# Questions on Convergence of $\int_1^\infty\left(\psi(x)-x\right)\ x^{-s-1}\ dx$ and $\int_1^\infty\left(\psi'(x)-1\right)\ x^{-s}\ dx$

Relationships (1) and (2) below are valid for $\Re(s)>1$.

(1) $\quad\int_1^\infty\psi(x)\ x^{-s-1}\ dx=-\frac{\zeta'(s)}{s\ \zeta(s)}\,,\quad \Re(s)>1$

(2) $\quad\int_1^\infty\psi'(x)\ x^{-s}\ dx=-\frac{\zeta'(s)}{\ \zeta(s)}\,,\quad \Re(s)>1$

For $\Re(s)>1$, relationship (3) below follows from relationship (1) above and relationship (4) below. I've been told the convergence of (3) for $Re(s)=1$ is equivalent to the Prime Number Theorem, and convergence of (3) for $Re(s)\in\left(\frac{1}{2},1\right)$ is consistent with the Riemann Hypothesis.

(3) $\quad\int_1^\infty(\psi(x)-x)\ x^{-s-1}\ dx=-\frac{\zeta'(s)}{s\ \zeta(s)}+\frac{1}{1-s}\,,\quad \Re(s)>\frac{1}{2}\ (assuming\ RH)$

(4) $\quad\int_1^\infty (-x)\ x^{-s-1}\ dx=\frac{1}{1-s}\,,\quad \Re(s)>1$

For $\Re(s)>1$, relationship (5) below follows from relationship (2) above and relationship (6) below. I believe the convergence of (5) for $Re(s)=1$ is equivalent to the Prime Number Theorem, and convergence of (5) for $Re(s)\in\left(\frac{1}{2},1\right)$ is consistent with the Riemann Hypothesis.

(5) $\quad\int_1^\infty(\psi'(x)-1)\ x^{-s}\ dx=-\frac{\zeta'(s)}{\ \zeta(s)}+\frac{1}{1-s}\,,\quad \Re(s)>\frac{1}{2}\ (assuming\ RH)$

(6) $\quad\int_1^\infty (-1)\ x^{-s}\ dx=\frac{1}{1-s}\,,\quad \Re(s)>1$

I believe that for $Re(s)\in\left(\frac{1}{2},1\right)$, relationship (3) above follows from relationships (7) and (8) below by splitting the integral in (7) at $x=1$ and moving the lower portion of the integral to the right side of (7).

(7) $\quad\int_0^\infty(\psi(x)-x)\ x^{-s-1}\ dx=-\frac{\zeta'(s)}{s\ \zeta(s)}\,,\quad \Re(s)\in\left(\frac{1}{2},1\right)\ (assuming\ RH)$

(8) $\quad\int_0^1 (-x)\,x^{-s-1}\,dx=-\frac{1}{1-s}\,,\quad \Re(s)<1$

I believe that for $Re(s)\in\left(\frac{1}{2},1\right)$, relationship (5) above follows from relationships (9) and (10) below by splitting the integral in (9) at $x=1$ and moving the lower portion of the integral to the right side of (9).

(9) $\quad\int_0^\infty(\psi'(x)-1)\ x^{-s}\ dx=-\frac{\zeta'(s)}{\ \zeta(s)}\,,\quad \Re(s)\in\left(\frac{1}{2},1\right)\ (assuming\ RH)$

(10) $\quad\int_0^1 (-1)\,x^{-s}\ dx=-\frac{1}{1-s}\,,\quad \Re(s)<1$

I'm been attempting to validate (3) and (5) above using (11) and (12) below and this seems to work well for $Re(s)\ge 1$, but not so well for $Re(s)\in\left(\frac{1}{2},1\right)$. I've noticed (11) and (12) seem to converge for $Re(s)\ge1$, but I don't believe either (11) or (12) converge for $Re(s)\in\left(\frac{1}{2},1\right)$.

