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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3$\begingroup$ If $x^\rho$ denotes $x^{\mbox{Re}\;\rho}$ why not writing it $x^{\mbox{Re}\;\rho}$? $\endgroup$ – Julien Mar 24 '13 at 15:28
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$\begingroup$ 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
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$\begingroup$ 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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5$\begingroup$ 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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4$\begingroup$ 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.
answered Nov 15 '15 at 4:56
