# Upper bound for $|\zeta'(s)|$ near the line $\sigma=1$, a detailed proof

In page 285 Apostol leaves as a reader's asigment the proof that $|\zeta'(s)|=O(\log^{2}t)$, this is for every $T>0$ there exists a positive constant $K$ (depending on T) such that $$|\zeta'(s)|\leq K\log^{2}t$$ for all $s=\sigma + it$ with $\sigma \geq 1/2$ satisfying $$\sigma>1-\frac{T}{\log t}$$ and $t\geq e$.

With Apostol's hint I start assuming that $\sigma\geq 2$ that provide $|\zeta'(s)|\leq|\zeta'(2)|=\sum_{n=1}^{\infty}\log n/n^{2}$, but I don't understand that this fulls this case, it is for every $T>0$, for $\sigma\geq 2$, $\exists K>0$ such that $|\zeta'(s)|\leq K\log^{2}t$. I don't kwnow how obtain the bound from previous series.

After following (even I follow Apostol' scheme that shows corresponding question handling for $|\zeta(s)|$) I assume $\sigma<2$ and compute $|s|=\sqrt {\sigma^2 +t^2}\leq 2+t\leq e+t\leq 2t$, $|s-1|\geq\sqrt{0+t^2}=t$, so, from $|s-1|\geq t$ I compute $1/|s-1|\leq 1/t$.

Now, I use a known (a previously proved inequality, formula (18) page 284 of , below I don't copy this, only I bound following Apostol's proof to estimate $|\zeta(s)|$, now to estimate $|\zeta'(s)|$). I essays the method to write $$|\zeta'(s)|\leq \sum_{n=1}^{N}\frac{\log n}{n^\sigma}+t\int_{N}^{\infty}\frac{\log x}{x^{\sigma+1}}dx+\int_{N}^{\infty}\frac{1}{x^{\sigma+1}}dx+\frac{N^{1-\sigma}\log N}{t}+\frac{N^{1-\sigma}}{t^2}$$

By an integration by parts I compute $$\int_{N}^{\infty}\frac{\log x}{x^{\sigma+1}}dx=\frac{\log N}{\sigma N^\sigma}+\frac{N^\sigma}{\sigma^2}$$ Now following Apostol I take $N=[t]$, then $\log n\leq \log t$ if $n\geq N$ and the hypothesis $1-\sigma<T/\log t$ implies $1/n^\sigma=O(1/n)$ see page 285 of  for a full detailed proof, and then I compute $(\log n)/n^\sigma$ as $O((\log n)/n)$.

Near the line $\sigma=1$, using $N<t+1$, I compute $t(\log N)/\sigma N^\sigma=O(\log N)=O(1)$, I believe that this holds because $N$ is fixed (next times I ask to you if big oh computations are true near $\sigma=1$), too $1/(\sigma N^\sigma)=O(1/N)=O(1)$, $N^{1-\sigma}\log N/\log t=O((\log N)/N)=O(1)$, too (as previous I've computed) $1/(N^\sigma)=O(1/N)=O(1)$, and finally $N^{1-\sigma}/t^2=O(1/N^2)=O(1)$ (I believe, another time because $N$ is constant).

By partial summation $$\sum_{n\leq x}\frac{\log n}{n}=\frac{\log x}{x}-\int_{1}^{x}(t+O(1))\frac{1-\log t}{t^2}dt$$ and I believe that I remains equals to $(\log x)/x+\log x-(\log^2 x)/2+O((\log t)/t^2)$. Thus I compute finally

$$|\zeta'(s)|\leq \left(\frac{\log t}{t}+\log t+\frac{\log^2 t}{2}\right)+O\left(\frac{\log t}{t^2}\right)+O(1)+O(1)+O(1)+O(1)+O(1)$$ and this equals to $\leq K\log^2 t$, if I can prove that error term is less than $\log^2 t$, but I don't know.

In , page 314 Murty gives as hint that we need take the inverse of square of companion condition, the inequality that appears at first paragraph in this post, that safisfies $\sigma$, to work the exercise.

I assume that this exercise could was written in some course notes. My question, if you don't find a reference with a detailed proof of this is

Question. A proof verification of statement in first paragraph concerning an upper bound for $|\zeta'(s)|$ near the line $\sigma=1$.

Thanks in advance. If you want edit this post, because this is so much large, you can show a summary when the points that possibly are wrong are clear, as you see.

References:

 Tom M. Apostol, Introduction to Analytic Number Theory, UTM Springer (1976).

 M. Ram Murty, Problems in Analytic Number Theory, Graduate Texts in Mathematics RIM 206, Springer, Second Edition (2008).

There are some mistakes in your computations. For example if $N=\left[t\right]$ then $t\log\left(N\right)/\left(\sigma N^{\sigma}\right)=O\left(\log\left(t\right)\right)$ and not $O\left(1\right)$. Another mistake is in the partial summation. Recalling the formula $$\sum_{n\leq N}a_{n}\phi\left(n\right)=A_{N}\phi\left(N\right)-\int_{1}^{N}A_{t}\phi'\left(t\right)dt$$ where $A_{N}=\sum_{n\leq N}a_{n}$ we have in your case $a_{n}=1$ and $\phi\left(n\right)=\log\left(n\right)/n$ (I'm assuming that because you wrote in the integral $t\left(1+O\left(1\right)\right)\frac{1-\log\left(t\right)}{t}$) and so the first term is $\log\left(N\right)$. But the general idea is right. I write the solution so you can check by yourself if this match with yours or if there are other errors. We have, $$\zeta'\left(s\right)=O\left(\sum_{n\leq N}\frac{\log\left(n\right)}{n}\right)+O\left(\int_{N}^{\infty}\frac{dx}{x^{\sigma+1}}\right)+O\left(t\int_{N}^{\infty}\frac{\log\left(x\right)}{x^{\sigma+1}}dx\right)+O\left(\frac{N^{1-\sigma}\log\left(N\right)}{t}\right)+O\left(\frac{N^{1-\sigma}}{t^{2}}\right)+O\left(\frac{\log\left(N\right)}{N^{\sigma}}\right).$$ For the first sum there is no need to use partial summation; we can observe that $$\sum_{n\leq N}\frac{\log\left(n\right)}{n}\leq\log\left(N\right)\sum_{n\leq N}\frac{1}{n}\ll\log^{2}\left(N\right)$$ for the first integral we have $$\int_{N}^{\infty}\frac{dx}{x^{\sigma+1}}=O\left(\frac{1}{N^{\sigma}}\right)=O\left(1\right)$$ for the second, recalling that $N=\left[t\right]$ $$t\int_{N}^{\infty}\frac{\log\left(x\right)}{x^{\sigma+1}}dx=O\left(\frac{t\log\left(N\right)}{\sigma N^{\sigma}}\right)=O\left(\log\left(t\right)\right)$$ and the other terms are trivially all $O\left(1\right)$. Then $$\zeta'\left(s\right)=O\left(\log^{2}\left(t\right)\right).$$