| 1. | Lecture 1 |
| 2. | Lecture 2 |
| 3. | Lecture 3 |
| 4. | Lecture 4 |
| 5. | Lecture 5 |
| 6. | Lecture 6 |
| 7. | Lecture 7 |
| 8. | Lecture 8 |
| 9. | Lecture 9 |
| 10. | Lecture 10 |
| 11. | Lecture 11 |
| 12. | Lecture 12 |
| 13. | Lecture 13 |
Recall that \(\Xi(s)=\pi^{-s / 2} \Gamma(s / 2) \zeta(s)\) can be extended analytically in the whole complex plane to a meromorphic function having simple poles at \(s=0\) and \(s=1\), and satisfies the functional equation \(\Xi(s)=\Xi(1-s)\). Let \[\begin{aligned} \xi(s)&=s(s-1)\pi^{-s / 2} \Gamma(s / 2) \zeta(s)\\ &=1+\frac{s(s-1)}{2}\int_{1}^{\infty} (\theta(x)-1)\left(x^{s / 2}+x^{(1-s) / 2}\right) \frac{dx}{x}, \end{aligned}\] where \(\theta\) is the Theta function \[\theta(x)=\sum_{n \in \mathbb{Z}} e^{-\pi n^{2}x}.\]
Then the function \(\xi(s)\) can be extended analytically in the whole complex plane to an entire function that satisfies the functional equation \(\xi(s)=\xi(1-s)\).
Thus the Riemann zeta-function can be extended analytically in the whole complex plane to a meromorphic function having a simple pole at \(s=1\) with residue \(1\). Furthermore, for all \(s \in \mathbb{C} \backslash\{1\}\), we have \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s).\]
The function \(\xi(s)\) is an entire function of order \(1\). Furthermore, \[\begin{aligned} \limsup _{|s| \rightarrow \infty} \frac{\log |\xi(s)|}{|s| \log |s|}=\frac{1}{2}. \end{aligned}\]
Proof.
Since \(\xi(s)=\xi(1-s)\), it suffices to bound \(|\xi(s)|\) in the half-plane \(\operatorname{Re}(s) \geq 1 / 2\).
We can estimate \(\xi(s)\) by invoking Stirling’s formula, and elementary upper bounds for \(\zeta(s)\). When \(\operatorname{Re}(s)>1\), we will use the identity \[\zeta(s)=\frac{s}{s-1}-s \int_{1}^{\infty}\frac{\{x\}}{x^{s+1}} d x.\]
However, since the last integral represents a holomorphic function in \(\operatorname{Re}(s)>0\), the identity holds in this larger domain.
In particular, we have \[|(s-1) \zeta(s)| =O(|s|^{2}) \quad \text { whenever } \quad \operatorname{Re}(s) \geq 1 / 2.\]
Moreover, since \(\log |\Gamma(s)| \leq|\log \Gamma(s)|\), Stirling’s formula yields \[\log |\Gamma(s)| \leq|s| \log |s|+O(|s|) \quad \text { whenever } \operatorname{Re}(s) \quad \geq 1 / 2.\]
Combining the last two estimates, since \(\xi(s)=s(s-1)\pi^{-s / 2} \Gamma(s / 2) \zeta(s)\), we obtain \[\log |\xi(s)| \leq \frac{1}{2}|s| \log |s|+O(|s|) \quad \text { whenever } \quad \operatorname{Re}(s) \geq 1 / 2.\] which establishes the first claim of the lemma.
For the second claim, we note that \[\log |\Gamma(|s|)|=\log \Gamma(|s|)=|s| \log |s|+O(|s|),\] whence \[\log \xi(|s|)=\frac{1}{2}|s| \log |s|+O(|s|) \quad \text { as } \quad |s| \rightarrow \infty.\]
This completes the proof of the lemma.$$\tag*{$\blacksquare$}$$
The function \(\xi(s)\) has infinitely many zeros in the strip \(0 \leq \operatorname{Re}(s) \leq 1\) and no zeros outside that strip. It can be written as
\[\xi(s)=e^{B s} \prod_{\rho}\left(1-\frac{s}{\rho}\right) e^{s / \rho}, \tag{*}\] where \(\rho\) runs through the zeros of \(\xi(s)\) counted according to their multiplicities and \[B=1+\frac{\gamma}{2}-\log(2\sqrt{\pi}).\]
Proof.
