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% The following slideshow was used in a presentation to the Analysis Seminar at Missouri S&T 9 December 2011
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% Title: Zagier's Proof of the Prime Number Theorem
% Speaker: Tom Cuchta
% Abstract: The prime number theorem states that the number of primes not exceeding x is asymptotic to x/log(x) as x tends to infinity.  This theorem was first proved independently in 1896 by Hadamard and de la Vallee Poussin.  Their proof had two elements: showing that Riemann's zeta function has no zeros on the line Re(x) = 1 and deducing the prime number theorem from this.  A short proof of the first assertion has been known for a long time but the deduction of the prime number theorem continued to involve difficult analysis.  In 1980 D.J. Newman discovered a very simple version of the Tauberian argument needed for an analytic proof of the prime number theorem.  We describe the resulting proof which has a simple structure and uses hardly anything beyond Cauchy's integral theorem.
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% Copyright (C) 2011 Tom Cuchta
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\begin{document}
\title{Zagier's Proof of the Prime Number Theorem}
\author[Tom Cuchta]{ Tom Cuchta }
\date[]{9 December 2011}
\makebeamertitle

\AtBeginSubsection[]{
  \frame<beamer>{ 
    \frametitle{Outline}   
    \tableofcontents[currentsection,currentsubsection] 
  }
}
\lyxframeend{}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{Definitions from last time}
Convention: $p$ is always a prime number

\begin{itemize}

\item The prime counting function, $\pi(x) = \displaystyle\sum_{p \leq x} 1$
\item $f \sim g$ if and only if $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1$
\item \textbf{The Prime Number Theorem:} $\pi(x) \sim \frac{x}{\log(x)}$

\end{itemize}

\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{Definitions from last time}
The Laplace transform of a piecewise continuous function $f$ is the function $\mathcal{L}f \colon \{s \in \mathbb{C} \colon \Re(s) > 0\} \rightarrow \mathbb{R}$ defined by

$$(\mathcal{L}f)(s) = \displaystyle\int_0^{\infty} f(t)e^{-st}.$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
The Riemann zeta function is the function defined by the formula

$$\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}.$$

Clearly it is convergent for $\Re(s) > 1$, but we need a bigger domain. We need to extend it to $\Re(s) > 0$.
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
To that end, we need a lemma.

\textbf{Euler Product Lemma:} For $\Re(s) > 1$, $\zeta(s) = \displaystyle\prod_p \frac{1}{1-p^{-s}}$.

\textbf{Proof:} Using absolute convergence to rearrange terms and the fundamental theorem of arithmetic,

$$\begin{array}{ll}
\displaystyle\sum_{n \geq 1} \frac{1}{n^s} &= \displaystyle\lim_{k \rightarrow \infty} \displaystyle\sum_{e_1 \geq 0} \displaystyle\sum_{e_2 \geq 0} \ldots \displaystyle\sum_{e_k \geq 0} \frac{1}{(p_1^{e_1}p_2^{e_2} \ldots p_k^{e_k})^s} \\
&= \displaystyle\lim_{k \rightarrow \infty} \left( \displaystyle\sum_{e_1 \geq 0} p_1^{-se_1} \right) \left( \displaystyle\sum_{e_2 \geq 0} p_2^{-se_2} \right) \ldots \left( \displaystyle\sum_{e_k \geq 0} p_k^{-se_k} \right) \\
&= \displaystyle\lim_{k \rightarrow \infty} \left( \frac{1}{1-p_1^{-s}} \right) \left( \frac{1}{1-p_2^{-s}} \right) \ldots \left( \frac{1}{1-p_3^{-s}} \right) \\
&= \displaystyle\prod_p \frac{1}{1-p^{-s}}.
\end{array}$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
\textbf{Corollary:} $\zeta(s) \neq 0$ for $\Re(s) > 1$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
\textit{\textbf{Lemma:} $\zeta(s) - \frac{1}{s-1}$ (defined for $\Re(s) > 1$) extends to an analytic function defined on $\Re(s) > 0$.}

\textbf{Claim 1:} $\sum_{n=1}^{\infty} \int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx = \zeta(s) - \frac{1}{s-1}$

