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\begin{lemma*}[Functional equation for $\Gamma$]
  The $\Gamma$ function extends meromorphically to
  $\Cbb$, with the only poles being simple poles
  at $s = 0, -1, -2, \ldots$.

  Moreover:
  \begin{enumerate}[(i)]
    \item $\Gamma(s + 1) = s\Gamma(s)$ for $s \in
      \Cbb$.
    \item $\Gamma(s) \Gamma(1 - s) =
      \frac{\pi}{\sin(\pi s)}$ for all $s \in
      \Cbb$ (Euler reflection formula).
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item For $\Re(s) > 0$, by integration by
      parts,
      \[
        \int_0^\infty t^s e^{-t} \dd t
        = s \int_0^\infty t^{s - 1} e^{-t} \dd t
      .\]
      This proves (i) for $\Re(s) > 0$.

      Now for any $k \in \Nbb$, for $\Re(s) > 0$
      we have
      \[
        \Gamma(s) = \frac{\Gamma(s + k)}{(s + k -
        1) \cdots (s + 1) s}
      .\]
      The RHS is analytic for $\Re(s) > -k$, so
      can use analytic continuation to extend
      $\Gamma(s)$ meromorphically to $\Re(s) >
      -k$, with the only poles being simple poles
      at $s = 0, -1, \ldots, -k, -1$.

      Let $k \to \infty$.
    \item Since both sides are analytic in $\Cbb
      \setminus \Zbb$, by analytic continuation,
      it suffices to prove the formula for $0 < s
      < 1$.

      Now, for any $t > 0$,
      \begin{align*}
        t^{s - 1} \Gamma(s - 1)
        &= t^{s - 1} \int_0^\infty u^{-s} e^{-u}
        \dd u \\
        &= \int_0^\infty v^{-s} e^{-vt} \dd t
        &&(u = vt)
      \end{align*}
      Multiply by $e^{-t}$ and integrate, and use
      Fubini to get
      \begin{align*}
        \Gamma(s) \Gamma(s - 1)
        &= \int_0^\infty \int_0^\infty v^{-s}
        e^{-vt} \dd v e^{-t} \dd t \\
        &= \int_0^\infty \int_0^\infty e^{-(v +
        1)t} \dd t v^{-s} \dd v \\
        &= \int_0^\infty \frac{v^{-s}}{1 + v} \dd
        v \\
        &= \int_{-\infty}^\infty \frac{e^{(1 -
        s)x}}{1 + e^x} \dd x
        &&(v = e^x)
      \end{align*}
      Hence, the remaining task is to show
      \[
        \int_{-\infty}^\infty \frac{e^{(1 -
        s)x}}{1 + e^x} \dd x
        = \frac{\pi}{\sin(\pi s)}
      .\]
      We will do this next time.
  \end{enumerate}
\end{proof}
