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\subsubsection*{The Brun-Titchmarsh Theorem}

\begin{fcthm}[Brun-Titchmarsh Theorem]
  \label{thm:bt}
  Assuming:
  - $x \ge 0$, $y \ge 2$
  - $\eps > 0$ and $y$ is large in terms of $\eps$
  Then:
  \[
    \pi(x + y) - \pi(x)
    \le \frac{(2 + \eps)y}{\log y}
  .\]
\end{fcthm}

\begin{remark*}
  We expect
  \[
    \pi(x + y) - \pi(x)
    = \frac{(n + o(1))y}{\log x}
  \]
  in a wide range of $y$ (e.g. $y \ge x^\eps$ for
  some $\eps > 0$ fixed). The prime number theorem
  gives this for $y \vgg x$. The \nameref{thm:bt}
  gives an upper bound of the expected order for
  $y \ge x^\eps$.
\end{remark*}

\begin{proof}
  Apply the \nameref{thm:ss} with $A = [x, x + y]
  \cap \Nbb$, $\mathcal{P} = \Pbb$. Note that for
  any $d \ge 1$,
  \[
    |\{a \in A : a \equiv 0 \pmod{d}\} =
    \frac{y}{d} + O(1)
  .\]
  Hence, the \gls{sievehyp} holds with $g(d) =
  \frac{1}{d}$, $R_d = O(1)$.

  Now, the function $h$ in \nameref{thm:ss} is
  given on primes by
  \[
    h(p) = \frac{g(p)}{1 - g(p)}
    = \frac{1}{p - 1}
    = \frac{1}{\varphi(p)}
  ,\]
  where $\varphi$ is the Euler totient function.
  In general,
  \[
    h(d) = \mob(d)^2 \cdot \frac{1}{\varphi(d)}
  \]
  (since $\varphi$ is multiplicative). Now, for
  any $z \ge 2$, \nameref{thm:ss} yields
  \[
    S(A, \Pbb, z) \le \frac{y}{\sum_{d \le z}
    \frac{\mob(d)^2}{\varphi(d)}} + O \left(
    \sum_{d \le z^2} \tauf_3(d) \right)
  .\]
  By Problem 11 on \es{1}, the error term is
  $O(z^2 (\log z)^2)$.

  Take $z = y^{\half - \frac{\eps}{10}}$. Then
  \[
    z^2(\log z)^2 \vll (y^{\half -
    \frac{\eps}{10}})^2 (\log y)^2
    \vll y^{1 - \frac{\eps}{20}}
  \]
  (for $y \ge y_0(\eps)$). We estimate
  \begin{align*}
    \sum_{d \le z} \frac{\mob(d)^2}{\varphi(d)}
    &= \sum_{d \le z} \frac{\mob(d)^2}{d}
    \frac{d}{\varphi(d)} \\
    &= \sum_{d \le z} \frac{\mob(d)^2}{d} \cdot
    \prod_{p \mid d} \ub{\left( 1 + \frac{1}{p} +
    \frac{1}{p^2} + \cdots \right)}_{= \frac{p}{p
    - 1} = \frac{p}{\varphi(p)}} \\
    &\ge \sum_{n \le z} \frac{1}{n}
  \end{align*}
  since any $n \le z$ has at least one
  representation as
  \[
    n
    = d p_1^{a_1} \cdots p_k^{a_k}
  ,\]
  where $d \le z$ is square-free and $p_i \mid d$
  are primes and $a_1 \ge 0$.

  We have proved
  \[
    \sum_{n \le z} \frac{1}{n}
    = \log z + O(1)
    \ge \left( 1 - \frac{1}{10} \right) \log z
  \]
  for $z \ge z_0(\eps)$.

  Putting everything together gives us
  \begin{align*}
    \pi(x + y) - \pi(x)
    &\le S(A, \Pbb, z) + z \\
    &\le \frac{y}{\left( 1 - \frac{\eps}{10}
    \right) \log z} + z + y^{1 - \frac{\eps}{20}}
    \\
    &\le \frac{(2 + \eps) y}{\log y}
  \end{align*}
  for $y \ge y_0(\eps)$.
\end{proof}

\newpage
\section{The Riemann Zeta Function}

Recall that the Riemann zeta function is defined
by
\[
  \zeta(s)
  = \sum_{n = 1}^{\infty} \frac{1}{n^s}
\]
for $\Re(s) > 1$.

Remarkable properties:
\begin{enumerate}[(1)]
  \item $\zeta(s)$ extends meromorphically to
    $\Cbb$.
  \item A functional equation relating $\zeta(s)
    \leftrightarrow \zeta(1 - s)$.
  \item All the (non-trivial) zeroes appear to be
    on the line $\Re(s) = \half$ (Riemann
    hypothesis).
  \item $\zeta(s)$ closely relates to the
    distribution of primes.
\end{enumerate}

\begin{notation*}
  $f(x) \asymp g(x)$ means $f(x) \vll g(x) \vll
  f(x)$. $\asymp_\sigma$ means that the constant
  in $\vll$ can depend on $\sigma$.
\end{notation*}

\begin{fclemma}[]
  Assuming:
  - $\sigma > 1$
  - $t \in \Rbb$
  Then:
  $|\zeta(\sigma + it)| \asymp_\sigma 1$.
\end{fclemma}

\begin{proof}
  By the Euler product formula,
  \[
    \zeta(\sigma + it)
    = \prod_p (1 - p^{-\sigma - it})^{-1}
  ,\]
  hence
  \[
    |\zeta(\sigma + it)|
    = \prod_p |1 - p^{-\sigma - it}|^{-1}
  .\]
  By the triangle inequality,
  \[
    1 - p^{-\sigma} \le |1 - p^{-\sigma - it}|
    \le 1 + |p^{-\sigma - it}| = 1 + p^{-\sigma}
  .\]
  Hence,
  \[
    \prod_p (1 - p^{-\sigma})^{-1}
    \ge |\zeta(\sigma + it)|
    \ge \prod_p (1 = p^{-\sigma})^{-1}
  .\]
  Note that these products converge if and only if
  $\sum_p p^{-\sigma}$ converges, and this sum
  converges by the comparison test.
\end{proof}

\begin{fclemma}[Polynomial growth of $\zeta$ in
  half-planes]
  \phantom{}
  \begin{cenum}[(i)]
    \item $f(\zeta)$ extends to a meromorphic
      function on $\Cbb$, with the only pole being
      a simple pole which is at $s = 1$.
    \item Let $k \ge 0$ be an integer. Then, for
      $\Re(s) \ge -k$ and $|s - 1| \ge \frac{1}{10}$
      we have $|\zeta(s)| \vll_k |s|^{k + 2} +
      1$.
  \end{cenum}
\end{fclemma}