It should be the case that, in some appropriate sense $$\pi (x)\sim \operatorname{Ri}(x)-\sum_{\rho}\operatorname{Ri}(x^{\rho}) \tag*{(4)}$$ with $\operatorname{Ri}$ denoting the Riemann function defined: $$\operatorname{Ri}(x)=\sum_{m=1}^\infty \frac{\mu (m)}{m}\operatorname{li}\left(x^{\frac{1}{m}}\right). \tag*{(5)}$$ This relation $(4)$ has been called "exact" [in Ribenboim's The New Book of Prime Number Records], yet we could not locate a proof in the literature; such a proof should be nontrivial, as the conditionally convergent series involved are problematic. In any case relation $(4)$ is quite accurate, and furthermore the Riemann function $\operatorname{Ri}$ can be calculated efficiently (...) The sum in $(4)$ over critical zeros is not absolutely convergent, and furthermore the phases of the summands interact in a frightfully complicated way.

—from Journal of Computational and Applied Mathematics by Borwein et al.

Of profound importance, Bernhard Riemann proved that the prime-counting function is exactly $$\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})$$ where $$\operatorname{R}(x)=\sum_{n=1}^\infty \frac{\mu (n)}{n}\operatorname{li}\left(x^{\frac{1}{n}}\right),$$ (...) $\rho$ indexes every zero of the Riemann zeta function, and $\operatorname{li}\left(x^{\frac{1}{n}}\right)$ is not evaluated with a branch cut but instead considered as $\operatorname{Ei}\left(\frac{\rho}{n}\ln x\right)$. Equivalently, if the trivial zeros are collected and the sum is taken only over the non-trivial zeros $\rho$ of the Riemann zeta function, then $\pi (x)$ may be written $$\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}.$$

—from Wikipedia's Prime counting function article


  1. According to Borwein and Ribenboim, the index in $(4)$ should run only over non-trivial zeros. According to Wikipedia, the index in $(4)$ should run over all zeros. Wikipedia states that if the sum runs only over non-trivial zeros, then we add the $\ln$ and $\arctan$ terms, which is even more confusing. So what's true?

  2. I'm pretty sure that Riemann did not prove the formula $(4)$. He only proposed a "weaker" form of it, namely $$\pi (x)=\sum_{m=1}^\infty \frac{\mu (m)}{m}J\left(x^{\frac{1}{m}}\right)$$ where $$J(x)=\operatorname{li}(x)-\sum_{\rho}\operatorname{li}\left(x^{\rho}\right)+\int_x^\infty \frac{dt}{t(t^2-1)\ln t}-\ln 2$$ and where $\rho$ runs over all non-trivial zeros. The formula was proven by Mangoldt, not Riemann. The Wikipedia article is wrong in that historical fact, isn't it?

  3. Even though the proof of $(4)$ is nowhere to be found in the literature, Raymond Manzoni provided a partial proof here: Two Representations of the Prime Counting Function. I call it partial because it is unknown whether the series converges at all: How could that be settled down?

Note: When I refer to the formula $(4)$ in this question, I assume $=$ instead of $\sim$.

  • $\begingroup$ What is your definition of $li$ ? With the function non-zero only for $x \ge 2$ the formula is true because it is a finite sum over $m$. The alternative is to consider the distribution (on $(0,\infty)$) whose Mellin transform is $-\log(s-1)/s$ in which case I doubt it converges even in the sense of distributions. $\endgroup$
    – reuns
    Apr 8, 2020 at 18:45
  • $\begingroup$ $\operatorname{li}x=\operatorname{Ei}\ln x$ where $\operatorname{Ei}$ is the exponential integral function. $\endgroup$
    – Wane
    Apr 8, 2020 at 22:25
  • $\begingroup$ It should be interpreted as the Cauchy principal value. $\endgroup$
    – Wane
    Apr 8, 2020 at 22:30
  • 1
    $\begingroup$ So that's the second one I mentioned : the distribution whose Mellin transform is $-\log(s-1)/s$. The explicit formula for $J(x)$ converges in the sense of distribution because its derivative does (this is the Riemann explicit formula), there is one of the books on the RH proving many facts about it (Montogmery, Patterson, Edwards..) nor of them appear to prove the above formula, not a good sign. $\endgroup$
    – reuns
    Apr 8, 2020 at 23:09
  • 1
    $\begingroup$ Now posted to MO, mathoverflow.net/questions/386213/… $\endgroup$ Mar 12, 2021 at 6:40

1 Answer 1


Now posted on MathOverflow by vitamin d: https://mathoverflow.net/questions/386213/is-pi-x-operatornamerx-sum-rho-operatornamerx-rho-correct-a/398456#398456


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