\textbf{Proof:} $$\begin{array}{ll}
 \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx &= \displaystyle\sum_{n=1}^{\infty} \left[ \frac{1}{n^s} x - \frac{x^{-s+1}}{-s+1} \right|_n^{n+1} \\
&= \displaystyle\sum_{n=1}^{\infty} \left[ \left( \frac{n+1}{n^s} - \frac{(n+1)^{-s+1}}{-s+1} \right) - \left( \frac{n}{n^s} - \frac{n^{-s+1}}{-s+1} \right) \right] \\
&= \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s} + \frac{(n+1)^{-s+1} - n^{-s+1}}{s-1} \\
&= \zeta(s) + \frac{1}{s-1} \displaystyle\sum_{n=1}^{\infty} (n+1)^{-s+1} - n^{-s+1} \\
&= \zeta(s) - \frac{1}{s-1}
\end{array}$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
\textit{\textbf{Lemma:} $\zeta(s) - \frac{1}{s-1}$ (defined for $\Re(s) > 1$) extends to an analytic function defined on $\Re(s) > 0$.}

Direct computation establishes $\int_n^x \frac{s}{u^{s+1}} du = \frac{1}{n^s} - \frac{1}{x^s}$.

\textbf{Claim 2:} $\displaystyle\sum_{n=1}^{\infty} \displaystyle\int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx$ converges absolutely for $\Re(s) > 0$.

\textbf{Proof:} $$\begin{array}{ll}
\left| \displaystyle\int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx \right| &= \left| \displaystyle\int_{n}^{n+1} \left( \displaystyle\int_{n}^x \frac{s}{u^{s+1}} du \right) dx \right| \\
&\leq \left( \displaystyle\max_{n \leq x \leq n+1} \left| \displaystyle\int_n^x \frac{s}{u^{s+1}} du \right| \right)(1) \\
&= \displaystyle\max_{n \leq x \leq n+1} \left| \displaystyle\int_n^x \frac{s}{u^{s+1}} du \right| \\
&\leq  \displaystyle\max_{n \leq x \leq n+1} \displaystyle\max_{n \leq u \leq x} \left| \frac{s}{u^{s+1}} \right| (x-n) \\
&= \displaystyle\max_{n \leq u \leq x} \left| \frac{s}{u^{s+1}} \right| = \left| \frac{s}{n^{s+1}} \right| = \frac{|s|}{n^{\Re(s) + 1}}
\end{array}$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$}
\textit{\textbf{Corollary:} $\zeta(s)$ extends to a meromorphic function on $\Re(s) > 0$ with simple pole of residue $1$ at $s=1$ and no other poles.}

\textbf{Proof:} Last lemma established $\zeta(s) = \sum_{n=1}^{\infty} \int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx + \frac{1}{s-1}$. This shows that 

$$\begin{array}{ll}
\frac{d}{ds} [ \zeta(s) ] &= \frac{d}{ds} \left[ \displaystyle\sum_{n=1}^{\infty} \int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx + \frac{1}{s-1} \right] \\
&= \displaystyle\sum_{n=1}^{\infty} \int_n^{n+1} \frac{d}{ds} \left[ \frac{1}{n^s} - \frac{1}{x^s} \right] dx + \frac{d}{ds} \left[ \frac{1}{s-1} \right] \\
&= \displaystyle\sum_{n=1}^{\infty} \int_n^{n+1} \left( \frac{\log(n)}{n^s} - \frac{\log(x)}{x^s} \right) dx - \frac{1}{(s-1)^{2}} \\
&= \displaystyle\sum_{n=1}^{\infty} \left[ \frac{\log(n)}{n^s}x - \frac{x^{1-s}((s-1)\log(x)+1}{(s-1)^2} \right|_n^{n+1} dx - \frac{1}{(s-1)^{2}} 
\end{array}$$

\lyxframeend{$\zeta$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\textit{\textbf{Corollary:} $\zeta(s)$ extends to a meromorphic function on $\Re(s) > 0$ with simple pole of residue $1$ at $s=1$ and no other poles.}
$$\begin{array}{ll}
&= \displaystyle\sum_{n=1}^{\infty} \frac{\log(n)}{n^s} + \displaystyle\sum_{n=1}^{\infty} \left[ \frac{(s-1)\log(n)+1}{(s-1)^s}n^{1-s} \right. \\
&\left. - \frac{(s-1)\log(n+1)+1}{(s-1)^2}(n+1)^{1-s} \right] - \frac{1}{(s-1)^{2}}  \\
&= \zeta'(s) + \frac{1}{(s-1)^2} - \frac{1}{(s-1)^2} 
\end{array}$$
\hspace*{-5pt}
This calculation showed that our formula for extending $\zeta$ is differentiable for $\Re(s) > 0$ and equals what we expect it to. Thus the extension is meromorphic since it is holomorphic on $\mathbb{C} \setminus \{0\}$. The residue of $s=1$ is $1$ because $1$ is the numerator of the term $\frac{1}{s-1}$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
Define $\Phi(s) = \displaystyle\sum_p \frac{\log(p)}{p^s}$ (defined for $\Re(s) > 1$).

