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The Wolfram page http://mathworld.wolfram.com/RiemannZetaFunction.html states that "Derivatives $\zeta^{(n)}(1/2)$ can also be given in closed form", but apart from an explicit formula for $\zeta'(1/2)$ provides neither any such formula, nor any reference. Can anyone point me to an appropiate reference?

John Fredsted
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    $\psi(s)= \frac{\Gamma'(s)}{\Gamma(s)} = -\gamma + \sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{s+n}$ so $\psi^{(k)}(1/2) = \sum_{n=0}^\infty \frac{(-1)^{k+1} k!}{(1/2+n)^{k+1}} = (-1)^{k+1} k! 2^k(1-2^{-k})\zeta(k+1)$ having no closed-form for $k \ge 2$ even. But (with the functional equation $\zeta(s) = \zeta(1-s)2^{s-1} \pi^s \sin(\pi s/2)\Gamma(1-s) \implies \frac{\zeta'(s)}{\zeta(s)}+ \frac{\zeta'(1-s)}{\zeta(1-s)} = \ldots -\psi(1-s)$ it means $\zeta^{(k)}(1/2)$ has an expression in term of $\zeta(m), 2 \le m\le k+1$ and the constants $\gamma,\pi$. Maybe that's what they meant – reuns Oct 09 '16 at 10:58
  • Thanks a lot for that very interesting input. It has given me quite something to ponder. – John Fredsted Oct 09 '16 at 11:09
  • well I forgot that $\frac{d^k}{ds^k} (\frac{\zeta'(s)}{\zeta(s)}+\frac{\zeta'(1-s)}{\zeta(1-s)}) = \frac{d^k \zeta'/\zeta}{ds^k}(s)+(-1)^k\frac{d^k \zeta'/\zeta}{ds^k}(1-s)$ so we have a problem for $k$ odd – reuns Oct 09 '16 at 11:32
  • Thanks for that update. It is a pitty, though, because I need all the derivatives. PS: Your formula for $\psi^{(k)}(1/2)$ should, I believe, have $2^{k+1}$ rather than $2^{k}$; otherwise, I cannot verify it using Maple. And the functional equation you give should, I believe, have $2^{s}\pi^{s-1}$ rather than $2^{s-1}\pi^{s}$. – John Fredsted Oct 09 '16 at 12:11

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By the reflection formula: $$ \Gamma\left(\tfrac{s}{2}\right)\pi^{-s/2}\zeta(s) = \Gamma\left(\tfrac{1-s}{2}\right)\pi^{(s-1)/2}\zeta(1-s) $$ and by considering $\frac{d}{ds}\log(\cdot)$ of both sides $$ \tfrac{1}{2}\psi\left(\tfrac{s}{2}\right)-\tfrac{\log\pi}{2}+\tfrac{\zeta'}{\zeta}(s) = -\tfrac{1}{2}\psi\left(\tfrac{1-s}{2}\right)+\tfrac{\log\pi}{2}-\tfrac{\zeta'}{\zeta}(1-s) $$ or $$\begin{eqnarray*} \tfrac{\zeta'}{\zeta}(s)+\tfrac{\zeta'}{\zeta}(1-s)&=&\log\pi-\tfrac{1}{2}\left[\psi\left(\tfrac{s}{2}\right)+\psi\left(\tfrac{1-s}{2}\right)\right]\\&=&\log\pi+\gamma+\sum_{n\geq 1}\left[\frac{1}{2n-1-s}+\frac{1}{2n-2+s}-\frac{1}{n}\right].\tag{A}\end{eqnarray*} $$ By evaluating both sides of $(A)$ at $s=\frac{1}{2}$ we get: $$ 2\cdot\tfrac{\zeta'}{\zeta}\left(\tfrac{1}{2}\right) = \log \pi+\gamma+\tfrac{\pi}{2}+3\log 2 \tag{B}$$ which agrees with the result $(41)$ in the mentioned MathWorld page.

