Derivatives of the Riemann zeta function at $s = 1/2$ 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?
 A: 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 $k=2m$.