\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is analytic for for $\Re(s) \geq 1$ and meromorphic for $\Re(s) > \frac{1}{2}$.}

By taking logarithm and differentiating $\zeta(s) = \displaystyle\prod_p \frac{1}{1-p^{-s}}$, 

$$\begin{array}{ll}
\log(\zeta(s)) &= -\displaystyle\sum_p \log(1-p^{-s}) \\
-\frac{\zeta'(s)}{\zeta(s)} &= \displaystyle\sum_p \frac{1}{1-p^{-s}} (p^{-s}\log(p)) \\
&= \displaystyle\sum_p \frac{\log(p)}{p^{-s}(1-p^{-s})} \\
&= \displaystyle\sum_p \frac{\log(p)}{p^s - 1}
\end{array}$$

\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is analytic for for $\Re(s) \geq 1$ and meromorphic for $\Re(s) > \frac{1}{2}$.}

But since $\frac{1}{p^s} + \frac{1}{p^s(p^s-1)} = \frac{1}{p^s-1}$, we see

$$-\frac{\zeta'(s)}{\zeta(s)} = \Phi(s) + \displaystyle\sum_p \frac{\log(p)}{p^s(p^s-1)}.$$

This sum converges for $\Re(s) > \frac{1}{2}$, so $\Phi(s)$ extends meromorphically ($\frac{\zeta'}{\zeta}$ and the sum are differentiable almost everywhere!) to $\Re(s) > \frac{1}{2}$ with poles only at $s=1$ and the zeros of $\zeta$. 
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is analytic for for $\Re(s) \geq 1$ and meromorphic for $\Re(s) > \frac{1}{2}$.}

Let $\alpha \in \mathbb{R}$. Assume $s_0 = 1 + i\alpha$ is a zero of $\zeta$ of order $\mu \geq 0$. Then by definition $\zeta(s) = (s-s_0)^{\mu} g(s)$ for some function $g$ that is analytic and nonzero at $s_0$. Then $\zeta'(s) = \mu(s-s_0)^{\mu-1}g(s) + (s-s_0)^{\mu}g'(s)$, so

$$\begin{array}{ll}
-\frac{\zeta'(s)}{\zeta(s)} &= -\frac{\mu(s-s_0)^{\mu-1}g(s) + (s-s_0)^{\mu}g'(s)}{g(s)(s-s_0)^{\mu}} \\
&= -\frac{\mu}{s-s_0} - \frac{g'(s)}{g(s)}
\end{array}$$

\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is analytic for for $\Re(s) \geq 1$ and meromorphic for $\Re(s) > \frac{1}{2}$.}

The previous slide yields the formula $\Phi(s) = - \left[ \frac{g'(s)}{g(s)} + \displaystyle\sum_p \frac{\log(p)}{p^s(p^s-1)} + \frac{\mu}{s-s_0} \right]$. 

Define $F(s) = -\frac{g'(s)}{g(s)} - \displaystyle\sum_p \frac{\log(p)}{p^s(p^s-1)}$.

Now consider

$$\Phi(1+ \epsilon + ia) = F(1+ \epsilon +ia) - \frac{\mu}{(1+ \epsilon +ia) - (1+ia)} = F(1 + \epsilon + ia) - \frac{\mu}{\epsilon}.$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is analytic for for $\Re(s) \geq 1$ and meromorphic for $\Re(s) > \frac{1}{2}$.}

$$\Phi(1+ \epsilon + ia) = F(1+ \epsilon +ia) - \frac{\mu}{(1+ \epsilon +ia) - (1+ia)} = F(1 + \epsilon + ia) - \frac{\mu}{\epsilon}.$$

We now see that $\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon + ia) = \displaystyle\lim_{\epsilon \searrow 0} \epsilon F(1+\epsilon+ia) - \mu = - \mu.$

\underline{Note:} Since the reflection principle implies $\overline{\zeta(s)} = \zeta(\bar{s})$, we have $\zeta(\bar{s_0})=0$, so we can derive 

$$\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon - ia) = \displaystyle\lim_{\epsilon \searrow 0} \epsilon F(1+\epsilon-ia) - \mu = - \mu.$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
We need to evaluate another limit.

