# Riemann explicit formula for $\pi(x)$ and its evaluation

We have for $x>1$:

$$\tag{1}\pi^*(x)=R(x)-\sum_{\rho} R(x^{\rho})$$

$\rho$-s are zeros of Riemann zeta function (trivial and nontrivial). I know how to derive $(1)$ and also know how to derive: $$\tag{2}\psi^*(x)=x-\sum_{\rho} \frac {x^{\rho}}{\rho}-\frac{\zeta'(0)}{\zeta(0)}$$ This question concerns only about $(1)$ so please do not answer like "instead of $(1)$ look at $(2)$". (I know it is somehow easier in understanding and evaluation.)

Back to $(1)$...

$R(x)$ is Rieamann R-function and is defined as:

$$\tag{3}R(x)=1+\sum _{n=1}^{\infty } \frac{\big(\log(x)\big)^n}{nn! \zeta (n+1)}$$ Now, whether I use $$\tag{3.1}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\log(x^\rho)\big)^n}{nn! \zeta (n+1)}$$ or $$\tag{3.2}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\rho \log(x)\big)^n}{nn! \zeta (n+1)}$$ in $(1)$, in both cases I do not get $\pi^*(x)$. (Series seems to diverge/oscillate.)

Can you tell me what a definition of $R(x)$ to use to get the correct results? I am interested only in exact formulas, approximations are out of interest!

I have seen approximations using Möbius function $\mu(x)$ and exponential integral $Ei\big(\rho \log(x)\big)$.

• @reuns How does this help me to evaluate $(1)$? – azerbajdzan Nov 18 '17 at 0:27
• A typo. $\text{Li}(x)$ appears from $\Pi(x) = \sum_{p^k \le x} \frac{1}{k} = \int_{2-\epsilon}^x \frac{\psi'(y)}{\log y}dy$. Then $R(x)$ appears from $\pi(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})$. – reuns Nov 18 '17 at 16:56
• I've been looking long and hard for a good reference for this, but it seems oddly under-researched in the literature... :-( – Markus Shepherd Nov 19 '17 at 7:24
• @Markus Shepherd As far as I know the expression $(1)$ has not yet been proved by anybody. And also I have not seen anywhere, that somebody used it for evaluation of $\pi(x)$, they always use a different/modified formula, that is just an approximation. The formula uses Moebius function and exponential integral. So $(1)$ is to me just hypothetical. – azerbajdzan Nov 19 '17 at 10:18

There is the easiest way to obtain convergent explicit formulas :

Since $$\psi(x) =\sum_{p^k \le x} \log p=( x-\sum_\rho \frac{x^\rho}{\rho}-\log 2\pi)1_{x > 2}$$ (sum over trivial and non-trivial zeros)

Letting $\Pi(x) = \sum_{p^k \le x} \frac{1}{k}= \int_{2-\epsilon}^x \frac{\psi'(x)}{\log x}dx$ and $1_{x > 2}\int_2^x ( \frac{d}{dt}\frac{t^\rho}{\rho}) \frac{dx}{\log x} = 1_{x > 2}\int_2^x \frac{\rho t^{\rho-1}}{\rho\log t} dt=1_{x > 2}\int_{2^\rho}^{x^\rho} \frac{u}{\log u} du= (\text{li}(x^\rho)-\text{li}(2^\rho))1_{x > 2}$

then $$\Pi(x) = (\text{li}(x)-\text{li}(2))1_{x > 2}-\sum_\rho (\text{li}(x^\rho)-\text{li}(2^\rho))1_{x > 2}$$

Now let $\pi(x) =\sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})=\sum_{n=1}^{\log_2(x)} \frac{\mu(n)}{n} \Pi(x^{1/n})1_{x^{1/n}> 2}$ and $\text{R}_\rho(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} (\text{li}(x^{\rho/n})-\text{li}(2^{\rho/n}))1_{x^{1/n} > 2}$ (those are finite sums)

therefore $$\pi(x) = \text{R}_1(x)-\sum_\rho \text{R}_\rho(x)$$

• It does not work even with your definition of $R_\rho(x)$. – azerbajdzan Nov 19 '17 at 18:41
• I agree that it is finite, but the result of such computation is not $\pi(x)$. – azerbajdzan Nov 19 '17 at 18:45
• @azerbajdzan What do you mean ? Of course it is, by definition. – reuns Nov 19 '17 at 18:50
• there was a typo I meant $1_{x^{1/n}>2}$ – reuns Nov 19 '17 at 21:03
• Here the definition is $1_{x > 2}\int_2^x ( \frac{d}{dt}\frac{t^\rho}{\rho}) \frac{dt}{\ln t}$ with $t^\rho = e^{\rho \ln(t)}$, there is no ambiguity about it. Also you don't have so many good reasons to not say $\Pi(x) = \sum_{n \le x} \frac{\psi(n+1/2)-\psi(n-1/2)}{\ln n}, \pi(x) = \sum_{n=1}^{\log_2(x)} \frac{\mu(n)}{n}\Pi(x^{1/n})$ is not an explicit formula for $\pi(x)$. – reuns Nov 19 '17 at 23:10

$$\psi^*(x) =\sum_{p^k \le x} \log p$$ $$\Pi^*(x) = \sum_{p^k \le x} \frac{1}{k}$$ There are indeed many traps while evaluating $\pi(x)$ using the $\zeta$ zeros
(I will neglect here the smooth and quickly decreasing with $x$ contribution of the trivial zeros).
This answer detailed most of these traps but let's examine your specific points :

Now, whether I use $$\tag{3.1}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\log(x^\rho)\big)^n}{nn! \zeta (n+1)}$$ or $$\tag{3.2}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\rho \log(x)\big)^n}{nn! \zeta (n+1)}$$ in $(1)$, in both cases I do not get $\pi^*(x)$. (Series seems to diverge/oscillate.)

