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Riemann R-function is defined as: $$R (x)=\sum _{n=1}^{\infty } \frac{\mu (n) \text{li}\left(x^{1/n}\right)}{n}$$

I know how it appears in Riemann explicit formula for prime counting function $\pi (x)$ and that it is (with some minor terms) the so called best continuous estimator for $\pi (x)$

But what it really represents?

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  • $\begingroup$ Maybe you can search in your library Prime Numbers and the Riemann Hypothesis, by Barry Mazur and William Stein, Cambridge University Press (2016). I am saying Chapter 36. Also maybe you can be interested in some pages from Tadej Kotnik, The prime-counting function and its analytic approximations, $\pi(x)$ and its approximations, Advances in Computational Mathematics Vol. 29, Issue 1 (July 2008). I am an aficionado thus I don't know if it is that you want to read, good luck. $\endgroup$ – user243301 Oct 31 '17 at 12:55
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Look instead at

$$Li(x) = \int_2^x\frac{1_{t > 2}}{\log t}dt, \qquad R(x) =\sum_{n=1}^\infty \frac{\mu(n)}{n} Li(x^{1/n})$$

  • The Riemann explicit formula is $$\psi(x)= \sum_{p^k \le x} \log p = x - \sum_\rho \frac{x^\rho}{\rho} - \log 2\pi -\sum_{k =1}^\infty \frac{x^{-2k}}{-2k}$$

    (for $x > 1$)

  • Which gives $$\Pi(x) = \sum_{p^k \le x} \frac{1}{k} = \int_{2-\epsilon}^x \frac{\psi'(x)}{\log x} dx = Li(x) - \sum_\rho Li(x^\rho) -\sum_{k =1}^\infty Li(x^{-2k})$$

  • And hence $$\pi(x) = \sum_{p \le x} 1 = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})= R(x) - \sum_\rho R(x^\rho) -\sum_{k =1}^\infty R(x^{-2k})$$

    Thus $R(x)$ is the main term in the explicit formula for $\pi(x)$.

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  • $\begingroup$ Thank you for your reply. But that is what I knew, that is why I wrote: "I know how it appears in Riemann explicit formula". But that does not explain what it represents relating to the primes. I mean there can be plenty continuous approximations to $\pi (x)$, or even such ones that they are equal to $\pi (x)$ at integer x. Simple example can be interpolating polynomial. $\endgroup$ – azerbajdzan Oct 31 '17 at 12:27
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    $\begingroup$ @azerbajdzan Look at the Mellin transform of all those things. $Li(x)$ is not defined in term of the primes, your interpolating polynomial is. $x,Li(x),R(x)$ tell you how the zeros of $\zeta(s)$ (and the RH) affect $\psi(x),\Pi(x),\pi(x)$. $\endgroup$ – reuns Oct 31 '17 at 12:33
  • $\begingroup$ I would say the opposite. The zeros of $\zeta (s)$ tell how to affect $R(x)$ to get $\pi (x)$. I give you one point for your answer at least for writing the formulas. Thank you. $\endgroup$ – azerbajdzan Oct 31 '17 at 12:39
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    $\begingroup$ I think there is so little written about $R(x)$. Have you seen some books or articles that focus more on $R(x)$ than on zeros of $\zeta (s)$? $\endgroup$ – azerbajdzan Oct 31 '17 at 12:44

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