Since $-\frac{\zeta'}{\zeta}$ has a simple pole of residue $1$ at $s=1$, we can write 

$$-\frac{\zeta'}{\zeta} = G(s) + \frac{1}{s-1},$$

where $G(s)$ is the the power series defining $-\frac{\zeta'}{\zeta}$ minus $\frac{1}{s-1}$. Since $\frac{\zeta'}{\zeta}$ has a pole at $s=1$, so does $\Phi$ (since the sum converges at $s=1$ by limit comparison test with $\displaystyle\sum_{n=1}^{\infty} \frac{\log n}{n^2}$), so we get

$$\Phi(s) = \frac{1}{s-1} + G(s) + \displaystyle\sum_p \frac{\log p}{p^s(p^s - 1)}.$$
\lyxframeend{}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
Therefore,

$$\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon) = \epsilon \frac{1}{\epsilon} + \epsilon G(1+\epsilon) + \epsilon \displaystyle\sum_p \frac{\log p}{p^{1+\epsilon}(p^{1+\epsilon}-1)} = 1.$$
\lyxframeend{}

\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is holomorphic for $\Re(s) \geq 1$.}

All terms of $\Phi(1+\epsilon) = \displaystyle\sum_p \frac{\log(p)}{p^{1+\epsilon}}$ are greater than $0$ and since for any prime $p$,

$$(p^{\frac{ia}{2}} + p^{-\frac{ia}{2}})^2 = (p^{ia} + 2 + p^{-ia}) = 2 + 2 \cos(a \log(p)) \geq 0,$$

we see that

$$\displaystyle\sum_p \frac{\log(p)}{p^{1+\epsilon}} (p^{\frac{ia}{2}} + p^{\frac{-ia}{2}})^2 \geq 0.$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
\textit{\textbf{Lemma:} $\zeta(s) \neq 0$ and $\Phi(s) - \frac{1}{s-1}$ is holomorphic for $\Re(s) \geq 1$.}

By algebra, from $\displaystyle\sum_p \frac{\log(p)}{p^{1+\epsilon}} (p^{\frac{ia}{2}} + p^{\frac{-ia}{2}})^2 \geq 0$, we get

$$\Phi(1+\epsilon+ia) + \Phi(1 + \epsilon - ia) + 2 \Phi(1+\epsilon) \geq 0.$$

By the limits we just computed,

$$\displaystyle\lim_{\epsilon \searrow 0} \epsilon[ \Phi(1+\epsilon+ia) + \Phi(1 + \epsilon - ia) + 2 \Phi(1+\epsilon) ] = -2\mu + 2 \geq 0$$

so we conclude $\mu \leq 1$. But we need $\mu=0$ if we are to show there are no zeros for $\Re(s)=1$!

\lyxframeend{}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\zeta$ and $\Phi$}
To that end, let $1 + 2ia$ as a zero of $\zeta$ of order $\nu$, now consider

$$\displaystyle\sum_p \frac{\log(p)}{p^{1+\epsilon}} (p^{\frac{ia}{2}} + p^{\frac{-ia}{2}})^2 \geq 0.$$

The same exact calculation we performed above with this new zero yields

$$6 - 8\mu - 2\nu \geq 0.$$

But $\mu, \nu \geq 0$, so we have $6 - 8\mu \geq 6 - 8\mu - 2\nu \geq 0$ implying

$$\mu \leq \frac{3}{4}.$$

This implies that $\mu=0$.
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
Now we define the function $\vartheta(x) = \displaystyle\sum_{p \leq x} \log p$.

This is a nice function to study because

\textit{\textbf{Lemma:} The prime number theorem holds if and only if $\vartheta(x) \sim x$.}

\textbf{Proof:} Since

$$\begin{array}{ll}
\vartheta(x) &= \displaystyle\sum_{p \leq x} \log p \\
&\leq \log(x) \displaystyle\sum_{p \leq x} 1 \\
&= \log(x) \pi(x).
\end{array}$$

Division by $x$ yields $0 \leq \frac{\vartheta(x)}{x} \leq \pi(x) \frac{\log x}{x}$

\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} The prime number theorem holds if and only if $\vartheta(x) \sim x$.}

\begin{itemize}
\item Clearly, for $1 > \epsilon > 0$, $\vartheta(x) \geq \displaystyle\sum_{x^{1-\epsilon} < p \leq x} \log(p)$.

