0
$\begingroup$

Riemann and von Mangoldt derived explicit formulas for the Riemann prime power counting function $J(x)$ and the second Chebyshev function $\psi(x)$ respectively via the following relationships.

(1) $\quad J(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\log\zeta(s)\,\frac{x^s}{s}\ ds=li(x)-\sum _\rho Ei\left(\log(x)\ \rho\right)-\log (2)+\int_x^{\infty } \frac{1}{t \left(t^2-1\right) \log (t)} \, dt$

(2) $\quad\psi(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\left(−\frac{\zeta′(s)}{\zeta(s)}\right)\frac{x^s}{s}\ ds=x-\sum_\rho\frac{x^\rho}{\rho}-\log(2\ \pi)-\frac{1}{2}\log(1-x^{-2})$

Note that the integral in (1) and (2) above also applies to the staircase function $S(x)$ as illustrated in (3) below which seems to imply the possible existence of an explicit formula for the staircase function expressed in terms of the zeta zeros.

(3) $\quad S(x)=\frac{1}{2\ \pi\ i}\int_{a-\infty\ i}^{a+\infty\ i}\zeta(s)\,\frac{x^s}{s}\ ds$

Question 1: Is it possible to derive an explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?

Question 2: Assuming the answer to question 1 above is yes, what is the explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?

$\endgroup$
3
$\begingroup$

The short answer is no. The reason that the zeros of $\zeta(s)$ show up in the first two formulas is because those zeros are actually poles of the integrands. Moving the contour of integration results in contribution from the poles of the integrand, not its zeros. But $\zeta(s) \frac{x^s}s$ doesn't have poles at the zeros of $\zeta(s)$; indeed, its only poles are at $x=1$ and $x=0$, and so the formula for $S(x)$ will be the sum of those two residues, which is $x-\frac12$, plus the contribution from the shifted contours (which will not tend to $0$ in this case).

$\endgroup$
  • $\begingroup$ Great, if you prove me wrong then I will have learned something! $\endgroup$ – Greg Martin Feb 17 '17 at 0:35
  • $\begingroup$ Greg Martin: I don't believe your answer above is entirely correct. For example, $\zeta'(s)=-s\int_0^\infty T(x)\,dx$ for $\Re(s)>1$ where $T(x)=\sum_{n=1}^{\lfloor x\rfloor}\log(n)$ has no poles at the zeta zeros. But note that $\psi(x)$ can be written as a function of $T(x)$ and vice versa. Since $T(x)$ can be written as a function of $\psi(x)$, and $\psi(x)$ can be written as a function of the zeta zeros, it follows that $T(x)$ can be written as a function of the zeta zeros. $\endgroup$ – Steven Clark May 19 '17 at 15:36
  • $\begingroup$ I suspect it might be the same for $S(x)$ as $\vartheta(x)$, $\psi(x)$, $\pi(x)$ and $J(x)$ can all be written as functions of $S(x)$, but I haven't yet derived an inversion formula for $S(x)$ as a function of $\psi(x)$ or $J(x)$. $\endgroup$ – Steven Clark May 19 '17 at 15:38
  • $\begingroup$ @Steven Clark Can you show us how you write $T(x)$ as a function of $\psi(x)$? $\endgroup$ – azerbajdzan Dec 19 '17 at 9:58
  • $\begingroup$ @azerbajdzan Sorry it took so long to respond but I just noticed your request. The relationships are as follows: $T(x)=\sum_{n=1}^{x-1}\psi\left(\frac{x}{n}\right)$ and $\psi(x)=\sum _{n=1}^{x-1}\mu(n)\,T\left(\frac{x}{n}\right)$. The function $T(x)=\sum_{n=1}^x \log(n)$ has a very precise asymptotic $T(x)\approx T_{a}(x)=x\,(\log(x)-1)+\frac{1}{2}\log(2\pi)$ with an error bound very close to $\pm\frac{1}{2}\log(x)=\log(\sqrt x)$. I've wondered for quite some time if the relationship between $\psi(x)$ and $T(x)$ and the precision of $T_a(x)$ can somehow be used to prove the Riemann hypothesis. $\endgroup$ – Steven Clark Feb 14 '18 at 20:29
1
$\begingroup$

I've derived an explicit formula for $S(x)$ from the explicit formula for $Q(x)$ based on the relationship between $S(x)$ and $Q(x)$. The $c(n)$ coefficient referenced in (3) below involves both a Dirichlet convolution and a Dirichlet inverse, but I believe it simplifies such that $c(n)=1$ when $n$ is a square integer ($n\in\{1,4,9,16,...\}$) and $c(n)=0$ when $n$ is not a square integer.

(1) $\quad Q(x)=\sum\limits_{n=1}^x a(n)\,,\quad a(n)=\left|\mu(n)\right|$

(2) $\quad Q_o(x)=\frac{6\,x}{\pi^2}+\sum\limits_{k=1}^K\left(\frac{x^{\frac{\rho_k}{2}}\zeta\left(\frac{\rho_k}{2}\right)}{\rho_k \zeta'\left(\rho_k\right)}+\frac{x^{\frac{\rho _{-k}}{2}}\zeta \left(\frac{\rho_{-k}}{2}\right)}{\rho_{-k}\zeta'\left(\rho_{-k}\right)}\right)+1+\sum\limits_{n=1}^N\frac{x^{-n}\,\zeta(-n)}{(-2\,n)\,\zeta'(-2\,n)}\,,\\$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad K\to\infty\land N\to\infty$

(3) $\quad S_o(x)=\sum_\limits{n=1}^x c(n)\,Q_o\left(\frac{x}{n}\right)$

The following plot illustrates $S_o(x)$ (orange) evaluated at $K=N=200$. $S(x)$ is illustrated in blue as a reference but is mostly hidden under the evaluation of $S_o(x)$. The red discrete portion of the plot illustrates the evaluation of $S_o(x)$ at integer values of $x$.

Illustration of $S_o(x)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.