The von Mangoldt function \(\Lambda\) is defined by \[\Lambda(n):= \begin{cases}\log p, & \text { if } n=p^{k} \text { for some prime } p \text { and } k \in \mathbb{Z}_+, \\ 0, & \text { otherwise }\end{cases}.\]
The first Chebyshev function \(\vartheta\) is defined for \(x \geqslant 2\) by \[\vartheta(x):=\sum_{p \in\mathbb P_{\le x}} \log p,\] while it is convenient to set \(\vartheta(x):=0\) for \(0<x<2\), where \[\mathbb P_{\le x}:=\{p\in\mathbb P: p\le x\}=\mathbb P\cap[0, x].\]
The second Chebyshev function \(\psi\) is defined for \(x \geqslant 2\) by \[\psi(x):=\sum_{n \in[x]} \Lambda(n),\] while it is convenient to set \(\psi(x):=0\) for \(0<x<2\).
The prime counting function \(\pi\) is defined by \[\pi(x):=\sum_{p \in\mathbb P_{\le x}} 1=\#\mathbb P_{\le x},\] while it is convenient to set \(\pi(x):=0\) for \(0<x<2\).
For \(x \geq 2\) we have \[\vartheta(x)=\pi(x) \log x-\int_{2}^{x} \pi(t) \frac{dt}{t},\] and \[\pi(x)=\frac{\vartheta(x)}{\log x}+\int_{2}^{x} \frac{\vartheta(t)}{t \log ^{2} t} d t.\]
Proof: We have \[\pi(x)=\sum_{p \in \mathbb P_{\le x}} 1=\sum_{1<n \leq x} {\bf 1}_{\mathbb P}(n),\] and \[\vartheta(x)=\sum_{p \in \mathbb P_{\le x}} \log p=\sum_{1<n \leq x} {\bf 1}_{\mathbb P}(n) \log n.\]
If \(x, y\in \mathbb R_+\) with \(\lfloor y\rfloor<\lfloor x\rfloor\), and \(g\in C^1([y, x])\), then we know
\[\sum_{y<n \leq x} f(n) g(n)=F(x) g(x)-F(y) g(y)-\int_{y}^{x} F(t) g^{\prime}(t) d t,\] where \(F(t):=\sum_{1\le n \leq x} f(n).\)
Taking \(f(n)={\bf 1}_{\mathbb P}(n)\) and \(g(x)=\log x\) with \(y=1\) we obtain \[\vartheta(x)=\sum_{1<n \leq x} {\bf 1}_{\mathbb P}(n) \log n=\pi(x) \log x-\pi(1) \log 1-\int_{1}^{x} \pi(t)\frac{dt}{t},\] which proves the first identity since \(\pi(t)=0\) for \(t<2\).
Next, let \(f(n)={\bf 1}_{\mathbb P}(n) \log n\) and \(g(x)=1 / \log x\) and write \[\pi(x)=\sum_{3 / 2<n \leq x} f(n) \frac{1}{\log n}, \quad \vartheta(x)=\sum_{1<n \leq x} f(n)\]
Using the summation by parts formula with \(y=3 / 2\) we obtain \[\pi(x)=\frac{\vartheta(x)}{\log x}-\frac{\vartheta(3 / 2)}{\log 3 / 2}+\int_{3 / 2}^{x} \frac{\vartheta(t)}{t \log ^{2} t} d t\] which proves the second identity, since \(\vartheta(t)=0\) if \(t<2\). $$\tag*{$\blacksquare$}$$
For all \(x\in\mathbb R_+\), we have \[\vartheta(x) \leqslant \psi(x) \leqslant \vartheta(x)+\pi(\sqrt{x}) \log x.\]
For all \(x \geqslant 2\) and all \(a>1\), we have \[\frac{\vartheta(x)}{\log x} \leqslant \pi(x) \leqslant \frac{a \vartheta(x)}{\log x}+\pi\big(x^{1 / a}\big).\]
Proof of (i): One may suppose \(x \geqslant 2\). We first have \[\psi(x)-\vartheta(x)=\sum_{p^{k} \leqslant x} \log p-\sum_{p \leqslant x} \log p =\sum_{p \leqslant \sqrt{x}} \sum_{k=2}^{\lfloor\frac{\log x}{\log p}\rfloor} \log p,\] so that \(\psi(x) \geqslant \vartheta(x).\)
