For $\mathrm{Re}s>0$, zeta admits the fractional-part integral formula with a simple residue-one pole at $1$

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 $1$.