By the previous lemma, we know that \(\xi(s)\) has order \(1\).
Noting that \(\xi(0)=1\), and using Hadamard’s theorem we obtain (*) for some \(B\).
If \(\xi(s)\) had only a finite number of zeros, (*) would imply the estimate \(\log |\xi(s)|=O(|s|)\), which contradicts the fact that \(\limsup _{|s| \rightarrow \infty} \frac{\log |\xi(s)|}{|s| \log |s|}=\frac{1}{2}\).
For \(\operatorname{Re} s>1\), the zeta function \(\zeta(s)\), and, consequently, \(\xi(s)\), have no zeros in this range. By using the equation \(\xi(s)=\xi(1-s)\) it follows that \(\xi(s) \neq 0\) for \(\operatorname{Re} s<0\). Since \(1=\xi(0)=\xi(1) \neq 0\), the zeros of \(\xi(s)\) lie in the strip \(0 \leq \operatorname{Re}(s) \leq 1\).
Show that \(B=1+\frac{\gamma}{2}-\log(2\sqrt{\pi})\). $$\tag*{$\blacksquare$}$$
The zeta function \(\zeta(s)\) has simple zeros at \(s=-2,-4,-6,-8, \ldots\). These zeros are called trivial zeros.
For \(\operatorname{Re} s>1\), the zeta function \(\zeta(s)\) has no zeros.
In the critical strip \(0\le \operatorname{Re}s\le 1\) the zeros of \[\xi(s)=s(s-1)\pi^{-s / 2} \Gamma(s / 2) \zeta(s),\] are precisely the zeros of \(\zeta(s)\). These zeros are called nontrivial zeros.
We also know that \(\zeta(\sigma)\neq 0\) whenever \(0<\sigma<1\).
In addition to the trivial zeros, the zeta function has infinitely many nontrivial zeros lying in the critical strip \(0 \leq \operatorname{Re} s \leq 1\).
The nontrivial zeros of the zeta function are distributed symmetrically with respect to the lines \(\operatorname{Re} s=1 / 2\) and \(\operatorname{Im} s=0\), which follows from the functional equation \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s).\]
The functional equation is very important, but also may be insufficient in some applications for it does not express \(\zeta(s)\) explicitly.
The following tool will be useful to get some estimates of \(\zeta(s)\) in the critical strip, especially when \(\sigma\) is close to \(1\).
We have uniformly for \(x \geqslant 1\) and \(s \in \mathbb{C} \backslash\{1\}\) such that \(\sigma>0\), we have \[\zeta(s)=\sum_{n \leqslant x} \frac{1}{n^{s}}+\frac{x^{1-s}}{s-1}+R_{0}(s ; x)\] with \[R_{0}(s ; x)=\frac{\psi(x)}{x^{s}}-s \int_{x}^{\infty} \frac{\psi(u)}{u^{s+1}} d u,\] and hence \[\left|R_{0}(s ; x)\right| \leqslant \frac{|s|}{\sigma x^{\sigma}}.\]
Proposition. Let \(s=\sigma+\) it with \(-1 \leqslant \sigma \leqslant 2\) and \(t\) not equal to an ordinate of a zero of \(\zeta(s)\). Set \(\tau=|t|+3\). Then we have \[-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\substack{\rho \\|t-{\rm Im}\rho| \leqslant 1}} \frac{1}{s-\rho}+O(\log \tau).\]
Proof.
Recall that \(\xi(s)=s(s-1) \pi^{-s / 2} \Gamma(s / 2) \zeta(s)\).
