Let \(0<\delta<\pi\), then for any \(z\in\mathbb C\) so that \(|\arg z|<\pi-\delta\), we have \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+O(|z|^{-1}),\] and also \[\frac{\Gamma'(z)}{\Gamma(z)}=\log z-\frac{1}{2z}+O(|z|^{-2}),\] uniformly as \(|z|\to\infty\), where logarithm has principal value, and the implicit constant depend at most on \(\delta\).
Let \(s=\sigma+it\) with \(-1 \leqslant \sigma \leqslant 2\) and \(t\) not equal to an ordinate of a zero of \(\zeta(s)\). Then we have \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \frac{1}{s-\rho}+O(\log (|t|+3) ),\] where the summation runs through all the non-trivial zeros of \(\zeta(s)\).
Proof: Recall that \(\xi(s)=s(s-1) \pi^{-s / 2} \Gamma(s / 2) \zeta(s)\). Set \(\tau=|t|+3\).
The logarithmic differentiation yields \[\frac{\xi^{\prime}(s)}{\xi(s)}=b+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right), \quad \text{ where }\quad b=\log(2\sqrt{\pi})-1-\frac{\gamma}{2},\] where the summation runs through all zeros \(\rho=\beta+i \gamma\) of \(\xi(s)\), which are exactly the non-trivial zeros of \(\zeta(s)\).
Moreover, the above sum is absolutely convergent, since \[\sum_{\rho}|\rho|^{-2}<\infty,\] as \(\xi\) is an entire function of order \(1\).
Now using the definition of \(\xi(s)=s(s-1) \pi^{-s / 2} \Gamma(s / 2) \zeta(s)\), we have \[\frac{\xi^{\prime}(s)}{\xi(s)}=\frac{1}{s}+\frac{1}{s-1}-\frac{\log \pi}{2}+\frac{\zeta^{\prime}(s)}{\zeta(s)}+\frac{1}{2} \frac{\Gamma^{\prime}(s/2)}{\Gamma(s/2)}.\]
By the logarithmic differentiation of \(\Gamma(s)\), we obtain \[-\frac{\Gamma^{\prime}(s)}{\Gamma(s)}=\frac{1}{s}+\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n+s}-\frac{1}{n}\right),\] since \[\Gamma(s)=s^{-1} e^{-\gamma s} \prod_{n=1}^{\infty}\left(1+\frac{s}{n}\right)^{-1} e^{s / n}.\]
Therefore, we may write \[\frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s-1}+\frac{\log \pi}{2}+\frac{\gamma}{2}+\sum_{n=1}^{\infty}\left(\frac{1}{2n+s}-\frac{1}{2n}\right)+b+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).\]
Note that \[\begin{aligned} \sum_{1\le n\le\tau}\left|\frac{1}{2n+s}-\frac{1}{2n}\right|=O(\log \tau), \ \text{ and } \ \sum_{n\ge\tau}\left|\frac{1}{2n+s}-\frac{1}{2n}\right|=O\bigg(\frac{|s|}{\tau}\bigg). \end{aligned}\]
Therefore, we can write that \[\begin{aligned} \frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)+O(\log \tau). \end{aligned}\tag{*}\]
Notice that \[\bigg|\frac{\zeta^{\prime}(2+it)}{\zeta(2+it)}\bigg|\le \sum_{n=1}^\infty\frac{\Lambda(n)}{n^2}=O(1).\]
Applying (*) with \(s=2+it\) and using the previous bound, we obtain \[\bigg|\sum_{\rho}\left(\frac{1}{2+it-\rho}+\frac{1}{\rho}\right)\bigg|=O(\log\tau).\]
Adding and subtracting the sum \(\sum_{\rho}\big(\frac{1}{2+it-\rho}+\frac{1}{\rho}\big)\) from (*), we obtain \[\begin{aligned} \frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_{\rho}\left(\frac{1}{s-\rho}-\frac{1}{2+it-\rho}\right)+O(\log \tau). \end{aligned}\]
Note that the zero \(\rho=\beta+i \gamma\), satisfies \[\begin{aligned} \operatorname{Re}\bigg(\frac{1}{2+it-\rho}\bigg)&=\frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2} \ge\frac{1}{4+4(t-\gamma)^2}\ge0,\\ \operatorname{Re}\bigg(\frac{1}{\rho}\bigg)&=\frac{\beta}{\beta^2+\gamma^2}\ge0. \end{aligned}\]
Therefore, we obtain \[\begin{aligned} \sum_{\rho}\frac{1}{4+4(t-\gamma)^2}\le \operatorname{Re}\bigg(\sum_{\rho}\left(\frac{1}{2+it-\rho}+\frac{1}{\rho}\right)\bigg)=O(\log\tau). \end{aligned}\]
This immediately implies that \[\sum_{|\operatorname{Im}\rho-t|\le 1}1\le \sum_{|\operatorname{Im}\rho-t|\le 1}\frac{2}{1+(t-\operatorname{Im}\rho)^2}=O(\log\tau).\]
In other words, the number of zeros \(\rho\) in the strip \(t\le \operatorname{Im}\rho \le t+1\) is at most \(O(\log\tau)\) for any \(t\ge 2\).
By the previous observation we see that \[\begin{aligned} \sum_{\rho: | t-\gamma \mid \leqslant 1} \frac{1}{|2+i t-\rho|} =O \Big(\sum_{\rho: | t-\gamma \mid \leqslant 1}1\Big)=O (\log \tau). \end{aligned}\tag{**}\]
By the previous observation we also have \[\begin{aligned} \sum_{\rho: | t-\gamma \mid > 1}\left|\frac{1}{s-\rho}-\frac{1}{2+it-\rho}\right|=O(\log\tau) \end{aligned}\tag{***}\]
In order to see (***), we split \(\sum_{\rho: | t-\gamma \mid > 1}=\sum_{k\in\mathbb Z_+}\sum_{\rho: k<| t-\gamma |\le k+1}\), and observe, arguing as in (**), that for each \(k\in\mathbb Z_+\) the number of zeros \(\rho\) obeying \(k<| t-\gamma |\le k+1\) is at most \(O(\log (\tau+k))\). Now let \(k \in \mathbb{Z}_+\) and consider the zeros \(\rho\) satisfying \(k<|\gamma-t| \leqslant k+1\). Since \[\left|\frac{1}{s-\rho}-\frac{1}{2+i t-\rho}\right|=\frac{2-\sigma}{|(s-\rho)(2+i t-\rho)|} \leqslant \frac{3}{|\gamma-t|^{2}} \leqslant \frac{3}{k^{2}}\] we infer that the contribution from the sum \(\sum_{\rho: k<| t-\gamma |\le k+1}\) is at most \(O(k^{-2} \log (\tau+k))\). Summing over \(k\in\mathbb Z_+\) we obtain (***).
Finally, combining (**) and (***) with \[\begin{aligned} \frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_{\rho}\left(\frac{1}{s-\rho}-\frac{1}{2+it-\rho}\right)+O(\log \tau), \end{aligned}\] we obtain \[\begin{aligned} -\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\substack{\rho \\|t-\operatorname{Im}\rho| \leqslant 1}} \frac{1}{s-\rho}+O(\log \tau), \end{aligned}\] as desired. $$\tag*{$\blacksquare$}$$
From the proof of the previous lemma, we obtain the following result.
For every real number \(T\ge2\) the number of nontrivial zeros \(\rho\) of the zeta function \(\zeta\) satisfying \(T\le \operatorname{Im} \rho\le T+1\) is at most \(O(\log T)\).
Let \(s=\sigma+it\) and assume that \(|s+2m|\ge1/2\) for every \(m\in\mathbb N\). Then \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \frac{1}{s-\rho}+O(\log (|t|+3) ),\] where the summation runs through all the non-trivial zeros of \(\zeta(s)\).
Proof: By the functional equation \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s),\] and the logarithmic differentiation, we have \[\frac{\zeta'(s)}{\zeta(s)}=\log(2\pi)+\frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right)-\frac{\Gamma'(1-s)}{\Gamma(1-s)}-\frac{\zeta'(1-s)}{\zeta(1-s)}.\]
Except for the logarithmic derivative of the gamma function, the terms on the right-hand side are uniformly bounded in the half plane \(\operatorname{Re}s\le-1\) after removing neighborhoods is \(|s-2m| < 1/2\) of the even integers \(2m\), these being poles of \(\cot(\pi s/2)\).