There is no hope of computing $\zeta''\left(\frac{1}{2}\right)$ from the reflection formula (if we have a symmetric function with respect to $s=\frac{1}{2}$, its even derivatives at such point equal zero), but by considering the Weierstrass product for the $\zeta$ function we have

$$ \frac{d^2}{ds^2}\log\zeta(s) = \frac{1}{(s-1)^2}-\sum_{n\geq 1}\frac{1}{(2n+s)^2}-\sum_{\rho}\frac{1}{(s-\rho)^2}\tag{C} $$ hence $$ \tfrac{\zeta''}{\zeta}\left(\tfrac{1}{2}\right) = \tfrac{1}{4}\left(\log \pi+\gamma+\tfrac{\pi}{2}+3\log 2\right)^2+8-2G-\tfrac{\pi^2}{4}-\sum_{\rho}\frac{1}{\left(\tfrac{1}{2}-\rho\right)^2}\tag{D}$$ In general $\zeta^{(k)}\left(\frac{1}{2}\right)$ can be computed by differentiating $(A)$ or $(C)$ an even number of times, then evaluating at $s=\frac{1}{2}$. We may also notice that $$ \sum_{\rho}\frac{1}{\left(\frac{1}{2}-\rho\right)^{2m}}\stackrel{\text{Cauchy}}{=}\lim_{T\to +\infty}\frac{1}{2\pi i}\oint_{R_T}\frac{\zeta'(s)}{\zeta(s)}\left(s-\tfrac{1}{2}\right)^{-2m}\,ds, \tag{E}$$ where $R_T$ is the rectangle having vertices at $-Ti,1-Ti,1+Ti,Ti$ (with an indentation avoiding the pole at $s=1$), can be simply estimated by summation by parts and the Riemann-Von Mangoldt theorem. The non trivial-zeros of $\zeta$ have a distance from $\frac{1}{2}$ which is always $\geq 14$, hence the residual series provide a negligible contribution for large and even values of $k=2m$.

Jack D'Aurizio
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  • Thanks for your extensive answer which, I think, really deserves more attention than this short comment of mine gives it (but at least, I have given it a thumbs up). Unfortunately, none of the formulas you provide constitute what I consider to be a closed form expression for the derivatives of $\zeta(s)$ at $s = 1/2$, that is, an expression in terms of, say, $\pi$, $e$, Catalan's constants, and the Stieltjes constants (here included Euler's constant), and importantly containing no infinite sums. That was what I thought "closed form" at the MathWorld page referred to. – John Fredsted Oct 31 '17 at 07:19
  • @JohnFredsted: well, but you can clearly see that the formula (41) on MathWorld only gives $\zeta'(1/2)$ in terms of $\zeta(1/2)$, so what were you expecting? – Jack D'Aurizio Oct 31 '17 at 10:05
  • I was expecting a closed form expression for the derivatives (plural) of $\zeta(s)$ at $s = 1/2$, not just the first derivative. – John Fredsted Oct 31 '17 at 11:55
  • Yes, certainly, if the exponent is odd, then the sum over the nontrivial zeros vanish identically due to symmetry stemming from the 'shift' of the nontrivial zeros to the left by $1/2$. But that is not my point: Taking the second derivative of Eq. (A), evaluating it at $s = 1/2$, and isolating $\zeta^{(3)}(1/2)$ will make the latter depend on $\zeta^{(2)}(1/2)$ which in turn depends on the sum over the nontrivial zeros, your Eq. (D). This, it seems, is inconsistent with your above statement "the derivatives of odd order just depend on $\pi, \gamma, \log2, \log\pi, G,\zeta(1/2)$". – John Fredsted Nov 01 '17 at 08:45
  • @JohnFredsted: you are right, I am retracting my previous comments (except the point about the dependency on $\zeta(1/2)$). On the other hand, I have another approach to suggest: $\zeta(1/2)$ can be computed from the regularization of a divergent series, and it is clearly negative by the behaviour of such series (https://math.stackexchange.com/a/2500083/44121). If the same applies to the derivatives of $\zeta$ at $\frac{1}{2}$, we are done. – Jack D'Aurizio Nov 01 '17 at 19:46
  • Happy to hear that we now agree on the dependency of the higher order derivatives on sums over the nontrivial zeros. Your point about dependency on $\zeta(1/2)$ is fully justified, and has been noted. I guess, I would have been very happy if the derivatives of $\zeta$ would be expressible in terms of some fundamental constants, and $\zeta$ evaluated at some finite set of points. I do not understand your new approach; how is it relevant to the evaluation of (higher order) derivatives? [PS: Why has some of the comments above been deleted?] – John Fredsted Nov 02 '17 at 07:28