# An upper bound of $\left| \frac{1}{s}\log\zeta(s) \right|$ for $\Re s=\sigma>1$, from this integral formula and a related comparison

For $\Re s=\sigma>1$ one has the following known formula $$\frac{1}{s}\log\zeta(s)=\int_1^\infty \Pi(x)x^{-s-1}dx,$$ then if we take the derivative we can write $$\frac{1}{s}\log\zeta(s)=s(s+1)\int_1^\infty \Pi(x)x^{-s-2}\log x dx-s\int_1^\infty\frac{\psi(x)}{x^{s+1}}dx.$$

I would like to do a comparison between my computations with the best/relatively good upper bound of $$\left| \frac{1}{s}\log\zeta(s) \right|$$ for $\Re s=\sigma>1$.

I say with these computations, where I use the Prime Number Theorem or the fact, that is a consequence of this theorem, that there exists superior limit of $\frac{\pi(x)}{x/\log x}$ as $x\to\infty$, we take this as a constant about $1.11$. We can use the triangle inequality to get $$\left| \frac{1}{s}\log\zeta(s) \right| \leq 1.11\cdot\left| (s+1) \right| + \left| s(s+1) \int_1^\infty\sum_{k=2}^\infty\frac{\pi(x^{1/k})}{kx/\log x}x^{-s-1}dx\right|$$ $$\qquad\qquad\qquad\qquad\qquad\qquad+\left| s\int_1^{\infty} \frac{\psi(x)}{x^{s+1}}dx\right| .$$

Question. Can you compute upper bounds for the second and the third summand, using only this kind of statements, I say the Prime Number Theorem or its consequences?

I know that for example for the third summand one has $$\left| s\int_1^{\infty} \frac{\psi(x)}{x^{s+1}}dx\right| \leq |s|\int_1^{\infty} \frac{2\psi(x)}{x^{\sigma+1}}dx,$$ and $\psi(x)=x+o(x)$. Can you finish my computations from this way, and refers us on the other hand a good upper bound of $\left| \frac{1}{s}\log\zeta(s) \right|$ for $\Re s=\sigma>1$? Many thanks.

How about this? Let $s = \sigma + it$ with $\sigma > 1$. $$\zeta(\sigma) = \sum_{n=1}^\infty \frac{1}{n^\sigma} \leq 1 + \int_1^\infty \frac{1}{x^\sigma} \, dx = 1 + \frac{1}{\sigma - 1}.$$ Then $$|\log \zeta(s)| \leq \log \zeta(\sigma) \leq \log\left(1 + \frac{1}{\sigma-1}\right) \leq \frac{1}{\sigma-1}.$$