2
$\begingroup$

Which books cover the proof that Riemann Hypothesis is equivalent to the best error bound for the Prime Number Theorem?

My understanding is that Riemann Hypothesis is equivalent to the best bound of prime number theorem. Von Koch (1901) proved that the Riemann hypothesis is equivalent to the "best possible" bound for the error of the prime number theorem, but Koch's paper is in German, I could not read it.

Can anyone recommend books or English articles which cover this proof?

Also, did Schoenfeld give an improved version of this argument?

$\endgroup$
3
  • 2
    $\begingroup$ See also math.stackexchange.com/questions/614009/…. $\endgroup$ Dec 12, 2014 at 23:03
  • $\begingroup$ Have you tried picking up any book about Analytic Number Theory? These questions are discussed in many textbooks. $\endgroup$ Dec 13, 2014 at 12:01
  • $\begingroup$ I think it's the one thing to prove that RH is equivalent to Koch's error bound, but I would also be interested in the proof that no bound could be optimal - that one has been proved by Littlewood I believe. $\endgroup$ Jan 5, 2015 at 7:45

2 Answers 2

5
$\begingroup$

The method to prove the Prime Number Theorem can be quickly summarized, and further details can be found in any introductory book on analytic number theory.

The proof relies on understanding the analytic properties of the Riemann zeta function $\zeta(s)$. In particular, the classical approach to proving the Prime Number Theorem is to understand the growth of $\Lambda(n)$, the von Mangoldt function, which is $\log p$ if $n = p^k$ for $p$ a prime and $0$ otherwise.

One can show that $\displaystyle \psi(x) = \sum_{n \leq x} \Lambda(n)$ behaves like $\pi(x) \log x$, and so proving the Prime Number Theorem is akin to proving that $\psi(x) \sim x$. The relation to $\zeta(s)$ is the fact that $$ -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \geq 1} \frac{\Lambda(n)}{n^s},$$ so the von Mangoldt function appears as the coefficients of a Dirichlet series whose analytic information is contained in the Riemann zeta function. To get that information, you perform an integral transform whose effect is to pull out the first $x$ coefficients. This is the Mellin transform $$ \psi(x) = \frac{1}{2\pi i} \int_{2 - i \infty}^{2 + i\infty} \left( - \frac{\zeta'(s)}{\zeta(s)} \right)\frac{x^s}{s} ds,$$ for $x$ non-integers.

Now it boils down to using the residue theorem and understanding convergence. We'll consider the former and waive the latter.

As we move the line of integration to the left, we pick up residues from the zeroes of $\zeta(s)$. The largest zero is at $s = 1$, which has the effect of contributing an $x$. For each additional zero of $\zeta(s)$, which I'll denote by $\rho$, we get additional contribution. In total, we would get $$ \psi(x) = x - \sum_{\rho} \frac{x^\rho}{\rho} - \frac{\zeta'(0)}{\zeta(0)} - \frac{1}{2} \log( 1 - x^{-2}).$$

So if there is a zero of real part $\rho$, then there is a contribution to the growth of $\pi(x)\log x$ of order $x^\rho$. The Riemann hypothesis states that the only zero of real greater than $1/2$ is from $1$, which means that the secondary growth terms contribute only things on the order of $x^{1/2}$. In particular, we would have that $\lvert\pi(x)\log x - x\rvert \ll x^{1/2 + \epsilon}$ for any $\epsilon > 0$, and where the $\epsilon$ comes from the fact that we are summing very many zeroes (and the sum contributes something like logarithmic growth).

That's the idea: relate it to analytic information about the zeta function, relate it to the zeroes of the zeta function, handle convergence of everything involved (very technical), and bask in success.

$\endgroup$
0
$\begingroup$

You can find a proof in -H. M. Edwards, "Riemann's Zeta Function", Dover publications, 1974.-

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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