$\zeta(s)=\frac{s}{s-1}-s\int_1^\infty\{x\}x^{-s-1}\,dx$ for $\Re(s)>0$

1.1. For \(\Re(s) > 1\) and \(M \in \mathbb{N}\), apply Abel summation (integration by parts):\[\sum_{n=1}^M n^{-s} = \int_{1^-}^M x^{-s}\,d\lfloor x \rfloor = \frac{\lfloor M \rfloor}{M^s} + s \int_1^M \frac{\lfloor x \rfloor}{x^{s+1}}\,dx = M^{1-s} + s \int_1^M \frac{x - \{x\}}{x^{s+1}}\,dx.\]

1.2. Split the integral:\[s \int_1^M \frac{x}{x^{s+1}}\,dx = s \int_1^M x^{-s}\,dx = s \left[ \frac{x^{1-s}}{1-s} \right]_1^M = \frac{s}{s-1} (1 - M^{1-s}).\]

1.3. Thus\[\sum_{n=1}^M n^{-s} = \frac{s}{s-1} + M^{1-s} \left(1 - \frac{s}{s-1}\right) - s \int_1^M \frac{\{x\}}{x^{s+1}}\,dx = \frac{s}{s-1} - \frac{M^{1-s}}{s-1} - s \int_1^M \frac{\{x\}}{x^{s+1}}\,dx.\]

1.4. For \(\Re(s) > 1\), as \(M \to \infty\), \(|M^{1-s}| = M^{1-\Re(s)} \to 0\).

1.5. Because \(0 \le \{x\} < 1\), the integral \(\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx\) converges absolutely and uniformly on compact subsets of \(\Re(s) > 0\).

1.6. Taking \(M \to \infty\) gives \(\zeta(s) = \frac{s}{s-1} - s \int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx\) for \(\Re(s) > 1\), and by analytic continuation this holds for all \(\Re(s) > 0\) with \(s \neq 1\).