Below is a plot of $\exp \sum _p^x -\frac{\cos \left(x \log \ p\right)}{\sqrt{p}}$, where $p$ runs over the primes, and the $x$-values of the Riemann $\zeta$ zeros are marked with dashed lines:

enter image description here

Below is the plot $\sum _\rho^{100} -\frac{\cos \left(\rho \log x \ \right)}{\rho }$, where $\rho$ is the imaginary part of the non-trivial $\zeta$ zeros, and this time the $x$-values of the primes are marked with dashed lines:

enter image description here

In what sense is $\sum _p \frac{\cos \left(x \log p\right)}{\sqrt{p}}$ the inverse of $\sum _\rho \frac{\cos \left(\rho \log x \right)}{\rho }$ (despite the former not converging on the infinite sum)?

Also, does the first sum spike at all zeros of the Riemann $\zeta$ function, or does the scaling factor $\sqrt p$ depend on the RH?


1 Answer 1


This is not a complete answer but hopefully provides some insight.

The function $$\exp\left(-\sum\limits_{p\le x},\frac{\cos(x\, \log(p))}{\sqrt{p}}\right)\tag{1}$$

seems to approximate

$$\left|\frac{1}{\zeta\left(\frac{1}{2}+i x\right)}\right|\tag{2}$$

which is illustrated in Figure (1) below.

Illustration of formulas (1) and (2) in orange and blue

Figure (1): Illustration of formulas (1) and (2) in orange and blue

The sum

$$-\sum\limits_{\rho} \frac{\cos\left(\Im(\rho)\, \log(x)\right)}{\Im(\rho)}\tag{3}$$

seems to be more related to the derivative of the explicit formula for

$$\psi(x)=\sum\limits_{n=1}^x \lambda(n)\tag{4}$$

which is related to the Dirichlet series

$$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{n=1}^x \frac{\lambda(n)}{n^s}\,,\quad\Re(s)>1\tag{5}.$$

More specifically the explicit formula for $\psi(x)$ contains the following sum over the non-trivial zeta zeros

$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \left(\frac{x^{\rho_{-k}}}{\rho_{-k}}+\frac{x^{\rho_k}}{\rho_k}\right)\right)\tag{6}$$

with derivative

$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \left(x^{\rho_{-k}-1}+x^{\rho_k-1}\right)\right)\tag{7}$$

which assuming the Riemann hypothesis can be evaluated as

$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \frac{2 \cos\left(\Im\left(\rho_k\right)\, \log(x)\right)}{\sqrt{x}}\right)\tag{8}.$$

If you look at your second plot carefully you'll see there seems to be peaks at the prime-powers as well as the primes which is consistent with the integers where $\psi(x)$ takes a step.

  • $\begingroup$ Many thanks for your answer. I assume that the $\sqrt x$ in $(8)$ is the scaling factor along the critical line, which answers my question about dependence on the RH. The resemblance of the first function to the reciprocal of Hardy's Z function makes sense too, since it comes from observations of the behaviour of partial Euler product along the line $\frac{1}{2}+i x$. $\endgroup$
    – martin
    Dec 17, 2023 at 20:39
  • 2
    $\begingroup$ @martin You're correct the scaling factor is based on the RH: $$x^{\rho_k-1}+x^{\rho_{-k}-1}=x^{\frac{1}{2}+i\, \Im\left(\rho_k\right)-1}+x^{\frac{1}{2}-i\, \Im\left(\rho_k\right)-1}=\frac{x^{i\, \Im\left(\rho_k\right)}+x^{-i\, \Im\left(\rho_k\right)}}{\sqrt{x}}=\frac{2 \cos\left(\Im\left(\rho_k\right)\, \log(x)\right)}{\sqrt{x}}$$ $\endgroup$ Dec 17, 2023 at 23:42

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