Let $J(x)$ be Riemann's prime counting function given by $\frac{1}{2}w(x) + \sum_{n < x} w(n)$, where $w(p^k) = \frac{1}{k}$ when $p$ is a prime number and $k$ is a positive integer, and $w$ vanishes everywhere else.

Let the logarithmic integral $\newcommand{\li}{\mathrm{li}} \li(x)$ be some antiderivative of $\frac{1}{\log(x)}$.

I have often seen it asserted that for $x > 1$ or perhaps for $x > 2$, we have $J(x) = \li(x) - \sum_{\rho} \li(x^{\rho})$ or perhaps $J(x) = \li(x) - \sum_{\rho} \li(x^{\rho}) - \log(2)$, or some such variant. Here, the sum is over all zeros (trivial and nontrivial) of the Riemann zeta function, in some suitable order. The contribution over just the trivial zeros is some antiderivative of $\frac{1}{t (1 - t^2) \log(t)}$. I understand how to morally (non-rigorously) derive these formulas up to additive constants, but the additive constant specifics elude me.

I have written this without pinning down $\li$ beyond up to an additive constant. Clearly, adding any nonzero constant to $\li$ will destroy the convergence of any such series with infinitely many terms, so if there is some particular fixed choice of $\li$ to be used for all the terms which makes this converge, there is a unique such choice.

Is it indeed the case, as I often see suggested, that using the particular offset logarithmic integral with $\li(2) = 0$ makes these series converge? If so, I am very curious, what is so special about $2$ here?

If not, is there some other fixed choice of logarithmic integral I should be using for all these terms, or should I be using varying choices of $\li$, or in general, how should I interpret these often seen formulas?


1 Answer 1


See Edwards' book Riemann's Zeta Function. He introduces $J(x)$ on p. 22 (in fact Edwards created the notation $J(x)$, which Riemann had written as $f(x)$). On p. 26 he defines ${\rm Li}(x)$ to be $\int_0^x dt/\log t$, with the integral defined across the point 1 using a Cauchy principal value. On p. 33, equation (3) is the formula $$ J(x) = {\rm Li}(x) - \sum_{{\rm Im}(\rho)> 0} ({\rm Li}(x^\rho) + {\rm Li}(x^{1-\rho})) + \int_x^\infty \frac{dt}{t(t^2-1)\log t} - \log 2 $$ for $x > 1$. The terms in the sum over nontrivial zeros $\rho$ with positive imaginary part pair together $\rho$ and $1-\rho$, where the latter are nontrivial zeros with negative imaginary part. These two terms never agree since $\rho = 1-\rho$ only when $\rho = 1/2$ and $\zeta(1/2) \not= 0$.

The sum counts each nontrivial zero with its multiplicity, and note $\rho$ and $1-\rho$ have the same multiplicity as zeros of the zeta-function. The $-\log 2$ at the end is $\log(1/2) = \log(\xi(0))$ where $\xi(s) = (s-1)\pi^{-s/2}\Gamma(s/2 + 1)\zeta(s)$, so $\xi(0) = -\zeta(0) = 1/2$.

  • $\begingroup$ I see. I take it that each zero with positive imaginary component is included once as an exponent in the sum as a $\mathrm{Li}(x^{\rho})$ term and once again as an exponent in the sum as a $\mathrm{Li}(x^{1 - \rho})$ term, and that this double counting of the zeros with positive imaginary component is why we can get away with leaving out the conjugate zeros with negative imaginary component? $\endgroup$ Oct 12, 2021 at 3:54
  • $\begingroup$ The sum is over zeros where ${\rm Im}(\rho) > 0$ with multiplicity, just like in the Hadamard factorization of an entire function, each term indexed by a zero occurs with the multiplicity of that zero. As $\rho$ runs over those zeros, $1-\rho$ runs over the zeros with negative imaginary part. You could think of the sum as running over pairs of nontrivial zeros $\{\rho, 1-\rho\}$, one with positive imaginary part and the other with negative imaginary part (and both of them with multiplicity). $\endgroup$
    – KConrad
    Oct 12, 2021 at 5:30
  • $\begingroup$ Oh, of course, I was being foolish, I forgot that $1 - \rho$ would itself have negative imaginary part; I was somehow thinking of its conjugate instead. Great, thank you, this has cleared things up for me. $\endgroup$ Oct 12, 2021 at 6:39
  • 1
    $\begingroup$ For $\rho$ on the critical line, $1-\rho$ is $\overline{\rho}$. $\endgroup$
    – KConrad
    Oct 12, 2021 at 8:11

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.