[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*}
[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]
[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*}
[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]
[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 $D$.
[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]
[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 $\operatorname{Re}(s)>1$.
[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]
[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*}
[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]
[/step]
