Is there any 'guess' as to how the Riemann zeta function $\zeta(\sigma+it)$ (or its modulus) behaves to leading order as $t\rightarrow\infty$, for fixed $\sigma$ in the critical strip? Obviously this can't be known for sure until at least the Lindelof Hypothesis is solved, but have people come up with either a good guess based on numerical data or a definite answer with the assumption of the Riemann Hypothesis?

I know the Lindelof Hypothesis (which is a consequence of RH) would give $\zeta(\sigma+it)=O(t^{0.5-\sigma+\epsilon})$ for $\sigma\le0.5$ and $\zeta(\sigma+it)=O(t^{\epsilon})$ for $\sigma\ge0.5$, but I'm looking for something like $|\zeta(\sigma+it)|\sim Kt^{0.5-\sigma+\epsilon}$ or $K(\log t)^{\alpha}t^{0.5-\sigma+\epsilon}$ for some constants $K$ and $\alpha$.

Many thanks in advance for any help with this!

  • $\begingroup$ Here is the good document related to your question home.olemiss.edu/~mbmilino/MicahThesis.pdf $\endgroup$
    – user21574
    Apr 10, 2014 at 17:22
  • $\begingroup$ I am not getting it? Is your "I'm looking for something...." not stronger then LH, which is certainly not known? $\endgroup$
    – Marc Palm
    Apr 10, 2014 at 17:37
  • $\begingroup$ @plusepsilon.de: Yes, it is stronger than LH. I know no such result has been proved; I was just looking for either numerical data or consequence of RH (which implies LH). $\endgroup$ Apr 10, 2014 at 17:44
  • $\begingroup$ $\zeta$ gets arbitrary small in $1/2 \leq \Re s <1$. I am not sure about $0< \Re s <1/2$. $\endgroup$
    – Marc Palm
    Apr 10, 2014 at 17:50

1 Answer 1


An asymptotic result is much stronger than a Big Oh bound, and no results like the ones you hope for can be true: In Titchmarsh's "Theory of the Riemann Zeta Function", Theorem 11.9 shows that for fixed $\sigma_0$ in the interval $(1/2,1]$, the values of $\log(\zeta(\sigma_0+it))$, $t>0$ are dense in the complex plane. In particular, the real part $\ln|\zeta(\sigma_0+i t)|$ is dense in $\mathbb R$.

Update in response to comment below: The theorem in Titchmarsh referenced above was generalized by Voronin: Let $D_r$ be the closed disk of radius $r<1/4$ centered at $3/4$, and let $f$ be any function holomorphic on the interior of $D_r$, and continuous and non-vanishing on $D_r$. For any $\epsilon>0$ there exists $t$ such that $$ \max_{s\in D_r}\left|\zeta(s+it)-f(s)\right|<\epsilon. $$

This is Voronin's "Universality Theorem." http://en.wikipedia.org/wiki/Zeta_function_universality refers to this as "the remarkable ability of the Riemann zeta-function to approximate arbitrary non-vanishing holomorphic functions arbitrarily well."

Another point of view for the behavior of $\zeta(s)$ on lines in the critical strip is given by the Theorem of Bohr and Jessen: There is a continuous function $F(z,\sigma)$ such that for any rectangle $R$ with sides parallel to the real and imaginary axes, as $T\to\infty$, $$ \frac{1}{2T}m\{t\in [-T,T]:\log(\zeta(\sigma+it)\in R\}\to\iint_RF(x+iy,\sigma)\, dxdy. $$ Here $m$ denotes Lebesgue measure.

The Riemann zeta function is too complicated an object for its absolute value to have an asymptotic on lines in the critical strip.

  • $\begingroup$ OK, so that means we can't get leading-order asymptotics for $\zeta$ itself. But what about $|\zeta|$? Even if the values it takes are dense in $\mathbb{R}^+$, something like $|\zeta(\sigma_0+it)|\sim\log(t)$ would be possible without violating this. $\endgroup$ Apr 8, 2014 at 21:04
  • $\begingroup$ Thanks for the updated answer. So the only value of $\sigma$ where there might possibly be a result like what I'm looking for is $\sigma=0.5$, and even that's unlikely! $\endgroup$ Apr 10, 2014 at 17:42
  • $\begingroup$ Not at $\sigma=0.5$, either. See this answer math.stackexchange.com/questions/268268/… $\endgroup$
    – Stopple
    Apr 10, 2014 at 22:44

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