The logarithmic differentiation provides \[\frac{\xi^{\prime}(s)}{\xi(s)}=b+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right),\] 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 \[\sum_{\rho}|\rho|^{-2}<\infty.\]
Now using the definition of \(s(s-1) \pi^{-s / 2} \Gamma(s / 2) \zeta(s)\) gives \[\frac{\xi^{\prime}(s)}{\xi(s)}=\frac{1}{s}+\frac{1}{s-1}-\frac{\log \pi}{2}+\frac{\zeta^{\prime}(s)}{\zeta(s)}+\frac{1}{2} \frac{\Gamma^{\prime}(s/2)}{\Gamma(s/2)}.\]
By logarithmic differentiation of \(\Gamma(s)\), we obtain \[-\frac{\Gamma^{\prime}(s)}{\Gamma(s)}=\frac{1}{s}+\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n+s}-\frac{1}{n}\right).\]
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 \[\sum_{1\le n\le\tau}\left|\frac{1}{2n+s}-\frac{1}{2n}\right|=O(\log \tau), \quad \text{ and } \quad \sum_{n\ge\tau}\left|\frac{1}{2n+s}-\frac{1}{2n}\right|=O\bigg(\frac{|s|}{\tau}\bigg).\]
Therefore, we can write that \[\begin{align*} \frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)+O(\log \tau).\tag{*} \end{align*}\]
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 \[\begin{aligned} {\rm Re}\bigg(\frac{1}{2+it-\rho}\bigg)&=\frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2} \ge\frac{1}{4+4(t-\gamma)^2}\ge0,\\ {\rm Re}\bigg(\frac{1}{\rho}\bigg)&=\frac{\beta}{\beta^2+\gamma^2}\ge0. \end{aligned}\]
Therefore, we obtain \[\begin{aligned} \sum_{\rho}\frac{1}{4+4(t-\gamma)^2}\le {\rm Re}\bigg(\sum_{\rho}\left(\frac{1}{2+it-\rho}+\frac{1}{\rho}\right)\bigg)=O(\log\tau). \end{aligned}\]
This immediately implies that \[\sum_{|{\rm Im}\rho-t|\le 1}1\le \sum_{|{\rm Im}\rho-t|\le 1}\frac{2}{1+(t-{\rm Im}\rho)^2}=O(\log\tau).\]
In other words, the number of zeros \(\rho\) in the strip \(t\le {\rm Im}\rho \le t+1\) is at most \(O(\log\tau)\) for any \(t\ge 2\).
By the previous observation we see that \[\begin{align*} \sum_{\rho: | t-\gamma \mid \leqslant 1} \frac{1}{|2+i t-\rho|} =O \Big(\sum_{\rho: | t-\gamma \mid \leqslant 1}1\Big)=O (\log \tau).\tag{**} \end{align*}\]
By the previous observation we also have \[\begin{align*} \sum_{\rho: | t-\gamma \mid > 1}\left|\frac{1}{2+it-\rho}+\frac{1}{\rho}\right|=O(\log\tau) \tag{***} \end{align*}\]
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 \[\begin{aligned} -\frac{\zeta^{\prime}(s)}{\zeta(s)}=\frac{1}{s-1}-\sum_{\substack{\rho \\|t-{\rm Im}\rho| \leqslant 1}} \frac{1}{s-\rho}+O(\log \tau), \end{aligned}\] as desired. $$\tag*{$\blacksquare$}$$
From the proof of the previous proposition, we obtain the following important result.
For every real number \(T\ge2\) the number of nontrivial zeros \(\rho\) of the zeta function \(\zeta\) satisfying \(T\le {\rm Im} \rho\le T+1\) is at most \(O(\log T)\).
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 proposition 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 \(|s-1|\le 2c_1\). Thus if \(\rho=\beta+i\gamma\) is a nontrivial zero of \(\xi(s)\) then \(|\rho-1|\ge 2c_1\). If we assume that \(|\gamma|\le c_1\), 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 proposition. 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 \(\left|\gamma_{0}\right| \geq c_{1}\). $$\tag*{$\blacksquare$}$$
Consider the following function \[\begin{aligned} h(x)= \begin{cases} 1 & \text{ if } x\in(1, \infty),\\ \frac{1}{2} &\text{ if } x=1, \\ 0 & \text{ if } x\in(0, 1). \end{cases} \end{aligned}\]
Let \(\kappa, T, T^{\prime}\in\mathbb R_+\) be given.
