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In the explicit expression for $$\psi_0(x) = x - \sum_{\rho} \frac{x^{\rho}}{\rho} - \frac{\zeta'(0)}{\zeta(0)} - \frac{1}{2} \log (1-x^{-2}) $$ $ x^\rho$ denotes $x^{\mathrm{Re} \rho}$. I wanted to know if there is some formula for $$\sum_{\rho} \frac{x^{\rho}}{\rho}$$ and whether that divisor also denotes the real part of $\rho$ or not.

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    $\begingroup$ If $x^\rho$ denotes $x^{\mbox{Re}\;\rho}$ why not writing it $x^{\mbox{Re}\;\rho}$? $\endgroup$ – Julien Mar 24 '13 at 15:28
  • $\begingroup$ because imaginary powers do not matter. $\rho$ is zero of riemann zeta function which is always imaginary. $\endgroup$ – user58491 Mar 24 '13 at 15:32
  • $\begingroup$ This does not answer my question. $e^{i\pi}=-1\neq 1=e^0=e^{\mbox{Re}\;(i\pi)}$. So if you mean the latter, you don't write the former, no? $\endgroup$ – Julien Mar 24 '13 at 15:36
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    $\begingroup$ I can't see anything in this link that says $x^\rho$ is $x^{\mbox{Re}\;\rho}$. Maybe what you mean is, since the function is real-valued, we can replace $x^\rho/\rho$ by $\mbox{Re}(x^\rho/\rho)$ to keep the real part. This is still different from what you say. $\endgroup$ – Julien Mar 24 '13 at 15:46
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    $\begingroup$ No, $x^\rho$ does not denote $x^{{\rm Re}\,\rho}$, any more than $x$ denotes $x^2$. $\endgroup$ – anon Mar 25 '13 at 1:46
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No, there is no particularly good formula for $$ \sum_\rho \frac{x^\rho}{\rho}, $$ except of course for the explicit formula itself. But this is enough for many applications, such as proving an explicit prime number theorem with error-term (along with the zero-free region for the zeta function).

In this answer, I give a quick sketch of how one gets the explicit formula. In short, one uses the residue theorem on the integral $$ \int_{(2)} -\frac{\zeta'(s)}{\zeta(s)} \frac{X^s}{s} ds,$$ and there are poles at zeroes of the zeta function. These poles have residues $x^\rho/\rho$. This does not mean the real part. And contrary to your claim, the imaginary parts are very important, both for the actual value and for convergence.

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