[proofplan] We subtract the elementary singular term from $\zeta(s)$ by first using the identity \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x)=\frac{1}{s-1}. \end{align*} The difference becomes a sum of local errors comparing the left endpoint value $n^{-s}$ with the integral average of $x^{-s}$ over $[n,n+1]$. Each error is controlled by integrating the derivative of $x^{-s}$, giving a locally uniformly convergent series near $s=1$. This series defines a holomorphic bounded correction term, so the displayed singular term is the only singular part at $s=1$. [/proofplan]

[step:Rewrite the singular term as an integral over $[1,\infty)$]Let \begin{align*} H := \{s \in \mathbb{C} : \operatorname{Re}(s) > 1\}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$, restricted to the Borel subsets of the intervals under consideration. For $s \in H$, define the function \begin{align*} f_s : [1,\infty) &\to \mathbb{C} \\ x &\mapsto x^{-s}, \end{align*} where $x^{-s} := e^{-s \log x}$ with the real logarithm on $(0,\infty)$. Since $\operatorname{Re}(s) > 1$, the improper [Lebesgue integral](/page/Lebesgue%20Integral) converges and \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x) &= \lim_{R \to \infty} \int_1^R x^{-s}\,d\mathcal{L}^1(x) \\ &= \lim_{R \to \infty} \frac{R^{1-s}-1}{1-s}. \end{align*} Since $\operatorname{Re}(s)>1$, we have \begin{align*} |R^{1-s}|=R^{1-\operatorname{Re}(s)} \to 0 \end{align*} as $R \to \infty$, and hence \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x) = \frac{1}{s-1}. \end{align*} Moreover, \begin{align*} \int_1^\infty |x^{-s}|\,d\mathcal{L}^1(x) = \int_1^\infty x^{-\operatorname{Re}(s)}\,d\mathcal{L}^1(x) <\infty, \end{align*} so countable additivity of the Lebesgue integral for absolutely integrable functions gives \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x) = \sum_{n=1}^{\infty}\int_n^{n+1}x^{-s}\,d\mathcal{L}^1(x). \end{align*} Therefore, for every $s \in H$, \begin{align*} \zeta(s)-\frac{1}{s-1} = \sum_{n=1}^{\infty} n^{-s} - \sum_{n=1}^{\infty}\int_n^{n+1} x^{-s}\,d\mathcal{L}^1(x). \end{align*} Both series on the right converge absolutely, so termwise subtraction gives \begin{align*} \zeta(s)-\frac{1}{s-1} = \sum_{n=1}^{\infty} \left( n^{-s} - \int_n^{n+1} x^{-s}\,d\mathcal{L}^1(x) \right). \end{align*}[/step]

[guided]The goal is to isolate the singular part of $\zeta(s)$ by comparing the Dirichlet series with an integral that can be evaluated explicitly. Define \begin{align*} H := \{s \in \mathbb{C} : \operatorname{Re}(s) > 1\}. \end{align*} Let $\mathcal{L}^1$ be one-dimensional Lebesgue measure on $\mathbb{R}$, restricted to the Borel subsets of the intervals being integrated over. For $s \in H$, define \begin{align*} f_s : [1,\infty) &\to \mathbb{C} \\ x &\mapsto x^{-s}, \end{align*} where $x^{-s}:=e^{-s\log x}$ and $\log x$ is the real logarithm on $(0,\infty)$. Since $\operatorname{Re}(s)>1$, the improper Lebesgue integral of $f_s$ over $[1,\infty)$ converges. For $R>1$, the [Fundamental Theorem of Calculus](/theorems/632) applied to the continuously differentiable function $x \mapsto x^{1-s}/(1-s)$ on $[1,R]$ gives \begin{align*} \int_1^R x^{-s}\,d\mathcal{L}^1(x) = \frac{R^{1-s}-1}{1-s}. \end{align*} Because $\operatorname{Re}(s)>1$, \begin{align*} |R^{1-s}|=R^{1-\operatorname{Re}(s)} \to 0 \end{align*} as $R\to\infty$. Hence \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x)=\frac{1}{s-1}. \end{align*} The same hypothesis gives absolute integrability: \begin{align*} \int_1^\infty |x^{-s}|\,d\mathcal{L}^1(x) = \int_1^\infty x^{-\operatorname{Re}(s)}\,d\mathcal{L}^1(x) <\infty. \end{align*} Therefore countable additivity of the Lebesgue integral for absolutely integrable functions permits the decomposition over the intervals $[n,n+1]$: \begin{align*} \int_1^\infty x^{-s}\,d\mathcal{L}^1(x) = \sum_{n=1}^{\infty}\int_n^{n+1}x^{-s}\,d\mathcal{L}^1(x). \end{align*} Since the Dirichlet series defining $\zeta(s)$ also converges absolutely on $H$, termwise subtraction is justified and yields \begin{align*} \zeta(s)-\frac{1}{s-1} = \sum_{n=1}^{\infty} \left( n^{-s} - \int_n^{n+1}x^{-s}\,d\mathcal{L}^1(x) \right). \end{align*} This is the key reduction: the singular integral has been removed, and what remains is a sum of local approximation errors.[/guided]

