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\begin{proof}
  Let $k \ge 0$ be an integer.

  We claim that there exist polynomials $P_k$,
  $Q_k$ of degree $\le k + 1$ and such that for
  $\Re(s) > 1$,
  \[
    \zeta(s) = \frac{1}{s - 1} + Q_k(s) + s(s + 1)
    \cdots (s + k) \int_1^\infty
    \frac{P_k(\{t\})}{t^{s + k + 1}} \dd t
    \tag{$*$}
    \label{lec12eq1}
  .\]
  First assume \eqref{lec12eq1} holds.

  Then, since $P_k(\{t\}) \vll_k 1$, for $\Re(s) >
  -k - \half$, $|s - 1| \ge \frac{1}{10}$,
  \eqref{lec12eq1} gives $|\zeta(s)| \vll_k |s|^{k
  + 2} + 1$. So (ii) follows.

  For (i), using analytic continuation and
  \eqref{lec12eq1}, it suffices to show that the
  RHS of \eqref{lec12eq1} is meromorphic for
  $\Re(s) > -k$, with the only pole a simple one
  at $s = 1$.

  Suffices to show that
  \[
    \int_1^\infty \frac{P_k(\{t\})}{t^{s + k + 1}}
    \dd t
  \]
  is analytic for $\Re(s) > -k$.

  This follows from the following criterion: If $U
  \subseteq \Cbb$ is open, $f : U \times \Rbb \to
  \Cbb$ is piecewise continuous, and if $s \mapsto
  f(s, t)$ is analytic in $U$ for any $t \in
  \Rbb$, then $\int_{\Rbb} f(s, t) \dd t$ is
  analytic in $U$, provided that $\int_{\Rbb}
  |f(s, t)| \dd t$ is bounded on compact subsets
  of $U$.

  Applying this with $f(s, t) =
  \frac{P_k(\{t\})}{t^{s + k + 1}} \indicator{[1,
  \infty)}$ concludes the
  proof of (i) assuming \eqref{lec12eq1}.

  We are left with proving \eqref{lec12eq1}. We
  use induction on $k$.

  Case $k = 0$: By \nameref{lemma:partsum},
  \begin{align*}
    \zeta(s)
    &= \sum_{n = 1}^\infty \frac{1}{n^s} \\
    &= s \int_1^\infty \frac{\left\lfloor u
    \right\rfloor}{u^{s + 1}} \dd u
    &&\text{(apply partial summation to $\sum_{n
    \le x} n^{-s}$ and let $x \to \infty$)} \\
    &= s \int_1^\infty \frac{1}{u^s} \dd u
    - \int_1^\infty \frac{\{u\}}{u^{s + 1}} \dd u
    \\
    &= \frac{1}{s - 1} + 1 - s \int_1^\infty
    \frac{\{u\}}{u^{s + 1}} \dd u
  \end{align*}
  Take $P_0(u) = u$, $Q_0(u) = 1$.

  Case $k + 1$ assuming $k$: Let
  \[
    c_k = \int_0^1 P_k(\{t\}) \dd t
  .\]
  For $\Re(s) > 1$, we have
  \[
    \zeta(s) = \frac{1}{s - 1} + Q_k(s) + c_k s(s
    + 1) \cdots (s + k - 1) + s(s + 1) \cdots (s +
    k) \int_1^\infty \frac{P_k(\{t\}) - c_k}{t^{s
    + k + 1}} \dd t
  .\]
  Let
  \[
    P_{k + 1}(u)
    = -\int_0^u (P_k(t) - c_k) \dd t
  .\]
  This is a polynomial of degree $\le k + 2$. By
  integration by parts,
  \[
    \int_1^\infty \frac{P_t(\{t\}) - c_k}{t^{s + k
    + 1}} \dd t
    = (s + k + 1) \int_1^\infty \frac{P_{k +
    1}(\{u\})}{u^{s + k + 1}} \dd u
  .\]
  Substituting this into the previous equation, we
  get that case $k + 1$ follows.
\end{proof}

\subsubsection*{The Gamma function}

\begin{fcdefnstar}[Gamma function]
  For $\Re(s) > 0$, let
  \[
    \Gamma(s)
    = \int_0^\infty t^{s - 1} e^{-t} \dd t
  .\]
\end{fcdefnstar}

\begin{lemma}[]
  $\Gamma(s)$ is analytic for $\Re(s) > 0$.
\end{lemma}

\begin{proof}
  Apply the same criterion for integral of
  analytic function being analytic as in the
  previous lemma, taking $f(s, t) = t^{s - 1}
  e^{-t} \indicator{[0, \infty)}(t)$.

  Note that for $0 < \sigma_1 \le \Re(s) \le
  \sigma_2$,
  \[
    \int_{\Rbb} |f(s, t)| \dd t
    \le \int_0^1 t^{\sigma_1 - 1} e^{-t} \dd t
    + \int_1^\infty t^{\sigma_2 - 1} e^{-t} \dd t
    < \infty
    .
    \qedhere
  \]
\end{proof}