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 \[\begin{aligned} \left|h(1)-\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} \frac{\mathrm{~d} s}{s}\right| & \leq \frac{\kappa}{T+\kappa}. \end{aligned}\]
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, valid for all \(y>0\), \[0 \leq \frac{\pi}{2}-\arctan y=\int_{y}^{\infty} \frac{d t}{1+t^{2}} \leq \frac{2}{1+y}\]
This concludes the proof of the lemma.$$\tag*{$\blacksquare$}$$
Let \[\begin{aligned} F(s):=\sum_{n=1}^{\infty} a_{n} n^{-s}, \end{aligned}\] be a Dirichlet series with abscissa of convergence \(\sigma_{c}\) and abscissa of absolute convergence \(\sigma_{a}\).
Its truncated variant is denoted by \[F_{\le x}(s):=\sum_{n\in[x]} a_{n} n^{-s}, \quad \text{ for } \quad x\in\mathbb R_+.\]
For \(\kappa>\max \left\{0, \sigma_{a}\right\}\), \(T \geq 1\) and \(x \geq 1\), we have \[\begin{aligned} F_{\le x}(s)=\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s) x^{s} \frac{\mathrm{~d} s}{s}+O\left(x^{\kappa} \sum_{n=1}^{\infty} \frac{\left|a_{n}\right|}{n^{\kappa}(1+T|\log (x / n)|)}\right). \end{aligned}\]
It suffices to show that, for any fixed \(\kappa>0\), and uniformly for \(y>0, T>0\), we have that \[\begin{align*} \left|h(y)-\frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} y^{s} s^{-1} d s\right| = O\big( y^{\kappa} /(1+T|\log y|)\big). \tag{*} \end{align*}\]
Indeed, applying this estimate with \(y=x / n\) and summing over \(n \in\mathbb Z_+\) (after multiplication by \(a_{n}\) ) we obtain precisely the stated formula.
When \(T|\log y|>1\), the estimate (*) follows from the first inequality of the previous lemma. Otherwise, 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 \[O \bigg(\int_{-T}^{T}|(t\log y) s^{-1}| d t\bigg) =O (T|\log y|) = O(1),\] and consequently, by the second inequality of the previous lemma, we see that the left-hand side of (*) is \(O(y^{\kappa})\). This concludes the proof. $$\tag*{$\blacksquare$}$$
Let \(F(s):=\sum_{n=1}^{\infty} a_{n} n^{-s}\) be a Dirichlet series with finite abscissa of absolute convergence \(\sigma_{a}\).
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
\[\begin{aligned} \sum_{n \leq x} \frac{a_{n}}{n^{s}}= & \frac{1}{2 \pi i} \int_{\kappa-i T}^{\kappa+i T} F(s+w) x^{w} \frac{dw}{w}\\ & +O\left(x^{\sigma_{a}-\sigma} \frac{(\log x)^{\alpha}}{T}+\frac{B(2 x)}{x^{\sigma}}\left(1+\frac{x\log x}{T}\right)\right). \end{aligned}\]
Apply formula (*) to the series \(\sum_{n=1}^{\infty} b_{n} n^{-w}\) with \(b_{n}:=a_{n} n^{-s}\).
The contribution of integers \(n\in\mathbb Z_+\) which do not belong to \(\left[\frac{1}{2} x, 2 x\right]\) is \[\begin{aligned} \sum_{n\not\in[x/2, 2x]}\frac{|x^{\kappa}||a_{n}|}{n^{\kappa}(1+T|\log (x / n)|)}&=O\bigg(x^{\kappa} T^{-1} \sum_{n=1}^{\infty}\left|a_{n}\right| n^{-\kappa-\sigma}\bigg)\\ &=O\big( x^{\sigma_{a}-\sigma} T^{-1}(\log x)^{\alpha}\big). \end{aligned}\]
When \(\frac{1}{2} x \leq n \leq 2 x\), using inequality \(\log y\ge 1-\frac{1}{y}\) for \(y\in\mathbb R_+\) we have \(|\log (x / n)|\ge \frac{|x-n|}{2x}\). This leads to the following estimate \[\begin{gathered} x^{-\sigma} \sum_{x / 2 \leq n \leq 2 x} \frac{\left|a_{n}\right|}{1+T|\log (x / n)|} =O\bigg( \frac{B(2 x)}{x^{\sigma}} \sum_{x / 2 \leq n \leq 2 x} \min\bigg\{1, \frac{x}{T|x-n|}\bigg\}\bigg). \end{gathered}\]
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. $$\tag*{$\blacksquare$}$$
For any \(2\le T \le x\), we have \[\psi(x)=x-\sum_{|{\rm Im}\rho| \leqslant T} \frac{x^{\rho}}{\rho}-\log 2 \pi+O\left(\frac{x(\log x)^{2}}{T}\right).\]
Proof.
