H. Edwards in his book on the zeta function says that $\sum\frac{x^{\rho}}{\rho}$ converges conditionally "even when $\rho ,1-\rho$ are paired." I tried calculating some terms (n = 500 or so) and noticed the result below. Could be bad programming on my part or an aspect of the calculation for low numbers.
The expressions below assume RH but I am asking about the calculation for small finite n. It may be that an answer has to assume RH.
$$\sum \frac{x^{\rho}}{\rho} = \sum_n \frac{x^{\rho_n}}{\rho_n}+\frac{x^{1-\rho_n}}{1-\rho_n}$$
seems to be zero or (to be piecewise continuous and) have discontinuities at x = prime. And like (for example) square wave Fourier approximations there is a noticeable Gibbs-like effect near these discontinuities.
$\rho_n$ are non-trivial zeros of the Riemann zeta function. I should add the possibility that what appears to be prime may be nearly prime.
I checked through $x = 100.$ 83 seemed doubtful until I used a better approximation and 97 seems to require more summands than I have time to calculate. Primes are not the only such points. The expression appears to show the same behavior for a lot of composites (4 is one) less than 20 but fewer as x grows.
Can anyone explain why this would be so? Thank you.
FWIW as a check on my work I think that the expression below in which $\alpha_{n}$ are imaginary part of $\rho_n$ is equivalent and (like the 2d one above) assumes RH. It seems to give the same approximation as the second expression above.
$$\sum_{\zeta(\rho)=0}\frac{x^{\rho}}{\rho} = \sqrt{x}\cdot\sum_{n =1}^{\infty}\frac{\cos(\alpha_{n}\ln x)+2\alpha_{n}\sin(\alpha_{n}\ln x)}{1/4+\alpha_{n}^2}\hspace{5mm}$$