The known non trivial zeros $\rho$ verify the RH and may thus be written $\,\displaystyle\rho:=\frac 12+it$.
Now $(3.1)$ can't indeed provide the right answer because the numerical evaluation of $x^\rho$, before applying the logarithm, will give you for $x$ real $>1\quad\displaystyle x^\rho=\exp\left(\left(\frac 12+it\right)\log(x)\right)$.
The problem is that the evaluation of the logarithm will not return $\,\left(\frac 12+it\right)\log(x)\;$ but something like $\left(\frac 12+it\right)\log(x)+2\pi ki\;$ with $k\in\mathbb{Z}$ depending of $t$ (the evaluation of the logarithm imposing a phase in $(-\pi,\pi]\;$)

Using a different phase reference (branch) for the terms of the series isn't a fine idea but what about the Gram series as used in $(3.2)$?
This formula is indeed correct and was the one I used to produce this animation for $x\in(1,100)$. But this formula too has to be handled with care because we have to evaluate something behaving like (since $\,\zeta(n)\sim 1\,$ as $\,n\to +\infty$) :$$S(z):=\sum_{n=1}^{\infty } \frac{z^n}{nn!},\quad z:=\left(\frac 12+it\right)\log(x)$$ For the first non trivial zero and (say) $\,x=100\,$ we gave $|z|>65$, for the tenth $|z|>229$, for the hundredth $|z|>1089$.
The problem is that the terms of $S(z)$ and thus $R(x^\rho)\,$ will become very large (about $\dfrac {e^z}{z\sqrt{2 \pi z}}$ near $n=z$) before decreasing again to get under unity after about $\,\lceil e\,z\rceil\,$ terms. In practice for $\,x=100\,$ and the $100$-th non trivial zero you will need a working precision of $480$ digits!

Another idea proposed by reuns is to use the Möbius $\mu$ function formula (no high precision here) :

$$\tag{1}R(x)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{li}\bigl(x^{1/n}\bigr)$$ (since $\ \displaystyle\pi^{*}(x)=\sum_{n=1}^{\infty} \frac{\mu(n)}n \Pi^*\bigl(x^{1/n}\bigr)\;$ with $\Pi^*(z)=0\;$ for $\,z<2\;$ we need only the $\,\lceil\log_2(x)\rceil\;$ first terms for all the $R(x^\rho)$ expressions ... not just $R(x)$)

But from the logarithm in the $\operatorname{li}$ function we get exactly the same problem of "phase reduction" as previously. Fortunately a simply remedy exists from the definition of the exponential integral $\operatorname{Ei}$ by using $\,\operatorname{li}\left(e^x\right)=\operatorname{Ei}(x)$ and replacing $(1)$ with : $$\tag{2}R(x)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{Ei}\left(\frac 1n\log(x)\right)$$ and $$\tag{3}R\left(x^\rho\right)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{Ei}\left(\frac {\rho}n\log(x)\right)$$

The result for $\;\pi(x),\;x\in(2,100)$ (starting with $R(x)$ evaluated using the Gram series, subtracting the exact $\;\displaystyle\frac 1{\log(x)}-\frac 1{\pi}\arctan\left(\frac{\pi}{\log(x)}\right)\;$ contribution for the trivial zeros and two times the real part of the sum of the $\,200\,$ first non trivial zeros $\,R\left(x^\rho\right)\,$ terms using $(3)\;$) should be : Concerning the use of the $\zeta$ zeros to evaluate the prime counting function $\pi(x)$ as in your expression $(1)$ this answer may he helpful. Shortly the combinatorial Meissel-Lehmer method allowed to compute $\pi(x)$ up to $10^{23}$ while the analytic method allowed to go up to $10^{25}$ (in $2016$, I don't know the state of the art). Note that more efficient expressions than $(1)$ were used as you may find in the history by Büthe and references.

• The problem with formulas involving Möbius μ function is that the plotted graph does not look nice, it is not smooth (like are mine two submitted graphs of $\psi^*(x)$ and $\Pi^*(x)$). And it also seems there is some other problem with your graph - for $2<x<3$ it is a way out of correct values. – azerbajdzan Jun 18 '18 at 13:56
• Also your animation has some problems. For example for $13<x<17$ the approximation with 100 zeros is worse than with n=20. – azerbajdzan Jun 18 '18 at 14:03
• @azerbajdzan: the problem in the $(2,3)$ interval was corrected by replacing the $R(x)$ Moebius sum I used by the (exact) Gram series. Concerning your other remarks : Well, we get what we get ! (I created my animation around year $2000$ and reverified the picture for $n=100$ it appears right : the Gram series seem to produce these "wide scale waves". I still don't know a convergence proof for your $(1)$ using Gram or Möbius sums but D. Platt studied the convergence of his sums without supposing R.H. in my second link). Excellent and nicer animations, – Raymond Manzoni Jun 21 '18 at 9:33