On the other hand, we have \[\begin{aligned} \psi(x)-\vartheta(x) & =\sum_{p \leqslant \sqrt{x}} \sum_{k=2}^{\lfloor\frac{\log x}{\log p}\rfloor} \log p \leqslant \sum_{p \leqslant \sqrt{x}} \log p\bigg\lfloor\frac{\log x}{\log p}\bigg\rfloor \leqslant \sum_{p \leqslant \sqrt{x}} \log x \\ & =\pi(\sqrt{x}) \log x. \qquad \end{aligned}\tag*{$\blacksquare$}\]
Proof of (ii): We have \[\pi(x)=\sum_{p \leqslant x} 1=\sum_{p \leqslant x} \frac{\log p}{\log p} \geqslant \frac{1}{\log x} \sum_{p \leqslant x} \log p=\frac{\vartheta(x)}{\log x}.\]
For \(2 \leqslant T<x\), we also have \[\pi(x)=\sum_{p \leqslant T} 1+\sum_{T<p \leqslant x} 1=\pi(T)+\sum_{T<p \leqslant x} \frac{\log p}{\log p} \leqslant \pi(T)+\frac{\vartheta(x)}{\log T}.\] and the choice of \(T=x^{1 / a}\) implies the asserted estimate. $$\tag*{$\blacksquare$}$$
The following relations are equivalent: \[\lim _{x \rightarrow \infty} \frac{\pi(x) \log x}{x}=1. \tag{A}\] \[\lim _{x \rightarrow \infty} \frac{\vartheta(x)}{x}=1. \tag{B}\] \[\lim _{x \rightarrow \infty} \frac{\psi(x)}{x}=1. \tag{C}\]
Proof: We know that \[\frac{\vartheta(x)}{x} \leqslant \frac{\psi(x)}{x} \leqslant \frac{\vartheta(x)}{x}+\frac{\pi(\sqrt{x}) \log x}{x}.\] Hence the equivalence between (B) and (C) follows, since \[\lim_{x\to \infty}\frac{\pi(\sqrt{x}) \log x}{x}=0.\]
For every \(a>1\) know that \[\frac{\vartheta(x)}{x} \leqslant \frac{\pi(x)\log x}{x} \leqslant \frac{a \vartheta(x)}{x}+\frac{\pi\big(x^{1 / a}\big)\log x}{x}.\]
For every \(a>1\) we also know that \[\lim_{x\to \infty}\frac{\pi\big(x^{1 / a}\big)\log x}{x}=0.\]
Then \[\lim_{x\to \infty}\frac{\vartheta(x)}{x} \leqslant \lim_{x\to \infty}\frac{\pi(x)\log x}{x} \leqslant\lim_{x\to \infty} \frac{a \vartheta(x)}{x}.\]
Since \(a>1\) is arbitrary, we obtain equivalence between (A) and (B).
This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
For all \(x \geqslant 1\), we have \[\vartheta(x) \leqslant x \log 4.\]
Proof: We first prove by induction that, for all \(n\in\mathbb Z_+\), we have \[\vartheta(n) \leqslant n \log 4.\]
This inequality is clearly true for \(n \in[3]\). If \(n \geqslant 4\) is even, we have \[\vartheta(n)=\vartheta(n-1) \leqslant (n-1)\log 4<n\log 4^{n}.\]
Suppose now that \(n \geqslant 5\) is odd and set \(n=2 m+1\) with \(m\in\mathbb Z_+\). The idea is to use the fact that the product \[\prod_{m+1<p \leqslant 2 m+1} p \qquad \text{divides the binomial coefficient} \qquad \binom{2 m+1}{m}.\]
To see this, observe that \(p\in\mathbb P\) such that \(m+1<p \leqslant 2m+1\) divides \((2m+1)!\) because of \(p \leqslant 2 m+1\), but does not divide \(m!(m+1)!\) because of \(p>m+1\), so that \[\prod_{m+1<p \leqslant 2 m+1} p \quad \text { divides } \quad (2 m+1)!=m!(m+1)!\binom{ 2 m+1}{m}\] and since the product is coprime to \(m!(m+1)!\) the claim follows.