\item Now since $\pi(x) - \pi(x^{1 - \epsilon}) = \displaystyle\sum_{x^{1-\epsilon} < p \leq x} 1$ we see
\end{itemize}
$$\displaystyle\sum_{x^{1-\epsilon} < p \leq x} \log(p) \geq \log(x^{1-\epsilon}) \displaystyle\sum_{x^{1-\epsilon} < p \leq x} 1 = (1-\epsilon) \log(x) (\pi(x) - \pi(x^{1-\epsilon})).$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} The prime number theorem holds if and only if $\vartheta(x) \sim x$.}

Clearly, $\pi(x^{1-\epsilon}) \leq x^{1-\epsilon}$, so using the two inequalities on the previous slide,

$$\vartheta(x) \geq (1-\epsilon) \log(x) (\pi(x) - \pi(x^{1-\epsilon}) \geq (1-\epsilon) \log(x) (\pi(x) - x^{1-\epsilon}).$$

Algebraic rearrangement yields

$$\pi(x) \leq \frac{1}{1-\epsilon} \frac{\vartheta(x)}{x} + \frac{\log(x)}{x^{\epsilon}}.$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} The prime number theorem holds if and only if $\vartheta(x) \sim x$.}

We combine these things we have discovered:

\begin{itemize}
\item $0 \leq \frac{\vartheta(x)}{x} \leq \pi(x) \frac{\log x}{x}$

\item $\pi(x) \leq \frac{1}{1-\epsilon} \frac{\vartheta(x)}{x} + \frac{\log(x)}{x^{\epsilon}}$
\end{itemize}

to get

$$\begin{array}{ll}
\frac{\vartheta(x)}{x} &\leq \pi(x) \frac{\log(x)}{x} \\
&\leq \left( \frac{1}{1-\epsilon} \frac{\vartheta(x)}{\log(x)} + x^{1 - \epsilon} \right) \frac{\log(x)}{x} \\
&= \frac{1}{1-\epsilon} \frac{\vartheta(x)}{x} + \frac{\log(x)}{x^{\epsilon}}.
\end{array}$$
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} The prime number theorem holds if and only if $\vartheta(x) \sim x$.}

We know $\frac{\vartheta(x)}{x} \leq \pi(x) \frac{\log(x)}{x} \leq \frac{1}{1-\epsilon} \frac{\vartheta(x)}{x} + \frac{\log(x)}{x^{\epsilon}}$, so first take the limit as $x \rightarrow \infty$ and then the limit as $\epsilon \rightarrow 0$ to get

$$\lim_{x \rightarrow \infty} \frac{\vartheta(x)}{x} \leq \lim_{x \rightarrow \infty} \pi(x) \frac{\log(x)}{x} \leq \lim_{x \rightarrow \infty} \frac{\vartheta(x)}{x}.$$

This shows that $\frac{\vartheta(x)}{x} \rightarrow 1$ if and only if $\pi(x) \frac{\log x}{x} \rightarrow 1$.
\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textbf{New notation:} If $f$ and $g$ are functions, we say $f = O(g)$ if and only if $|f(x)| \leq c |g(x)|$ for all sufficiently large $x$.

\textit{\textbf{Lemma:} $\vartheta(x) = O(x)$}

\textbf{Proof:} First, $2^{2n} = (1+1)^{2n} = \displaystyle\sum_{k=1}^{2n} {2n \choose k} \geq {2n \choose n} = \frac{(2n)!}{(n!)^2}$. 

Now notice that every prime $p$ in the set $\{n+1, n+2, \ldots, 2n\}$ must divide $(2n)!$ but cannot divide $n!$. So all such $p$ divide $\frac{(2n)!}{(n!)^2}$. Since all the $p$'s are relatively prime, $\frac{(2n)!}{(n!)^2}$ is also divisible by $\displaystyle\prod_{n < p \leq 2n} p$. So we get the inequality

$$2^{2n} \geq \frac{(2n)!}{(n!)^2} \geq \displaystyle\prod_{n < p \leq 2n} p.$$

\lyxframeend{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} $\vartheta(x) = O(x)$}

\textbf{Claim:} $e^{\vartheta(2n) - \vartheta(n)} = \displaystyle\prod_{n < p \leq 2n} p$

\textbf{Proof of claim:} 

$$\begin{array}{ll}
e^{\vartheta(2n) - \vartheta(n)} &= \exp \left( {{\displaystyle\sum_{p \leq 2n} \log(p) - \displaystyle\sum_{p \leq n} \log(p)}} \right) \\
&= \exp \left( {{\displaystyle\sum_{n < p \leq 2n} \log(p)}} \right) \\
&= \displaystyle\prod_{n < p \leq 2n} e^{\log(p)} \\
&= \displaystyle\prod_{n < p \leq 2n} p
\end{array}$$

\lyxframeend{}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} $\vartheta(x) = O(x)$}

