# von Mangoldt's formula for Chebyshev $\psi$ function

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?