We may suppose that \(x \notin \mathbb{Z}\).
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 \[\kappa-i T^{\prime},\quad \kappa+i T^{\prime}, \quad -1/2+i T^{\prime}, \quad \text{ and } \quad -1/2-i T^{\prime}.\]
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 \[\frac{1}{2 \pi i} \int_{\mathcal{R}}-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s} \mathrm{~d} s=x-\sum_{|\gamma| \leqslant T^{\prime}} \frac{x^{\rho}}{\rho}-\frac{\zeta^{\prime}(0)}{\zeta(0)},\] since \[\begin{gathered} {\rm res}_0\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=-\frac{\zeta^{\prime}(0)}{\zeta(0)}, \quad \text{ and } \quad {\rm res}_1\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=x,\\ \quad \text{ and } \quad {\rm res}_{\rho}\bigg(-\frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^{s}}{s}\bigg)=\frac{x^{\rho}}{\rho}. \end{gathered}\]
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 \(\sigma_a=\alpha=1\), \(\sigma=0={\rm Re} s\), \(\kappa=1+1/\log x\) and \(a_n=\Lambda(n)\) and \(B(n)=\log n\) to obtain \[\begin{aligned} \psi(x)= & x-\sum_{|\gamma| \leqslant T^{\prime}} \frac{x^{\rho}}{\rho}-\log 2 \pi-\sum_{j=1}^{2} I_{\mathcal{H}_{j}}-I_{\mathcal{V}} \\ & +O\left(\frac{x(\log x)^{2}}{T^{\prime}}+\log x\right). \end{aligned}\] where \(I_{\mathcal{H}_{j}}\) denotes the integrals taken over the two horizontal sides and \(I_{\mathcal{V}}\) is the integral taken over the vertical side.
Since \(T^{\prime} \simeq T\), we obtain \[\begin{aligned} I_{\mathcal{H}_{j}} & =O \bigg(\int_{-1/2}^{\kappa}\left|-\frac{\zeta^{\prime}(\sigma\pm i T^{\prime})}{\zeta(\sigma\pm i T^{\prime})}\right| \frac{x^{\sigma}}{\left|\sigma\pm i T^{\prime}\right|} d\sigma \bigg)\\ & =O\bigg( (\log T)^{2} \int_{-1/2}^{\kappa} \frac{x^{\sigma}}{\left|\sigma\pm i T^{\prime}\right|} d\sigma\bigg) \\ & = O\bigg(\frac{(\log T)^{2}}{T} \int_{-1/2}^{\kappa} x^{\sigma} d\sigma\bigg) =O\bigg( \frac{x(\log T)^{2}}{T}\bigg), \quad \text{ for } \quad j \in[2]. \end{aligned}\]
Moreover, we have \[\begin{aligned} I_{\mathcal{V}} &=O\bigg( \int_{-T^{\prime}}^{T^{\prime}}\left|-\frac{\zeta^{\prime}(-1/2+i t)}{\zeta(-1/2+i t)}\right| \frac{x^{-1/2}}{|-1/2+i t|} dt\bigg)\\ &=O\bigg( x^{-1/2}\int_{-T^{\prime}}^{T^{\prime}}\log(2+|t|) \frac{dt}{|-1/2+it|}\bigg)\\ &=O(x^{-1/2}(\log T)^2)=O\bigg( \frac{x(\log T)^{2}}{T}\bigg), \end{aligned}\] since \(2\le T\le x\). This completes the proof. $$\tag*{$\blacksquare$}$$
There exists an absolute constant \(c\in(0, 1)\) such that as \(x \to \infty\), one has \[\begin{align*} \psi(x)&=x+ O\big(xe^{-c \sqrt{\log x}}\big), \tag{*}\\ \pi(x) &= {\rm Li}(x) + O\big(xe^{-c \sqrt{\log x}}\big),\tag{**} \end{align*}\] where \[{\rm Li}(x):=\int_{2}^x\frac{dt}{\log t}.\] Moreover, one has \[\operatorname{Li}(x)=\frac{x}{\log x}+x \sum_{k=1}^{N-1} \frac{k!}{(\log x)^{k+1}}+O\left(\frac{x}{(\log x)^{N+1}}\right).\] For instance, for \(x \to \infty\), the estimate \[\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{(\log x)^{2}}\right)\] is useful in many applications.
