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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?
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