\textbf{Claim:} $\vartheta(2^m) < 2^{m+1} \log 2$

\textbf{Proof of claim:} First notice that by telescoping and the fact that $\vartheta(1) = 0$ we see 

$$\displaystyle\sum_{n=1}^m \vartheta(2^n) - \vartheta(2^{n-1}) = \vartheta(2^m) - \vartheta(2^0) = \vartheta(2^m).$$

Taking a logarithm in the relationship $\displaystyle\prod_{n < p \leq 2n} p = e^{\vartheta(2n) - \vartheta(n)} \leq 2^{2n}$ yields

$$\displaystyle\sum_{n < p \leq 2n} \log(p) = \vartheta(2n) - \vartheta(n) \leq 2n \log(2).$$

\lyxframeend{}
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\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} $\vartheta(x) = O(x)$}

\textit{\textbf{Claim:} $\vartheta(2^m) < 2^{m+1} \log 2$}

So by the above inequality and the formula for a partial geometric series,

$$\begin{array}{ll}
\vartheta(2^m) &= \displaystyle\sum_{n=1}^m \vartheta(2^n) - \vartheta(2^{n-1}) \\
&\leq \displaystyle\sum_{n=1}^m 2^n \log(2) \\
&= (2^{m+1} - 2) \log(2) \\
&< 2^{m+1} \log(2)
\end{array}$$
\lyxframeend{}
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\lyxframe{$\vartheta$}
\textit{\textbf{Lemma:} $\vartheta(x) = O(x)$}

So we know $\vartheta(2^m) < 2^{m+1} \log 2$, use that fact in the following to finish the proof: 

\textbf{Claim:} $\vartheta(x) \leq (4 \log 2) x$

\textbf{Proof:} For any $x$ choose $m$ so that $2^{m-1} \leq x < 2^m$, then

$$\begin{array}{ll}
\vartheta(x) &\leq \vartheta(2^m) \\
&\leq 2^{m+1} \log(2) \\
&=(4 \log(2)) 2^{m-1} \\
&\leq (4 \log(2))x.
\end{array}$$	
\lyxframeend{}
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\lyxframe{Bring it all together}
Let $\psi(x) \colon \mathbb{R} \rightarrow \mathbb{R}$ be a function.

\textbf{Lemma:} $\psi(x) \sim x$ if and only if for every $\epsilon > 0$ we have $(1 - \epsilon)x < \psi(x) < (1+\epsilon)x$ for sufficiently large $x$.

\textbf{Proof:} $(\rightarrow)$ Assume that $\displaystyle\lim_{x \rightarrow \infty} \frac{\psi(x)}{x} = 1$. Then, formally, we say that for every $\epsilon > 0$ there exists $M>0$ such that if $x > M$, then $|\frac{\psi(x)}{x} - 1| \leq \epsilon$. The result follows by definition of absolute value.

\hspace{20pt}

$(\leftarrow)$ Assume for all $\epsilon > 0$, that $(1 - \epsilon)x < \psi(x) < (1+\epsilon)x$. Then, $1 - \epsilon < \frac{\psi(x)}{x} < 1 + \epsilon$, but epsilon is arbitrary, so the result immediately follows.
\lyxframeend{}
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\lyxframe{Bring it all together}
\textbf{Lemma:} If $\psi(x) \colon [1,\infty) \rightarrow \mathbb{R}$ is any non-decreasing function of $x$ such that $\displaystyle\int_1^{\infty} \frac{\psi(u) - u}{u^2} du$ converges, then $\psi(x) \sim x$.

We do this by contradiction.

Assume there exists $\epsilon > 0$ such that $\psi(x) \geq (1+\epsilon)x$ for arbitrarily large $x$. Since $\psi$ non-decreasing, we have for such $x$

$$\begin{array}{ll}
\displaystyle\int_x^{(1+\epsilon)x} \frac{\psi(u) - u}{u^2} du &= \displaystyle\int_x^{(1+\epsilon)x} \left( \frac{\psi(u)}{u} - 1 \right) \frac{du}{u} \\
&\geq \displaystyle\int_x^{(1+\epsilon)x} \left( \frac{(1+\epsilon)x}{u} - 1 \right) \frac{du}{u} \\
\end{array}$$
\lyxframeend{}
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\lyxframe{Bring it all together}
\textbf{Lemma:} If $\psi(x) \colon [1,\infty) \rightarrow \mathbb{R}$ is any non-decreasing function of $x$ such that $\displaystyle\int_1^{\infty} \frac{\psi(u) - u}{u^2} du$ converges, then $\psi(x) \sim x$.