Then by Stirling’s formula for \(\operatorname{Re}s\le-1\), we obtain \[\bigg|\frac{\zeta'(s)}{\zeta(s)}\bigg|=O(\log(1+|s|)), \quad \text{ since } \quad \bigg|\frac{\Gamma'(1-s)}{\Gamma(1-s)}\bigg|=O(\log(1+|s|)).\]
For \(\operatorname{Re}s\ge2\), we have \[\bigg|\frac{\zeta'(s)}{\zeta(s)}\bigg|=O(1).\]
Also if \(\operatorname{Re}s\le-1\) or \(\operatorname{Re}s\ge2\), then we have \[\frac{1}{s-1}-\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \frac{1}{s-\rho}=O(\log (|t|+3)).\]
If we combine this with the previous lemma the proof follows.$$\tag*{$\blacksquare$}$$
For every real number \(T \geqslant 2\), there exists \(T^{\prime} \in[T, T+1]\) such that, uniformly for \(-1 \leqslant \sigma \leqslant 2\), we have \[\bigg|\frac{\zeta^{\prime}(\sigma+i T^{\prime})}{\zeta(\sigma+i T^{\prime})}\bigg|=O(\log ^{2} T).\]
Proof: We subdivide \([T, T+1]\) into \(O(\log T)\) equal parts of length \(c / \log T\), where \(c>0\) is chosen so that the number of parts exceeds the number of zeros.
By the Dirichlet pigeonhole principle, we deduce that there is a part that contains no zeros. Hence for \(T^{\prime}\) lying in this part, we must have \(\left|T^{\prime}-\gamma\right| \ge c' / \log T\) for some \(c'>0\).
We infer that each summand in the previous lemma is \(O(\log T)\) and since there are \(O(\log T)\) summands by the previous corollary, we obtain the desired estimate. This completes the proof.$$\tag*{$\blacksquare$}$$
There exists an absolute constant \(C>0\) such that \(\zeta(s)\) has no zero \(\rho=\beta+i \gamma\) satisfying \[\beta \geq 1-\frac{C}{\log (|\gamma|+2)}. \tag{*}\]
Proof: At the point \(s = 1\) the zeta function \(\zeta(s)\) has a pole, and so there exists \(c_1\in\mathbb R_+\) so that \(\zeta(s)\) has no zeros in the domain \(|s-1|< 2c_1\). Thus if \(\rho=\beta+i\gamma\) is a nontrivial zero of \(\xi(s)\) then \(|\rho-1|\ge 2c_1\). If \(|\gamma|\le c_1\), then \(1-\beta\ge c_1\ge \frac{c_1}{4\log 2}\ge \frac{c_1}{4\log(|\gamma|+2)}\) implying (*) with \(C=c_1/4\).
We now fix a particular zero \(\rho_0=\beta_{0}+i \gamma_{0}\) of \(\zeta(s)\) such that \(|\gamma_0|>c_1\).
Suppose that \(s=\sigma+it\) with \(\sigma>1\). Taking real parts we obtain \[-\operatorname{Re}\left(\frac{\zeta^{\prime}(s)}{\zeta(s)}\right)=\sum_{n=1}^{\infty} \Lambda(n) n^{-\sigma} \cos (t \log n).\]
Since \(3+4 \cos \theta+\cos 2 \theta=2(1+\cos \theta)^{2} \geq 0\) for any \(\theta\in\mathbb R\), we have \[-3 \frac{\zeta^{\prime}(\sigma)}{\zeta(\sigma)}-4 \operatorname{Re}\left(\frac{\zeta^{\prime}(\sigma+i t)}{\zeta(\sigma+it)}\right)-\operatorname{Re}\left(\frac{\zeta^{\prime}(\sigma+2 it)}{\zeta(\sigma+2 it)}\right) \geq 0. \tag{**}\]
Since \(\zeta(s)\) has a pole of residue \(1\) at \(s=1\), we have \[-\frac{\zeta^{\prime}(\sigma)}{\zeta(\sigma)}=\frac{1}{\sigma-1}+O(1).\]
We consider \(s=\sigma+it\) with \(t=\gamma_{0}\). Since \(\left|\gamma_{0}\right| \geq c_{1}>0\), we have \[\begin{aligned} -\operatorname{Re}\left(\frac{\zeta^{\prime}\left(\sigma+i \gamma_{0}\right)}{\zeta\left(\sigma+i \gamma_{0}\right)}\right) & \leq-\operatorname{Re} \sum_{\left|\gamma-\gamma_{0}\right| \leq 1} \frac{1}{(\sigma-\beta)+i\left(\gamma_{0}-\gamma\right)}\\ &+c_{2} \log \left(\left|\gamma_{0}\right|+2\right) \\ & \leq \frac{-1}{\left(\sigma-\beta_{0}\right)}+c_{2} \log \left(\left|\gamma_{0}\right|+2\right), \end{aligned}\] by proceeding as in the previous lemma.
Similarly, we have \[-\operatorname{Re}\left(\frac{\zeta^{\prime}\left(\sigma+2 i \gamma_{0}\right)}{\zeta\left(\sigma+2 i \gamma_{0}\right)}\right) \leq c_{3} \log \left(\left|\gamma_{0}\right|+2\right).\]
Inserting these three estimates into (**), we deduce that for \(\sigma\) close to 1, \[4\left(\sigma-\beta_{0}\right)^{-1}-3(\sigma-1)^{-1} \leq c_{4} \log \left(\left|\gamma_{0}\right|+2\right)\]
Choosing \(\sigma=1+\frac{1}{2 c_{4} \log \left(\left|\gamma_{0}\right|+2\right)}\), we obtain \[\beta_{0} \leq 1-\frac{1}{14 c_{4} \log \left(\left|\gamma_{0}\right|+2\right)},\] which establishes (*) when \(\left|\gamma_{0}\right| \geq c_{1}\). $$\tag*{$\blacksquare$}$$
Let \(\kappa, T, T^{\prime}\in\mathbb R_+\) be given, and consider the following function \[\begin{aligned} h(x)= \begin{cases} 1 & \text{ if } x\in(1, \infty),\\ \frac{1}{2} &\text{ if } x=1, \\ 0 & \text{ if } x\in(0, 1). \end{cases} \end{aligned}\]
If \(x \neq 1\), then \[\left|h(x)-\frac{1}{2 \pi i} \int_{\kappa-i T^{\prime}}^{\kappa+i T} x^{s} \frac{\mathrm{~d} s}{s}\right| \leq \frac{x^{\kappa}}{2 \pi|\log x|}\left(\frac{1}{T}+\frac{1}{T^{\prime}}\right).\]
If \(x=1\), then \[\begin{aligned} \left|h(1)-\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} \frac{\mathrm{~d} s}{s}\right| & \leq \frac{\kappa}{T+\kappa}. \end{aligned}\]
Proof: Consider first the case when \(x>1\).
Let \(k\) be a sufficiently large integer and let \(\mathcal{R}_{k}\) denote the rectangle with vertices \(\kappa-i T^{\prime}, \kappa+i T\), \(\kappa-k+i T, \kappa-k-i T^{\prime}\).
Since \(0\) belongs to the interior of \(\mathcal{R}_{k}\). By the Cauchy theorem, we may write \[\frac{1}{2 \pi i} \int_{\mathcal{R}_{k}} x^{s} \frac{d s}{s}=1=h(x).\]
Now we have the following upper bounds \[\begin{gathered} \left|\int_{\kappa+i T}^{\kappa-k+i T} x^{s} s^{-1} d s\right|\le \int_{\kappa-k}^\kappa\frac{x^udu}{(u^2+T^2)^{1/2}} \leq \frac{x^{\kappa}}{T|\log x|}, \\ \left|\int_{\kappa-k-i T^{\prime}}^{\kappa-i T^{\prime}} x^{s} s^{-1} d s\right| \le \int_{\kappa-k}^\kappa\frac{x^udu}{(u^2+(T')^2)^{1/2}}\leq \frac{x^{\kappa}}{T^{\prime}|\log x|}, \\ \left|\int_{\kappa-k+i T}^{\kappa-k-i T^{\prime}} x^{s} s^{-1} d s\right| \leq \frac{x^{\kappa-k}}{k-\kappa}\left(T+T^{\prime}\right). \end{gathered}\]
We deduce the stated result by letting \(k\) tend to infinity.
