# Approximation of Products of Truncated Prime $\zeta$ Functions

The problem arose, while I was looking at products of power prime zeta functions $$P_x(ks)=\sum_{p\,\in\mathrm{\,primes}\leq x} p^{-ks},$$ with $k\in \mathbb{N}$ and $s=it$ with real $t$. By using (see here) $$\sum_{p\leq x}p^{-s}= \mathrm{li}(x^{1-s}) + O \left( {2|s|x^{1/2}}\log x \right)$$ I get (omitting the error terms for the moment) $$\begin{eqnarray*} P_x(ks)P_x(ms) &\sim& \text{li}(x^{1-ks})\text{li}(x^{1-ms}) &=&\int_0^1 \int_0^1 \frac{1}{\ln( x^{1-ks} u_k) \ln(x^{1-ms}u_m)} du_k du_m\\ &=&{\rm Ei}(1-ks){\rm Ei}(1-ms)&&\\ \end{eqnarray*}$$ How can I simplify this expression? Non-trivial approximations are also welcome, when bounds are given.

Convolution: Since $\text{li}(x^{1-s})=\int_0^{x^{1-s}}(\ln u)^{-1}du$, I thought Integration of Convolutions, with something like $$\int(f*g)(x) \, dx=\left(\int f(x) \, dx\right)\left(\int g(x) \, dx\right)$$ might help, since my problem also deals with products of integrals?

Special Case: When I allow negative values for $k$ and $m$ and set $m=-k$ and get the following $$\begin{eqnarray*} P_x(ks)P_x(-ks)&=&\pi(x) + \sum_{j<m}2\cos (\ln \frac{p_j}{p_k})\\ &\sim& \frac{x^{2}}{(1-ks)(1+ks)(\log x)^2}=\frac{1}{(1-(k|s|)^2)}\frac{x^{2}}{(\log x)^2}\\ \end{eqnarray*}$$ The approximation has only real value, as the original expression had. The following function has the same approximation $$\begin{eqnarray*} \frac{1}{\left(1-(k|s|\right)^2)} \text{li}^2(x),\\ \end{eqnarray*}$$ which would very wealky support something like $$P_x(ks)P_x(-ks) \sim \int_0^{x^{1-ks}} \frac{1}{\ln u} du\int_0^{x^{1+ks}} \frac{1}{\ln u} du \sim K(k,s)\times \left(\int_0^{x} \frac{1}{\ln u} du\right)^2 .$$ Does this give anybody a hint?

Intuitively (from the viewpoint of Fourier Analysis), I would expect the prime zeta function $P(s)=\zeta_P(s)$ to be undefined on the imaginary axis, as argued below. Wikipedia says that "singularities cluster near all points of" the line $\Re s=0$, which also bounds the critical strip of the (full) Riemann zeta function on the left in the complex plane.

Perhaps you should call the partial sums (over $p\leq x$), which you are treating, the "incomplete" or "truncated" prime zeta function.

For $s=it$ with $0\neq t\in\mathbb{R}$ and $0\neq k\in\mathbb{Z}$, $$p^{-ks}=e^{-ikt\log p}=\cos(kt\log p)-i\sin(kt\log p)$$ so that for a fixed $k$, the sum over primes $p$ $$P(ks) =\sum_{p\text{ prime}}p^{-ks} =\sum_{p}e^{-ikt\log p} =\sum_{p}e^{-ik\theta_p}$$ should behave like a "random walk" with unit (or "quantum") steps but whose directions $-k\theta_p$ are not discrete but, rather, continuous, "incoherent" or "decoherent" (in the spectral/polarization/quantum physical sense, in the sense of compressed sensing, where its significance connects with the concept of basis functions in Fourier analysis, or in the general sense of meaning uncorrelated), aperiodic and decreasing, since $$\theta_p=2\pi \left( \frac{t\log p}{2\pi}-\left\lfloor \frac{t\log p}{2\pi}\right\rfloor \right)$$ where the rate of change of direction (modulo $2\pi$ we can ignore the jumps) $\frac{\partial}{\partial p}\theta_p=\frac{t}{p}$ decreases in magnitude (deccelerates) slowly (and hence the autocorrelation between terms increases) for $p$ sufficiently large. Since each $\log p$ is irrational (presumably also transcendental) and each $\theta_p$ is probably also irrational (except perhaps for one value of $p$, due to $t$), the directions never repeat and are therefore dense modulo $2\pi$, so that the sum or expected value should be undefined.

Compare this with a Fourier Series or the Jacobi Theta function. I would not expect convergence from the sequence of partial sums, unless some kind of symmetry were exploited (for example between $k$ & $m$) or imposed in some manner, for example analogously to Cauchy principle values for improper integrals, or perhaps some other sort of regularization. Because of the irrationality of $\log p$ for each $p$, I also would not expect any such symmetry trick for $m,n$ as one sees for example when computing that $\{e^{inx}\}_{n\in\mathbb{Z}}$ form an orthonormal basis in Hilbert space.

