Choose some $x > 1$. Then
$$
\lim_{T\to\infty} \sum_{\Im(\rho)<T}\cos(\ln(x)\Im(\rho))=-\infty
$$ where $\rho$ ranges over all zeros of the zeta function iff $x$ is prime or the power of some prime. This implies that the zeta function's zeros are in a sort of arithmetic progression, and are more likely to appear when $\cos(\ln(x)\rho)$ is negative.
This can be (non-rigorously) obtained using the explicit formula for Chebyshev's function:
$$
\psi(x)=x-\sum_{\rho} \frac{x^{\rho}}{\rho}-\ln 2\pi-\ln(1-\frac{1}{x^{2}}),
$$
so by taking the derivative, we get that
$$
\psi'(x)=1-\sum_{\rho} x^{\rho}+\frac{2}{x-x^{3}}.
$$
Since $\psi'(x)$ is $\infty$ if $x$ is prime or the power of a prime, $\sum_{\rho} x^{\rho}$ must be $-\infty$. By using Euler’s formula, canceling sines, and dividing by $\sqrt{x}$, we get that
$$
\lim_{T\to\infty}\sum_{\Im(\rho)<T}\cos(\ln(x)\Im(\rho))=-\infty
$$
when $x$ is prime or a power of a prime.
I have also done computations for various values of $x$. If $x=2$ for example, $\sum_{\Im(\rho)<100}\cos(\ln(2)\Im(\rho)) = -7.078$ (first 29 zeros summed), and $\sum_{\Im(\rho)<1000}\cos(\ln(2)\Im(\rho)) = -76.524$ (first 649 zeros summed).
The rate of divergence seems to be almost linear with respect to the number of zeros included in the summation, but it slows down very gradually.
Is this a known result, and is there a formal proof? Also, is there a good approximation for the rate at which this sum goes to $-\infty$?
Best Answer
This is called Landau's formula. More precisely, if we extend the von Mangoldt function $\Lambda(n)$ to the function $\Lambda:\mathbb R_+\to \mathbb R$ by $\Lambda(x)=0$ for non-integer $x$, then $$ \sum_{|\mathrm{Im}\,\rho|\leq T}x^{\rho}=-\frac{T}{2\pi}\Lambda(x) +O_x(\ln T) $$ For $x=2$ and $T=1000$, as the Riemann hypothesis is true for zeros with small imaginary part, you get $$ \sum_{0\leq\mathrm{Im}\,\rho\leq T}\cos(\ln(2)\mathrm{Im}\,\rho) \approx -\frac{1000}{2\pi\sqrt{2}}\ln(2)=-78.0064... $$