The Riemann Zeta Function

Because $K$ is compact and lies in the open half-plane $\mathrm{Re}s>1$, there is a real number $\sigma >1$ with $\mathrm{Re}s\ge \sigma$ for every $s\in K$. For $n\ge 1$ and $s\in K$, [L1] and [L2] give $n^{-s}=\exp (-s\log n)$, so $\mathrm{\mid }{n}^{-s}\mathrm{\mid }=\exp (-\mathrm{Re}s\log n)\le \exp (-\sigma \log n)=n^{-\sigma }\mathrm{.}$

Taking $\sigma >1$ rational if necessary, [L4] makes ${\sum }_{n\ge 1}{n}^{-\sigma }$ convergent. Step 1.1 and [L3] therefore give uniform convergence of $\sum n^{-s}$ on $K$. Since $\mathrm{\mid }{n}^{-s}\mathrm{\mid }$ is bounded by the same summable majorant, the series also converges absolutely at each point of $K$, hence at each $s$ with $\mathrm{Re}s>1$.

Write $\sigma :=\mathrm{Re}s$. Since the primes are among the integers at least $2$, ${\sum }_p\mathrm{\mid }{p}^{-s}\mathrm{\mid }={\sum }_p{p}^{-\sigma }\le {\sum }_{n=2}^{\mathrm{\infty }}{n}^{-\sigma }<\mathrm{\infty }$ by [L1]. Therefore [L2] applies to $a_p:=-p^{-s}$, so the product ${\prod }_p(1-p^{-s})$ converges and is nonzero.

For a finite set $P$ of primes, ${\prod }_{p\in P}\frac{1}{1-p^{-s}}={\prod }_{p\in P}{\sum }_{k\ge 0}{p}^{-ks}\mathrm{.}$ Multiplying out this finite product lists exactly the terms $n^{-s}$ for those integers $n\ge 1$ whose prime divisors all lie in $P$, and [L3] with [L4] shows that each such integer appears exactly once. Hence ${\prod }_{p\in P}\frac{1}{1-p^{-s}}={\sum }_{\begin{array}{c}n\ge 1\\ p\mid n\Rightarrow p\in P\end{array}}{n}^{-s}\mathrm{.}$

Let $P_N$ be the set of primes at most $N$. By step 1.2 the corresponding partial products are the partial sums over integers all of whose prime divisors lie in $P_N$. Every fixed integer eventually has this property, and the omitted terms are bounded in absolute value by the tail of the absolutely convergent series in [L1]. Therefore these partial products converge pointwise to ${\sum }_{n\ge 1}{n}^{-s}=\zeta (s)$.

Let $K\subseteq \{s:\mathrm{Re}s>1\}$ be compact, and choose $\sigma >1$ with $\mathrm{Re}s\ge \sigma$ on $K$. If an integer $n$ is omitted from the partial product over $P_N$, then $n$ has some prime divisor greater than $N$, hence $n>N$. So for $s\in K$, $\mid \zeta (s)-{\prod }_{p\le N}\frac{1}{1-p^{-s}}\mid \le {\sum }_{n>N}{n}^{-\sigma }\mathrm{.}$ The tail on the right tends to $0$ independently of $s$, so the partial products converge uniformly on $K$. Thus the Euler product converges locally uniformly on $\mathrm{Re}s>1$. Combining this with steps 1.1 and 2.1 proves the stated formula together with its absolute and locally uniform convergence.

By [L1], $\zeta (s)$ is the reciprocal of the absolutely convergent product ${\prod }_p(1-p^{-s})$.

By [L2], that product is nonzero, so its reciprocal is also nonzero. Therefore $\zeta (s)\mathrm{\ne }0$.

For $N\ge 2$, $s{\int }_1^N\lfloor x\rfloor x^{-s-1}dx={\sum }_{n=1}^{N-1}n\text{\hspace{0.17em}⁣}{\int }_n^{n+1}s{x}^{-s-1}dx={\sum }_{n=1}^{N-1}n(n^{-s}-(n+1{)}^{-s})\mathrm{.}$ Expanding the last sum gives $s{\int }_1^N\lfloor x\rfloor x^{-s-1}dx={\sum }_{n=1}^{N-1}{n}^{-s}-(N-1)N^{-s},$ and therefore ${\sum }_{n=1}^N{n}^{-s}=N^{1-s}+s{\int }_1^N\lfloor x\rfloor x^{-s-1}dx\mathrm{.}$

Since $\mathrm{Re}s>1$, the term $N^{1-s}$ tends to $0$ as $N\to \mathrm{\infty }$. Letting $N\to \mathrm{\infty }$ in step 1.1 and using [L1] yields $\zeta (s)=s{\int }_1^{\mathrm{\infty }}\lfloor x\rfloor x^{-s-1}dx=s{\int }_1^{\mathrm{\infty }}(x-\{x\})x^{-s-1}dx\mathrm{.}$ Also $s{\int }_1^{\mathrm{\infty }}{x}^{-s}dx=\frac{s}{s-1},$ so $\zeta (s)=\frac{s}{s-1}-s{\int }_1^{\mathrm{\infty }}\{x\}{x}^{-s-1}dx$ on $\mathrm{Re}s>1$.