(11) $\quad\int_1^\infty(\psi(x)-x)\ x^{-s-1}\ dx=\sum _{n=1}^N \frac{\Lambda (n) \left(n^{-s}-N^{-s}\right)}{s}+\frac{1-N^{1-s}}{1-s}\,,\quad\Re(s)\ge 1\ \&\ N\to\infty$

(12) $\quad\int_1^\infty(\psi'(x)-1)\ x^{-s}\ dx=\sum_{n=1}^N\Lambda (n)\,n^{-s}+\frac{1-N^{1-s}}{1-s}\,,\quad \Re(s)\ge 1\ \&\ N\to\infty$

The following three plots illustrate the error between the right side of (11) and the right side of (3) for $\Re(s)=2$, $\Re(s)=1$, and $\Re(s)=\frac{1}{2}+0.01$ using evaluation limits of $N=100$ (blue) and $N=1000$ (orange). Note as $N$ increases from $100$ (blue) to $1000$ (orange), the right side of (11) seems to be converging to the right side of (3) in the first plot for $\Re(s)=2$ and in the second plot for $\Re(s)=1$, but not in the third plot for $\Re(s)=\frac{1}{2}+0.01$.

The following three plots illustrate the error between the right side of (12) and the right side of (5) for $\Re(s)=2$, $\Re(s)=1$, and $\Re(s)=\frac{1}{2}+0.01$ using evaluation limits of $N=100$ (blue) and $N=1000$ (orange). Note as $N$ increases from $100$ (blue) to $1000$ (orange), the right side of (12) seems to be converging to the right side of (5) in the first plot for $\Re(s)=2$ and in the second plot for $\Re(s)=1$, but not in the third plot for $\Re(s)=\frac{1}{2}+0.01$.

Question 1: Assuming the Riemann Hypothesis, is the right side of (11) supposed to converge to the right side of (3) as $N\to\infty$ for $Re(s)\in\left(\frac{1}{2},1\right)$?

Question 2: If the answer to question 1 is no, what formula can be used to evaluate the integral on the left side of (11) for $Re(s)\in\left(\frac{1}{2},1\right)$?

Question 3: Assuming the Riemann Hypothesis, is the right side of (12) supposed to converge to the right side of (5) as $N\to\infty$ for $Re(s)\in\left(\frac{1}{2},1\right)$?

Question 4: If the answer to question 3 is no, what formula can be used to evaluate the integral on the left side of (12) for $Re(s)\in\left(\frac{1}{2},1\right)$?

With respect to questions 2 and 4 above, I'm interested in formulas derived from the $\Lambda(n)$, $UnitStep(x-n)$, and $\delta(x-n)$ functions versus formulas derived from von Mangoldt's explicit formula involving a sum over the zeta zeros.

Theorem 20 : If $\zeta(s)$ has no zeros on $Re(s) > \sigma$ then $\forall \epsilon > 0, \psi(x) = x+O(x^{\sigma+\epsilon})$
Exercice 21 : Use the theorem 20, summation by parts or Abel's summation formula and the theory of Dirichlet series for showing that the RH is equivalent to $-\frac{\zeta'(s)}{\zeta(s)}-\zeta(s) = \sum_{n=1}^\infty (\Lambda(n)-1)n^{-s}$ for $Re(s) > 1/2$
• The formula $-\frac{\zeta'(s)}{\zeta(s)}-\zeta(s)=\sum_{n=1}^N(\Lambda(n)-1)\,n^{-s}$ seems to have the same problem as (11) and (12). It seems to converge as $N$ increases for $Re(s)\ge 1$, but not for $Re(s)\in\left(\frac{1}{2},\,1\right)$. I'm looking for a formula which can be used to illustrate convergence as $N$ increases for $Re(s)\in\left(\frac{1}{2},\,1\right)$. Is there no such formula? I'm wondering if there's perhaps a conditional convergence requirement that needs to be considered when using a finite value of $N$ which is not captured by any of these formulas. Commented Mar 24, 2017 at 3:03
• @StevenClark That's the exercice I proposed : prove $\sum_{n=1}^\infty (\Lambda(n)-1)n^{-s}$ converges for $Re(s) > 1/2$ iff the RH is true Commented Mar 24, 2017 at 3:12