We now make a change of variables with $w = \frac{u}{x}$. Then $dw = \frac{1}{x} du$. Dividing both sides by $u$ yields $\frac{1}{x} \frac{du}{u} = \frac{dw}{u}$ which implies $\frac{du}{u} = x \frac{dw}{u} = x \frac{dw}{wx} = \frac{dw}{w}$ and we have

$$\begin{array}{ll}
\displaystyle\int_x^{(1+\epsilon)x} \left( \frac{(1+\epsilon)x}{u} - 1 \right) \frac{du}{u} &= \displaystyle\int_1^{1+\epsilon} \left( \frac{1+\epsilon}{w} - 1 \right) \frac{dw}{w} \\
&= \displaystyle\int_1^{1+\epsilon} \left( \frac{1+\epsilon}{w} - 1) \frac{dw}{w} \right) \\
&= ( -1 - \log(1+\epsilon)) - (-(1+\epsilon) - 0)  \\
&= \epsilon - \log(1+\epsilon),
\end{array}$$

which does not depend on $x$. 
\lyxframeend{}
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\lyxframe{Bring it all together}
\textbf{Lemma:} If $\psi(x) \colon [1,\infty) \rightarrow \mathbb{R}$ is any non-decreasing function of $x$ such that $\displaystyle\int_1^{\infty} \frac{\psi(u) - u}{u^2} du$ converges, then $\psi(x) \sim x$.

Since

$$\displaystyle\int_x^{(1+\epsilon)x} \frac{\psi(u) - u}{u^2} du \geq \epsilon - (1+\epsilon)\log(1+\epsilon)$$

for all sufficiently large $x$, we see that $\displaystyle\int_1^{\infty} \frac{\psi(u) - u}{u^2} du$ does not converge, a contradiction.

Same argument if there is some $\epsilon > 0$ with $\psi(x) \leq (1-\epsilon)x$ for arbitrarily large $x$. We must conclude now that $\psi(x) \sim x$.
\lyxframeend{}
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\lyxframe{Bring it all together}
\textbf{Corollary:} If $\displaystyle\int_0^{\infty} (\vartheta(e^t)e^{-t} - 1) dt$ converges, then the prime number theorem is true.

\textbf{Proof:} If $u=e^t$ then,

$$\begin{array}{ll}
\displaystyle\int_0^{\infty} \frac{\vartheta(u) - u}{u^2} du &= \displaystyle\int_0^{\infty} \frac{\vartheta(e^t) - e^t}{e^t} dt \\
&= \displaystyle\int_0^{\infty} (\vartheta(e^t)e^{-t} - 1) dt.
\end{array}$$

Then since we are assuming that $\displaystyle\int_0^{\infty} (\vartheta(e^t)e^{-t} - 1) dt$ converges and we know that $\vartheta \colon [1, \infty) \rightarrow \mathbb{R}$ is a non-decreasing function of $x$, by previous lemma, we can say that $\vartheta(x) \sim x$. We had a result earlier that showed $\vartheta(x) \sim x$ if and only if the prime number theorem, so applying that yields this corollary.
\lyxframeend{}
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\lyxframe{Bring it all together}
Define $H \colon [0, \infty) \rightarrow \mathbb{R}$ by $H(x) = \vartheta(e^x)e^{-x} - 1$.

\textbf{Lemma:} The function $H$ is bounded and piecewise continuous.

\textbf{Proof:} We know $\vartheta$ is piecewise continuous (in fact, piecewise constant), so $H$ must be. Recall that $\vartheta(x) = \mathcal{O}(x)$ so $\vartheta(x) \leq c|x|$ for some $c$ and all sufficiently large $x$. So, 

$$\begin{array}{ll}
H(x) &= \vartheta(e^x)e^{-x} - 1 \\
&\leq c|x|e^{-x} - 1 \\
&= cxe^{-x} - 1 \\
\end{array}$$

which approaches $-1$ as $x \rightarrow \infty$. Since the limit exists, $H$ is bounded.
\lyxframeend{}
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\lyxframe{Bring it all together}
\textit{\textbf{Lemma:} $(\mathcal{L}\vartheta(e^t))(s) = \frac{\Phi(s)}{s}$ for $\Re(s) > 1$.}

\textbf{Proof:} We know:

\begin{itemize} 
\item $\vartheta(e^t) \leq Ce^t$ for all sufficiently large $t \geq 0$ 

\item $(\mathcal{L}\vartheta(e^t))$ converges absolutely for $\Re(s) > 1$ (by first Laplace transform proof from last week)
\end{itemize}

