# More elegant $\zeta(s)$ zeros counting function than $N(T)$

The explicit formula expresses the deep connection between the primes $p$ and the non-trivial zeros $\rho$ of $\zeta(s)$. The prime-counting function is given by the following formula giving primes in terms of $\rho$ (the notation $*$ means that at any point of discontinuity we take the average of the left and right limits):

$$\pi^*(x)=R(x)-\sum_\rho R(x^\rho)-\frac{1}{\ln{x}}+\frac{1}{\pi}\arctan{\frac{\pi}{\ln{x}}}$$

where $R(x)=\sum\limits_{n=1}^\infty \frac{\mu(n)}{n}\operatorname{li}(x^{\frac{1}{n}})$ where $\operatorname{li}(x)$ is the logarithmic integral. This is complicated, and the Mobius function $\mu(n)$ contains information about primes anyway. However we have an explicit formula for the alternative counting function $\psi^*(x)$ (counting logarithms of $p$ at prime powers):

$$\psi^*(x)=x-\sum_\rho \frac{x^\rho}{\rho}-\ln{2\pi}-\frac{1}{2}\ln{\left(1-\frac{1}{x^2}\right)}$$

which is a simpler and more elegant expression for how $p$ depend on $\rho$. I always assumed this would have a companion formula giving a counting function for $\rho$ in terms of $p$, and wondered what it was. This formula is not an example since the RHS refers to $\zeta(s)$ and doesn't capture the Fourier-type duality between $p$ and $\rho$. However I recently came across a formula due to Guinand (paper here; formula also mentioned here) for the counting function $N(T)=\frac{1}{2\pi i}\int_{\delta R}\frac{\zeta'(s)}{\zeta(s)}\;ds$ where $R=[-\varepsilon,1+\varepsilon]\times[0,T]$ (i.e. $\left|\{\rho\;|\;\Im(\rho)\in[0,T]\}\right|$; see this) assuming the Riemann hypothesis:

$$N^*(T)=\frac{T}{2\pi}\ln{\frac{T}{2\pi}}-\frac{T}{2\pi}-{\frac{1}{\pi}\lim_{N\rightarrow\infty}\left[\sum_{n=1}^N \Lambda(n)\frac{\sin{(T\ln{n})}}{\sqrt{n}\ln{n}}-\int_1^N \frac{\sin{(T\ln{t})}}{\sqrt{t}\ln{t}}\;dt-\frac{\sin{(T\ln{N})}}{\ln{N}}\left(\sum_{n=1}^N \frac{\Lambda(n)}{\sqrt{n}}-2\sqrt{N}\right)\right]}+\frac{1}{2\pi}\left(\arg{\Gamma\left(\frac{1}{2}+iT\right)}-T\ln{T}+T\right)+\frac{1}{\pi}\arctan{2T}-\frac{1}{4\pi}\arctan{\sinh{\pi T}}$$

This is what I was looking for since it effectively gives $N(T)$ in terms of the primes. However it is very complicated, making me wonder if there is an alternative way to define a counting function on $\rho$ for which a more elegant expression in terms of $p$ holds.

My Question: Does anyone know of an alternative way to $N(T)$ of counting $\rho$ (analogous to $\psi^*(x)$ for the primes) for which there is a simpler or more elegant expression than that for $N(T)$ in terms of $p$?

• I don't see what you mean. Assuming the RH, the Fourier transform of $h(u) = \frac{\psi(e^u)-e^u-\frac{\ln(1-e^{-2u})}{2}}{e^{u/2}}$ is $\hat{h}(\xi) = \sum_\lambda \delta(\xi-\lambda)$ where $\lambda= Im(\rho)$. This is what says en.wikipedia.org/wiki/… by looking at $\langle h,f \rangle =\langle \hat{h},\hat{f} \rangle$ where $f$ is a test function. Of course $h$ is given by the primes, and $N(T) = \langle \hat{h},\hat{f} \rangle$ for some appropriate test function $\hat{f}$. – reuns Jan 14 '17 at 19:05
• And the other way for expressing $N(T)$ is with the argument principle – reuns Jan 14 '17 at 19:11
• @user1952009 I am fully aware that the argument principle can be used (I mentioned that in my question); however, the resulting formula is not the sort of formula I was asking for (being simply the definition of $N(T)$). As to your other point; I am not sure I quite get your point. I know that the Weil formula can be used, but your method has left $f$ undefined; is there a formula that can be deduced by this method that is any simpler than the Guinand formula I quoted? I suspect that this is not the case. – Anon Jan 17 '17 at 5:45
• @Anon The answers to a similar question at the following link provide more information on this topic: mathoverflow.net/q/82635 – Steven Clark Mar 18 '18 at 17:31
• @StevenClark Thank you for the link; I'd actually already seen this question and linked to it in my question (the formula for $N^*(T)$ I gave is given in the answers to that question). What I was specifically looking for was a different way of counting $\rho$ that gave a more elegant and perhaps intuitive formula than the formula for $N^*(T)$ I've given (just as we can count the primes by $\psi^*(x)$ rather than by $\pi^*(x)$ and get a simpler explicit formula). – Anon Mar 23 '18 at 22:32