The case \(0<x<1\) can be dealt with in a symmetric way, applying the same argument with \(k\) replaced by \(-k\). We omit the details.
When \(x=1\), we simply note that \[\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} s^{-1} ds=\frac{1}{2 \pi}(\arg (\kappa+i T)-\arg (\kappa-i T))=\frac{1}{\pi} \arctan (T / \kappa).\]
The stated upper bound is now immediate from the following bounds \[0 \leq \frac{\pi}{2}-\arctan y=\int_{y}^{\infty} \frac{d t}{1+t^{2}} \leq \frac{2}{1+y},\] which is valid for all \(y>0\).
This concludes the proof of the lemma.$$\tag*{$\blacksquare$}$$
Let \[\begin{aligned} F(s):=\sum_{n=1}^{\infty} b_{n} n^{-s}, \end{aligned}\] be a Dirichlet series with abscissa of convergence \(\sigma_{c}\) and abscissa of absolute convergence \(\sigma_{a}\).
For \(\kappa>\max \left\{0, \sigma_{a}\right\}\), \(T \geq 1\) and \(x \geq 1\), we have \[\begin{aligned} \sum_{1\le n \le x}b_n=&\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s) x^{s} \frac{ds}{s}\\ &+O\left(x^{\kappa} \sum_{n=1}^{\infty} \frac{\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}\right). \end{aligned}\]
It suffices to show that, for any fixed \(\kappa>0\), and uniformly for \(y>0, T>0\), we have that \[\begin{aligned} \left|h(y)-\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} y^{s} s^{-1} d s\right| = O\big( y^{\kappa} /(1+T|\log y|)\big). \end{aligned}\tag{*}\]
Indeed, for \(\kappa>\max \left\{0, \sigma_{a}\right\}\), \(T \geq 1\) and \(x \geq 1\), we have \[\begin{aligned} \frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s)x^{s} s^{-1} d s =\sum_{n=1}^\infty b_n\bigg(\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} \bigg(\frac{x}{n}\bigg)^{s} \frac{ds}{s}\bigg). \end{aligned}\]
Hence, applying (*) with \(y=x / n\) we obtain \[\begin{aligned} \sum_{n=1}^\infty& b_n\bigg(\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} \bigg(\frac{x}{n}\bigg)^{s} \frac{ds}{s}\bigg)\\ &=\sum_{1\le n \le x}b_n + O\left(x^{\kappa} \sum_{n=1}^{\infty} \frac{\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}\right). \end{aligned}\]
It remains to prove (*): \[\begin{aligned} \left|h(y)-\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} y^{s} s^{-1} d s\right| = O\big( y^{\kappa} /(1+T|\log y|)\big). \end{aligned}\tag{*}\]
When \(T|\log y|>1\), the estimate (*) follows from the first inequality of the previous lemma. Otherwise, when \(T|\log y|\le 1\), we can write
\[\int_{\kappa-i T}^{\kappa+i T} y^{s} s^{-1} ds=y^{\kappa} \int_{\kappa-i T}^{\kappa+i T} s^{-1} ds +y^{\kappa} \int_{\kappa-i T}^{\kappa+i T}\left(y^{i t}-1\right) s^{-1} ds.\]
The second integral is
\[O \bigg(\int_{-T}^{T}|(t\log y) s^{-1}| d t\bigg) =O (T|\log y|) = O(1).\]
Consequently, by the second inequality of the previous lemma, we see that the left-hand side of (*) is \(O(y^{\kappa})\) as desired. $$\tag*{$\blacksquare$}$$
Let \(F(s):=\sum_{n=1}^{\infty} a_{n} n^{-s}\) be a Dirichlet series with \(\sigma_{a}<\infty\).
Suppose that there exists some real number \(\alpha \geq 0\) such that \[\sum_{n=1}^{\infty}\left|a_{n}\right| n^{-\theta} =O\big(\left(\theta-\sigma_{a}\right)^{-\alpha}\big) \quad \text{ for } \quad \theta>\sigma_{a}.\]
Assume that that \(B\) is a non-decreasing function satisfying \[\left|a_{n}\right| \leq B(n)\quad \text{ for all } \quad n\in\mathbb Z_+.\]
Then for \(x \geq 2, T \geq 2, \sigma \leq \sigma_{a}\), and \(\kappa:=\sigma_{a}-\sigma+1 / \log x\), we have \[\begin{aligned} \sum_{1\le n \leq x} \frac{a_{n}}{n^{s}}= & \frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s+w) x^{w} \frac{dw}{w}\\ & +O\left(x^{\sigma_{a}-\sigma} \frac{(\log x)^{\alpha}}{T}+\frac{B(2 x)}{x^{\sigma}}\left(1+\frac{x\log x}{T}\right)\right). \end{aligned}\]
Proof: Apply the first effective Perron formula to the series \(\sum_{n=1}^{\infty} b_{n} n^{-w}\) with \(b_{n}:=a_{n} n^{-s}\), we obtain that \[\begin{aligned} \sum_{1\le n \leq x} \frac{a_{n}}{n^{s}}=&\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s+w) x^{w} \frac{dw}{w}\\ &+O\left(x^{\kappa} \sum_{n=1}^{\infty} \frac{\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}\right), \end{aligned}\] since \(\kappa+\sigma>\sigma_a+1/\log x>\sigma_a\) and \(F(s+w)\) converges absolutely.
We now write \[\begin{aligned} \sum_{n=1}^{\infty} \frac{x^{\kappa}\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}= &\sum_{n\not\in[x/2, 2x]} \frac{x^{\kappa}\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}\\ &+ \sum_{n\in[x/2, 2x]} \frac{x^{\kappa}\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}. \end{aligned}\]
Recalling that \(|b_{n}|=|a_{n}| n^{-\sigma}\), and \(\kappa=\sigma_a-\sigma+1/\log x\), and \[\sum_{n=1}^{\infty}\left|a_{n}\right| n^{-\theta} =O\big(\left(\theta-\sigma_{a}\right)^{-\alpha}\big) \quad \text{ for } \quad \theta>\sigma_{a},\] and using this bound with \(\theta:=\kappa+\sigma=\sigma_a+1/\log x\), we see that the contribution of the first sum is: \[\begin{aligned} \sum_{n\not\in[x/2, 2x]}\frac{x^{\kappa}|b_{n}|}{n^{\kappa}(1+T|\log (x / n)|)}&=O\bigg(x^{\kappa} T^{-1} \sum_{n=1}^{\infty}\left|a_{n}\right| n^{-\kappa-\sigma}\bigg)\\ &=O\big( x^{\sigma_{a}-\sigma} T^{-1}(\log x)^{\alpha}\big), \end{aligned}\] since \(|\log (x / n)|\ge 2\), and \(x^\kappa=x^{\sigma_{a}-\sigma}\).
