least near... ;-) ) $\endgroup$ Commented Apr 12, 2023 at 22:39

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.


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?

  • $\begingroup$ No, I just didn't find reliably a value for the half derivative, so I couldn't say whether it is close to something. What is its value? $\endgroup$ Commented Mar 12, 2014 at 22:27
  • $\begingroup$ the 2nd derivative of zeta @ 0 ~ -2.006356455 (per mathworld) . Was thinking there could be a connection between the nth and the 1/nth derivative of zeta $\endgroup$
    – o-90
    Commented Mar 12, 2014 at 22:41
  • $\begingroup$ I'd highly doubt of it. $\endgroup$ Commented Nov 27, 2016 at 17:10

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