Taking logarithms, we then obtain \[\vartheta(2 m+1)-\vartheta(m+1)=\sum_{m+1<p \leqslant 2 m+1} \log p \leqslant \log \binom{2 m+1}{m}.\]
Using Stirling’s formula we have \(\binom{2 m+1}{m}\le \frac{2m+1}{m+1}\frac{4^m}{\sqrt{\pi m}}\le 4^m\), thus \[\vartheta(2 m+1)\le m\log 4+\vartheta(m+1)\le m\log 4+ (m+1)\log 4=(2m+1)\log 4,\] where we have used the induction hypothesis applied to \(\vartheta(m+1)\).
The lemma follows from \[\vartheta(x)=\vartheta(\lfloor x\rfloor) \le \lfloor x\rfloor \log 4 \leqslant x \log 4.\]
The proof for the upper bound is now completed.$$\tag*{$\blacksquare$}$$
For all \(x \geqslant 1537\), we have \[\vartheta(x)>\frac{x}{\log 4}.\]
Proof: We first notice that the function \(f\) defined by \[f(x)=\lfloor x\rfloor -\bigg\lfloor\frac{x}{2}\bigg\rfloor-\bigg\lfloor\frac{x}{3}\bigg\rfloor-\bigg\lfloor\frac{x}{5}\bigg\rfloor +\bigg\lfloor\frac{x}{30}\bigg\rfloor,\] is periodic of period \(30\), since \(\lfloor x+n\rfloor =\lfloor x \rfloor+n\) for any \(n\in\mathbb Z\).
Moreover, for \(x \notin \mathbb{Z}\), we have \[f(30-x)=1-f(x),\] since \(\lfloor -x \rfloor=-\lfloor x \rfloor-1\) for \(x \notin \mathbb{Z}\).
An inspection of its values when \(x \in[1,15)\) allows us to infer that \(f(x)\) only takes the values \(0\) or \(1\) if \(x \notin \mathbb{Z}\).
Since \(f\) is continuous on the right, we also have \(f(x)=0\) or \(1\) when \(x \in \mathbb{Z}\). By periodicity, we infer that \(f(x)=0\) or \(1\) for all \(x \in \mathbb{R}\).
Since \(\lfloor x / k n\rfloor=0\) whenever \(n>x / k\) for \(k \in\{2,3,5,30\}\), we obtain \[\begin{aligned} \psi(x) \geqslant & \sum_{n \leqslant x} \Lambda(n) f\left(\frac{x}{n}\right) = \sum_{n \leqslant x} \Lambda(n)\left(\bigg\lfloor\frac{x}{n}\bigg\rfloor -\bigg\lfloor\frac{x}{2n}\bigg\rfloor-\bigg\lfloor\frac{x}{3n}\bigg\rfloor-\bigg\lfloor\frac{x}{5n}\bigg\rfloor +\bigg\lfloor\frac{x}{30n}\bigg\rfloor\right) \\ = & \sum_{n \leqslant x} \Lambda(n)\bigg\lfloor\frac{x}{n}\bigg\rfloor-\sum_{n \leqslant x / 2} \Lambda(n)\bigg\lfloor\frac{x}{2n}\bigg\rfloor-\sum_{n \leqslant x / 3} \Lambda(n)\bigg\lfloor\frac{x}{3n}\bigg\rfloor\\ & -\sum_{n \leqslant x / 5} \Lambda(n)\bigg\lfloor\frac{x}{5n}\bigg\rfloor+\sum_{n \leqslant x / 30} \Lambda(n)\bigg\lfloor\frac{x}{30n}\bigg\rfloor. \end{aligned}\]
By a simplified form of Striling’s formula we know for all \(x\ge 1\) that \[\sum_{n\in[x]}\log n=x\log x-x+1+R(x), \quad \text{ where } \quad 0\le R(x)\le \log x.\]
Since \(\Lambda\star{\bf 1}=\log\), then we conclude that \[\begin{aligned} \sum_{n\in[x]}\log n=\sum_{n\in[x]}\sum_{d\mid n}\Lambda(d)=\sum_{d\in[x]}\Lambda(d)\sum_{k\in[x/d]}=\sum_{d\in[x]}\Lambda(d)\bigg\lfloor\frac{x}{d}\bigg\rfloor. \end{aligned}\]