For \(\frac{1}{2} x \leq n \leq 2 x\), using inequality \(\log y\ge 1-\frac{1}{y}\) for \(y\in\mathbb R_+\) we have \[|\log (x / n)|\ge \frac{|x-n|}{x}.\]
This leads to the following estimate \[\begin{aligned} \sum_{n\in[x/2, 2x]}& \frac{x^{\kappa}\left|b_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}= O\bigg(x^{-\sigma} \sum_{x / 2 \leq n \leq 2 x} \frac{\left|a_{n}\right|}{1+T|\log (x / n)|}\bigg)\\ &=O\bigg( \frac{B(2 x)}{x^{\sigma}} \sum_{x / 2 \leq n \leq 2 x} \min\bigg\{1, \frac{x}{T|x-n|}\bigg\}\bigg). \end{aligned}\]
Splitting \(\sum_{x / 2 \leq n \leq 2 x}=\sum_{x / 2 \leq n \leq x-1}+ \sum_{x-1<n<x+1}+\sum_{x +1 \leq n \leq 2 x}\) we obtain \[\begin{aligned} O\bigg( \frac{B(2 x)}{x^{\sigma}}& \sum_{x / 2 \leq n \leq 2 x} \min\bigg\{1, \frac{x}{T|x-n|}\bigg\}\bigg)\\ &=O\bigg(\frac{B(2 x)}{x^{\sigma}}\bigg(1+\frac{x\log x}{T}\bigg)\bigg). \end{aligned}\]
This completes the proof. $$\tag*{$\blacksquare$}$$
There exists \(T_0\ge2\) such that for any \(T\ge T_0\) and for any \(x\ge 2\), we have \[\begin{aligned} \psi(x)=&x-\sum_{|\operatorname{Im}\rho| \leqslant T}\frac{x^{\rho}}{\rho}-\log (2 \pi)-\frac{1}{2} \log \left(1-\frac{1}{x^{2}}\right)\\ &+O\left(\frac{x(\log x T)^{2}}{T}+\log x\right), \end{aligned}\] where \[\psi(x)=\sum_{n\le x}\Lambda(n).\] Observe that for \(T_0\le T\le x\), we have \[O\left(\frac{x(\log x T)^{2}}{T}+\log x\right)=O\left(\frac{x(\log x)^{2}}{T}\right).\]
Proof: We may suppose that \(x \notin \mathbb{Z}\).
Recall that for every real number \(T \geqslant 2\), there exists \(T^{\prime} \in[T, T+1]\) such that, uniformly for \(-1 \leqslant \sigma \leqslant 2\), we have \[\bigg|\frac{\zeta^{\prime}(\sigma+i T^{\prime})}{\zeta(\sigma+i T^{\prime})}\bigg|=O(\log ^{2} T).\]
Let \(T^{\prime}\) be the number supplied by the above item. Let \(\mathcal{R}\) be the rectangle with vertices \[\kappa-i T^{\prime},\quad \kappa+i T^{\prime}, \quad -(2K+1)+i T^{\prime}, \quad \text{ and } \quad -(2K+1)-i T^{\prime},\] where \(K\in\mathbb N\) is large.
We know that \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}+B-\sum_{n=1}^{\infty}\left(\frac{1}{2n+s}-\frac{1}{2n}\right)-\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).\]
This implies that \(-\zeta^{\prime}(s) / \zeta(s)\) has simple poles at \(s=-2 k\) for \(k\in\mathbb Z_+\) with residue \(-1\), and the residue at \(s=1\), which is equal to \(1\).
Therefore, by the residue theorem we obtain \[\frac{1}{2 \pi i} \int_{\mathcal{R}}-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s} \mathrm{~d} s=x-\sum_{|\operatorname{Im} \rho| \leqslant T^{\prime}} \frac{x^{\rho}}{\rho}-\sum_{1\le k<K+1/2 }\frac{x^{-2k}}{-2k}-\frac{\zeta^{\prime}(0)}{\zeta(0)},\] since \[\begin{aligned} \operatorname{res}_0\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=-\frac{\zeta^{\prime}(0)}{\zeta(0)}, \quad &\text{ and } \quad \operatorname{res}_1\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=x,\\ \operatorname{res}_{-2k}\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=\frac{x^{-2k}}{2k}, \quad &\text{ and } \quad \operatorname{res}_{\rho}\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=-\frac{x^{\rho}}{\rho}. \end{aligned}\]
It can be shown that \(\zeta^{\prime}(0) / \zeta(0)=\log (2 \pi)\).
Note that \(\psi(x)=\sum_{n\le x}\Lambda(n)=\sum_{n\le x}\frac{\Lambda(n)}{n^s}\) with \(s=0\).
We know that \[\begin{aligned} \bigg|-\frac{\zeta^{\prime}(1+\sigma+it)}{\zeta(1+\sigma+it)}\bigg| =-\frac{\zeta^{\prime}(1+\sigma)}{\zeta(1+\sigma)}<\frac{1}{\sigma} \end{aligned}\] for any \(\sigma>0\).
Using the second Perron formula with \(\sigma_a=\alpha=1\), \(\sigma=0=\operatorname{Re} s\), \(\kappa=1+1/\log x\) and \(a_n=\Lambda(n)\) and \(B(n)=\log n\), we obtain \[\begin{aligned} \psi(x)= &\frac{1}{2 \pi i} \int_{\kappa-i T'}^{\kappa+i T'} \bigg(-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}\bigg) x^{w} \frac{dw}{w}\\ & +O\left(\frac{x\log x}{T'}+\left(\log x+\frac{x(\log x)^2}{T'}\right)\right). \end{aligned}\]
Therefore, we may write
\[\begin{aligned} \psi(x)= & x-\sum_{|\operatorname{Im}\rho| \leqslant T^{\prime}} \frac{x^{\rho}}{\rho}-\log (2 \pi) +\sum_{1\le k<K+1/2 }\frac{x^{-2k}}{2k}-I_{\mathcal H_{\pm}}-I_{\mathcal{V}} \\ & +O\left(\frac{x(\log x)^{2}}{T^{\prime}}+\log x\right), \end{aligned}\] where \(I_{\mathcal H_{\pm}}=I_{\mathcal{H}_{-}}+I_{\mathcal{H}_{+}}\) and
\(I_{\mathcal{H}_{-}}\) denotes the integral taken over the bottom horizontal side connecting \(-(2K+1)- i T^{\prime}\) with \(\kappa- i T^{\prime}\),
\(I_{\mathcal{H}_{+}}\) denotes the integral taken over the top horizontal side connecting \(-(2K+1)+ i T^{\prime}\) with \(\kappa+ i T^{\prime}\),
\(I_{\mathcal{V}}\) is the integral taken over the left vertical side connecting \(-(2K+1)-i T^{\prime}\) with \(-(2K+1)+i T^{\prime}\).
Since \(T^{\prime} \simeq T\), we obtain \[\begin{aligned} I_{\mathcal{H}_{\pm}} =O \bigg(\int_{-(2K+1)}^{\kappa}\left|\frac{\zeta^{\prime}(\sigma\pm i T^{\prime})}{\zeta(\sigma\pm i T^{\prime})}\right| \frac{x^{\sigma}}{\left|\sigma\pm i T^{\prime}\right|} d\sigma \bigg). \end{aligned}\]
We know that for \(s=\sigma+it\) with \(\sigma\le -1\) one has \[\left|\frac{\zeta^{\prime}(s)}{\zeta(s)}\right|=O(\log(1+|s|)),\] provided that circles of radii \(1/2\) around the trivial zeros \(s = -2k\) are excluded. See the corollary after the first lemma.
