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]. For every prime $p$, $\mathrm{\mid }{p}^{-s}\mathrm{\mid }=p^{-\sigma }<1$, so $1-p^{-s}\mathrm{\ne }0$. Therefore [L2] applies to $a_p:=-p^{-s}$ and makes the product ${\prod }_p(1-p^{-s})$ convergent and 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 ${\prod }_p(1-p^{-s})$.

Each factor $1-p^{-s}$ is nonzero because $\mathrm{\mid }{p}^{-s}\mathrm{\mid }=p^{-\mathrm{Re}s}<1$. By [L2], the absolutely convergent 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 $\mathrm{Re}s>0$.

Repeated differentiation of $g_t(x)=e^{-\pi t{x}^2}$ gives a polynomial times the same Gaussian. For every polynomial $P$, $P(x)e^{-\pi t{x}^2}\to 0$ faster than any reciprocal power as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$, since $t>0$. Thus $g_t$ is Schwartz and [L3] applies.

By [L1], $\theta (t)={\sum }_{n\in \mathbb{Z}}{g}_t(n)$. Poisson summation from [L3], step 1.1 and the exact Gaussian transform [L2] 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 $1$. Under [A2], 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 $s=1$.

Under [A2], 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 $F(s):=1\mathrm{/}(s(s-1))+H(s)$ on $\mathbb{C}$, using the globally convergent integral for $H$ in [L4]. For every $s\in \mathbb{C}$, its two powers of $t$ exchange under $s\mapsto 1-s$, and $1\mathrm{/}(s(s-1))=1\mathrm{/}((1-s)(-s))$. Therefore $F(s)=F(1-s)$ as meromorphic functions on $\mathbb{C}$. This conclusion uses the global integral identity, not merely its initial validity on $\mathrm{Re}s>1$.

On $\mathrm{Re}s>1$, [L1] names the completed function as $\mathrm{\Lambda }(s)$ and step 1.1 identifies it with $F$. By [L4] this is its meromorphic continuation, so [A1] identifies $F$ with $\mathrm{\Lambda }$ on all of $\mathbb{C}$. Step 1.2 now gives $\mathrm{\Lambda }(s)=F(s)=F(1-s)=\mathrm{\Lambda }(1-s)$. Since $H$ is entire, the nonzero simple principal parts of $1\mathrm{/}(s(s-1))$ give simple poles at $0$ and $1$.

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.

Split the Mellin integral [L3] at $1$ and apply the theta relation [L2] on $(0,1)$. Changing variables $u=1\mathrm{/}t$ gives $\mathrm{\Lambda }(s)=\frac{1}{s(s-1)}+H(s),H(s)=\frac{1}{2}{\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-s\mathrm{/}2-1\mathrm{/}2})dt\mathrm{.}$ For $t\ge 1$, [L2] gives $0\le \theta (t)-1\le 2e^{-\pi t}\mathrm{/}(1-e^{-\pi t})$, so the integral converges locally uniformly for all $s\in \mathbb{C}$ and $H$ is entire. The formula continues $\mathrm{\Lambda }$ meromorphically with simple poles at $0,1$ and is invariant under $s\mapsto 1-s$, since its two integrand powers are exchanged. By [L1], $\xi (s)=\frac{1}{2}s(s-1)\mathrm{\Lambda }(s)$ therefore extends to an entire function and satisfies $\xi (s)=\xi (1-s)$.