Let $K\subseteq \{s\in \mathbb{C}:\mathrm{Re}s>0\}$ be compact, and choose $\sigma >0$ with $\mathrm{Re}s\ge \sigma$ on $K$. Because $0\le \{x\}<1$, $\mid \{x\}{x}^{-s-1}\mid \le x^{-\sigma -1}(x\ge 1,s\in K)\mathrm{.}$ Taking $\sigma$ rational with $\sigma >0$, [L2] implies ${\int }_1^{\mathrm{\infty }}{x}^{-\sigma -1}dx<\mathrm{\infty }$, so the integral in step 2.1 converges absolutely and locally uniformly on $\mathrm{Re}s>0$. Hence it defines a holomorphic function there. Therefore the displayed formula continues meromorphically to $\mathrm{Re}s>0$, and the only singularity is the simple pole of $s\mathrm{/}(s-1)$ at $1$, whose residue is $1$.

Choose $\sigma >0$ and $M\ge 0$ so that $\mathrm{Re}s\ge \sigma$ and $\mathrm{\mid }s\mathrm{\mid }\le M$ on $K$. Pair the alternating series as $\eta (s)={\sum }_{n=1}^{\mathrm{\infty }}((2n-1{)}^{-s}-(2n{)}^{-s})\mathrm{.}$ Using [A1] and [L4], $\mathrm{\mid }(2n-1{)}^{-s}-(2n{)}^{-s}\mathrm{\mid }=\mid s{\int }_{2n-1}^{2n}{x}^{-s-1}dx\mid \le M(2n-1{)}^{-\sigma -1}\mathrm{.}$ Taking $\sigma$ rational, [L5] makes ${\sum }_n(2n-1{)}^{-\sigma -1}$ convergent. Therefore [L3] gives local uniform convergence of the paired series on $K$, so $\eta$ is holomorphic on $\mathrm{Re}s>0$.

On $\mathrm{Re}s>1$, the zeta series of [L1] converges absolutely, so regrouping odd and even terms gives $\eta (s)={\sum }_{n\ge 1}{n}^{-s}-2{\sum }_{n\ge 1}(2n{)}^{-s}=(1-2^{1-s})\zeta (s)\mathrm{.}$

By step 1.1, $\eta$ is holomorphic on $\mathrm{Re}s>0$. By [L2], the function $(1-2^{1-s})\zeta (s)$ is also holomorphic there: the factor $1-2^{1-s}$ vanishes at $s=1$ and removes the only pole of $\zeta$. Step 1.2 shows that the two holomorphic functions agree on the nonempty open set $\mathrm{Re}s>1$, so [A2] gives the identity $\eta (s)=(1-2^{1-s})\zeta (s)$ throughout $\mathrm{Re}s>0$.

Evaluating the transform in [L3] at $\xi =0$ gives $C_t={\widehat{g}}_t(0)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\pi t{x}^2}dx\mathrm{.}$ With the change of variables $u=\sqrt{\pi t}x$ and [L2], this integral equals $t^{-1\mathrm{/}2}$. Therefore ${\widehat{g}}_t(\xi )=t^{-1\mathrm{/}2}{e}^{-\pi {\xi }^2\mathrm{/}t}\mathrm{.}$

By [L1], $\theta (t)={\sum }_{n\in \mathbb{Z}}{g}_t(n)$. Poisson summation from [L3] and step 1.1 therefore give $\theta (t)={\sum }_{m\in \mathbb{Z}}{\widehat{g}}_t(m)=t^{-1\mathrm{/}2}{\sum }_{m\in \mathbb{Z}}{e}^{-\pi m^2\mathrm{/}t}=t^{-1\mathrm{/}2}\theta (1\mathrm{/}t)\mathrm{.}$

Write $\sigma :=\mathrm{Re}s$. By [L2], $\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\theta (t)-1)t^{s\mathrm{/}2-1}dt={\sum }_{n\ge 1}{\int }_0^{\mathrm{\infty }}{e}^{-\pi n^{2}t}{t}^{s\mathrm{/}2-1}dt,$ provided the interchange is justified. Since $\mathrm{\mid }{t}^{s\mathrm{/}2-1}\mathrm{\mid }=t^{\sigma \mathrm{/}2-1}$ and $\sigma >1$, the summands are absolutely integrable and nonnegative after taking absolute values, so [L4] applies to the absolute-value kernel.