Moreover, uniformly for \(-1 \leqslant \sigma \leqslant 2\), we have \[\bigg|\frac{\zeta^{\prime}(\sigma+i T^{\prime})}{\zeta(\sigma+i T^{\prime})}\bigg|=O(\log ^{2} T).\]
Hence, we may conclude that \[\begin{aligned} I_{\mathcal{H}_{\pm}} &=O \bigg(\int_{-(2K+1)}^{\kappa}\left|\frac{\zeta^{\prime}(\sigma\pm i T^{\prime})}{\zeta(\sigma\pm i T^{\prime})}\right| \frac{x^{\sigma}}{\left|\sigma\pm i T^{\prime}\right|} d\sigma \bigg)\\ &=O \bigg(\int_{-(2K+1)}^{-1} \frac{x^{\sigma}\log(1+|\sigma\pm i T^{\prime}|)}{\left|\sigma\pm i T^{\prime}\right|} d\sigma +\int_{-1}^{\kappa} \frac{x^{\sigma}(\log T)^2}{\left|\sigma\pm i T^{\prime}\right|} d\sigma \bigg)\\ &=O\bigg( \frac{x(\log T)^{2}}{T}\bigg). \end{aligned}\]
Moreover, we have \[\begin{aligned} I_{\mathcal{V}} &=O\bigg( \int_{-T^{\prime}}^{T^{\prime}}\left|-\frac{\zeta^{\prime}(-2K-1+i t)}{\zeta(-2K-1+i t)}\right| \frac{x^{-2K-1}}{|-2K-1+i t|} dt\bigg)\\ &=O\bigg( \frac{x^{-2K-1}T\log(KT)}{2K+1}\bigg). \end{aligned}\]
Letting \(K\to\infty\) in the last formula we obtain \[\begin{aligned} \psi(x)= & x-\sum_{|\operatorname{Im}\rho| \leqslant T} \frac{x^{\rho}}{\rho}-\log (2 \pi) +\sum_{k=1}^\infty\frac{x^{-2k}}{2k} \\ & +O\left(\frac{x(\log x)^{2}}{T}+\frac{x(\log T)^{2}}{T}+\log x\right), \end{aligned}\]
This gives the desired formula, since \[\sum_{k=1}^\infty\frac{x^{-2k}}{2k}=-\frac{1}{2} \log \left(1-\frac{1}{x^{2}}\right).\]
This complete the proof. $$\tag*{$\blacksquare$}$$
There exists an absolute constant \(c\in(0, 1)\) such that as \(x \to \infty\), one has \[\begin{align*} \psi(x)&=x+ O\big(xe^{-c \sqrt{\log x}}\big), \tag{*}\\ \pi(x) &= \operatorname{Li}(x) + O\big(xe^{-c \sqrt{\log x}}\big), \tag{**}\end{align*}\] where \[\operatorname{Li}(x):=\int_{2}^x\frac{dt}{\log t}.\] Moreover, one has \[\operatorname{Li}(x)=\frac{x}{\log x}+x \sum_{k=1}^{N-1} \frac{k!}{(\log x)^{k+1}}+O\left(\frac{x}{(\log x)^{N+1}}\right).\]
Proof: For any fixed \(N\in\mathbb Z_+\), by repeated integration by parts, we have \[\operatorname{Li}(x)=\frac{x}{\log x}+x \sum_{k=1}^{N-1} \frac{k!}{(\log x)^{k+1}}+O\left(\frac{x}{(\log x)^{N+1}}\right).\]
The second part (**) follows from the first part (*) by summation by parts. Therefore, it suffices to prove the first part (*).
By Landau’s theorem we obtain, for any \(2\le T \le x\), that \[|\psi(x)-x|\le \sum_{|\operatorname{Im}\rho| \leqslant T} \frac{x^{\operatorname{Re}\rho}}{|\rho|}+O\left(\frac{x(\log x)^{2}}{T}\right).\]
By the zero-free region estimates, there exists an absolute constant \(C\in\mathbb R_+\) such that for every nontrivial zero \(\rho=\beta+i\gamma\) of \(\zeta(s)\), we have \[\operatorname{Re}\rho=\beta\le 1-\frac{C}{\log(|\gamma|+2)}\le1-\frac{C}{\log T}.\]
Hence, inserting this bound into the previous one, we otain \[\sum_{|{\rm Im}\rho| \leqslant T} \frac{x^{{\rm Re}\rho}}{|\rho|}\le x^{1-\frac{C}{\log T}}(\log T)^2.\]
Taking \(T=e^{\sqrt{\log x}}\), we obtain the desired result and (*) follows.$$\tag*{$\blacksquare$}$$
As an immediate corollary, for \(x \to \infty\), we obtain the following estimate \[\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{(\log x)^{2}}\right),\] which is useful in many applications.
Riemann hypothesis (1859) All non-trivial zeros of \(\zeta(s)\) are on the critical line \(\operatorname{Re} s=\frac{1}{2}\).
The Riemann \(\zeta(s)\neq0\) for all \(\operatorname{Re}s>1/2\) if and only if \[\begin{aligned} \psi(x)=x+ O\big(\sqrt{x}(\log x)^2\big). \end{aligned}\tag{*}\]
Proof: (\(\Longrightarrow\)) For zeros \(\rho=\beta+i\gamma\) of \(\zeta(s)\) satisfying \(|\gamma|\le T\), we have \(\beta\le 1/2\). Choosing \(T=\sqrt{x}\) and applying Landau’s theorem we obtain \[\begin{aligned} |\psi(x)-x|&\le \sum_{|\operatorname{Im}\rho| \leqslant T} \frac{x^{\operatorname{Re}\rho}}{|\rho|}+O\left(\frac{x(\log x)^{2}}{T}\right)\\ &=O\left(\sqrt{x}(\log T)^{2}+\sqrt{x}(\log x)^{2}\right)\\ &=O\left(\sqrt{x}(\log x)^{2}\right). \end{aligned}\]
We now prove the reverse implication (\(\Longleftarrow\)). Assume that (*) holds.
For \(\operatorname{Re} s>1\), note that \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}=s\int_1^{\infty}\frac{\psi(y)}{y^{s+1}}dy,\] where \(\psi(x)=\sum_{n\le x}\Lambda(n)\).
If we set \(\psi(x)=x+E(x)\), where \(E(x)=O\big(\sqrt{x}(\log x)^2\big)\), then \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{s}{s-1}+s\int_1^{\infty}\frac{E(y)}{y^{s+1}}dy,\] and clearly the integral defines a holomorphic function for \(\operatorname{Re} s>1/2\).
Consequently the Riemann function \(\zeta(s)\) cannot vanish in this region and the proof is completed. $$\tag*{$\blacksquare$}$$
For every real number \(T\ge2\) the number of nontrivial zeros \(\rho\) of the zeta function \(\zeta\) satisfying \(T\le \operatorname{Im} \rho\le T+1\) is at most \(O(\log T)\).
Suppose \(\gamma\) is a closed path in a region \(\Omega\subseteq \mathbb C\), such that \(\operatorname{Ind}_{\gamma}(\alpha)=0\) for every \(\alpha\not\in\Omega\). Suppose also that \(\operatorname{Ind}_{\gamma}(\alpha)=0\) or \(1\) for ever \(\alpha \in \Omega\setminus\gamma^{*}\), and let \(\Omega_{1}=\{\alpha\in\mathbb C:\operatorname{Ind}_{\gamma}(\alpha)=1\}\). For any \(f \in H(\Omega)\) let \(N_{f}\) be the number of zeros of \(f\) in \(\Omega_{1}\), counted according to their multiplicities. If \(f\in H(\Omega)\) and \(f\) has no zeros on \(\gamma^{*}\) then \[N_{f}=\frac{1}{2 \pi i} \int_{\gamma} \frac{f^{\prime}(z)}{f(z)} d z=\operatorname{Ind}_{\Gamma}(0),\] where \(\Gamma=f \circ \gamma\).
Let \(N(T)\) be the number of zeros of \(\zeta(s)\) with \(s=\sigma+it\) in the region \(0<\sigma<1\), \(0<t \leqslant T\). If \(T\) is not the ordinate of zero of \(\zeta(s)\), then \[N(T)=\frac{T}{2 \pi} \log \bigg(\frac{T}{2 \pi}\bigg)-\frac{T}{2 \pi}+O(\log T)\quad \text{ as } \quad T\to\infty. \tag{*}\]
Proof: Recall that \(\xi(s)=s(s-1) \pi^{-s / 2} \Gamma(s / 2) \zeta(s)\) and the zeros of \(\xi(s)\) are nontrivial zeros of \(\zeta(s)\). Moreover, \(\overline{\xi(s)}=\xi(\overline{s})\).
Let \(\mathcal R\) denote the rectangle with vertices \(2 \pm i T\), \(-1 \pm i T\). Then by the argument principle we have \[2N(T)=\frac{1}{2 \pi i} \int_{\mathcal R} \frac{\xi^{\prime}(s)}{\xi(s)} d s = \frac{1}{2 \pi } \operatorname{Im}\int_{\mathcal R} \frac{\xi^{\prime}(s)}{\xi(s)} d s,\] since \(\overline{\xi(s)}=\xi(\overline{s})\).