Hence for \(x\ge 1\), we have \[\sum_{d\in[x]}\Lambda(d)\bigg\lfloor\frac{x}{d}\bigg\rfloor=x\log x-x+1+R(x), \quad \text{ where } \quad 0\le R(x)\le \log x.\]
Inserting this bounds to the previous formula we obtain \[\begin{aligned} \psi(x) \geqslant & x \log x-x+1-\left(\frac{x}{2} \log \frac{x}{2}-\frac{x}{2}+1+\log \frac{x}{2}\right) \\ & -\left(\frac{x}{3} \log \frac{x}{3}-\frac{x}{3}+1+\log \frac{x}{3}\right)-\left(\frac{x}{5} \log \frac{x}{5}-\frac{x}{5}+1+\log \frac{x}{5}\right) \\ & +\left(\frac{x}{30} \log \frac{x}{30}-\frac{x}{30}+1\right) \\ = & x \log \left(2^{7 / 15} 3^{3 / 10} 5^{1 / 6}\right)-3 \log x+\log 30-1 . \end{aligned}\]
Using the estimate \(\psi(x)\le \vartheta(x)+\pi(\sqrt{x})\log x\) we obtain the desired lower bound \[\vartheta(x)\ge x \log \left(2^{7 / 15} 3^{3 / 10} 5^{1 / 6}\right)-(\sqrt{x}+3) \log x\ge \frac{x}{\log 4},\] whenever \(x\in\mathbb R_+\) is sufficiently large. This completes the proof.$$\tag*{$\blacksquare$}$$
Note that \[\log \left(2^{7 / 15} 3^{3 / 10} 5^{1 / 6}\right) \approx 0.92129 \ldots,\] which is a very good lower bound. It was sufficient to allow Chebyshev to prove Bertrand’s famous postulate.
Let \(n\in\mathbb Z_+\). Then the interval \((n, 2n]\) contains a prime number.
Proof: We check numerically the result for \(n \in[768]\).
Suppose \(n \geqslant 769\), using Chebyshev’s upper and lower bounds we deduce \[\sum_{n<p \leqslant 2 n} \log p=\vartheta(2 n)-\vartheta(n)>n\left(\frac{2}{\log 4}-\log 4\right)>0.\]
This shows that \((n, 2n]\cap \mathbb P\neq \varnothing\), which implies the desired result.$$\tag*{$\blacksquare$}$$
Let \(b>a\) and \(q \geq 1\) be integers. Let \(f\in C^q([a, b])\), then \[\sum_{n=a+1}^{b} f(n)=\int_{a}^{b} f(x) d x+\sum_{r=1}^{q}(-1)^{r} \frac{B_{r}}{r!}\left(f^{(r-1)}(b)-f^{(r-1)}(a)\right)+R_{q},\] where \[R_{q}=\frac{(-1)^{q+1}}{q!} \int_{a}^{b} B_{q}(x-\lfloor x\rfloor) f^{(q)}(x) d x.\]
If \(f \in C^{1}([a, b])\), and \(\psi(x)=\{x\}-1/2\) for \(x\in\mathbb R\), then \[\sum_{a<n \leqslant b} f(n)=\int_{a}^{b} f(x) dx+f(a) \psi(a)-f(b) \psi(b) +\int_{a}^{b} f^{\prime}(x) \psi(x) dx.\]
Proof: We apply the Euler–Maclaurin–Jacobi summation formula with \(q=1\), then \[B_{1}(x)=B_{0} x+B_{1}=x-\frac{1}{2},\quad \text{ and } \quad B_{1}^{\prime}(x)=1,\] and consequently, we have \[R_{1}=\int_{a}^{b} f^{\prime}(x) \psi(x) dx,\] which completes the proof. $$\tag*{$\blacksquare$}$$
By the absolute convergence for all complex numbers \(s=\sigma+i t\) such that \(\sigma>1\) we also have the Euler product formula \[\zeta(s)=\prod_{p\in\mathbb P}\left(1-\frac{1}{p^{s}}\right)^{-1}.\]
The Euler product formula enables us to see that \(\zeta(s) \neq 0\) in the half-plane \(\sigma>1\).