If we were given a function $f\in L^2([-\pi,\pi])$ so that the sequence of Fourier coefficients $\{c_k=\hat{f}\}$ in $\mathbb{C}$ tended toward zero as $k\rightarrow\pm\infty$, then we could form a convergent Fourier series and interchange the order of summations in the double summation below to obtain $$\sum_{k\in\mathbb{Z}}c_kP(ks)= \sum_{k\in\mathbb{Z}} \sum_{p}c_ke^{-ik\theta_p}= \sum_{p} \sum_{k\in\mathbb{Z}} c_ke^{-ik\theta_p}= \sum_{p}f(-\theta_p),$$ but this would not be allowed for $c_k\equiv1$ since then $f$ would not be integrable.

The truncated sums of $P(ks)$, except for being deterministic (and the angles $\theta_p$ being aperiodic), also bear some resemblance to quantum random walks (see the discussion around equation 44 in the linked introductory paper). A characteristic difference between classical and quantum random walks is that the expected value $\mathbb{E}|X_n|$ is $O(\sqrt n)$ for classical, but $O(n)$ for quantum random walks, i.e. QRWs are ballistic rather than diffusive. In this context, it would be interesting to see if your results show ballistic-diffusive "crossover".

• Thanks for your answer, but maybe I didn't point out that the sum runs just over all primes less than $x$. But nevertheless it sounds interesting. What do you mean by "$P(s)$ behaves like a random walk with incoherent direction"? – draks ... Feb 7 '12 at 18:10
• Thanks, I did not see that the sum was truncated (did you add this subsequently??). I added some explanation and links relating "coherence" to "correlation" and "incoherence" to what we observe here as well as to its analagous roles in compressed sensing (where it plays a similar role to orthogonal bases in Fourier analysis) and quantum physics (where Quantum decoherence has to do with predictability of a system on a macroscopic level when the quantum effects "cancel out" -- exactly what we might hope of this infinite sum). – bgins Feb 7 '12 at 21:01
• After your answer, I added the $\leq x$ to the very first equation, but it was always in the second line. Sorry for the confusion and thanks for the suggested title. – draks ... Feb 7 '12 at 21:23
• Sorry for not noticing it then, and kudos for some cool math! I would also recommend indicating in some way that it is not $P(ks)$, but also depends on $x$: $P_x(ks)$? – bgins Feb 7 '12 at 21:47

After all I can at least give a partial result and a possible path way to a full solution. I think the question could be settled if this Question on (Semi) Prime Counting Functions could be settled. Let's start with the simple case $k=m=1$ and assume $x$ very large:

$P_x(s)^2$ runs over all semi-primes. So along to the line of thought here, I continue like that: $$P_x(s)^2 = \int_2^x t^{-s} d(2\pi_2(t)-\pi(t^{1/2}))=2\left(t^{-s}\pi_2(t)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\pi_2(t)dt\right)-\left(t^{-s}\pi(t^{1/2})\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\pi(t^{1/2})dt\right),$$ where $\pi_2(t)$ counts all 2-almost primes. The second (-) part is already solved in the linked answer. For the first part there are 2 options:

1. Putting in the asymptotic $\pi_2(t)\sim \left( \frac{t}{\log t} \right) \log\log t$, I get $$\begin{eqnarray*} P_x(s)^2&\sim& C_1 +s\int_{2}^{x}\left( \frac{t^{-s}\log\log t}{\log t} \right) dt\\ &=& C_1 + s\int_{2}^{x^{1-s}}\left( \frac{\log\frac{\log u}{1-s}}{\log u} \right) du\\ &=& C_1 + s\int_{2}^{x^{1-s}}\left( \frac{\log\log u}{\log u} \right) du-s\int_{2}^{x^{1-s}}\left( \frac{\log(1-s)}{\log u} \right) du\\ \end{eqnarray*}$$ with $t=u^{1/(1-s)}$. Here I got stuck with the first integral.
2. Using $\pi_2(t)=\sum_{i=1}^{\pi(t^{1/2})}\left[\pi\left(\frac{t}{p_i}\right)-i+1\right]$ to get $$P_x(s)^2 = C_2 + s\int_{2}^{x}t^{-s-1} \sum_{i=1}^{\pi(t^{1/2})}\left[\pi\left(\frac{t}{p_i}\right)-i+1\right] dt .$$ Also here I'm unsure how to continue: Is it possible to switch summation and integration?

As a non-expert, I'm not sure, if I can adapt the line of thought here that easy. But if I'm not wrong, I think this may be extended to all cases of interest by choosing the correct counting function. Another question concerned with that problem can be found here.

...to be continued...