[step:Express each local error through the derivative of $x^{-s}$]For $n \in \mathbb{N}$, define the map \begin{align*} a_n:\mathbb C&\to\mathbb C\\ s&\mapsto n^{-s} - \int_n^{n+1} x^{-s}\,d\mathcal{L}^1(x). \end{align*} Equivalently, \begin{align*} a_n(s) = \int_n^{n+1} \bigl(n^{-s}-x^{-s}\bigr)\,d\mathcal{L}^1(x). \end{align*} For fixed $s \in \mathbb{C}$, the map $x \mapsto x^{-s}$ is continuously differentiable on $[n,n+1]$ and has derivative $-s x^{-s-1}$. Hence, the [Fundamental Theorem of Calculus](/theorems/632) gives, for $x \in [n,n+1]$, \begin{align*} n^{-s}-x^{-s} = s\int_n^x t^{-s-1}\,d\mathcal{L}^1(t). \end{align*} Substituting this identity into the formula for $a_n(s)$ and using [Fubini's theorem](/page/Probability%20and%20Measure%20Mathematical%20Tripos%20Part%20II%20Cambridge) on the compact rectangle $\{(t,x): n \leq t \leq x \leq n+1\}$ gives \begin{align*} a_n(s) &= s\int_n^{n+1}\int_n^x t^{-s-1}\,d\mathcal{L}^1(t)\,d\mathcal{L}^1(x) \\ &= s\int_n^{n+1}(n+1-t)t^{-s-1}\,d\mathcal{L}^1(t). \end{align*}[/step]

[guided]The point of this step is to turn the difference between a summand and an integral into a derivative estimate. Fix $n \in \mathbb{N}$, and define \begin{align*} a_n:\mathbb C&\to\mathbb C\\ s&\mapsto n^{-s} - \int_n^{n+1} x^{-s}\,d\mathcal{L}^1(x). \end{align*} Now fix $s\in\mathbb C$. Since $n^{-s}$ is constant as a function of $x$, we may write \begin{align*} a_n(s) = \int_n^{n+1} \bigl(n^{-s}-x^{-s}\bigr)\,d\mathcal{L}^1(x). \end{align*} Now we compute the difference $n^{-s}-x^{-s}$. The function $x \mapsto x^{-s}$ is continuously differentiable on $[n,n+1]$, because $n \geq 1$ keeps the interval away from $0$. Its derivative is \begin{align*} \frac{d}{dx}x^{-s}=-s x^{-s-1}. \end{align*} The [Fundamental Theorem of Calculus](/theorems/632) on $[n,x]$ gives \begin{align*} x^{-s}-n^{-s} = \int_n^x -s t^{-s-1}\,d\mathcal{L}^1(t), \end{align*} and therefore \begin{align*} n^{-s}-x^{-s} = s\int_n^x t^{-s-1}\,d\mathcal{L}^1(t). \end{align*} Substituting this into $a_n(s)$ yields \begin{align*} a_n(s) = s\int_n^{n+1}\int_n^x t^{-s-1}\,d\mathcal{L}^1(t)\,d\mathcal{L}^1(x). \end{align*} The integrand is continuous on the compact triangular region \begin{align*} \{(t,x) \in \mathbb{R}^2 : n \leq t \leq x \leq n+1\}, \end{align*} so [Fubini's theorem](/page/Probability%20and%20Measure%20Mathematical%20Tripos%20Part%20II%20Cambridge) applies. Reversing the order of integration changes the inner variable $x$ to range from $t$ to $n+1$, hence \begin{align*} a_n(s) &= s\int_n^{n+1}\int_t^{n+1} t^{-s-1}\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) \\ &= s\int_n^{n+1}(n+1-t)t^{-s-1}\,d\mathcal{L}^1(t). \end{align*} This formula is useful because $0 \leq n+1-t \leq 1$, leaving a summable factor of size approximately $n^{-\operatorname{Re}(s)-1}$.[/guided]