Indeed, for \(\sigma>1\) we have \[\begin{aligned} \frac{1}{|\zeta(s)|}&=\prod_{p\in\mathbb P}\left|1-\frac{1}{p^{s}}\right| \le \prod_{p\in\mathbb P}\left(1+\frac{1}{p^{\sigma}}\right)\\ &\le \sum_{n=1}^\infty\frac{1}{n^\sigma}\le 1+\int_1^\infty\frac{dt}{t^\sigma}=\frac{\sigma}{\sigma-1}. \end{aligned}\] Thus \(|\zeta(s)|\ge\frac{\sigma-1}{\sigma}>0\).
Euler’s summation formula If \(f \in C^{1}([a, b])\), and \(\psi(x)=\{x\}-1/2\) for \(x\in\mathbb R\), then \[\sum_{a<n \leqslant b} f(n)=\int_{a}^{b} f(x) dx+f(a) \psi(a)-f(b) \psi(b) +\int_{a}^{b} f^{\prime}(x) \psi(x) dx.\]
Let \(x \geqslant 1\) be a real number and \(s=\sigma+i t\) with \(\sigma>1\). By the Euler summation formula with \(a=1, b=x\) and \(f(x)=x^{-s}\), we can write \[\sum_{n \leqslant x} \frac{1}{n^{s}}=\frac{1}{2}+\frac{1-x^{1-s}}{s-1}-\frac{\psi(x)}{x^{s}}-s \int_{1}^{x} \frac{\psi(u)}{u^{s+1}} d u.\]
Taking \(x \to \infty\) we obtain \[\zeta(s)=\frac{1}{2}+\frac{1}{s-1}-s \int_{1}^{\infty} \frac{\psi(u)}{u^{s+1}} du.\tag{*}\]
Since \(|\psi(x)| \leqslant \frac{1}{2}\), the integral converges for \(\operatorname{Re}s>0\) and is uniformly convergent in any compact set to the right of the line \(\operatorname{Re}s=0\).
This implies that it defines an analytic function in the half-plane \(\operatorname{Re}s>0\), and therefore (*) extends \(\zeta\) to a meromorphic function in this half-plane, which is analytic except for a simple pole at \(s=1\) with residue \(1\).
Replacing \(x\) by \(\pi n^{2} x\) in the integral defining \(\Gamma(s/2)\) gives \[\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right) n^{-s}=\int_{0}^{\infty} x^{s / 2-1} e^{-\pi n^{2} x} d x \quad \text{ for all } \quad \sigma>0.\]
The purpose is to sum both sides of this equation. To this end, we define the following two Theta functions. For all \(x>0\), we set \[\omega(x)=\sum_{n=1}^{\infty} e^{-\pi n^{2} x} \quad \text { and } \quad \theta(x)=2 \omega(x)+1=\sum_{n \in \mathbb{Z}} e^{-\pi n^{2}x}.\]
Then \(g(t) =e^{-\pi t^{2}}\) satisfies \(\int_{\mathbb{R}} g(t) dt=1\), and \[\widehat{g}(u)=e^{-\pi u^{2}}.\]
For a Schwartz function \(f\), by the Poisson summation formula, we have \(\sum_{n\in \mathbb Z}\widehat{f}(n)=\sum_{n\in \mathbb Z}f(n)\), hence \[\theta(x)=\sum_{n\in\mathbb Z}g(\sqrt{x}n)=x^{-1/2}\theta(x^{-1}) \quad \text{ for all } \quad x>0.\]
Summing this equation over \(n\in \mathbb Z_+\) and interchanging the sum and integral, we obtain for all \(\sigma>1\) that \[\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right) \zeta(s)=\int_{0}^{\infty} x^{s / 2-1} \omega(x) d x,\] since the sum and integral converge absolutely for \(\sigma>1\).
