Sum over zeta zeroes - how to handle conjugates?

35 Views Asked by At

Consider the prime counting function:

$$\pi(n)=R(x)-\sum\limits_\rho R\left(x^{\rho}\right)-\frac{1}{\text{ln } x}+\frac{1}{\pi}\text{ arctan }\frac{\pi }{\text{ln } x}$$

where $R$ is Riemann's function (link) and the $\rho$ are the zeroes of the zeta function, with the trivial zeroes easily summed (because they're the even negative integers), and the non-trivial zeroes taken in conjugate pairs in order of increasing imaginary part. Am I right in thinking that

$$\sum_{\rho} R(n^{\rho})=\sum _{\Im(\rho )\leq t} \left(R\left(n^{\bar{\rho }}\right)+R\left(n^{\rho }\right)\right)$$

Is this the right way to deal with the conjugate pairs?