# Counting prime powers $p^{\frac{k}{t}} \left( t \in \mathbb{R}{+}, k \in \mathbb{N} \right)$ by changing $\rho$'s in $\psi(x)$?

With $\rho=\beta+ \gamma \,i$ being a non-trivial zero of $\zeta(s)$, the logarithmic prime counting function is:

$$\psi(x) = x - \log(2\pi) - \frac12 \log\left(1- \frac{1}{x^2}\right) - \sum_{\rho} \left(\dfrac{x^{\beta+ \gamma \,i}}{\beta+ \gamma \,i}+\dfrac{x^{1-\beta- \gamma \,i}}{1-\beta- \gamma \,i}\right)$$

This function counts all prime powers of the type $p^k, k \in \mathbb{N}$ residing $\le x$. I wondered what would happen to the $\rho$'s when I would divide $k$ by a real number $t \in \mathbb{R}^{+}$. Numerical evidence suggests that the following formula holds for all $t$ (i.e. counting all prime powers of the form $p^{\frac{k}{t}})$:

$$\psi(x,t) = x - \log(2\pi) - \frac12 \log\left(1- \frac{1}{x^2}\right) - \sum_{\gamma} \left(\dfrac{x^{1-\frac{t}{2}+ t\,\gamma \,i}}{1-\frac{t}{2}+ t\,\gamma \,i}+\dfrac{x^{1-\frac{t}{2}- t\,\gamma \,i}}{1-\frac{t}{2}- t\,\gamma \,i}\right)$$

The graph below illustrates the point (for three values of $t$ and using the first 1000 $\rho$'s):

The adjusted real part $\beta = 1-\frac{t}{2}$ can now become any value $<1$. For instance: at $t=2$ then $\beta=0$ and $\psi(x,2)$ counts all primes, prime powers but now also its squared roots $\le x$.

Could $\psi(x,t)$ also be formally derived 'bottom-up' through Fourier analysis (and assuming the RH)?

• What do you mean by "primes of the type $p^k$"? There are no primes of the type $p^k$, not unless $k=1$. Apr 25, 2015 at 23:32
• Thanks Gerry. That is indeed a sloppy formulation. Have changed it in the post and in its title.
– Agno
Apr 26, 2015 at 8:06

Consider, for $\sigma>1$, the function $$\psi(x,t):=\frac{1}{2\pi i}\int_{\sigma-\infty}^{\sigma+\infty}\frac{x^s}{s}\cdot\frac{-\zeta'}{\zeta}\left(1+\frac{s-1}{t}\right)ds.$$ On the one hand, shifting the contour indefinitely to the left, the residue theorem yields $$\psi(x,t)=x-\frac{\zeta'}{\zeta}\left(1-\frac{1}{t}\right)-\sum_{n=1}^\infty \frac{x^{1-(2n+1)t}}{1-(2n+1)t}-\sum_\rho \frac{x^{1-t+t\rho}}{1-t+t\rho},$$ where the $\rho$-sum is over the zeros of $\zeta$ in the critical strip, with $\rho$ paired up with $1-\rho$. The second term comes from the residue at $s=0$, while the $n$-sum is the contribution of the $\zeta$-zeros at negative even integers $s=-2n$. That is, the function $\psi(x,t)$ defined above is essentially the same as your $\psi(x,t)$ under the Riemann Hypothesis, except that the constant $\log(2\pi)$ and the tiny term $\frac12 \log\left(1- \frac{1}{x^2}\right)$ in your formula for $\psi(x,t)$ should be adjusted as given above for an exact match. On the other hand, by an inverse Mellin transform we obtain, for $x$ not an integer, $$\psi(x,t)=\sum_{n^{1/t}\leq x}\Lambda(n)n^{\frac{1}{t}-1},$$ where $\Lambda(n)$ is the von Mangoldt function supported on the prime powers $n=p^k$. (For $x$ an integer the same holds except that the term for $n=x^t$ must be halved.)

In particular, $\psi(x,t)$ jumps exactly at the numbers of the form $n^{1/t}=p^{k/t}$, and your observation has been explained rigorously.

P.S. Note that for $t=1$ we have $$\frac{\zeta'}{\zeta}\left(1-\frac{1}{t}\right)=\log(2\pi)\qquad\text{and}\qquad \sum_{n=1}^\infty \frac{x^{1-(2n+1)t}}{1-(2n+1)t}=\frac12 \log\left(1- \frac{1}{x^2}\right).$$

• Many thanks GH. Much appreciated! Just to be sure: what you call "the tiny next term in your formula" refers to $\frac12 \log\left(1- \frac{1}{x^2}\right)$? If so, isn't that term equal to the series of the trivial zeros at the negative even integers (i.e. already included in your $\psi(x,t)$)?
– Agno
Apr 26, 2015 at 8:14
• @Agno: I updated my response by making those comments of mine explicit. In particular, see the second and third terms in my second display, as well as the P.S. section at the end. Apr 26, 2015 at 15:00
• Thanks GH. Really like your comprehensive answer. Wondered whether it also works for $t < 0$, i.e. counting reciprocals of prime powers like $\frac{1}{p^{\frac{k}{t}}}$. The terms in $\psi(x,t)$ still converge except for the $n$-sum term that goes to $\infty$ when $t < 0$. This could be solved by extending $\zeta(s)$ to something like $\Gamma(\frac{s}{2})\,\zeta(s)$ to annihilate the trivial zeros (or use the Riemann $\xi(s)$-function). Tried your formula for $t=-1$ (without the $n$-term) and it yields a 'saw tooth' graph that is $0$ at the primes (but also at some other non-integer values).
– Agno
Apr 28, 2015 at 13:25
• @Agno: For $t<0$ several more fundamental problems arise, e.g. the first display for $\sigma>1$ does not yield the third display, and the third display converges to a finite value as $x\to\infty$. At any rate, $\sum_n\Lambda(n)W(n)$ with any smooth compactly supported weight function $W:(0,\infty)\to\mathbb{C}$ can be expressed in terms of the $\zeta$-zeros, this is a standard technique in analytic number theory. Apr 28, 2015 at 22:49