With $\zeta$ denoting the Riemann zeta function, prove that

$$ |\zeta(1+it)| \gg \frac{1}{\log^7 t} $$

uniformly for all $t\geq 10 $.

As a hint, it is suggested to use a part of the proof that $\zeta$ has no roots on the line $\Re(s)=1$.

I guess they refer to the inequality $$\zeta(\sigma)^3|\zeta(\sigma + it)|^4|\zeta(\sigma + 2it)| \geq 1 $$

Also, if you want, you can use the bounds $|\zeta(\sigma + it)| = O(\log t)$ and $|\zeta'(\sigma + it)| = O(\log^2 t)$ uniformly on all $1\leq \sigma \leq 2$, $ t \geq 10$ (though I am pretty sure they work for $t\geq \frac{5}{2}$, as well!).

My thought was to (if necessary) replace $t$ by $2t$ in the above inequality to get a new one, and then multiply both to get big powers and use the $\zeta$ estimate above, but in any case I do not know how to get rid of $\zeta(\sigma))$ which is problematic since the argument inside has imaginary part $0$ and $\sigma =1$ is a simple pole. Perhaps an inequality like $\zeta(\sigma) \leq \frac{\sigma}{\sigma-1}$ might help?! No idea.

Any help appreciated!


1 Answer 1


From the inequalities you posted we have $$\left|\frac{1}{\zeta\left(\sigma+it\right)}\right|\leq\zeta\left(\sigma\right)^{3/4}\left|\zeta\left(\sigma+2it\right)\right|^{1/4}\ll\frac{\log^{1/4}\left(t\right)}{\left(\sigma-1\right)^{3/4}},\,\sigma>1$$ and $$\zeta\left(1+it\right)-\zeta\left(\sigma+it\right)=-\int_{1}^{\sigma}\zeta^{\prime}\left(s+it\right)ds\ll\left(\sigma-1\right)\log^{2}\left(t\right),\,\sigma>1-A/\log\left(t\right),\,A>0$$ hence $$\left|\zeta\left(1+it\right)\right|\gg\frac{\left(\sigma-1\right)^{3/4}}{\log^{1/4}\left(t\right)}-\left(\sigma-1\right)\log^{2}\left(t\right),\sigma>1-A/\log\left(t\right),\,A>0$$ then tacking $$\sigma-1=A_{1}\log^{-9}\left(t\right)$$ where $A_{1}>0$ is sufficiently small we get $$\left|\zeta\left(1+it\right)\right|\gg\frac{1}{\log^{7}\left(t\right)}$$ since all terms have the same order. This proof is taken from Titchmarsh, the theory of the Riemann Zeta function, p. $50-51$.


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