Is it true that

$$|\zeta(\frac{1}{2}+it)|< \frac{1}{4}+t^2$$ for all $t\geq 0$, where $\zeta$ denotes the Riemann zeta function ?

It is known that the left hand-side is $O(t^{0.25})$, and on the the Lindelof Hypothesis, it is actually $O(t^\epsilon)$ for any positive $\epsilon$. But since these results are only valid for large $|t|$, thyey dont seem to answer my question.

Note: I have added abs value for The LHS of inequality because it is not clear if $\zeta(\frac{1}{2}+it)$ is always real number

  • $\begingroup$ In the paper: Hiary, Ghaith A. An explicit van der Corput estimate for ζ(1/2+it). Indag. Math. (N.S.) 27 (2016), no. 2, 524–533. It is proved that $|\zeta(1/2+it)|\le 0.63 t^{1/6}\log t$ for $t\ge 3$ and we have $|\zeta(1/2+it)|\le 1.461$ for $0\le t\le 3$. Of course the proof is more complicate than that in the answer of Carlo Beenakker. $\endgroup$
    – juan
    Feb 26, 2018 at 16:18
  • $\begingroup$ You can find further useful bounds here: iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html $\endgroup$
    – GH from MO
    Feb 26, 2018 at 23:02

2 Answers 2


The inequality at the top of page 99 in Lectures on The Riemann Zeta–Function gives $$|\zeta(\tfrac{1}{2}+it)|<9t^{1/2},\;\;t\geq 1.$$ To cover the whole the interval $t\geq 0$, just add $3/2$ to the right-hand-side of the inequality.

The inequality $|\zeta(\tfrac{1}{2}+it)|<\tfrac{1}{4}+t^2$ holds for $t> 0.8$.



enter image description here

where black dots are values of $\zeta\left(\frac12+it\right)$ and the red ones are at the distance $\frac14+t^2$ from the origin.


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.