# Question on Integral Transform Related to Riemann Zeta Function $\zeta(s)$

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.

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

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

• Note that the statement that the integral in (5) converges for $\Re(x)>\frac12$ is unknown—it's equivalent to the Riemann hypothesis. – Greg Martin Mar 14 '17 at 5:45
• Greg: Thanks for reminding me this a conjecture versus a proven fact. – Steven Clark Mar 14 '17 at 5:48
• 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. – Greg Martin Mar 14 '17 at 6:00
• 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. – Steven Clark Mar 14 '17 at 6:27
• 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. – Greg Martin Mar 14 '17 at 6:43