[step:Prove local uniform convergence of the correction series near $s=1$]Let \begin{align*} D := \{s \in \mathbb{C} : |s-1| < 1/2\}. \end{align*} For $s \in D$, we have $|s| \leq 3/2$ and $\operatorname{Re}(s) > 1/2$. Thus, for every $n \in \mathbb{N}$ and $s \in D$, \begin{align*} |a_n(s)| &\leq |s|\int_n^{n+1}(n+1-t)\,|t^{-s-1}|\,d\mathcal{L}^1(t) \\ &\leq \frac{3}{2}\int_n^{n+1} t^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(t) \\ &\leq \frac{3}{2}\int_n^{n+1} t^{-3/2}\,d\mathcal{L}^1(t) \\ &\leq \frac{3}{2} n^{-3/2}. \end{align*} The numerical series \begin{align*} \sum_{n=1}^{\infty} \frac{3}{2}n^{-3/2} \end{align*} converges. Define \begin{align*} A:D&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty} a_n(s). \end{align*} The [Weierstrass $M$-test](/page/Sequence%20of%20Functions) implies that this series converges uniformly on $D$. For each fixed $n$, the map $(s,t) \mapsto s(n+1-t)t^{-s-1}$ is continuous on $D \times [n,n+1]$ and holomorphic in $s$ for each $t$. On every compact set $K \subset D$, the integrand is bounded uniformly on $K \times [n,n+1]$, so holomorphic parameter integration implies that $a_n: D \to \mathbb{C}$ is holomorphic. The [locally uniform limit theorem for holomorphic functions](/page/Holomorphic%20Function) then implies that the uniform limit $A: D \to \mathbb{C}$ is holomorphic on $D$.[/step]

[guided]We now prove the estimate that makes the correction term harmless near $s=1$. Define the open disc \begin{align*} D := \{s \in \mathbb{C} : |s-1| < 1/2\}. \end{align*} For $s \in D$, the triangle inequality gives $|s| \leq |s-1|+1 < 3/2$, and the real part satisfies $\operatorname{Re}(s) > 1/2$. Using the formula from the previous step, \begin{align*} a_n(s) = s\int_n^{n+1}(n+1-t)t^{-s-1}\,d\mathcal{L}^1(t), \end{align*} we estimate its absolute value. For $t \in [n,n+1]$, the factor $n+1-t$ lies between $0$ and $1$. Also, \begin{align*} |t^{-s-1}| = t^{-\operatorname{Re}(s)-1}, \end{align*} because $t>0$ and $t^{-s-1}=e^{-(s+1)\log t}$. Therefore \begin{align*} |a_n(s)| &\leq |s|\int_n^{n+1}(n+1-t)t^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(t) \\ &\leq \frac{3}{2}\int_n^{n+1} t^{-\operatorname{Re}(s)-1}\,d\mathcal{L}^1(t). \end{align*} Since $\operatorname{Re}(s)>1/2$, we have $-\operatorname{Re}(s)-1<-3/2$, and for $t\geq 1$ this implies \begin{align*} t^{-\operatorname{Re}(s)-1} \leq t^{-3/2}. \end{align*} Thus \begin{align*} |a_n(s)| &\leq \frac{3}{2}\int_n^{n+1} t^{-3/2}\,d\mathcal{L}^1(t) \\ &\leq \frac{3}{2}n^{-3/2}. \end{align*} The comparison series \begin{align*} \sum_{n=1}^{\infty} \frac{3}{2}n^{-3/2} \end{align*} converges. Define \begin{align*} A:D&\to\mathbb C\\ s&\mapsto \sum_{n=1}^{\infty}a_n(s). \end{align*} The [Weierstrass $M$-test](/page/Sequence%20of%20Functions) shows that this series converges uniformly on $D$. Each $a_n$ is holomorphic on $D$ because it is given by an integral over the fixed compact interval $[n,n+1]$ of the function $(s,t) \mapsto s(n+1-t)t^{-s-1}$, which is continuous jointly and holomorphic in $s$, with uniform bounds on compact subsets of $D$. The [locally uniform limit theorem for holomorphic functions](/page/Holomorphic%20Function) then implies that $A: D \to \mathbb{C}$ is holomorphic.[/guided]

