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According to mathworld 41,42. "Derivatives $\zeta^{(n)}(1/2)$ can also be given in closed form" with example for the first derivative.

What is the closed form? References?

The motivation is that this question expresses $\zeta(3)$ in terms of $\zeta(1/2)$ and the first 3 derivatives, so closed form possibly might result in closed form for zeta(3) (unless the closed form is derived by the linked question).

Particaluraly intersted in the second and third derivatives.

On what the derivatives would depend helps too.

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  • $\begingroup$ Let's hope we get an answer here. Things on Mathworld (and Wikipedia, and so on) stated without citation are not entirely reliable... $\endgroup$
    – Gerald Edgar
    Commented May 10, 2013 at 13:50
  • $\begingroup$ Hm, does mathematica.stackexchange.com answer questions about mathworld? $\endgroup$
    – joro
    Commented May 10, 2013 at 14:18

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Edit: My original answer was incorrect.

You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation:

$$ \zeta(s)\Gamma(\tfrac{s}{2}) \pi^{-s/2} = \zeta(1{-}s)\Gamma(\tfrac{1-s}{2}) \pi^{(s-1)/2}. $$

Differentiating both sides sides of the equation, plugging in $s=\frac{1}{2}$, and then solving for $\zeta'(\frac{1}{2})$, I get the value listed on the MathWorld website.

As Noam Elkies points out, taking higher derivatives, this process allows you to write $\zeta^{(2n+1)}(\frac{1}{2})$ in terms of the smaller even derivatives $\zeta(\frac{1}{2}),\zeta''(\frac{1}{2}), \zeta^{(4)}(\frac{1}{2}), \ldots, \zeta^{(2n)}(\frac{1}{2})$.

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    $\begingroup$ Actually you only get every other derivative for free this way. The functional equation says $\xi(s) = \pi^{s/2} \Gamma(s/2) \zeta(s)$ is symmetric about $s = 1/2$, so its odd-order derivatives vanish there, which gives linear equations on the $\zeta^{(n)}(1/2)$ that let you solve for $\zeta^{(2m+1)}(1/2)$ as a linear combination of $\zeta(1/2)$, $\zeta''(1/2)$, $\zeta^{(4)}(1/2)$, ..., $\zeta^{(2m)}(1/2)$. But you still can't solve for the even-order derivatives in terms of derivatives of lower order. $\endgroup$
    – Noam D. Elkies
    Commented May 10, 2013 at 19:59
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    $\begingroup$ Thank you Micah. Are you sure you won't have additional terms like $\zeta(2k+1)$? (My approach gets such). Would please give an example for $\zeta'''(1/2)$? $\endgroup$
    – joro
    Commented May 11, 2013 at 5:41

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