For each $n\ge 1$, substitute $u=\pi n^{2}t$. Then ${\int }_0^{\mathrm{\infty }}{e}^{-\pi n^{2}t}{t}^{s\mathrm{/}2-1}dt={\pi }^{-s\mathrm{/}2}{n}^{-s}{\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{s\mathrm{/}2-1}du={\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2)n^{-s}$ by [L3].

Summing the identity of step 1.2 over $n$ and using [L1] yields $\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\theta (t)-1)t^{s\mathrm{/}2-1}dt={\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2){\sum }_{n\ge 1}{n}^{-s}={\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2)\zeta (s)\mathrm{.}$ This is the claimed Mellin representation.

On $\mathrm{Re}s>1$, split the integral in [L3] at $1$. Using [L2] on $(0,1)$ and the change of variables $u=1\mathrm{/}t$ gives $\mathrm{\Lambda }(s)=\frac{1}{s(s-1)}+\frac{1}{2}{\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-(s+1)\mathrm{/}2})dt\mathrm{.}$

For $t\ge 1$, [L2] and the definition of $\theta$ give $0<\theta (t)-1=2{\sum }_{n\ge 1}{e}^{-\pi n^{2}t}\le 2{\sum }_{n\ge 1}{e}^{-\pi nt}=\frac{2e^{-\pi t}}{1-e^{-\pi t}}\mathrm{.}$ Hence the integral in step 1.1 converges absolutely and locally uniformly for every $s\in \mathbb{C}$, because the powers of $t$ contribute only polynomial growth while the right-hand side decays exponentially. Therefore $H(s):=\frac{1}{2}{\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-(s+1)\mathrm{/}2})dt$ is entire, and step 1.1 shows that $\mathrm{\Lambda }(s)=1\mathrm{/}(s(s-1))+H(s)$ is meromorphic on $\mathbb{C}$ with at most simple poles at $0$ and $1$.

By [L5] and [L6], $1\mathrm{/}\mathrm{\Gamma }(s\mathrm{/}2)$ is entire, with a simple zero at $s=0$ and zeros only at the negative even integers. Thus $\tilde{\zeta }(s):={\pi }^{s\mathrm{/}2}\mathrm{\Lambda }(s)\mathrm{/}\mathrm{\Gamma }(s\mathrm{/}2)$ is meromorphic on $\mathbb{C}$. On $\mathrm{Re}s>1$, [L4] makes $\tilde{\zeta }(s)=\zeta (s)$. By [A1], this is the unique meromorphic continuation of zeta. The zero of $1\mathrm{/}\mathrm{\Gamma }(s\mathrm{/}2)$ cancels the pole of $\mathrm{\Lambda }$ at $0$, and no further poles are introduced at the negative even integers. Since [L1] already shows that zeta is holomorphic on $\mathrm{Re}s>0$ away from $1$, the only pole of the continuation is the simple residue-one pole at $s=1$.

Repeating the split-at-$1$ calculation from the Mellin integral in [L3] and using [L2] on $(0,1)$ gives $\mathrm{\Lambda }(s)=\frac{1}{s(s-1)}+\frac{1}{2}{\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-s\mathrm{/}2-1\mathrm{/}2})dt$ for $\mathrm{Re}s>1$.

Define $F(s):=\frac{1}{s(s-1)}+H(s),$ where $H$ is the entire function from [L4]. By step 1.1, on $\mathrm{Re}s>1$ this equals the explicit right-hand side there. That explicit formula is unchanged when $s$ is replaced by $1-s$, because $1\mathrm{/}(s(s-1))=1\mathrm{/}((1-s)(-s))$ and the two powers of $t$ are exchanged. Hence $F(s)=F(1-s)(\mathrm{Re}s>1)\mathrm{.}$

On $\mathrm{Re}s>1$, [L1] names the completed function as $\mathrm{\Lambda }(s)$, and step 1.1 identifies that same function with the explicit split formula. Combined with [L4], this shows that $F(s)=\mathrm{\Lambda }(s)$ there. Since both $F$ and $\mathrm{\Lambda }$ are meromorphic on $\mathbb{C}$, [A1] gives $F=\mathrm{\Lambda }$ on all of $\mathbb{C}$. Applying [A1] again to the meromorphic functions $F(s)$ and $F(1-s)$, which agree on $\mathrm{Re}s>1$ by step 2.1, yields $F(s)=F(1-s)$ on $\mathbb{C}$. Therefore $\mathrm{\Lambda }(s)=F(s)=F(1-s)=\mathrm{\Lambda }(1-s)$ for every $s\in \mathbb{C}$. The explicit pole term in step 1.1 shows that the poles at $0$ and $1$ are simple.

