[proofplan] We derive an integral representation for $\zeta(s)$ valid on $\{\operatorname{Re}(s) > 1\}$ via summation by parts, then recognise the representation as the sum of $\frac{1}{s-1}$ and a function that remains analytic on the larger half-plane $\{\operatorname{Re}(s) > 0\}$. The crucial analytic fact is that the integrand $(x - \lfloor x \rfloor) x^{-s-1}$ is dominated by $x^{-\sigma - 1}$, which is integrable over $[1, \infty)$ precisely when $\sigma > 0$. Locally uniform dominated convergence combined with Morera's theorem then upgrades the integral to a holomorphic function on $\{\operatorname{Re}(s) > 0\}$, while the term $\frac{1}{s-1}$ contributes a simple pole with residue $1$ at $s = 1$. [/proofplan] [step:Rewrite the zeta series by Abel summation to expose a telescoping structure] Fix $s \in \mathbb{C}$ with $\sigma := \operatorname{Re}(s) > 1$. By the [Convergence of the Zeta Series](/theorems/1746), the series $\zeta(s) = \sum_{n \geq 1} n^{-s}$ converges absolutely. We group terms to form a telescoping sum. Partial summation: \begin{align*} \sum_{n=1}^{N} \frac{1}{n^s} = \sum_{n=1}^{N-1} n \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right) + \frac{N}{N^s} \cdot \frac{1}{1}. \end{align*} More directly, using $1 = n - (n-1)$ for $n \geq 2$ and reindexing, \begin{align*} \zeta(s) = \sum_{n=1}^{\infty} n \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right), \end{align*} provided $N/N^s = N^{1-s} \to 0$ as $N \to \infty$, which holds because $|N^{1-s}| = N^{1-\sigma}$ and $\sigma > 1$. [guided] The series $\sum n^{-s}$ converges but does not reveal any structure at $s = 1$ (where it diverges) or continuation to smaller $\sigma$. The plan is to rewrite it as an integral, so we can separate an explicit pole term from a regular piece. The first step is Abel summation (summation by parts). We want to write $\sum_n n^{-s}$ as a sum of differences $n^{-s} - (n+1)^{-s}$ weighted by integers. Start from the partial sums \begin{align*} S_N := \sum_{n=1}^N \frac{1}{n^s}. \end{align*} Writing $n^{-s} = \sum_{m=1}^n (m^{-s} - (m-1+1)^{-s}) \cdot (\text{something})$ is unwieldy; it is cleaner to just verify the telescoping identity directly. Consider \begin{align*} T_N := \sum_{n=1}^{N-1} n \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right) = \sum_{n=1}^{N-1} \frac{n}{n^s} - \sum_{n=1}^{N-1} \frac{n}{(n+1)^s}. \end{align*} Reindex the second sum with $m = n + 1$: \begin{align*} \sum_{n=1}^{N-1} \frac{n}{(n+1)^s} = \sum_{m=2}^{N} \frac{m-1}{m^s} = \sum_{m=2}^{N} \frac{m}{m^s} - \sum_{m=2}^{N} \frac{1}{m^s}. \end{align*} So \begin{align*} T_N = \frac{1}{1^s} + \sum_{m=2}^{N-1} \frac{m - m}{m^s} + \left( - \frac{N - 1}{N^s} \text{ term handling} \right) + \sum_{m=2}^N \frac{1}{m^s}. \end{align*} Computing carefully, \begin{align*} T_N &= \sum_{n=1}^{N-1} \frac{n}{n^s} - \sum_{m=2}^N \frac{m}{m^s} + \sum_{m=2}^N \frac{1}{m^s} \\ &= \frac{1}{1^s} - \frac{N}{N^s} + \sum_{m=2}^N \frac{1}{m^s} = \sum_{m=1}^N \frac{1}{m^s} - N^{1-s} = S_N - N^{1-s}. \end{align*} Hence $T_N = S_N - N^{1-s}$. Since $|N^{1-s}| = N^{1-\sigma} \to 0$ as $N \to \infty$ (because $\sigma > 1$), taking $N \to \infty$ yields \begin{align*} \zeta(s) = \lim_{N \to \infty} S_N = \lim_{N \to \infty} T_N = \sum_{n \geq 1} n \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right). \end{align*} [/guided] [/step] [step:Convert the telescoping difference into a Riemann integral] For each integer $n \geq 1$, write the difference as a definite integral via the Fundamental Theorem of Calculus applied to $\phi(t) = t^{-s}$, which has derivative $\phi'(t) = -s t^{-s-1}$. Integrating over $[n, n+1]$: \begin{align*} \frac{1}{n^s} - \frac{1}{(n+1)^s} = -\int_n^{n+1} \phi'(t) \, d\mathcal{L}^1(t) = s \int_n^{n+1} \frac{1}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} Substituting into the identity from Step 1, \begin{align*} \zeta(s) = \sum_{n=1}^{\infty} n \cdot s \int_n^{n+1} \frac{d\mathcal{L}^1(t)}{t^{s+1}} = s \int_1^{\infty} \frac{\lfloor t \rfloor}{t^{s+1}} \, d\mathcal{L}^1(t), \end{align*} where the last equality identifies $n = \lfloor t \rfloor$ on the interval $[n, n+1)$ and combines the piecewise integrals into a single integral over $[1, \infty)$. The interchange of sum and integral is justified by absolute convergence: $\sum_n n \int_n^{n+1} t^{-\sigma-1} \, d\mathcal{L}^1(t) \leq \int_1^\infty t \cdot t^{-\sigma - 1} \, d\mathcal{L}^1(t) = \int_1^\infty t^{-\sigma} \, d\mathcal{L}^1(t) < \infty$ for $\sigma > 1$. [guided] We now convert the discrete difference $n^{-s} - (n+1)^{-s}$ into an integral so we can combine all terms into one integral over $[1,\infty)$. The function $\phi: (0, \infty) \to \mathbb{C}$, $t \mapsto t^{-s}$, is holomorphic on the complement of the origin (using the principal branch of logarithm), with derivative $\phi'(t) = -s t^{-s-1}$ for real $t > 0$. By the Fundamental Theorem of Calculus on $[n, n+1]$, \begin{align*} \frac{1}{(n+1)^s} - \frac{1}{n^s} = \int_n^{n+1} \phi'(t) \, d\mathcal{L}^1(t) = -s \int_n^{n+1} \frac{d\mathcal{L}^1(t)}{t^{s+1}}. \end{align*} Flipping signs, \begin{align*} \frac{1}{n^s} - \frac{1}{(n+1)^s} = s \int_n^{n+1} \frac{d\mathcal{L}^1(t)}{t^{s+1}}. \end{align*} Substituting into the identity of Step 1, \begin{align*} \zeta(s) = \sum_{n=1}^\infty n \left( \frac{1}{n^s} - \frac{1}{(n+1)^s} \right) = s \sum_{n=1}^\infty n \int_n^{n+1} \frac{d\mathcal{L}^1(t)}{t^{s+1}}. \end{align*} On the interval $[n, n+1)$ the floor function satisfies $\lfloor t \rfloor = n$, so $n = \lfloor t \rfloor$ is the correct coefficient. We combine the piecewise integrals into a single integral: \begin{align*} \sum_{n=1}^\infty n \int_n^{n+1} \frac{d\mathcal{L}^1(t)}{t^{s+1}} = \sum_{n=1}^\infty \int_n^{n+1} \frac{\lfloor t \rfloor}{t^{s+1}} \, d\mathcal{L}^1(t) = \int_1^\infty \frac{\lfloor t \rfloor}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} The interchange of summation and integration is justified by the [Monotone Convergence Theorem](/theorems/???) applied to the absolute value: since $|\lfloor t \rfloor / t^{s+1}| \leq t^{-\sigma}$ and $\int_1^\infty t^{-\sigma} \, d\mathcal{L}^1(t) = \frac{1}{\sigma - 1} < \infty$ for $\sigma > 1$, the integral converges absolutely. Thus \begin{align*} \zeta(s) = s \int_1^\infty \frac{\lfloor t \rfloor}{t^{s+1}} \, d\mathcal{L}^1(t) \qquad (\sigma > 1). \end{align*} [/guided] [/step] [step:Split the floor into its smooth and fractional parts] Write $\lfloor t \rfloor = t - \{t\}$ where $\{t\} := t - \lfloor t \rfloor \in [0, 1)$ is the fractional part. Then \begin{align*} \zeta(s) = s \int_1^\infty \frac{t}{t^{s+1}} \, d\mathcal{L}^1(t) - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} The first integral is elementary: for $\sigma > 1$, \begin{align*} s \int_1^\infty \frac{d\mathcal{L}^1(t)}{t^s} = s \cdot \left[ \frac{t^{1-s}}{1-s} \right]_1^\infty = s \cdot \frac{0 - 1}{1 - s} = \frac{s}{s - 1}, \end{align*} where $t^{1-s} \to 0$ as $t \to \infty$ because $1 - \sigma < 0$. Therefore \begin{align*} \zeta(s) = \frac{s}{s-1} - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} Rewriting using $\frac{s}{s-1} = 1 + \frac{1}{s-1}$, \begin{align*} \zeta(s) - \frac{1}{s-1} = 1 - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t) \qquad (\sigma > 1). \end{align*} [guided] We have \begin{align*} \zeta(s) = s \int_1^\infty \frac{\lfloor t \rfloor}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} The obstacle to extending this to smaller $\sigma$ is that the integrand is only bounded by $t^{-\sigma}$, so integrability requires $\sigma > 1$ — the same constraint as the original series. We need to extract a regular piece; the idea is to subtract the "smooth part" of $\lfloor t \rfloor$, namely $t$ itself. Write $\{t\} := t - \lfloor t \rfloor$ for the fractional part, so $\lfloor t \rfloor = t - \{t\}$. Then \begin{align*} \zeta(s) = s \int_1^\infty \frac{t - \{t\}}{t^{s+1}} \, d\mathcal{L}^1(t) = s \int_1^\infty \frac{d\mathcal{L}^1(t)}{t^s} - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} The first integral is a power-law integral. For $\sigma > 1$, the antiderivative of $t^{-s}$ is $\frac{t^{1-s}}{1-s}$ (since $s \neq 1$), so \begin{align*} \int_1^\infty \frac{d\mathcal{L}^1(t)}{t^s} = \left[ \frac{t^{1-s}}{1-s} \right]_1^\infty = \frac{0 - 1}{1 - s} = \frac{1}{s-1}, \end{align*} using that $\lim_{t \to \infty} t^{1-s} = 0$ for $\sigma > 1$ (because $|t^{1-s}| = t^{1-\sigma} \to 0$). Multiplying by $s$, \begin{align*} s \int_1^\infty \frac{d\mathcal{L}^1(t)}{t^s} = \frac{s}{s-1} = 1 + \frac{1}{s-1}. \end{align*} Substituting, \begin{align*} \zeta(s) = 1 + \frac{1}{s-1} - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t), \end{align*} and rearranging, \begin{align*} \zeta(s) - \frac{1}{s-1} = 1 - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} This identity holds for $\sigma > 1$; the right-hand side will turn out to be analytic on the larger region $\sigma > 0$. [/guided] [/step] [step:Verify the fractional-part integral is holomorphic on $\{\operatorname{Re}(s) > 0\}$] Define \begin{align*} F: \{s \in \mathbb{C} : \operatorname{Re}(s) > 0\} &\to \mathbb{C} \\ s &\mapsto \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t). \end{align*} We verify that $F$ is well-defined and holomorphic on this half-plane. Well-definedness: for $\sigma := \operatorname{Re}(s) > 0$, the integrand is bounded in modulus by \begin{align*} \left| \frac{\{t\}}{t^{s+1}} \right| = \frac{\{t\}}{t^{\sigma + 1}} \leq \frac{1}{t^{\sigma+1}}, \end{align*} because $0 \leq \{t\} < 1$. The function $t \mapsto t^{-\sigma-1}$ is integrable on $[1, \infty)$ with respect to $\mathcal{L}^1$ since $\sigma + 1 > 1$: \begin{align*} \int_1^\infty t^{-\sigma - 1} \, d\mathcal{L}^1(t) = \frac{1}{\sigma} < \infty. \end{align*} Hence the integral defining $F(s)$ converges absolutely. Holomorphy: fix a compact subset $K \subset \{\sigma > 0\}$ and let $\sigma_0 := \min_{s \in K} \operatorname{Re}(s) > 0$. Then \begin{align*} \left| \frac{\{t\}}{t^{s+1}} \right| \leq \frac{1}{t^{\sigma_0 + 1}} \qquad \text{for all } s \in K \text{ and } t \geq 1, \end{align*} and $\int_1^\infty t^{-\sigma_0 - 1} \, d\mathcal{L}^1(t) = 1/\sigma_0 < \infty$. For each $t \geq 1$, the map $s \mapsto \{t\}/t^{s+1} = \{t\} \cdot e^{-(s+1) \log t}$ is holomorphic on $\mathbb{C}$. By the standard theorem on differentiation of parameter-dependent integrals (a consequence of [Morera's Theorem](/theorems/???) combined with Fubini), $F$ is holomorphic on any open set where such a uniform dominating function exists. Since $K$ was arbitrary, $F$ is holomorphic on $\{\sigma > 0\}$. [guided] We set \begin{align*} F: \{\sigma > 0\} \to \mathbb{C}, \qquad F(s) = \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t), \end{align*} and must show this is a well-defined holomorphic function. Well-definedness first. The integrand $\{t\}/t^{s+1}$ is bounded in modulus by $t^{-\sigma-1}$ (using $0 \leq \{t\} < 1$), and $\int_1^\infty t^{-\sigma-1} \, d\mathcal{L}^1(t) = [(-1/\sigma) t^{-\sigma}]_1^\infty = 1/\sigma$ is finite for $\sigma > 0$. So $F(s)$ converges absolutely. Holomorphy next. We invoke the standard criterion: if $f(s, t)$ is holomorphic in $s$ for each fixed $t$, and if on every compact $K \subset \{\sigma > 0\}$ there is a dominating function $g_K \in L^1([1,\infty), \mathcal{L}^1)$ with $|f(s,t)| \leq g_K(t)$ for all $s \in K$, then $F(s) = \int f(s,t) \, d\mathcal{L}^1(t)$ is holomorphic on $\{\sigma > 0\}$. We verify both conditions. For fixed $t \geq 1$, $s \mapsto \{t\}/t^{s+1} = \{t\} e^{-(s+1) \log t}$ is an entire function of $s$. Given any compact $K \subset \{\sigma > 0\}$, the real part $\sigma = \operatorname{Re}(s)$ attains its minimum on $K$, say $\sigma_0 > 0$. Then for all $s \in K$, \begin{align*} \left| \frac{\{t\}}{t^{s+1}} \right| = \frac{\{t\}}{t^{\sigma + 1}} \leq \frac{1}{t^{\sigma_0 + 1}}, \end{align*} and $g_K(t) := t^{-\sigma_0 - 1}$ is integrable on $[1, \infty)$ with integral $1/\sigma_0 < \infty$. The criterion applies (its proof reduces to [Morera's Theorem](/theorems/???) combined with [Fubini's Theorem](/theorems/???) applied to $\oint_\gamma F(s) \, ds = \int_1^\infty \oint_\gamma f(s,t) \, ds \, d\mathcal{L}^1(t) = 0$ for any small contour $\gamma$), so $F$ is holomorphic on $\{\sigma > 0\}$. This is where the transition "from $\sigma > 1$ to $\sigma > 0$" happens in the proof: the integral $\int_1^\infty \{t\} t^{-s-1} \, d\mathcal{L}^1(t)$ converges on a larger domain than $\int_1^\infty t \cdot t^{-s-1} \, d\mathcal{L}^1(t)$ because the fractional part is bounded, whereas the identity part contributes the polar