Let $\sigma >1-\dfrac{c}{2\log(|t|+3)},|t|>7/8,$ where $c$ is constant from Theorem about region without zeros of Riemann zeta function. Using the fact that $$\log \zeta(s) - \log \zeta(s_1) = \int_{s_1}^{s} \frac{\zeta'(z)}{\zeta(z)}dz$$ for $s=\sigma + it, s_1=1+\dfrac{1}{\log(|t|+3)}+it$, prove that for $s$ in this region is $$|\log \zeta(s)|\leq \log \log (|t|+3)+O(1)$$ and $$\frac{1}{\zeta(s)}=O(\log(|t|+3)).$$

My work: It's pretty useless, but I will write it.

We know that $\log \zeta(s) = \log \zeta(s_1)+\displaystyle\int_{s_1}^{s} \frac{\zeta'(z)}{\zeta(z)}dz$. Now, $$\begin{align*}|\log \zeta(s)| &\le |\log \zeta(s_1)|+\left|\displaystyle\int_{s_1}^{s} \frac{\zeta'(z)}{\zeta(z)}dz\right|\\ &=|\log \zeta(s_1)|+O(1)\\ &=\left|\log \zeta\left(1+\dfrac{1}{\log(|t|+3)}+it\right)\right|+O(1)\\ &\le |\log(1+\log(|t|+3)+it)|+O(1)\\ &\leq \log \log (|t|+3)+O(1)\end{align*}$$

I used that for $\Re(s)>1$ $$\zeta(s)\le 1+\frac{1}{s-1}.$$

What do you think?

I know (99%) that this work isn't correct.


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