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Chebyshev's $\psi$ function is defined for primes $p$ as

$$\psi(x)=\sum _{p^k\leq x} \log (p)$$

von Mangoldt found an explicit formula for this, with the exception that the function takes half-values at each 'step':

$${\psi}^{}_{0} (x)=x-\frac{\zeta '(x)}{\zeta(x)}-\frac{1}{2}\ln\bigl(1-x^2 \bigr)-\sum_{\rho}\frac{x^{\rho}}{\rho}$$

where $\rho$ denotes the non-trivial zeroes of the zeta function. I have two questions:

  1. Does this formula assume that the Riemann hypothesis is correct, or does it remain valid if instances of $\rho$ exist that lie away from the critical line but still within the critical strip?
  2. Given that $\frac{\zeta '(x)}{\zeta(x)}=\ln(2\pi)$, and given that $\frac{1}{2}\ln\bigl(1-x^2 \bigr)$ is defined only in the half-plane $\ge 1$, is always negative with a pole at $x=1$, and rapidly converges towards $0$ from below, the expression $x-\frac{\zeta '(x)}{\zeta(x)}-\frac{1}{2}\ln\bigl(1-x^2 \bigr)$ is asymptotic to $x-\ln(2\pi)$. Is it therefore possible to write von Mangoldt's formula using big or little O notation for the expression $x-\frac{\zeta '(x)}{\zeta(x)}-\frac{1}{2}\ln\bigl(1-x^2 \bigr)$? If so, how?
Richard Burke
  • 925

2 Answers2

1

The answer to your first question, in a nutshell, is no. Depending on whether the Riemann Hypothesis is true or not, $\psi$ will take certain or other values. In particular, if the Riemann Hypothesis is true, $|\psi(x)-x|$ will be bounded by $O(\sqrt(x)\log^2(x))$, and conversely. This was proved by Shoenfeld. In addition, if the Riemann Hypothesis is true, then $\psi(x)-x$ will “oscillate” logarithmically (that is, argument of a sine function describing the oscillation will be a function of log x). The “dominant” log frequency will the absolute value of the imaginary part of the first zeta zero. If the Riemann Hypothesis is false, then the dominant frequency will be the absolute value of the smallest imaginary part of a zeta zero out of those zeta zeros with the largest real part. All this can be derived from con Mangoldt’s formula. Two good references to get started with this are Edwards’ book on the Riemann Zeta function, or Mazur and Stein’s “Prime numbers and the Riemann Hypothesis”. But perhaps your best resource is Dittrich’s “Reassessing Riemann’s paper: on the number of primes less than a given magnitude”, chapter 6, which deals with the derivation of von Mangoldt’s formula.

EGME
  • 457
1

Starting with

$$ \zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} +...$$

$$= \left(1 + \frac{1}{2^s} + \frac{1}{2^{2s}} +...\right)\left(1 + \frac{1}{3^s} + \frac{1}{3^{2s}} +...\right)\left(1 + \frac{1}{5^s} + \frac{1}{5^{2s}} +...\right) + ...$$

$$=\left( \frac{1}{1-\frac{1}{2^s}}\right)\left( \frac{1}{1-\frac{1}{3^s}}\right)\left( \frac{1}{1-\frac{1}{5^s}}\right)...$$

Taking logarithm,

$$ \log (\zeta(s)) = - \sum_p \log(1 - \frac{1}{p^s}) $$

Differentiating both sides,

$$ \frac{\zeta^\prime (s)}{\zeta(s)} = - \sum_p \frac{-\frac{\log p}{p^s}}{\left(1 - \frac{1}{p^s} \right)} = - \sum_p \frac{\log p}{\left(1 - p^s\right)}$$

and so

$$ \frac{\zeta^\prime (s)}{\zeta(s)} \stackrel{?}{=} -\frac{\log 2}{2^s} - \frac{\log 3}{3^s} - \frac{\log 2}{4^{s}} - \frac{\log 5}{5^s} - \frac{\log 7}{7^s} - \frac{\log 2}{8^{s}} - \frac{\log 3}{9^{s}} - \frac{\log 11}{11^s}-...$$

$$= \sum_{p^k} \frac{\log p}{p^{ks}}$$

Multiplying both sides by $\frac{x^s}{s}$ and integrating w.r.t. s,

$$ \oint \frac{\zeta^\prime (s)}{\zeta(s)}\left(\frac{x^s}{s}\right)ds = -\oint \sum_{p^k} \left(\frac{\log p}{s} \right) \left(\frac{x}{p^k}\right)^s ds $$ $$= -\sum_{p^k}\log p \left(\oint \left(\frac{x}{p^k}\right)^s \left(\frac{1}{s} \right) ds \right)$$

$$ = -2\pi i \sum_{p^k}\log p $$

by the complex residue theorem at $s=0$.