[step:Identify the bounded holomorphic correction term]For $s \in H \cap D$, the identity from the first step and the definition of $A$ give \begin{align*} \zeta(s)-\frac{1}{s-1}=A(s). \end{align*} Since $A$ is holomorphic on $D$, it is continuous at $1$. Choose \begin{align*} r := \frac{1}{4}. \end{align*} The closed disc \begin{align*} \overline{D}_r := \{s \in \mathbb{C} : |s-1| \leq r\} \end{align*} is compact and contained in $D$, so the continuous function $A$ is bounded on $\overline{D}_r$. Define \begin{align*} C := \sup_{s \in \overline{D}_r} |A(s)|. \end{align*} Then $C < \infty$, and whenever $\operatorname{Re}(s)>1$ and $0<|s-1|<r$, \begin{align*} \left|\zeta(s)-\frac{1}{s-1}\right| = |A(s)| \leq C. \end{align*} Thus \begin{align*} \zeta(s)=\frac{1}{s-1}+O(1) \end{align*} as $s \to 1$ with $\operatorname{Re}(s)>1$.[/step]

[guided]The previous steps constructed a function \begin{align*} A: D \to \mathbb{C} \end{align*} which is holomorphic, hence continuous, on the disc $D=\{s\in\mathbb{C}:|s-1|<1/2\}$. For $s \in H\cap D$, the local-error identity gives \begin{align*} \zeta(s)-\frac{1}{s-1}=A(s). \end{align*} Thus the only remaining task is to translate continuity of $A$ near $1$ into the $O(1)$ assertion. Choose \begin{align*} r:=\frac{1}{4}, \end{align*} and define the closed disc \begin{align*} \overline{D}_r:=\{s\in\mathbb{C}:|s-1|\leq r\}. \end{align*} This set is compact and satisfies $\overline{D}_r\subset D$. Since $A$ is continuous, the extreme value theorem gives a finite bound \begin{align*} C:=\sup_{s\in\overline{D}_r}|A(s)|<\infty. \end{align*} Therefore, whenever $\operatorname{Re}(s)>1$ and $0<|s-1|<r$, we have $s\in H\cap D$ and hence \begin{align*} \left|\zeta(s)-\frac{1}{s-1}\right| = |A(s)| \leq C. \end{align*} This is precisely \begin{align*} \zeta(s)=\frac{1}{s-1}+O(1) \end{align*} as $s\to 1$ through the half-plane $\operatorname{Re}(s)>1$.[/guided]

[step:Read off the pole order and residue]The identity \begin{align*} \zeta(s)=\frac{1}{s-1}+A(s) \end{align*} on $H \cap D$ extends $\zeta$ meromorphically to $D$ by defining the [meromorphic function](/page/Meromorphic%20Function) \begin{align*} Z: D \setminus \{1\} &\to \mathbb{C} \\ s &\mapsto \frac{1}{s-1}+A(s). \end{align*} Since $A$ is holomorphic at $1$, the Laurent expansion of $Z$ at $1$ has principal part exactly \begin{align*} \frac{1}{s-1}. \end{align*} Therefore $s=1$ is a simple pole of the meromorphic continuation of $\zeta$, and the residue is the coefficient of $(s-1)^{-1}$, namely \begin{align*} \operatorname{Res}_{s=1}\zeta = 1. \end{align*}[/step]

[guided]On $H\cap D$ we have proved the identity \begin{align*} \zeta(s)=\frac{1}{s-1}+A(s), \end{align*} where $A:D\to\mathbb{C}$ is holomorphic. This identity defines a meromorphic function \begin{align*} Z:D\setminus\{1\}&\to\mathbb{C}\\ s&\mapsto \frac{1}{s-1}+A(s) \end{align*} which agrees with $\zeta$ on $H\cap D$. Since $A$ is holomorphic at $1$, its Laurent expansion at $1$ has no negative powers. Thus the entire principal part of $Z$ at $1$ is \begin{align*} \frac{1}{s-1}. \end{align*} There is exactly one negative-power term, so the pole is simple. The residue is the coefficient of $(s-1)^{-1}$ in the Laurent expansion, and that coefficient is \begin{align*} \operatorname{Res}_{s=1}\zeta=1. \end{align*}[/guided]