Thus we have \[N(T)=\frac{1}{2 \pi i} \int_{\mathcal R} \frac{\xi^{\prime}(s)}{\xi(s)} d s = \frac{1}{4 \pi } \operatorname{Im}\int_{\mathcal R} \frac{\xi^{\prime}(s)}{\xi(s)} d s.\]
With \(\eta(s)=\pi^{-s / 2} \Gamma(s / 2) \zeta(s)\) we therefore may write \[\frac{\xi^{\prime}(s)}{\xi(s)}=\frac{1}{s}+\frac{1}{s-1}+\frac{\eta^{\prime}(s)}{\eta(s)}.\]
Then \[\operatorname{Im}\left[\int_{\mathcal{R}}\left(\frac{1}{s}+\frac{1}{s-1}\right) d s\right]=4 \pi,\] while \(\eta(s)=\eta(1-s)\) and \(\eta(\sigma \pm i t)\) are conjugates, so that \[\operatorname{Im}\left(\int_{\mathcal{R}} \frac{\eta^{\prime}(s)}{\eta(s)} d s\right)=4 \operatorname{Im}\left(\int_{\mathcal{L}} \frac{\eta^{\prime}(s)}{\eta(s)} d s\right),\] where \(\mathcal{L}\) consists of the segments \([2,2+i T]\) and \(\left[2+i T, \frac{1}{2}+i T\right]\).
Therefore \[\begin{aligned} \operatorname{Im}\left(\int_{\mathcal{L}} \frac{\eta^{\prime}(s)}{\eta(s)} d s\right) & =\operatorname{Im}\left[\int_{\mathcal{L}}\left(-\frac{1}{2} \log \pi +\frac{1}{2} \frac{\Gamma^{\prime}(s / 2)}{\Gamma(s / 2)}+\frac{\zeta^{\prime}(s)}{\zeta(s)}\right) d s\right] \\ & =-\frac{1}{2}(\log \pi) T +\operatorname{Im}\left(\int_{\mathcal{L}} \frac{\Gamma^{\prime}(s / 2)}{2 \Gamma(s / 2)} d s +\int_{\mathcal{L}} \frac{\zeta^{\prime}(s)}{\zeta(s)} d s\right) \end{aligned}\]
Using Stirling’s formula and \[\operatorname{Im}\left(\int_{\mathcal{L}} \frac{\Gamma^{\prime}(s / 2)}{2 \Gamma(s / 2)} d s\right)= \operatorname{Im} \log \Gamma\left(\frac{1}{4}+\frac{1}{2} i T\right),\] we obtain \[\operatorname{Im}\left(\int_{\mathcal{L}} \frac{\Gamma^{\prime}(s / 2)}{2 \Gamma(s / 2)} d s\right)=\frac{1}{2} T \log \left(\frac{T}{2}\right)-\frac{T}{2}-\frac{\pi}{8}+O\left(\frac{1}{T}\right).\]
Thus from the above estimates we have \[N(T)=\frac{T}{2 \pi} \log \frac{T}{2 \pi}-\frac{T}{2 \pi}+\frac{7}{8}+\frac{1}{\pi} \operatorname{Im}\left(\int_{\mathcal{L}} \frac{\zeta^{\prime}(s)}{\zeta(s)} d s\right)+O\left(\frac{1}{T}\right).\]
To prove the theorem it remains to show that \[\operatorname{Im}\left(\int_{1 / 2+i T}^{2+i T} \frac{\zeta^{\prime}(s)}{\zeta(s)} d s\right)=O(\log T), \tag{*}\] since the integral over the other segment of \(\mathcal{L}\) is clearly bounded.
We also know that for \(s=\sigma+it\) with \(-1 \leqslant \sigma \leqslant 2\) and \(t\) not equal to an ordinate of a zero of \(\zeta(s)\), we have \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \frac{1}{s-\rho}+O(\log (|t|+3) ),\] where the summation runs through all the non-trivial zeros of \(\zeta(s)\).
Therefore, for \(-1 \leqslant \sigma \leqslant 2\) and \(t\ge 2\) not equal to an ordinate of a zero of \(\zeta(s)\), we have \[\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \frac{1}{s-\rho}+O(\log t),\] where the summation runs through all the non-trivial zeros of \(\zeta(s)\).
Now the proof easily follows, since \[\begin{aligned} \operatorname{Im}\left(\int_{1 / 2+i T}^{2+i T} \frac{\zeta^{\prime}(s)}{\zeta(s)} d s\right) =& O(\log T)+\operatorname{Im}\left(\int_{1 / 2+i T}^{2+i T} \sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1}\frac{d s}{s-\rho}\right) \\ =O(\log T)+&\sum_{\rho:|t-\operatorname{Im}\rho| \leqslant 1} \Delta \arg (s-\rho)=O(\log T), \end{aligned}\] since \(|\Delta \arg (s-\rho)|<\pi\) on \(\left[\frac{1}{2}+i T, 2+i T\right]\) and (*) holds.$$\tag*{$\blacksquare$}$$
Let \(z_{0} \in \mathbb{C}\), and \(r\in(0, R)\) and \(f(z)\) be analytic in \(D\left(z_{0}, R\right)\) given by \[f(z)=\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n} \quad \text { for } \quad z \in D\left(z_{0}, R\right).\] Let \(U\) be a real number such that \(\operatorname{Re}(f(z)) \leq U\) for \(z \in D\left(z_{0}, R\right)\). Then \[\left|c_{n}\right| \leq \frac{2\left(U-\operatorname{Re}\left(f\left(z_{0}\right)\right)\right)}{R^{n}} \quad \text { for } \quad n \in\mathbb N.\]
Further for \(z \in \overline{D}\left(z_{0}, r\right)\), we have \[\left|f(z)-f\left(z_{0}\right)\right| \leq \frac{2 r}{R-r}\left(U-\operatorname{Re}\left(f\left(z_{0}\right)\right)\right).\]
Proof: By considering the function \(f\left(z+z_{0}\right)\) in place of \(f(z)\), we may assume that \(z_{0}=0\).
For \(z\in D(0, R)\), we write \[f(z)=\sum_{n=0}^{\infty} c_{n} z^{n},\] and \[\phi(z)=U-f(z)=U-\sum_{n=0}^{\infty} c_{n} z^{n}=\sum_{n=0}^{\infty} b_{n} z^{n},\] where \[b_{0}=U-c_{0},\quad b_{n}=-c_{n} \quad \text { for } \quad n \in\mathbb N\quad \text { and }\quad \beta_{0}:=\operatorname{Re}\left(b_{0}\right).\]
Then for \(n \geq 0\), we see that \[b_{n}=\frac{1}{2 \pi i} \int_{|z|=r} \frac{\phi(z)}{z^{n+1}} d z.\]
Setting \(z=r e^{i \theta}\) with \(-\pi<\theta \leq \pi\), we obtain for \(n \geq 0\) that \[b_{n}=\frac{1}{2 \pi i} \int_{-\pi}^{\pi} \frac{\phi\left(r e^{i \theta}\right) i r e^{i \theta}}{r^{n+1} e^{i(n+1) \theta}}=\frac{r^{-n}}{2 \pi} \int_{-\pi}^{\pi} \phi\big(r e^{i \theta}\big) e^{-i n \theta} d \theta.\]
From now on, we write \(\phi\big(r e^{i \theta}\big)=P(r, \theta)+i Q(r, \theta):=P+i Q\), where \(P(r, \theta)\) and \(Q(r, \theta)\) are real-valued functions.
