Yesterday, a certain very talented and passionate young student from Southern Africa asked me the following question about the Riemann zeta function $\zeta(s)$. He says he "thinks" he knows the answer, but he just wants to hear my views. However, I'm not a number theorist, hence I couldn't answer him. So below is the question:

Consider the Riemann zeta function $\zeta(s)$, and let $\alpha$ be the supremum of the real parts of its zeros. Let $\mu$ denote the Möbius function. Define $S(x)= \sum_{n\leq x} \frac{\mu(n)\log n}{n}$.

Note that

$$\Big(\frac{1}{\zeta(s+1)}\Big)' = -s \int_{1}^{\infty} S(x)x^{-s-1} \mathrm{d}x$$ for $\Re(s)> \alpha-1$, where the prime denotes differentiation. It is known that $S(x)=-1 + o(1)$, thus the above integral converges if and only if $\Re(s)>0$. The student's question is: what does this tell us, if anything, about the value of $\alpha$ ?

PS: Personally, i couldn't verify the above identity, neither could I verify the "known" result that $S(x) = -1 + o(1)$, hence I couldn't answer his question.


The identity $$\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$ was conjectured by Möbius (1832) and proved by Landau (1899). It is a consequence of the prime number theorem. Not surprisingly, the rate of convergence is determined by the (known) zero-free region of $\zeta(s)$. In particular, $$S(x)=-1+O_\epsilon(x^{\alpha-1+\epsilon})$$ holds for any $\epsilon>0$, and $\alpha$ in the exponent cannot be lowered.

Here is a sketch of the proof of the mentioned facts. By Perron's formula, we have (at least for $x\not\in\mathbb{N}$) $$S(x)=\frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty}\left(\frac{-1}{\zeta(s+1)}\right)' \frac{x^s}{s}\,ds.$$ The integration is meant over the vertical line with abscissa $1$. By truncating the integral at some height, and applying the residue theorem appropriately, we can move the line segment of integration to the left with the benefit of $x^s$ being much smaller there. This is the same technique by which the prime number theorem was originally proven. At $s=0$, the derivative inside the integral equals $-1$, while $x^s/s$ has a simple pole with residue $1$. Therefore, as we move the curve of integration to the left of $s=0$, we pick up the main term $-1$. The error term then depends on how far to the left we can move the curve of integration without encountering further poles, i.e. where the zeros of $\zeta(s+1)$ are located. The standard zero-free region already implies my first display. If $\alpha<1$, then we have a much wider zero-free region, and the second display follows. The fact that the exponent is optimal follows by reversing this logic, namely by examining the analytic continuation of the RHS of the OP's formula to the left of $s=0$.

I hope this helps your student, or perhaps this is exactly what she/he had in mind. It is standard material, but a good way to better understand the prime number theorem and its relation to the zeros of $\zeta(s)$.

  • 1
    $\begingroup$ @non-numbertheorist: It does not say anything about $\alpha$. The left hand side of your display is holomorphic in the half-plane $\Re(s)>\alpha-1$, and the RHS is a particular way to represent it in the half-plane $\Re(s)>0$. The fact that this particular representation does not converge beyond the smaller half-plane, says nothing about the LHS (or the zeros of $\zeta(s)$ for that matter). It is conjectured that $\alpha$ equals $1/2$, but as of now, no value in $[1/2,1]$ has been excluded for $\alpha$. As far as we know today, $\alpha$ can be anywhere between $1/2$ and $1$. $\endgroup$
    – GH from MO
    Apr 4 '20 at 8:46
  • $\begingroup$ @non-numbertheorist: How could the identity hold for $s=-0.1$ if the RHS does not converge for $s=-0.1$? The RHS has no value at $s=-0.1$, so there is no equality to talk about. $\endgroup$
    – GH from MO
    Apr 4 '20 at 8:57
  • $\begingroup$ @non-numbertheorist: The identity holds at every $s$ where both sides are defined, i.e. at every $s$ with positive real part. $\endgroup$
    – GH from MO
    Apr 4 '20 at 9:02
  • 1
    $\begingroup$ oh yes, now I see...thanks for the clear explanations ! The identity theorem (which facilitates the extension of the identity to a larger plane, requires both sides to be defined. But I will delete my comments in this long thread, since they seem to divert the subject of the question... $\endgroup$ Apr 4 '20 at 9:14
  • 2
    $\begingroup$ But am sure your answer would be much more helpful to the student than anything I would have told him. $\endgroup$ Apr 4 '20 at 9:22

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.