5
$\begingroup$

This is a reference-request about a very simple statement.

The Riemann hypothesis is well-known to be equivalent to $$(1)\ \ \ \pi(x) = \mathrm{Li}(x)+O(x^{1/2} \log x)$$ and to $$(2)\ \ \theta(x)=x+O(x^{1/2} \log^2 x).$$ Here as usual, $\pi(x) = \sum_{p \leq x} 1$ and $\theta(x) = \sum_{p \leq x} \log p$.

I am looking for a reference giving the proof of the equivalence between (1) and (2).

All the analytic number theory textbooks I have looked at gives at best a proof that $\theta(x) \sim \pi(x) \log x$, which is clearly not enough.

$\endgroup$
2
  • 4
    $\begingroup$ Have you tried writing $\pi(x)=\int_1^x\frac 1{\log x}d\theta(x)$, using integration by parts on the right hand side, and then plugging in (2) for $\theta(x)$? For the other direction, do something similar. $\endgroup$ Feb 17, 2013 at 19:52
  • 1
    $\begingroup$ Thanks to unknown, Greg, and GH. I have accepted Greg's answer because what I mainly needed was an authoritative reference that I could cite in a paper that I am written (in a first version, I said "it is well-known that..." but I wanted the whole proof to be complete). $\endgroup$
    – Joël
    Feb 18, 2013 at 15:14

2 Answers 2

7
$\begingroup$

The argument that (2) implies (1) is given as equation (13.5) in Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory. A similar partial summation argument (probably two lines long instead of one) will establish that (1) implies (2).

$\endgroup$
6
$\begingroup$

To complement Greg Martin's response, here is a proof that (1) implies (2).

Write $\pi(t)=\mathrm{Li}(t)+f(t)$, so that $f(t)=O(t^{1/2}\log t)$ by assumption. Then $$ \theta(x) = \int_{2-}^x\log t\ d\pi(t)=\int_{2-}^x\log t\ d\mathrm{Li}(t)+\int_{2-}^x\log t\ d f(t). $$ Here the first integral is $$ \int_{2-}^x\log t\ d\mathrm{Li}(t) = \int_2^x\log t\frac{dt}{\log t} = x-2 $$ and the second integral is $$ \int_{2-}^x\log t\ df(t) = f(x)\log x - \int_2^x \frac{f(t)}{t}dt = O(x^{1/2}\log^2 x) + \int_2^x O(t^{-1/2}\log t)\ dt. $$ The last integral is clearly $$ O(\log x)\int_2^x t^{-1/2}\ dt = O(x^{1/2}\log x),$$ hence altogether $$ \theta(x) = x-2 + O(x^{1/2}\log^2 x)+O(x^{1/2}\log x) = x+O(x^{1/2}\log^2 x). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.