Rearranging [L1] gives $\zeta (s)={\pi }^{s-1\mathrm{/}2}\frac{\mathrm{\Gamma }((1-s)\mathrm{/}2)}{\mathrm{\Gamma }(s\mathrm{/}2)}\zeta (1-s)\mathrm{.}$

Apply [L3] with $z=(1-s)\mathrm{/}2$ to obtain $\mathrm{\Gamma }((1-s)\mathrm{/}2)\mathrm{\Gamma }(1-s\mathrm{/}2)=2^s\sqrt{\pi }\mathrm{\Gamma }(1-s)\mathrm{.}$ Apply [L2] with $z=s\mathrm{/}2$ to obtain $\mathrm{\Gamma }(s\mathrm{/}2)\mathrm{\Gamma }(1-s\mathrm{/}2)=\frac{\pi }{\sin (\pi s\mathrm{/}2)}\mathrm{.}$ Dividing the first identity by the second yields ${\pi }^{s-1\mathrm{/}2}\frac{\mathrm{\Gamma }((1-s)\mathrm{/}2)}{\mathrm{\Gamma }(s\mathrm{/}2)}=2^s{\pi }^{s-1}\sin (\pi s\mathrm{/}2)\mathrm{\Gamma }(1-s)\mathrm{.}$

Substitute the factor identity from step 1.2 into step 1.1. This gives $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s\mathrm{/}2)\mathrm{\Gamma }(1-s)\zeta (1-s),$ which is the classical functional equation.

By [L3], $\mathrm{\Lambda }$ has simple poles at $0$ and $1$. Multiplying by $\frac{1}{2}s(s-1)$ in [L1] cancels exactly those poles, so $\xi$ is entire. The same two facts give $\xi (1-s)=\frac{1}{2}(1-s)(-s)\mathrm{\Lambda }(1-s)=\frac{1}{2}s(s-1)\mathrm{\Lambda }(s)=\xi (s)\mathrm{.}$

For real $x>1$, the Dirichlet series in [L5] is a sum of positive real terms, so $\zeta (x)\in \mathbb{R}$. The remaining factors in [L1] are also real there, hence $\xi (x)\in \mathbb{R}$ for all $x>1$. Therefore the entire functions $s\mapsto \xi (s)$ and $s\mapsto \overline{\xi (\stackrel{‾}{s})}$ agree on $(1,\mathrm{\infty })$, so [A1] makes them equal on all of $\mathbb{C}$. In particular $\xi (x)$ is real for every real $x$.

On the half-plane $\mathrm{Re}s\ge 2$, [L5] gives $\mathrm{\mid }\zeta (s)\mathrm{\mid }\le {\sum }_{n\ge 1}{n}^{-2}$. Applying [L4] to $z=s\mathrm{/}2$ on that sector shows $\mathrm{\mid }\mathrm{\Gamma }(s\mathrm{/}2)\mathrm{\mid }\le \exp (C\mathrm{\mid }s\mathrm{\mid }\log (2+\mathrm{\mid }s\mathrm{\mid }))$ for some constant $C$, hence [L1] gives $\mathrm{\mid }\xi (s)\mathrm{\mid }\le \exp (C_1\mathrm{\mid }s\mathrm{\mid }\log (2+\mathrm{\mid }s\mathrm{\mid }))(\mathrm{Re}s\ge 2)\mathrm{.}$ By the symmetry from step 1.1, the same bound holds on $\mathrm{Re}s\le -1$. On the strip $-1\le \mathrm{Re}s\le 2$, the explicit split formula in [L3] gives $\xi (s)=\frac{1}{2}+\frac{1}{4}s(s-1){\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-s\mathrm{/}2-1\mathrm{/}2})dt\mathrm{.}$ For $t\ge 1$ and $-1\le \mathrm{Re}s\le 2$, both powers of $t$ have modulus at most $1$, while $\theta (t)-1$ decays exponentially in $t$. Hence the integral is uniformly bounded on the strip, so $\mathrm{\mid }\xi (s)\mathrm{\mid }\le C_2(1+\mathrm{\mid }s{\mathrm{\mid }}^2)$ there. Therefore the same exponential bound holds on all of $\mathbb{C}$, and $\xi$ has order at most $1$.