Then we have \[b_{n} r^{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}(P+i Q) e^{-i n \theta} d \theta \quad \text { for } \quad n \geq 0.\]
Next, by the Cauchy theorem, for \(r<R\) and \(n \geq 1\), we have \[\begin{aligned} 0=\frac{1}{2 \pi i} \int_{|z|=r} \phi(z) z^{n-1} d z&=\frac{r^{n}}{2 \pi i} \int_{-\pi}^{\pi} \phi\big(r e^{i \theta}\big) i e^{i n \theta} d \theta\\ &=\frac{r^{n}}{2 \pi} \int_{-\pi}^{\pi}(P+i Q) e^{i n \theta} d \theta. \end{aligned}\]
By taking conjugates on both sides, we have \[0=\frac{1}{2 \pi} \int_{-\pi}^{\pi}(P-i Q) e^{-i n \theta} d \theta,\] which, together with \(b_{n} r^{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}(P+i Q) e^{-i n \theta} d\), implies that \[b_{n} r^{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} P(r, \theta) e^{-i n \theta} d \theta \quad \text { for } \quad n \in\mathbb N.\]
Now we take absolute values on both sides to obtain \[\left|b_{n}\right| r^{n} \leq \frac{1}{\pi} \int_{-\pi}^{\pi}\big|P\big(r e^{i \theta}\big)\big| d \theta \quad \text { for } \quad n \in\mathbb N.\]
But we have \[P\big(r e^{i \theta}\big)=\operatorname{Re}\left(\phi\big(r e^{i \theta}\big)\right)=U-\operatorname{Re}\left(f\big(r e^{i \theta}\big)\right) \geq 0.\]
Therefore \[\left|b_{n}\right| r^{n} \leq \frac{1}{\pi} \int_{-\pi}^{\pi} P\big(r e^{i \theta}\big) d \theta \quad \text { for }\quad n \in\mathbb N.\]
We recall that \[\phi\big(r e^{i \theta}\big)=\sum_{n=0}^{\infty} b_{n} r^{n}(\cos n \theta+i \sin n \theta).\]
Therefore \[P(r, \theta)=\operatorname{Re}\left(\phi\big(r e^{i \theta}\big)\right)=\sum_{n=0}^{\infty} r^{n}\big(\operatorname{Re}\left(b_{n}\right) \cos n \theta-\operatorname{Im}\left(b_{n}\right) \sin n \theta\big)\] and hence, using \(\beta_{0}:=\operatorname{Re}\left(b_{0}\right)\), we have \[\frac{1}{2 \pi} \int_{-\pi}^{\pi} P(r, \theta) d \theta=\beta_{0},\] since \(\int_{-\pi}^{\pi} \cos n \theta d \theta=0\) for \(n \geq 1\) and \(\int_{-\pi}^{\pi} \sin n \theta d \theta=0\) for \(n \geq 0\).
Then by \(\left|b_{n}\right| r^{n} \leq \frac{1}{\pi} \int_{-\pi}^{\pi} P\big(r e^{i \theta}\big) d \theta\) we see that \[\left|b_{n}\right| r^{n} \leq 2 \beta_{0} \quad \text { for } \quad n \in\mathbb N.\]
Letting \(r\) tend to \(R\), we deduce that \[\left|c_{n}\right|=\left|b_{n}\right| \leq \frac{2 \beta_{0}}{R^{n}} \quad \text { for } \quad n \in\mathbb N.\]
Now for \(|z| \leq r<R\), we have \[|f(z)-f(0)|=\left|\sum_{n=1}^{\infty} c_{n} z^{n}\right| \leq \sum_{n=1}^{\infty}\left|b_{n}\right| r^{n} \leq 2 \beta_{0} \sum_{n=1}^{\infty}\left(\frac{r}{R}\right)^{n}=2 \beta_{0} \frac{r}{R-r}\]
Inserting \(\beta_{0}=\operatorname{Re}\left(b_{0}\right)=U-\operatorname{Re}(f(0))\) by (5.2.5) in the above inequalities, the lemma follows. $$\tag*{$\blacksquare$}$$
Let \(f(z)\) be holomorphic for \(\left|z-z_{0}\right| \leqslant r\), and \(f\left(z_{0}\right) \neq 0\) and suppose that \(\left|f(z) / f\left(z_{0}\right)\right| \leqslant M\) for \(\left|z-z_{0}\right| \leqslant r\). If \(f(z) \neq 0\) for \(\left|z-z_{0}\right| \leqslant r / 2\), and \(\operatorname{Re}\left(z-z_{0}\right) \geqslant 0\), then
\[\begin{align*} & \operatorname{Re} \bigg(\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}\bigg) \geqslant-\frac{4}{r} \log M \tag{*}\\ & \operatorname{Re} \bigg(\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}\bigg) \geqslant-\frac{4}{r} \log M+\operatorname{Re}\left(\frac{1}{z_{0}-\rho}\right), \tag{**} \end{align*}\] where \(\rho\) is an arbitrary zero of \(f(z)\) in the region \(\left|z-z_{0}\right| \leqslant r / 2\) with \(\operatorname{Re}\left(z-z_{0}\right) <0\).
Proof: Let \[g(z)=f(z) \prod_{\rho}\frac{1}{z-\rho}, \quad z \neq \rho, \quad g(\rho)=\lim _{z \rightarrow \rho} g(z),\] where \(\rho\) denotes zeros of \(f(z)\) in the circle \(\left|z-z_{0}\right| \leqslant r / 2\) counted with respective multiplicities.
Then \(g(z)\) is holomorphic for \(\left|z-z_{0}\right| \leqslant r / 2\), and for \(\left|z-z_{0}\right|=r\) \[\left|\frac{g(z)}{g\left(z_{0}\right)}\right|=\left|\frac{f(z)}{f\left(z_{0}\right)} \prod_{\rho} \frac{z_{0}-\rho}{z-\rho}\right| \leqslant M,\]
By the maximum modulus principle this holds also for \(\left|z-z_{0}\right| \leqslant r\).
Since \(g(z) \neq 0\) for \(\left|z-z_{0}\right| \leqslant r / 2\), then taking the principal branch of the logarithm we see that \(F(z)=\log g(z)-\log g\left(z_{0}\right)\) is regular for \(\left|z-z_{0}\right| \leqslant r / 2\) and \[\operatorname{Re} F(z)=\log \left|g(z) / g\left(z_{0}\right)\right| \leqslant \log M,\] and \(M \geqslant 1\), since \(g(z) / g\left(z_{0}\right)=1\) when \(z=z_{0}\).
Moreover \(\operatorname{Re} F\left(z_{0}\right)=0\), so by the Borel–Carathéodory lemma with \(R=r / 2\) we obtain \[\left|F^{\prime}\left(z_{0}\right)\right|=\left|g^{\prime}\left(z_{0}\right) / g\left(z_{0}\right)\right| \leqslant \frac{4}{r} \log M,\] while by logarithmic differentiation we have \[\left|\frac{g^{\prime}\left(z_{0}\right)}{g\left(z_{0}\right)}\right|=\left|\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}-\sum_{\rho} \frac{1}{z_{0}-\rho}\right| \leqslant \frac{4}{r} \log M,\]
Thus we obtain \[\left|\operatorname{Re}\bigg(\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}-\sum_{\rho} \frac{1}{z_{0}-\rho}\bigg)\right| \leqslant \frac{4}{r} \log M,\]
which implies \[\operatorname{Re}\left(\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}-\sum_{\rho} \frac{1}{z_{0}-\rho}\right) \geqslant-\frac{4}{r} \log M .\]
Finally, the condition \(\operatorname{Re}\left(z_{0}-\rho\right)>0\) ensures that the conclusion of the lemma follows from the last bound, and \[\begin{aligned} \operatorname{Re} \bigg(\frac{f^{\prime}\left(z_{0}\right)}{f\left(z_{0}\right)}\bigg) \geqslant-\frac{4}{r} \log M+\operatorname{Re}\left(\frac{1}{z_{0}-\rho}\right) \end{aligned}\] as desired. $$\tag*{$\blacksquare$}$$
Let \(\varphi(t)\) and \(1 / \theta(t)\) be two positive, nondecreasing functions of \(t\) for \(t \geqslant t_0\) such that \(\theta(t) \leqslant 1\), and \(\lim_{t \to \infty}\varphi(t) = \infty\) and \[\frac{\varphi(t)}{\theta(t)}=o\left(e^{\varphi(t)}\right) \quad\text{ as } \quad t \to \infty.\] If \(\zeta(s) =O \big(e^{\varphi(t)}\big)\) for \(1-\theta(t) \leqslant \sigma \leqslant 2\), and \(t \geqslant t_0\), then \(\zeta(s) \neq 0\) for \[\sigma \geqslant 1-C \frac{\theta(2 t+1)}{\varphi(2 t+1)} \quad\text{ and }\quad t \geqslant t_{0},\] where \(C>0\) is an absolute constant.
Proof: Let \(s=\sigma+it\). Let \(\zeta(\beta+i \gamma)=0\), with \(\beta \leqslant 1\), and \(\gamma \geqslant t_{0}\).