Splitting the integral \(\int_{0}^\infty=\int_0^1+\int_1^\infty\) and changing the variables \(x\mapsto 1 / x\) in the first integral yields \[\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right) \zeta(s)=\int_{1}^{\infty} x^{s / 2-1} \omega(x) d x+\int_{1}^{\infty} x^{-s / 2-1} \omega\left(\frac{1}{x}\right) dx.\]
Using \(\theta(x^{-1})=x^{1/2}\theta(x)\) we write \(\omega\left(\frac{1}{x}\right)=x^{1 / 2} \omega(x)+\frac{x^{1 / 2}-1}{2},\) and \[\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right) \zeta(s) =-\frac{1}{s}+\frac{1}{s-1}+\int_{1}^{\infty} \omega(x)\left(x^{s / 2}+x^{(1-s) / 2}\right) \frac{dx}{x},\] whenever \(\sigma>1\).
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 \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s).\]
Proof: For \(\sigma>1\) we have \[\begin{aligned} \Xi(s)=-\frac{1}{s}+\frac{1}{s-1}+\int_{1}^{\infty} \omega(x)\left(x^{s / 2}+x^{(1-s) / 2}\right) \frac{dx}{x}. \end{aligned}\tag{*}\] Then we see that \[\xi(s)=s(s-1) \Xi(s).\]
Since \(\omega(x) =O (e^{-\pi x})\) as \(x \to \infty\), we infer that the integral in (*) is absolutely convergent for all \(s \in \mathbb{C}\) whereas the left-hand side is a meromorphic function on \(\sigma>0\). This implies that:
The identity (*) is valid for all \(\sigma>0\).
The function \(\Xi(s)\) can be defined by this identity as a meromorphic function on \(\mathbb{C}\) with simple poles at \(s=0\) and \(s=1\).
Since the right-hand side of (*) is invariant under the substitution \(s \mapsto 1-s\), we obtain \(\Xi(s)=\Xi(1-s)\) and \(\xi(s)=\xi(1-s)\).
The function \(s \mapsto \xi(s)=s(s-1) \Xi(s)\) is entire on \(\mathbb{C}\). Indeed, if \(\sigma>0\), the factor \(s-1\) counters the pole at \(s=1\), and the result on all \(\mathbb{C}\) follows from the functional equation.
