Added Background: The pair correlation of the zeros of the Riemann zeta function is influenced by the the derivative of the logarithmic derivative $(\zeta^\prime(s)/\zeta(s))^\prime$; see for example the answers to 
[this question](https://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul)




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I'm looking for references for computation of $\zeta^\prime(s)/\zeta(s)$, as well as the its derivative, in the critical strip $0<\text{Re}(s)<1$. (I'm sure they're out there but google scholar/MathSciNet searches return way too many irrelevant hits.)


Of course, both $\zeta(s)$ and $\zeta^\prime(s)$ are implemented in packages like Sage, one can just take the quotient and then use this to numerically estimate the derivative via the difference quotient, but this seems computationally wasteful. We have that 
$$
\frac{\zeta^\prime(s)}{\zeta(s)}=\log(2\pi)-1-\gamma/2-\frac{1}{s-1}-\frac12\frac{\Gamma^\prime(s/2+1)}{\Gamma(s/2+1)}+\sum\_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).
$$


Similarly, one gets $(\zeta^\prime(s)/\zeta(s))^\prime$ upon differentiating term by term. 


The digamma function $\Gamma^\prime/\Gamma$ as well as the Riemann zeros $\rho$ are implemented in *Mathematica*. So I think what I'm asking is a reference to answer the following: 


Given $\epsilon$, how many zeros do I need to take as a as a function of $t=\text{Im}(s)$ so the error is bounded by $\epsilon$?




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Added: In Theorem 9.6(A) in Titchmarsh's "Theory of the Riemann Zeta Function", one can compute the relevant constants to show that
$$
\left|\frac{\zeta^\prime(s)}{\zeta(s)}-\sum\_{|\rho-s|\le 6}\frac{1}{s-\rho}\right|\le 4\log t.
$$
So in answer to Joro's question below, yes the sum is dominated by the zeros close to $s$.



Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$
I believe the comments of joro and Carlo Benakker right under your question is to the point. Since the zeroes close to $s$ will be the ones that contributes in the sum, the zeroes close to s must be computed first and in order to do that several values of the zeta-function must certainly be calculated and the time for each such computation (and zero) will certainly not be less than computing any other particular value of $\zeta(s)$ or $\zeta'(s)$.


As for calculating a particular value $\zeta(s)$ the recent method of Ghaith Hiary <http://arxiv.org/abs/0711.5005> "Fast methods to compute the Riemann zeta function" (published in Annals of Mathematics 2011) contains the fastest method known. He shows a method that calculates the value of $\zeta(1/2+it)$ with an error term less than $|t|^{-N}$ in time $O\_{\varepsilon,N}(t^{4/13+\varepsilon})$. The method of Odlyzko and Schönhage "Fast algorithms for multiple evaluations of the Riemann zeta function", doi:10.2307/2000939 is faster if sufficiently many values of $t$ needs to be calculated, i.e. it can calculate $\sqrt T$ values in the range $[T,T+\sqrt T]$ in time $O\_\varepsilon(T^{1/2+\varepsilon})$. 


Now, In Hiary's paper it might look like he only considers the critical line. However in fact his method works for any value in the critical strip. Indeed in part of his argument he considers just the line Re$(s)=0$, but then he writes "It is clear that the restriction $\sigma=0$ is not important and a similar conclusions can be drawn for other values of $\sigma$" (page 6, second to last paragraph in Hiary's paper). It is true that his main interest is the critical line, but the algorithm holds for any $s$ in any vertical strip such as the critical strip.


Now to calculate $\zeta'(s)$ up to error of order $t^{-N}$ it is sufficient to calculate $\zeta(s+ih)$ and $\zeta(s-ih)$ with error less than $t^{-3N/2-1/4}$ for $h=t^{-N/2-1/4}$ and consider the difference quotient $(\zeta(s+ih)-\zeta(s-ih))/h$ by also using that $\zeta'''(s) =O(\sqrt t)$ in the critical strip and some version of the mean value theorem inequality.
It is simple to use finite difference quotients to calculate $\zeta^{(k)}(s)$ up to any desired degree of accuracy. Thus we can certainly use this method to calculate for example
$$
 \frac d {ds} \frac{\zeta'(s)}{\zeta(s)}= \frac{\zeta''(s)\zeta(s)-\zeta'(s)^2}{\zeta(s)^2}
$$
by this method. This method should not be computionally wasteful.