For any fixed \(N\in\mathbb Z_+\), by repeated integration by parts, we may derive \[\operatorname{Li}(x)=\frac{x}{\log x}+x \sum_{k=1}^{N-1} \frac{k!}{(\log x)^{k+1}}+O\left(\frac{x}{(\log x)^{N+1}}\right).\]
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 \[|\psi(x)-x|\le \sum_{|{\rm Im}\rho| \leqslant T} \frac{x^{{\rm Re}\rho}}{|\rho|}+O\left(\frac{x(\log x)^{2}}{T}\right).\]
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 \[{\rm Re}\rho=\beta\le 1-\frac{C}{\log(|\gamma|+2)}\le1-\frac{C}{\log T}.\]
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.$$\tag*{$\blacksquare$}$$
Riemann hypothesis (1859). All non-trivial zeros of \(\zeta(s)\) are on the critical line \({\rm Re} s=\frac{1}{2}\).
Remark.
If the Riemann hypothesis is true, then we would have \[\begin{align*} \psi(x)=x+ O\big(\sqrt{x}(\log x)^2\big). \tag{*} \end{align*}\]
This follows immediately by adapting the argument from the previous slide with \(T=\sqrt{x}\), where the essential input comes from the fact that all nontrivial zeros \(\rho\) of the \(\zeta(s)\) satisfy \({\rm Re}\rho=1/2\).
A striking result is that the asymptotic (*) implies the Riemann hypothesis.
Exercise. Prove that the Riemann hypothesis is equivalent to the PNT as in (*).
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 \[\sigma \geqslant 1-\frac{c_{0}}{(\log |t|)^{2 / 3}(\log \log |t|)^{1 / 3}} \quad \text { and } \quad|t| \geqslant 3. \tag{**}\]
The estimate (**) is essentially the best zero-free region for \(\zeta(s)\) up to now, which was independently obtained by Korobov and Vinogradov.
There exists an absolute constant \(c\in(0, 1)\) such that as \(x \to \infty\), one has \[\begin{aligned} \psi(x)&=x+O\left(x \exp \left(-c(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right),\\ \pi(x)&=\operatorname{Li}(x)+O\left(x \exp \left(-c(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right). \end{aligned}\]
In fact, any order of magnitude of \(\zeta(s)\) in a certain domain implies a zero-free region as may be seen in the next result devised by Landau.
Let \(\theta(t)\) and \(\phi(t)\) be positive functions such that \(\theta(t)\) is decreasing, \(\phi(t)\) is increasing and \(e^{-\phi(t)} \leqslant \theta(t) \leqslant \frac{1}{2}\). Assume that \(\zeta(s) =O(e^{\phi(t)})\) in the region \(\sigma \geqslant 1-\theta(t)\) and \(t \geqslant 2\). Then the following assertions hold.