Let \(\sigma_{0}\) satisfy \[1+e^{-\varphi(2 \gamma+1)} \leqslant \sigma_{0} \leqslant 2.\]
Let further \[s_{0}=\sigma_{0}+i \gamma,\quad \text{ and } \quad s_{0}^{\prime}=\sigma_{0}+2 i \gamma,\quad \text{ and } \quad r= \theta(2 \gamma+1)\le 1.\]
Then both the circles \(\left|s-s_{0}\right| \leqslant r\) and \(\left|s-s_{0}^{\prime}\right| \leqslant r\) lie in the region \(\sigma \geqslant 1-\theta(t)\) and \(t \geqslant t_{0}\), since \(|\sigma-\sigma_0|\le r\), and \[\begin{aligned} \sigma\ge \sigma_{0}-r & =\sigma_{0}-\theta(2 \gamma+1) \geqslant 1+e^{-\varphi(2 \gamma+1)}-\theta(2 \gamma+1) \\ & \geqslant 1-\theta(2 \gamma+1)\ge 1-\frac{\theta(2 \gamma+1)}{\theta(t)}\theta(t)\ge 1-\theta(t). \end{aligned}\]
The last inequality follows from the fact that \(t\le 2\gamma+r\le 2\gamma+1\), and \(1/\theta(t)\) is nondecreasing. Hence \(1/\theta(t)\le 1/\theta(2\gamma+1)\) and consequently \(\theta(2\gamma+1)/\theta(t)\le 1\), giving the last lower bound.
For \(\sigma>1\) and some \(A>0\) we have \(|1 / \zeta(s)| \leqslant \zeta(\sigma)<A(\sigma-1)^{-1}\).
Hence \[\left|1 / \zeta\left(s_{0}\right)\right| \leqslant A e^{\varphi(2 \gamma+1)}, \quad \text{ and } \quad \left|1 / \zeta\left(s_{0}^{\prime}\right)\right| \leqslant A e^{\varphi(2 \gamma+1)},\] since \(1+e^{-\varphi(2 \gamma+1)} \leqslant \sigma_{0} \leqslant 2\).
By hypothesis \(\zeta(s) =O \big(e^{\varphi(t)}\big)\) for \(1-\theta(t) \leqslant \sigma \leqslant 2\), so that there must exist \(A_{2}>0\) such that \[\begin{aligned} \left|\zeta(s) / \zeta\left(s_{0}\right)\right|<e^{A_{2} \varphi(2 \gamma+1)} & \text { for } \quad \left|s-s_{0}\right| \leqslant r, \\ \left|\zeta(s) / \zeta\left(s_{0}^{\prime}\right)\right|<e^{A_{2} \varphi(2 \gamma+1)} & \text { for } \quad\left|s-s_{0}^{\prime}\right| \leqslant r . \end{aligned}\]
Using (*) from the previous lemma with \(M=e^{A_{2} \varphi(2 \gamma+1)}\), we obtain \[-\operatorname{Re} \frac{\zeta^{\prime}\left(\sigma_{0}+2 i \gamma\right)}{\zeta\left(\sigma_{0}+2 i \gamma\right)}<A_{3} \frac{\varphi(2 \gamma+1)}{\theta(2 \gamma+1)} \quad\text{ for some } \quad A_{3}>0.\tag{A}\]
We have \(\beta \leqslant 1<\sigma_{0}\), while for \(\beta>\sigma_{0}-r / 2\), inequality (**) of the previous lemma gives \[-\operatorname{Re} \frac{\zeta^{\prime}\left(\sigma_{0}+i \gamma\right)}{\zeta\left(\sigma_{0}+i \gamma\right)}<A_{3} \frac{\varphi(2 \gamma+1)}{\theta(2 \gamma+1)}-\frac{1}{\sigma_{0}-\beta}.\tag{B}\]
Also we have \[-\zeta^{\prime}\left(\sigma_{0}\right) / \zeta\left(\sigma_{0}\right)<B /\left(\sigma_{0}-1\right), \tag{C}\] where \(B \rightarrow 1^+\) as \(\sigma_{0} \rightarrow 1^+\), since \(s=1\) is a pole of first order of \(\zeta(s)\) with the residue \(1\).
Recall that \(3+4 \cos \theta+\cos 2 \theta=2(1+\cos \theta)^{2} \geq 0\) for any \(\theta\in\mathbb R\).
Hence, we have \[-3 \frac{\zeta^{\prime}(\sigma_0)}{\zeta(\sigma_0)}-4 \operatorname{Re}\left(\frac{\zeta^{\prime}(\sigma_0+i \gamma)}{\zeta(\sigma_0+i\gamma)}\right)-\operatorname{Re}\left(\frac{\zeta^{\prime}(\sigma_0+2 i\gamma)}{\zeta(\sigma_0+2 i\gamma)}\right) \geq 0. \tag{D}\]
Inserting inequalities (A), (B) and (C) to inequality (D), we obtain \[\frac{3 B}{\sigma_{0}-1}+5 A_{3} \frac{\varphi(2 \gamma+1)}{\theta(2 \gamma+1)}-\frac{4}{\sigma_{0}-\beta} \geqslant 0,\] or \[\sigma_{0}-\beta \geqslant\left(\frac{3 B}{4\left(\sigma_{0}-1\right)}+\frac{5}{4} A_{3} \frac{\varphi(2 \gamma+1)}{\theta(2 \gamma+1)}\right)^{-1},\] which gives then \[1-\beta \geqslant \frac{1-\frac{3}{4} B-\frac{5}{4} A_{3}(\varphi(2 \gamma+1) / \theta(2 \gamma+1))\left(\sigma_{0}-1\right)}{\left(3 B / 4\left(\sigma_{0}-1\right)\right)+\frac{5}{4} A_{3}(\varphi(2 \gamma+1) / \theta(2 \gamma+1))}.\]
Now we choose \(B=\frac{5}{4}\) and \(\sigma_{0}=1+\left(40 A_{3}\right)^{-1} \theta(2 \gamma+1) / \varphi(2 \gamma+1)\), and then regardless of \(A_{3}\) the condition \(\sigma_{0} \geqslant 1+\exp (-\varphi(2 \gamma+1))\) holds in view of \(\frac{\varphi(t)}{\theta(t)}=o(e^{\varphi(t)})\) as \(t\to\infty\).
With this choice of \(B\) and \(\sigma_{0}\) the last inequality reduces to \[1-\beta \geqslant \frac{\theta(2 \gamma+1)}{1240 A_{3} \varphi(2 \gamma+1)},\] which gives the desired estimate provided that \(\beta>\sigma_{0}-r / 2\).
It remains to consider the case \(\beta \leqslant \sigma_{0}-r / 2\), when \[\begin{aligned} \beta & \leqslant \sigma_{0}-r / 2=1+\left(40 A_{3}\right)^{-1} \frac{\theta(2 \gamma+1)}{\varphi(2 \gamma+1)}-\frac{1}{2} \theta(2 \gamma+1) \\ & \leqslant 1-\left(1240 A_{3}\right)^{-1} \frac{\theta(2 \gamma+1)}{\varphi(2 \gamma+1)}, \end{aligned}\] since \(\lim _{t \rightarrow \infty} \varphi(t)=\infty\). This completes the proof. $$\tag*{$\blacksquare$}$$
Remark.
We can take \(\theta(t)=1/2\) and \(\varphi(t)=\log(t+2)\) and by the previous theorem, we obtain that there exists an absolute constant \(C>0\) such that \(\zeta(s)\) has no zero \(\rho=\beta+i \gamma\) satisfying \[\beta \geq 1-\frac{C}{\log (|\gamma|+2)}.\]
For all \(s=\sigma+i t \in \mathbb{C}\) such that \(\frac{1}{2} \leqslant \sigma \leqslant 1\) and \(t \geqslant 3\), one has \[|\zeta(s)| \leqslant A t^{B(1-\sigma)^{3 / 2}}(\log t)^{2 / 3}.\tag{*}\]
Inequality (*) implies that there exists an absolute constant \(c_{0}>0\) such that \(\zeta(s)\) has no zero in the region \[\sigma \geqslant 1-\frac{c_{0}}{(\log |t|)^{2 / 3}(\log \log |t|)^{1 / 3}} \quad \text { and } \quad|t| \geqslant 3. \tag{**}\]
This combined with the previous theorem yields the following variant of the PNT with the best error term to date.
There exists an absolute constant \(c\in(0, 1)\) such that as \(x \to \infty\), one has \[\begin{aligned} \psi(x)&=x+O\left(x \exp \left(-c(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right),\\ \pi(x)&=\operatorname{Li}(x)+O\left(x \exp \left(-c(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right). \end{aligned}\]