Hence

$$ \sum_{p^k}\log p = -\frac{1}{2\pi i} \oint \frac{\zeta^\prime (s)}{\zeta(s)}\left(\frac{x^s}{s}\right)ds$$

The singularities of the RHS occur:

at $s=0$ with residue

$$\frac{\zeta^\prime (0)}{\zeta(0)}x^0=\frac{-\frac{1}{2}\log(2\pi)}{-\frac{1}{2}}(1) = \log(2\pi)$$

at $s=1$ with residue

$$ -\frac{x^1}{1} \stackrel{?}{=} - x $$

at $s=-2, -4, -6, ...$ with residue

$$ \frac{1}{2x^2} + \left(\frac{1}{2x^2}\right)^2 + + \left(\frac{1}{2x^2}\right)^3 + ... = \frac{1}{2} \log \left(1 - \frac{1}{x^2}\right)$$

at all other zero locations $\alpha$ with residue

$$\sum_\alpha \frac{x^\alpha}{\alpha}$$

Therefore,

$$\boxed{\sum_{p^k \le x} \log(p) = x - \log(2\pi) - \frac{1}{2} \log \left(1 - \frac{1}{x^2}\right) - \sum_\alpha \frac{x^\alpha}{\alpha}}$$

i.e. the von Mangoldt formula for $\psi(x) = \sum_{p^k \le x} \log(p)$ does not assume that $Re(\alpha) = \frac{1}{2}$ (Riemann hypothesis).

In order to connect the zeros of $\zeta(s)$ with the prime counting function $\pi(x) = \sum_{p<x} 1$, consider the quantity

$$ \sum_{p^k\le x} \frac{1}{k} = \sum_{p \le x} \left( 1 + \frac{1}{2} + ... + \frac{1}{\lfloor \frac{\log x}{\log p}\rfloor} \right)$$

$$ = \sum_{p \le x} \left( \lfloor \frac{\log x}{\log p}\rfloor + 0.57721 \right)$$

where $0.57721$ is the Euler-Mascheroni constant.

Then

$$ \boxed{\left(\log x \sum_{p \le x} \frac{1}{\log p} -0.42279 \sum_{p \le x} 1 \right) \le \sum_{p^k\le x} \frac{1}{k} \le \left(\log x \sum_{p \le x} \frac{1}{\log p} + 0.57721 \sum_{p \le x} 1 \right)}$$

On the other hand,

$$ \sum_{p^k\le x} \frac{1}{k} = \sum_{p^k\le x} \frac{\log p}{k \log p} = \sum_{p^k\le x} \frac{\log p}{\log p^k} $$

which bounds

$$ \boxed{\left( \sum_{p^k\le x} \log p \right) \le \sum_{p^k\le x} \frac{1}{k} \le \left( \frac{1}{\log x} \sum_{p^k\le x} \log p \right)}$$

implying that

$$\sum_{p^k\le x} \log p \le \log x \sum_{p \le x} \frac{1}{\log p} + 0.57721 \sum_{p \le x} 1 $$

and

$$ \log x \sum_{p \le x} \frac{1}{\log p} -0.42279 \sum_{p \le x} 1 \le \frac{1}{\log x} \sum_{p^k\le x} \log p $$

Hence the two bounds

$$\boxed{x - \log(2\pi) - \frac{1}{2} \log \left(1 - \frac{1}{x^2}\right) - \sum_\alpha \frac{x^\alpha}{\alpha} \le \log x \sum_{p \le x} \frac{1}{\log p} + 0.57721 \sum_{p \le x} 1 }$$

and

$$\boxed{\log x \sum_{p \le x} \frac{1}{\log p} -0.42279 \sum_{p \le x} 1 \le \frac{1}{\log x} \left( x - \log(2\pi) - \frac{1}{2} \log \left(1 - \frac{1}{x^2}\right) - \sum_\alpha \frac{x^\alpha}{\alpha} \right) }$$

connect the prime counting function $\pi(x)$ to the zeros of $\zeta(s)$.

James
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