It remains to show that the functional equation can be written as \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s).\]
Since \(\Xi(s)=\Xi(1-s)\), we have \[\Gamma(s / 2) \zeta(s)=\pi^{s/2}\Xi(s)=\pi^{s/2}\Xi(1-s)=\pi^{s-1 / 2} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s).\]
Multiplying both sides by \(\pi^{-1 / 2} 2^{s-1} \Gamma\left(\frac{1+s}{2}\right)\) and using the duplication formula, asserting that \(\Gamma(s)= \pi^{-1 / 2}2^{s-1} \Gamma\left(s/2\right) \Gamma\left((s+1)/2\right)\) we see \[\Gamma(s) \zeta(s)=(2 \pi)^{s-1} \Gamma\left(\frac{1-s}{2}\right) \Gamma\left(\frac{1+s}{2}\right) \zeta(1-s)\]
Now the reflection formula \(\frac{\sin \pi s}{\pi}=\frac{1}{\Gamma(s) \Gamma(1-s)}\), implies that \[\zeta(s)=(2 \pi)^{s-1}\left(\frac{\sin \pi s}{\sin (\pi(1+s) / 2)}\right) \Gamma(1-s) \zeta(1-s)\] and the result follows from the identity \[\sin \pi s=2 \sin \left(\frac{\pi s}{2}\right) \sin \left(\frac{\pi}{2}(1+s)\right).\]
The proof is complete. $$\tag*{$\blacksquare$}$$
\(\zeta(s)\) has simple zeros at \(s=-2,-4,-6,-8, \ldots\). Indeed, since the integral in (*) is absolutely convergent for all \(s \in \mathbb{C}\) and since \(\omega(x)>0\) for all \(x\in\mathbb R\), we have \[\Xi(-2 n)=\frac{1}{2 n}-\frac{1}{2 n+1}+\int_{1}^{\infty} \omega(x)\left(x^{-n}+x^{n+1 / 2}\right) \frac{dx}{x}>0\] for all \(n\in\mathbb Z_+\). The result follows from the fact that \(\Gamma(s / 2)\) has simple poles at \(s=-2 n\).
These zeros are the only ones lying in the region \(\sigma<0\). They are called trivial zeros of the Riemann zeta-function.
For all \(0<\sigma<1\), we have \(\zeta(\sigma) \neq 0\). Indeed, for all \(\sigma>0\) we see \[\zeta(s)=\frac{s}{s-1}-s \int_{1}^{\infty} \frac{\{x\}}{x^{s+1}} dx\] we infer that, for all \(0<\sigma<1\), we obtain \[\left|\zeta(\sigma)-\frac{\sigma}{\sigma-1}\right|<\sigma \int_{1}^{\infty} \frac{dx}{x^{\sigma+1}}=1,\] which implies that \(\zeta(\sigma)<1+\sigma /(\sigma-1)\) for all \(0<\sigma<1\).
Hence \(\zeta(\sigma)<0\) for all \(\frac{1}{2} \leqslant \sigma<1\), and the functional equation implies the asserted result.
Let \(0<\delta<\pi\), then for any \(z\in\mathbb C\) so that \(|\arg z|<\pi-\delta\), we have \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+O(|z|^{-1}),\] uniformly as \(|z|\to\infty\), where logarithm has principal value, and the implicit constant depend at most on \(\delta\).
Let \(a, b\in\mathbb R\) be fixed and \(a \leq b\).
Then for every \(z=\sigma+ it\) with \(\sigma\in[a, b]\) and \(|t| \geq 1\), we have \[\Gamma(z)=\sqrt{2 \pi} e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}} e^{i|t|(\log |t|-1)} e^{\frac{\pi i}{2}\left(\sigma-\frac{1}{2}\right)}\left(1+O\left(\frac{1}{|t|}\right)\right).\]
Moreover, \(|\Gamma(z)|=\sqrt{2 \pi} e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}}\left(1+O\left(\frac{1}{|t|}\right)\right)\).
This implies \(|\Gamma(z)|=O\left(e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}}\right)\) and
\(\frac{1}{|\Gamma(z)|}=O\left(e^{\frac{\pi}{2}|t|}|t|^{\frac{1}{2}-\sigma}\right)\).
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 \(|\xi(s)|\) for \(\operatorname{Re}(s) \geq 1 / 2\).
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 \[\log |\Gamma(s)| \leq|s| \log |s|+O(|s|) \quad \text { whenever } \quad\operatorname{Re}(s) \geq 1 / 2.\]
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 the trivial zeros.
For \(\operatorname{Re} s>1\), the zeta function \(\zeta(s)\) has no zeros.
In addition to the trivial zeros, the zeta function has infinitely many nontrivial zeros lying in the critical strip \(0 \leq \operatorname{Re} s \leq 1\).
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 \(\zeta(s)\).
We also know that \(\zeta(\sigma)\neq 0\) whenever \(0<\sigma<1\).
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 \[\zeta(s)=2^{s} \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s).\]