Change in argument of the Riemann zeta function near the critical line

Question

Let $L$ denote the path $\frac{1}{2}+\epsilon$ $\to$ $\frac{1}{2}+\epsilon+iT$ where $\epsilon>0$ is arbitrarily small and $T>3$ is not an ordinate of zero of $\zeta(s)$.

Question: If $\Delta$ denotes the variation in the argument then prove that$\Delta_{L} \arg\zeta(s)=\mathcal{O}(\log T)$?

We have by the argument principle $$\Delta_{L} \arg\zeta(s)= \Im\left\{\int_{L}\frac{\zeta'(s)}{\zeta(s)}ds\right\}$$ So substituting $s=\frac{1}{2}+\epsilon+it$ we have $ds=i \ dt$ and hence

$$\Delta_{L} \arg\zeta(s)= \Im\left\{i\int_{0}^{T}\frac{\zeta'(1/2+\epsilon+it)}{\zeta(1/2+\epsilon+it)}dt\right\}$$ I am thinking to use the Fundamental theorem of calculus but my Professor told it is a bit delicate and not applicable here.

Requesting to share your ideas.

Answer 1

Under the RH there is a branch of $F(s)=\log (s-1)\zeta(s)$ analytic for $\Re(s) > 1/2$. Considering some branches of $\log \zeta(s)$ (a primitive of $\frac{\zeta'(s)}{\zeta(s)}$) and $\log (s-1)$ analytic on $\Re(s)\in (1/2,1)$ you'll have that $$\log \zeta(s) = \log \zeta(1/2+\epsilon)+ F(s)-F(1/2+\epsilon)- \log (s-1)+\log (\epsilon-1/2)$$ whenever $\Re(s)\in (1/2,1)$.

$$\log \zeta(1/2+\epsilon+iT) =F(\zeta(1/2+\epsilon+iT)+O(\log T)= O(\log T)$$ is proven in Titchmarsh in the chapter on the density of zeros, noting that $$F(1/2+\epsilon+iT)=F(2+iT)+\int_{2+iT}^{1/2+\epsilon+iT} F'(s)ds = O(1)+O(\log T)$$

Be careful that the $O$ constants depend on $\epsilon$.

If the RH is untrue then you need to consider a branch of $\log \zeta(s)$ analytic on $\Re(s) > 1$ and on horizontal lines without zeros for $\log \zeta(1/2+\epsilon+iT) =O(\log T)$ to hold.

Answer 2

This is not true without RH or a strong hypothesis on the number of zeroes outside the critical line as if one takes the integral $\frac{1}{2\pi i}\int_{R}\frac{\zeta'(s)}{\zeta(s)}ds$ on the rectangle $R=1/2+\epsilon \to 1/2+\epsilon+iT \to 2+iT \to 2 \to 1/2 +\epsilon$ with an indent around $1$, this is precisely the number of zeroes to the right of $1/2 +\epsilon$ up to height $T$ and we know that the integral on $R-L$ is $O(\log T)$ by classical results (see any RZ book like Titchmarsh) so if the result would hold the number of said zeros would be $O(\log T)$ and that is a very strong result about noncritical line zeroes;

By indenting out zeroes to the right of $1/2+\epsilon$, one can easily see that going on the vertical line $L$ in the OP, the argument jumps by $2\pi$ at each zero so one can see directly that it will be essentially the number of zeroes up to a constant (of course here one has to be careful as the jumps can be apriori both positive and negative) but the first paragraph proves rigorously that $\Delta_{L} \arg\zeta(s)=2\pi N_{1/2+\epsilon}(T)+O(\log T)$ where as usual $N_{\sigma}(T)$ is the number of zeroes to the right of $\sigma$, above the real axis and up to height $T$