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?

  • 1
    $\begingroup$ 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, 2017 at 12:55

1 Answer 1


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 $\pi(x)$.

  • $\begingroup$ 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. $\endgroup$ Oct 31, 2017 at 12:27
  • 1
    $\begingroup$ @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, 2017 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. $\endgroup$ Oct 31, 2017 at 12:39
  • 1
    $\begingroup$ 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 $\zeta (s)$? $\endgroup$ Oct 31, 2017 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.