The question below assumes the following definitions.

$\quad\zeta(s)$ - Riemann zeta function

$\quad\psi(x)$ - second Chebyshev function

$\quad J(x)$ - Riemann prime-power counting function

The following relationships are valid for $\Re(s)>1$.

(1) $\quad\int_0^\infty J(x)\ x^{-s-1}\ dx=\frac{\log\zeta(s)}{s}$

(2) $\quad\int_0^\infty\psi(x)\ x^{-s-1}\ dx=-\frac{\zeta'(s)}{s\ \ \zeta(s)}$

Assuming the Riemann Hypothesis, relationship (5) below is a bit more special in that it's valid for $\Re(s)>\frac{1}{2}$ where $s\ne 1$.

(5) $\quad\int_1^\infty\left(\psi(x)-x\right)\ x^{-s-1}\ dx=-\frac{\zeta'(s)}{s\ \zeta(s)}+\frac{1}{1-s}$

Question:: Assuming the Riemann Hypothesis, is there a relationship for $J(x)$ (or $J'(x)$) that converges for $\Re(s)>\frac{1}{2}$ (perhaps with exceptions such as $s=1$) analogous to relationship (5) above for $\psi(x)$?

Note I changed enumeration of the last relationship above from (5) to (3) in one of my earlier edits. Since earlier comments refer to this relationship as (5), I changed the enumeration back to (5). Later comments that refer to relationship (3) are really referring to relationship (5). I apologize for the confusion which I've created.

Integral (A) below is the closest I've come to convergence for $Re(s)<1$. Integral (A) seems to approximate $\log\zeta(s)$ better for large imaginary values of $s$ than for small imaginary values of $s$.

(A) $\quad \int_{2}^{N}\frac{d\,(\text{J}(x)-\text{li}(x))}{dx}x^{-s}dx$

The following two plots illustrate integral (A) evaluated along the line $\Re(s)=\frac{1}{2}+0.01$ using an upper integration limit of $N=1000$. The real and imaginary parts of integral (A) are illustrated in blue, and the corresponding parts of $\log\zeta(s)$ are illustrated in orange as references. The red discrete portions of the two plots illustrate the evaluation of integral (A) at the first 10 zeta zeros.

enter image description here

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The two plots of integral (A) above were created using formula (B) below.

(B) $\quad\sum_{n=1}^N\text{If}\left[\text{PrimePowerQ}[n],\frac{n^{-s}}{\Omega (n)},0\right]-\text{Ei}((1-s)\log (N))+\text{Ei}((1-s)\log(\text{2}))$

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    $\begingroup$ Note that the statement that the integral in (5) converges for $\Re(x)>\frac12$ is unknown—it's equivalent to the Riemann hypothesis. $\endgroup$ – Greg Martin Mar 14 '17 at 5:45
  • $\begingroup$ Greg: Thanks for reminding me this a conjecture versus a proven fact. $\endgroup$ – Steven Clark Mar 14 '17 at 5:48
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    $\begingroup$ Assuming that your $J(x)$ counts each prime power $p^k$ with weight $\frac1k$: the integral you want is $\int_2^\infty (J(x)-\mathop{\rm li}(x))x^{-s-1}\,dx$. The right-hand side will then be $\frac{\log\zeta(s)}s - \int_2^\infty \mathop{\rm li}(x)x^{-s-1}\,dx$, of course, but I'm not sure whether this last integral has a nice form. $\endgroup$ – Greg Martin Mar 14 '17 at 6:00
  • $\begingroup$ Greg: Thanks. With an upper integration limit of $\infty$, Mathematica gives me an answer for $\Re(s)>1$, but not for $Re(s)>\frac{1}{2}$. But I believe I can at least evaluate $\int_2^X\text{li}(x)\,x^{-s-1}dx$ where X is finite for $\Re(s)>\frac{1}{2}$ which is better than nothing. $\endgroup$ – Steven Clark Mar 14 '17 at 6:27
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    $\begingroup$ No. (3) follows from (2) by just splitting up the left-hand side and integrating the two terms separately; $\int_1^\infty x\cdot x^{-s-1}\,dx = -\frac1{1-s}$ is elementary. Similarly, my comment follows from (1) by just splitting the left-hand integral into two pieces. $\endgroup$ – Greg Martin Mar 14 '17 at 6:43

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