There exists an absolute constant \(c_{0}>0\) such that \(\zeta(s)\) has no zero in the region \[\sigma \geqslant 1-c_{0} \frac{\theta(2 t+1)}{\phi(2 t+1)}.\]
In the region \(\sigma \geqslant 1-\left(c_{0} / 2\right) \theta(2 t+2) \phi(2 t+2)^{-1}\), we have \[\frac{1}{\zeta(s)} = O\bigg(\frac{\theta(2 t+2)}{\varphi(2 t+2)}\bigg) \quad \text { and } \quad \frac{\zeta^{\prime}(s)}{\zeta(s)} = O\bigg(\frac{\theta(2 t+2)}{\varphi(2 t+2)}\bigg).\]
The PNT enables us to improve on estimates for some functions of prime numbers by proceeding as follows. Let \(x \geqslant 2\) be a large real number and \(f \in C^{2}([2, x])\). Then one can write \[\begin{aligned} \sum_{p \leqslant x} f(p) = & f(x) \pi(x)-\int_{2}^{x} f^{\prime}(u) \operatorname{Li}(u) \mathrm{d} u-\int_{2}^{x} f^{\prime}(u)(\pi(u)-\operatorname{Li}(u)) \mathrm{d} u \\ = & \int_{2}^{x} \frac{f(u)}{\log u} \mathrm{~d} u+f(2) \operatorname{Li}(2)+f(x)(\pi(x)-\operatorname{Li}(x)) \\ & -\int_{2}^{x} f^{\prime}(u)(\pi(u)-\operatorname{Li}(u)) \mathrm{d} u. \end{aligned}\]
If the integral \(\int_{2}^{\infty} f^{\prime}(u)(\pi(u)-\operatorname{Li}(u)) \mathrm{d} u\) converges, then we obtain \[\begin{aligned} \sum_{p \leqslant x} f(p)=&\int_{2}^{x} \frac{f(u)}{\log u} \mathrm{~d} u+c_{f}+f(x)(\pi(x)-\operatorname{Li}(x))\\ &+\int_{x}^{\infty} f^{\prime}(u)(\pi(u)-\operatorname{Li}(u)) \mathrm{d} u \end{aligned}\] with \[c_{f}=f(2) \operatorname{Li}(2)-\int_{2}^{\infty} f^{\prime}(u)(\pi(u)-\operatorname{Li}(u)) \mathrm{d} u\]
For instance, with the latest version of the error term in the PNT we obtain \[\begin{aligned} \sum_{p \leqslant x} \frac{1}{p} & =\log \log x+B+O\left(\exp \left(-c_{1}(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right), \\ \sum_{p \leqslant x} \frac{\log p}{p} & =\log x-E+O\left(\exp \left(-c_{1}(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right), \\ \prod_{p \leqslant x}\left(1-\frac{1}{p}\right) & =\frac{e^{-\gamma}}{\log x}+O\left(\exp \left(-c_{1}(\log x)^{3 / 5}(\log \log x)^{-1 / 5}\right)\right), \end{aligned}\] for some absolute constant \(c_{1}>0\), where \[B \approx 0.261497212 \ldots\] is the Mertens constant and \[E \approx 1.332582275 \ldots.\]
Let \(a, q \in\mathbb Z_+\) be such that \((a, q)=1\). It is customary to define the function \[\pi(x ; q, a)=\sum_{\substack{p \leqslant x \\ p \equiv a(\bmod q)}} 1.\]
We know \(\lim _{x \rightarrow \infty} \pi(x ; q, a)=\infty\) and hence the question of its order of magnitude arises naturally.
We expect the prime numbers to be well distributed in the \(\varphi(q)\) reduced residue classes modulo \(q\). Applying naively the methods that we have already developed to the function \[\Psi(x, \chi)=\sum_{n \leqslant x} \Lambda(n) \chi(n),\] where \(\chi\) is a non-principal Dirichlet character modulo \(q\) and to its Dirichlet series \(-\frac{L^{\prime}}{L}(s, \chi)\), we obtain \[\pi(x ; q, a)=\frac{\operatorname{Li}(x)}{\varphi(q)}+O_{q}\left(x e^{-c_{0}(q) \sqrt{\log x}}\right)\] for some constant \(0<c_{0}(q)<1\) depending on \(q\), the constants implied in the error term depending also on \(q\).
This dependence makes this result useless in practice.
A great deal of effort has been made to prove some efficient estimates where the constants do not depend on the modulus.
One of the most important results in the theory is called the Siegel-Walfisz-Page theorem or Siegel–Walfisz theorem.
Let \(a, q \in\mathbb Z_+\) be coprime integers.