term $1/(s-1)$. [/guided] [/step] [step:Identify the analytic continuation and extract the pole structure] Define \begin{align*} G: \{s \in \mathbb{C} : \operatorname{Re}(s) > 0\} &\to \mathbb{C} \\ s &\mapsto 1 - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t) = 1 - s F(s). \end{align*} By Step 4, $F$ is holomorphic on $\{\sigma > 0\}$, and $s \mapsto 1 - sF(s)$ is a product and sum of holomorphic functions, so $G$ is holomorphic on $\{\sigma > 0\}$. By Step 3, $\zeta(s) - \frac{1}{s-1} = G(s)$ for all $s$ with $\sigma > 1$. Since $\zeta$ is holomorphic on $\{\sigma > 1\}$ and $G$ is holomorphic on $\{\sigma > 0\} \supset \{\sigma > 1\}$, the function \begin{align*} \zeta(s) := G(s) + \frac{1}{s-1}, \qquad s \in \{\sigma > 0\} \setminus \{1\}, \end{align*} agrees with the original $\zeta$ on $\{\sigma > 1\}$ and is holomorphic on $\{\sigma > 0\} \setminus \{1\}$ as a sum of holomorphic functions (with $\frac{1}{s-1}$ holomorphic away from $s = 1$). At $s = 1$, the term $G(s)$ is holomorphic and $\frac{1}{s-1}$ has a simple pole with residue \begin{align*} \lim_{s \to 1} (s - 1) \cdot \frac{1}{s-1} = 1. \end{align*} Since $(s-1)G(s) \to 0$ as $s \to 1$, the full function $\zeta$ has a simple pole at $s = 1$ with residue $1$. Elsewhere on $\{\sigma > 0\}$, $\zeta$ is holomorphic. This completes the proof that $\zeta(s) - \frac{1}{s-1}$ extends analytically to $\{\operatorname{Re}(s) > 0\}$, establishing the claim. [guided] From Step 3 we have the identity on $\{\sigma > 1\}$: \begin{align*} \zeta(s) - \frac{1}{s-1} = 1 - s \int_1^\infty \frac{\{t\}}{t^{s+1}} \, d\mathcal{L}^1(t) = 1 - s F(s) =: G(s). \end{align*} From Step 4, $F$ is holomorphic on $\{\sigma > 0\}$, and products and sums of holomorphic functions are holomorphic, so $G$ is holomorphic on $\{\sigma > 0\}$. Why does this prove analytic continuation? The left-hand side, $\zeta(s) - \frac{1}{s-1}$, is a priori only defined for $\sigma > 1$. The right-hand side, $G(s)$, is a holomorphic function on the strictly larger domain $\{\sigma > 0\}$. On the intersection $\{\sigma > 1\}$ they agree. Therefore $G$ is the unique analytic continuation of $\zeta(s) - \frac{1}{s-1}$ from $\{\sigma > 1\}$ to $\{\sigma > 0\}$ (by the [Identity Theorem](/theorems/???) for holomorphic functions). Equivalently, define \begin{align*} \tilde\zeta(s) := G(s) + \frac{1}{s-1} \qquad \text{for } s \in \{\sigma > 0\} \setminus \{1\}. \end{align*} Then $\tilde\zeta = \zeta$ on $\{\sigma > 1\}$, and $\tilde\zeta$ is holomorphic on $\{\sigma > 0\} \setminus \{1\}$. At $s = 1$, the summand $G(s)$ has a finite value $G(1) = 1 - 1 \cdot F(1)$, while $\frac{1}{s-1}$ has a simple pole with residue $1$: \begin{align*} \lim_{s \to 1} (s-1) \tilde\zeta(s) = \lim_{s \to 1} (s-1)G(s) + \lim_{s \to 1} (s-1) \cdot \frac{1}{s-1} = 0 + 1 = 1. \end{align*} Hence the continuation $\tilde\zeta$ has a simple pole with residue $1$ at $s = 1$, and is holomorphic elsewhere on $\{\sigma > 0\}$. This is the content of the theorem: $\zeta(s) - \frac{1}{s-1}$ (which is $G(s)$, the piece without the pole) extends to an analytic function on $\{\operatorname{Re}(s) > 0\}$. [/guided] [/step]