The Riemann zeta function extends meromorphically to the complex plane with its only pole at $1$

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 $s=1$.