For all \(A>0\), there exists \(c_{1}(A)\in\mathbb R_+\) not depending on \(q\) such that, for all \(q \leqslant(\log x)^{A}\), we have \[\pi(x ; q, a)=\frac{\operatorname{Li}(x)}{\varphi(q)}+O_{A}\left(x e^{-c_{1}(A) \sqrt{\log x}}\right).\]
For all \(A>0\) and all \(q \geqslant 1\), we have \[\pi(x ; q, a)=\frac{\operatorname{Li}(x)}{\varphi(q)}+O_{A}\left(\frac{x}{(\log x)^{A}}\right).\]
Obviously, if we define the Chebyshev type functions \[\theta(x ; q, a)=\sum_{\substack{p \leqslant x \\ p \equiv a(\bmod q)}} \log p \quad \text { and } \quad \psi(x ; q, a)=\sum_{\substack{n \leq x \\ n \equiv a(\bmod q)}} \Lambda(n),\] then similar estimates hold for these functions, namely \[\left.\begin{array}{c} \theta(x ; q, a) \\ \psi(x ; q, a) \end{array}\right\}=\frac{x}{\varphi(q)}+O_{A}\left(x e^{-c_{1}(A) \sqrt{\log x}}\right),\] and \[\left.\begin{array}{c} \theta(x ; q, a) \\ \Psi(x ; q, a) \end{array}\right\}=\frac{x}{\varphi(q)}+O_{A}\left(\frac{x}{(\log x)^{A}}\right).\]
The proof of the Siegel–Walfisz theorem rests on an explicit formula for \(\psi(x, \chi)\) similar to that of the PNT theorem, namely \[\psi(x, \chi)=E_{0}(\chi) x-\sum_{|\gamma| \leqslant T} \frac{x^{\rho}}{\rho}+O\left(\frac{x}{T}(\log q x)^{2}+x^{1 / 4} \log x\right),\] where \[E_{0}(\chi)= \begin{cases}1, & \text { if } \chi=\chi_{0}, \\ 0, & \text { otherwise. }\end{cases}\]
This in turn implies that \[\psi(x ; q, a)=\frac{x}{\varphi(q)}+\sum_{\chi(\bmod q)} \sum_{|\gamma| \leqslant T} \frac{x^{\rho}}{\rho}+O\left(\frac{x}{T}(\log q x)^{2}+x^{1 / 4} \log x\right).\]
The other important tool is the knowledge of a zero-free region for the function \(L(s, \chi)\). The arguments generalize those of the function \(\zeta(s)\), except that there is an unforeseen difficulty in connection with the possible existence, still unproven, of an exceptional zero \(\beta_{1} \in \mathbb{R}\) near the point \(1\) of a function \(L(s, \chi)\) attached to a quadratic Dirichlet character. More precisely, we have the following result.
Let \(q \in \mathbb{Z}_+\) and \(\tau=|t|+3\). There exists an absolute constant \(c_{0}>0\) such that if \(\chi\) is a Dirichlet character modulo \(q\), then the function \(L(s, \chi)\) has no zero in the region \[\sigma \geqslant 1-\frac{c_{0}}{\log (q \tau)}\] unless \(\chi\) is a quadratic character, in which case \(L(s, \chi)\) has at most one, necessarily real, zero \(\beta_{1}<1\) in this region. This zero is called exceptional.
At the present time, we do not know much more about this exceptional zero. Nevertheless, Landau, Page and Siegel provided some very important results, showing in particular that such a zero occurs at most rarely.
Let \(q\in\mathbb Z_+\) and \(\tau=|t|+3\).
(Landau). There exists an absolute constant \(c_{1}>0\) such that the function \(\prod_{\chi(\bmod q)} L(s, \chi)\) has at most one zero in the region \[\sigma \geqslant 1-\frac{c_{1}}{\log (q \tau)}. \] If such a zero \(\beta_{1}\) exists, then it is necessarily real and associated to a quadratic character \(\chi_{1}\). This character is called the exceptional character.
(Page). If \(\chi\) is a Dirichlet character modulo \(q\), then \(L(\sigma, \chi) \neq 0\) in the region \[\sigma \geqslant 1-\frac{c_{2}}{q^{1 / 2}(\log (q+1))^{2}},\] where \(c_{2}>0\) is an effectively computable absolute constant.
(Siegel). Let \(\chi\) be a quadratic Dirichlet character modulo \(q\). For all \(\varepsilon>0\), there exists \(a\) non-effectively computable constant \(c_{\varepsilon}>0\) such that \[L(1, \chi)>\frac{c_{\varepsilon}}{q^{\varepsilon}}.\] This implies that, if \(\chi\) is a quadratic character modulo \(q\), then \(L(\sigma, \chi) \neq 0\) in the region \[\sigma \geqslant 1-\frac{c_{\varepsilon}}{q^{\varepsilon}}.\]