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\begin{fcthm}[Functional equation for $zeta$]
  Assuming:
  - $\xi(s) = \half s(s - 1) \pi^{-s/2} \Gamma
  \left( \frac{s}{s} \right) \zeta(s)$ for $s \in
  \Cbb$
  Then:
  $\xi$ is an entire function and $\xi(s) = \xi(1
  - s)$ for $s \in \Cbb$. Hence
  \[
    \pi^{-s / 2} \Gamma \left( \frac{s}{2} \right)
    \zeta(s)
    = \pi^{-\frac{1 - s}{2}} \Gamma \left( \frac{1
    - s}{2} \right) \zeta(1 - s)
  \]
  for $s \in \Cbb \setminus \{0, 1\}$.
\end{fcthm}

\begin{proof}
  Let $\Re(s) > 1$. Then,
  \[
    \Gamma \left( \frac{s}{2} \right) =
    \int_0^\infty t^{\frac{s}{2} - 1} e^{-t} \dd t
  .\]
  Make the change of variables $f = \pi n^2 u$ to
  get
  \[
    \Gamma \left( \frac{s}{2} \right)
    = \pi^{s/2} n^s \int_0^\infty u^{\frac{s}{2} -
    1} e^{-\pi n^2 u} \dd u
  .\]
  Hence
  \[
    \pi^{-\frac{s}{2}} n^{-s} \Gamma \left(
    \frac{s}{2} \right)
    = \int_0^\infty u^{\frac{s}{2} - 1} e^{-\pi
    n^2 u} \dd u
  .\]
  Summing over $n \in \Nbb$ and using Fubini,
  \begin{align*}
    \pi^{-\frac{s}{2}} \Gamma \left( \frac{s}{2}
    \right) \zeta(s)
    &= \sum_{n = 1}^{\infty} \int_0^\infty
    u^{\frac{s}{2} - 1} e^{-\pi n^2 u} \dd u \\
    &= \half \int_0^\infty u^{\frac{s}{2} - 1}
    (\theta(u) - 1) \dd u
    &&\theta(u) = \sum_{n = -\infty}^\infty
    e^{-\pi n^2 u} \\
    &= \half \int_0^1 u^{\frac{s}{2} - 1}
    (\theta(u) - 1) \dd u
    + \half \int_1^\infty u^{\frac{s}{2} - 1}
    (\theta(u) - 1) \dd u
    \tag{$*$}
    \label{lec13eq1}
  \end{align*}
  By the functional equation
  \[
    \theta(u) = \frac{1}{\sqrt{u}} \theta \left(
    \frac{1}{u} \right)
  ,\]
  we have
  \begin{align*}
    \int_0^1 u^{\frac{s}{2} - 1} (\theta(u) - 1)
    \dd u
    &= \int_0^1 u^{\frac{s}{2} - \frac{3}{2}}
    \theta \left( \frac{1}{u} \right) \dd u
    - \int_0^1 u^{\frac{s}{2} - 1} \dd u \\
    &= \int_1^\infty v^{-\frac{s + 1}{2}}
    \theta(v) \dd v
    - \int_0^1 u^{\frac{s}{2} - 1} \dd u
    &&(u = \frac{1}{v}) \\
    &= \int_1^\infty v^{-\frac{s + 1}{2}}
    (\theta(v) - 1) \dd v
    + \ub{\int_1^\infty v^{-\frac{s + 1}{2}} \dd
    v}_{\frac{1}{\frac{s - 1}{2}}}
    - \ub{\int_0^1 u^{\frac{s}{2} - 1} \dd
    u}_{\frac{1}{\frac{s}{2}}}
  \end{align*}
  Plugging this into \eqref{lec13eq1}, we get
  \[
    \pi^{-\frac{s}{2}} \Gamma \left( \frac{s}{2}
    \right) \zeta(s)
    = \half \int_1^\infty (u^{-\frac{s + 1}{2}} +
    u^{\frac{s}{2} - 1})(\theta(u) - 1) \dd u -
    \frac{1}{s(s - 1)}
  .\]
  Hence
  \[
    \xi(s) = -\half + \frac{1}{4} s(s - 1)
    \int_1^\infty (u^{-\frac{s + 1}{2}} +
    u^{\frac{s}{2} - 1}) (\theta(u) - 1) \dd u
    \tag{$**$}
    \label{lec13eq2}
  .\]
  Since $|\theta(u) - 1| \vll e^{-\pi u}$,
  applying the criterion for integrals of analytic
  functions being analytic, we see that $\xi(s)$
  is entire. So by analytic continuation,
  \eqref{lec13eq2} holds for all $s \in \Cbb$.

