What is the explicit formula for the staircase function $S(x)$ defined in terms of the zeta zeros?

Question

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?

Answer 1

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).

Answer 2

The whole number staircase function (sometimes called the integer floor function, or natural number counting function) may be expressed as, $$ S(w) = \int_{r=1^-}^{w} \exp^*(dJ(e^v))[\ln r] dr, $$ where $\ln r$ is the independent variable of the convolution exponential, and the argument of the c.e. is the usual prime-power staircase function (steps of $1/k$ at each $p^k$, $p$ prime; see eg "Riemann's Zeta Function," H.M. Edwards), $$ dJ(e^u) = \left( \frac{1}{u} - \sum_{\rho} \frac{e^{(\rho-1)u}}{u} - \frac{1}{e^{u}(e^{2u}-1)u} \right), $$ where the $\rho$ are the non-trivial zeros of the Riemann zeta function.

(Note, for those unfamiliar the convolution exponential (when it converges) is defined as $$ \exp^*(f) = \sum_{k=0}^{\infty} \frac{f^{*k}}{k!} = \delta_0 + f + f*f/2! + f*f*f/3!+..., $$ where $\delta_0$ is the Dirac delta function. It has some of the expected properties, like for functions $f$ and $g$ under convergence conditions, $\exp^*(f+g) = \exp^*(f)*\exp^*(g)$, and if $0<f<g$, then $\exp^*(f)<\exp^*(g)$.)

For proof see this question.