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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?

2 Answers 2

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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 $S_o(x)$

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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).

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    Great, if you prove me wrong then I will have learned something! – 2017-02-17
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    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. – 2017-05-19
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    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 $J(x)$. – 2017-05-19
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    @Steven Clark Can you show us how you write $T(x)$ as a function of $\psi(x)$? – 2017-12-19
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    @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. – 2018-02-14
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    Also, note that $\zeta'(s)=-s\int_0^\infty T(x)\,x^{-s-1}\,dx$ for $\Re(s)>1$, and $T_a(\frac{x}{2\pi})$ provides a pretty good estimate of the number of non-trivial zeta zeros with positive imaginary part less than x. – 2018-02-14
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    With respect to my original question above, note that $S(x)=\sum_{n=1}^x 1=\lfloor x\rfloor$ and $T(x)$ are related via their first-order derivatives: $T'(x)=\log(x)\,S'(x)$. This suggests the possible derivation $S(y)=\sum_{n=1}^{y-1}\int_{y_0}^y\frac{\frac{\partial}{\partial x}\psi\left(\frac{x}{n}\right)}{\log(x)}\,dx$, but one problem I had with this derivation is I can't seem to evaluate the integral $\int_{y_0}^y\frac{\frac{n^2}{n^2 x-x^3}}{\log(x)}\,dx$ where $\frac{n^2}{n^2 x-x^3}=\frac{\partial\left(-\frac{1}{2}\log\left(1-\frac{n^2}{x^2}\right)\right)}{\partial x}$. – 2018-02-14
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    @Steven Clark It does not look for a nice formula for $T(x)$ in terms of non-trivial zeros. For $S(x)$ it would be even more cumbersome. I do not think it will lead to a proof of RH. – 2018-02-14
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    @azerbajdzan I agree the explicit formula for $T(x)$ is less elegant than the explicit formula for $\psi(x)$, and I'm not even sure its possible to derive one for $S(x)$, but the explicit formula for $T(x)$ derived from the explicit formula for $\psi(x)$ is analogous to the explicit formula for the fundamental prime counting function $\pi(x)$ derived from the explicit formula for Riemann's prime-power counting function $\Pi(x)$. The explicit formula for $T(x)$ also illustrates explicit formulas can be derived for some functions even though their transforms don't have poles at the zeta zeros. – 2018-02-15
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    With respect to a proof of the Riemann hypothesis, I wasn't suggesting use of an explicit formula for $T(x)$ or $\psi(x)$ for that matter. I was only pointing out that $T(x)$ and $\psi(x)$ are intimately related and $T(x)$ has a very precise error bound, and wondering if these two observations could somehow be leveraged for a proof of the Riemann hypothesis. – 2018-02-15
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    The relationship between $T(x)$ and $\psi(x)$ was noticed at least as far back as the 1870s; indeed, that relationship was used by Mertens to prove asymptotic formulas for sums like $\sum_{n\le x} \Lambda(n)/n$–results that are known to be weaker than the prime number theorem itself, much less the sharp version of PNT that the Riemann hypothesis would give. That's not a proof that such leveraging is impossible, but a lot of smart people have looked at these functions over the last century and a half.... – 2018-02-15
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    @GregMartin You're more than likely right that if such a leveraging was possible, someone much smarter than myself would have taken advantage of it by now. I still get confused every time I re-investigate the relationship between $T(x)$ and $\psi(x)$ and try to establish an error bound on $\psi(x)$ based on the narrow error bound on $T(x)$. – 2018-02-21