# Is the Mean Value of $|P_k(it)|$ equal to $\sqrt{k}$?

Let $P_k$ be the truncated Prime $\zeta$ function, like $$P_k(it)=\sum_{n=1}^k p_n^{it},$$ with $p_n$ being the $n$th prime. Numerics seem to indicate that the mean value of $|P_k|$ taken over all values of $t$ tends towards $\sqrt{k}$ when $k$ and $t$ gets large, e.g. with $k=1229$ and $t<10^5$

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The numerical mean is $31.4234$, which is still below $\sqrt{1229}=35.057$. Is it possible to prove that? If so how to do that?

And how to calculate something like the standard deviation in this case?

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– draks ... Nov 15 '12 at 0:02

Write $p_k^{it} = e^{i \lambda_k t}$ where $\lambda_k = \log p_k$. Since $\lambda_k$ are linearly independent over the rationals, for given $K$ the dynamical system $F_t(z_1,\ldots,z_K) = (z_1 e^{i \lambda_1 t}, \ldots, z_K e^{i \lambda_K t}) \in {\mathbb T}^K$ is ergodic. Taking $f(z_1,\ldots, z_K) = \left| \sum_{j=1}^K z_j\right|$, we have $$E[f^2] = \int_{[0,1]^K} f(e^{2 \pi i \theta_1},\ldots,e^{2\pi i \theta_K})^2 \ d\theta_1 \ldots d\theta_K = K$$
and $E[f]^2 \le E[f^2]$ by Jensen's inequality. By the Birkhoff ergodic theorem, for almost every $(z_1, \ldots, z_K) \in {\mathbb T}^K$ we have $$\lim_{T \to \infty} \frac{1}{T} \int_0^T\left| z_1 e^{i\lambda_1 t} + \ldots + z_K e^{i \lambda_K t}\right|\ dt = E[f] \le \sqrt{K}$$ for almost every $(z_1, \ldots, z_K) \in {\mathbb T}^K$.
Where did my $P_k(it)=\sum_{n=1}^k p_n^{it}$ go? Looks like I get get when all $z_k=1$ in your $F_t$. Further is there a $e^{i\lambda_jt}$ missing in $f(z_1,\ldots, z_K) = \left| \sum_{j=1}^K z_j\right|$? – draks ... Nov 15 '12 at 0:00
Your $|P_k(it)| = f(F_t(1,\ldots,1))$. No, there isn't an $e^{i\lambda_j t}$ in $f(z_1,\ldots, z_K)$. – Robert Israel Nov 15 '12 at 0:13