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The Mathworks page on Riemann's $\zeta$ function says:

Let $\rho_k$ denote the $k$th nontrivial zero of $\zeta(s)$, and write the sums of the negative integer powers of such zeros as $$ Z(n)=\sum_k\rho_k^{-n} $$ ... Such sums can be computed analytically, and the first few are $$ Z(1) = \frac12[2+\gamma-\ln(4\pi)] =0.0230957... $$ where $\gamma$ is the Euler-Mascheroni constant,...

How to prove that? It can be simplified (assuming RH) to $$ Z(1) = \sum_k \frac4{(1+4t_k^2)} $$ where $t_k$ is the imaginary part of $\rho_k$, but they are thought likely to be transcendental numbers (from here and references therein).

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$\gamma$ is thought to be transcendental, and having a 'closed form' does not change the fact that it is. (Just because something has a closed form, that does not make it algebraic or even rational). For example, $\ln 2$ is transcendental, but it does not look complex and has a simple closed form. Hence, having a closed form and being transcendental are not mutually exclusive. – George V. Williams Jan 8 '13 at 23:13
@GeorgeV.Williams thanks for pointing that out... – draks ... Jan 8 '13 at 23:17
up vote 1 down vote accepted

Start from $$ \xi(s)=s(s-1)\pi^{s/2}\Gamma(s/2)\zeta(s)=\frac{1}{2}\exp(b s)\prod_\rho\left(1-\frac{s}{\rho}\right)\exp(s/\rho) $$ with $b=\log(2\pi)-1-\gamma/2$. Take the logarithmic derivative and group $\rho$ with $1-\rho$ to get $$ \frac{\xi^\prime}{\xi}(s)=b+\sum_{\rho}\frac{1}{\rho-s}. $$ So $\xi^\prime/\xi(0)$ allows you to compute $Z(1)$. Higher derivatives give $Z(n)$ in general.

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Thanks, but one question if you don't mind: What does your Gravatar represent? – draks ... Jan 9 '13 at 8:05
It's a plot of $\arg(\zeta(s))$, where the interval $[0,2\pi]$ is mapped to the color wheel. See my webpage. – stopple Jan 9 '13 at 15:49

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