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On the one hand, the Prime Counting Function $\pi_0(x)$ maybe be written $$ \pi_0(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1} $$ with $ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$ and $\rho$ running over all the zeros of $\zeta$ function. The derivative of $(1)$ is $$ \pi_0'(x) = \operatorname{R}'(x) - \sum_{\rho}\operatorname{R}'(x^{\rho}) \tag{2} $$ with $\displaystyle \operatorname{R}'(x^k) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \frac{x^{k/n-1}}{\log x}$.


On the other hand, we have $$ \pi_1(x)=\sum_{k=1}^{N} H(x-p_k), \tag{3} $$ with $H(\cdot)$ being the Heaviside function. This would result in (a prime comb) $$ \pi_1'(x)=\sum_{k=1}^\infty \delta(x-p_k). $$


$(3)$ obviously diverges at a prime $p_k$. Is it possible to show that $(2)$ also diverges?

Thanks to Raymond there is also a nice gif:

$\hskip1.7in$what text

The lower animated graph is the derivative of the function above, and we see the positions of the primes emerging as Dirac delta-type spikes. The horizontal bar in the middle has been constructed so that the intensity of brightness is related to the absolute value of the derivative. In this way we see the positions of the primes gradually emerge as narrow bands of light. [for reference see Matthew Watkins homepage].

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Even deriving (2) from (1) requires an assumption - namely that the sequence of partial sums in (2) converges uniformly to the full series - and I don't think that holds here. (The same proviso holds for your expression for $R'$, though there the convergence might well be uniform.) See en.wikipedia.org/wiki/Uniform_convergence#To_differentiability –  Greg Martin Dec 26 '13 at 19:19

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