  Moreover, the expression for $\zeta(s)$ is
  symmetric with respect to $s \mapsto 1 - s$, so
  $\xi(s) = \xi(1 - s)$, $s \in \Cbb$.
\end{proof}

\begin{fccoro}[Zeroes and poles of $zeta$]
  The $\zeta$ function extends to a meromorphic
  function in $\Cbb$ and it has
  \begin{cenum}[(i)]
    \item Only one pole, which is a simple pole at
      $s = 1$, residue $1$.
    \item Simple zeroes at $s = -2, -4, -6,
      \ldots$.
    \item Any other zeroes satisfy $0 \le \Re(s)
      \le 1$.
  \end{cenum}
\end{fccoro}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(ii) -- (iii)]
    \item Follows from the lemma on polynomial
      growth of $\zeta$ on vertical lines.
    \item[(ii) -- (iii)] We know $\zeta(s) \neq 0$
      for $\Re(s) > 1$. We want to show that if
      $\zeta(s) = 0$ and $\Re(s) > 0$ then $s \in
      \{-2, -4, -6, \ldots\}$ and $s$ is a simple
      zero.

      Let $\Re(s) > 0$. By the functional equation
      for $\zeta$,
      \[
        \zeta(s)
        = \ub{\pi^{s - \half}}_{\neq 0}
        \frac{\Gamma \left( \frac{1 - s}{2}
        \right)}{\Gamma \left( \frac{s}{2}
        \right)} \zeta(1 - s)
      .\]
      We claim that $\Gamma(s) \neq 0$ for all $s
      \in \Cbb$. By the Euler reflection formula,
      \[
        \Gamma(s) \Gamma(1 - s) \sin(\pi s)
        = \pi
      ,\]
      for $s \in \Cbb$.

      If $\Gamma(s) = 0$, then $\Gamma$ has a
      pole at $1 - s$. Hence, $1 - s = -n$ for
      some $n \ge 0$ integer. But then $s = n +
      1$, and $\Gamma(n + 1) = n! \neq 0$.

      We conclude that for $\Re(s) < 0$,
      \begin{align*}
        \zeta(s) = 0
        &\iff \frac{s}{2} \text{ is a pole of
        $\Gamma$} \\
        &\iff s = -2n, n \in \Nbb
      \end{align*}
      Since the poles of $\Gamma$ are simple,
      $\zeta$ has a simple zero at $s = -2n$, $n
      \in \Nbb$.
      \qedhere
  \end{enumerate}
\end{proof}

\subsection{Partial fraction approximation of
$\zeta$}

This is a formula for
$\frac{\zeta'(s)}{\zeta(s)}$.

For the proof we need a lemma. We write, for $z
\in \Cbb$, $r > 0$,
\begin{align*}
  B(z_0, r)
  &= \{z \in \Cbb : |z - z_0| < r\} \\
  \ol{B(z_0, r)}
  &= \{z \in \Cbb : |z - z_0| \le r\} \\
  \partial B(z_0, r)
  &= \{z \in \Cbb : |z - z_0| = r\}
\end{align*}

\begin{fclemma}[Borel-Caratheodory Theorem]
  \label{thm:borel-cara}
  Assuming:
  - $0 < r < R$
  - $f$ analytic in $\ol{B(0, R)}$, with $f(0) = 0$
  Then:
  \[
    \sup_{|z| \le r} |f(z)|
    \le \frac{2r}{R - r} \sup_{|z| = R} \Re(f(z))
  .\]
\end{fclemma}

\noproof

\begin{proof}
  This is Exercise 10 on \es{2}.
\end{proof}
