I've been told the Riemann hypothesis implies (1) below.

(1)$\quad\psi\left(e^{\ u}\right)-e^{\ u}=o\left(e^{\ u\ (1/2+\epsilon)}\right)$

I've also been told (1) above implies (2) below converges and is $C^{\ \infty}$.

(2)$\quad\int_{-\infty}^{\infty} \frac{\psi\left(e^{\ u}\right)-e^{\ u}}{e^{\ u\ (1/2+\epsilon)}}\ e^{\ i\ \omega\ u}du$

Question 1: Does a demonstration of (1) above also imply the Riemann hypothesis?

Question 2: Does a demonstration of the convergence of (2) above also imply the Riemann hypothesis?

Question 3: Is there a way to derive an exact formula for (2) above?

Question 4: Can (1) and (2) above be generalized with respect to $\psi\left(e^{\ u}\right)-\left(a\ e^{\ u}+b\right)$?

Question 5: If the answer to question 4 above is yes, what are the generalized versions corresponding to (1) and (2) above, and what are the restrictions on a and b?

To clarify, when I say $\psi\left(e^{\ u}\right)$ I mean $\psi\left(e^{\ u}\right)=\sum_{n<e^u}\Lambda(n)$ versus von Mangoldt's formula for $\psi\left(e^{\ u}\right)$, and I'm looking for a generalization beyond one that is based on terms specific to von Mangoldt's formula.

  • $\begingroup$ I take it by $\psi(e^u)$ you mean $\sum_{n \leq e^u} \Lambda(n)$? $\endgroup$ – Peter Humphries Jan 31 '17 at 22:34
  • $\begingroup$ Yes, and I edited my question above to clarify that point and a bit more. $\endgroup$ – Steven Clark Jan 31 '17 at 23:04

Make the substitution $x = e^u$. Then it is well known that RH is equivalent to $\psi(x) = x + O_{\varepsilon}(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, which answers Question 1 affirmatively. (2) can be written as \[\int_{0}^{\infty} \frac{\psi(x) - x}{\sqrt{x}} x^{-s} \, \frac{dx}{x}\] with $s = -i\omega + \varepsilon$. Actually, it is more natural for this integral to be from $1$ to $\infty$. Then this is essentially the Mellin transform of $\frac{\psi(x) - x}{\sqrt{x}}$, and it can be shown with a little work (using partial summation) that \[\frac{\zeta'}{\zeta}(s) = -s \int_{1}^{\infty} \frac{\psi(x)}{x^s} \, \frac{dx}{x}.\] It follows that \[\int_{1}^{\infty} \frac{\psi(x) - x}{\sqrt{x}} x^{-s} \, \frac{dx}{x} = -\frac{1}{s} \frac{\zeta'}{\zeta}\left(s + \frac{1}{2}\right) + \frac{1}{\frac{1}{2} - s}\] whenever the integral converges. (If you add a bit from $0$ to $1$, then you additionally get a term of the form $-\frac{1}{\frac{1}{2} - s}$.) In particular, if the integral converges for $\Re(s) > 0$, then the right-hand side is a holomorphic function for $\Re(s) > 0$, which is equivalent to RH. This answers Questions 2 and 3.

Questions 4 and 5 seems silly to me because $\psi(x)$ is asymptotic to $x$, not $ax + b$ with $a \neq 1$, so $\psi(x) - (ax + b)$ is asymptotic to $(1 - a)x$ as $x$ tends to infinity. I don't see what you're trying to achieve here.

  • $\begingroup$ My motivation is my formula for $\psi\left(e^u\right)$ has a couple of terms that converge to $a\ e^u+b$, and I can evaluate $\psi\left(e^u\right)-\left(a\ e^u+b\right)$ by simply omitting those terms from my evaluation. As evaluation parameters approach $\infty$, $a$ oscillates around the value 1, and $b$ oscillates around the value $-\log 2\ \pi$, but I don't believe either $a$ or $b$ ever evaluate exactly to these values. $\endgroup$ – Steven Clark Jan 31 '17 at 23:48
  • 1
    $\begingroup$ @SteveC. As P.Humphries wrote it is obvious that $\psi(x) -x = \mathcal{O}(x^{a+\epsilon})$ implies that $\frac{\zeta'(s)}{s\zeta(s)}+\frac{1}{s-1}$ is holomorphic on $Re(s) > a$ (so that $\zeta(s)$ has no zero on $Re(s) > a$). The converse is more tricky, that $\zeta(s)$ has no zero on $Re(s) > a \implies \psi(x) -x = \mathcal{O}(x^{a+\epsilon})$. You'll need what is done for proving the prime number theorem : the residue theorem and some theorems about the density of zeros (i.e. showing the Riemann explicit formula converges) $\endgroup$ – reuns Jan 31 '17 at 23:48

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