Let $p_n$ denote the $n$th prime number and let $p_0=0$. Then $\vartheta(e^t)$ is constant for $\log(p_n) < t < \log(p_{n+1})$, so

$$\displaystyle\int_{\log(p_n)}^{\log(p_{n+1})} e^{-st}\vartheta(e^t)dt = \vartheta(p_n) \displaystyle\int_{\log(p_n)}^{\log(p_{n+1})} e^{-st} dt = \frac{1}{s} \vartheta(p_n)(p_n^{-s} - p_{n+1}^{-s}).$$
\lyxframeend{}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\lyxframe{Bring it all together}
\textit{\textbf{Lemma:} $(\mathcal{L}\vartheta(e^t))(s) = \frac{\Phi(s)}{s}$ for $\Re(s) > 1$.}

Sum this for all $n \in \mathbb{Z}^+$ and for $\Re(s) > 1$ you get,

$$\begin{array}{ll}
\displaystyle\int_0^{\infty} e^{-st}\vartheta(e^t) dt &= \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{\log(p_n)}^{\log(p_{n+1})} e^{-st}\vartheta(e^t)dt \\
&= \displaystyle\sum_{n=1}^{\infty} \frac{1}{s} \vartheta(p_n)(p_n^{-s} - p_{n+1}^{-s}) \\
&= \frac{1}{s} \left[ \vartheta(p_1)(p_1^{-s} - p_2^{-s}) + \vartheta(p_2)(p_2^{-s}-p_3^{-s}) + \ldots \right] \\
&= \frac{1}{s} \left[ (\vartheta(p_1)-\vartheta(p_0))p_1^{-s} + (\vartheta(p_2)-\vartheta(p_1))p_2^{-s} + \ldots \right] \\
&= \frac{1}{s} \displaystyle\sum_{n=1}^{\infty} (\vartheta(p_n)-\vartheta(p_{n-1}))p_n^{-s} \\
&= \frac{1}{s} \displaystyle\sum_{n=1}^{\infty} \left( \displaystyle\sum_{p_{n-1} < p \leq p_n} \log(p) \right) p_n^{-s} \\
&= \frac{1}{s} \displaystyle\sum_{n=1}^{\infty} \frac{\log(p_n)}{p_n^{s}} = \frac{1}{s} \Phi(s).
\end{array}$$

\lyxframeend{}
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\lyxframe{Bring it all together}
Recall, in an earlier lemma lemma, we had derived the following: $\Phi(s) = \frac{1}{s-1} + G(s) + \displaystyle\sum_p \frac{\log p}{p^s(p^s - 1)}.$ We now prove a simple corollary to that lemma.

\textbf{Corollary:} $(\mathcal{L}H)(s) = \frac{\Phi(s+1)}{s+1} - \frac{1}{s}$ extends to a meromorphic function on the half-plane $\Re(s) > -\frac{1}{2}$, and is analytic for $\Re(s) \geq 0$.

\textbf{Proof:} Follows from the fact that the residue of $\frac{\Phi(s+1)}{s+1}$ at $s=0$ is $1$.

\lyxframeend{}
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\lyxframe{Bring it all together}
\textbf{Lemma:} The improper integral $\displaystyle\int_0^{\infty} H(t) dt$ converges.

\textbf{Proof:} We know $H(t)$ is bounded and piecewise continuous. Apply the analytic theorem to $H(t)$, whose Laplace transform $(\mathcal{L}H)(s) = \frac{\Phi(s+1)}{s+1} - \frac{1}{s}$ extends to an analytic function for $\Re(s) \geq 0$.

Then $\displaystyle\int_0^{\infty} H(t) dt$ converges.
\lyxframeend{}
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\lyxframe{Thank you!}
Thank you for attending! \\
\vspace{20pt}
\appendix


\lyxframeend{}\subsection*{References}


\lyxframeend{}\lyxframe{[allowframebreaks]For Further Reading}

\footnotesize
\beamertemplatebookbibitems
\begin{thebibliography}{2}
\bibitem{ZAG} Don Zagier. Newman's Short Proof of the Prime Number Theorem. The American Mathematical Monthly, 104(1997), 705-708.
\bibitem{CHURCH} James Ward Brown, Ruel V. Churchill. Complex Variables and Applications (7th Edition). McGraw-Hill, 2004.
\bibitem{BAKCLARK} Matt Baker, Dennis Clark. The Prime Number Theorem. Lecture notes. (http://people.math.gatech.edu/~mbaker/pdf/pnt2.pdf)
\bibitem{aaa} Sigurdur Helgason, Zuoqin Wang. Lecture 21 and 22: The Prime Number Theorem. MIT OpenCourseWare lecture notes. (	http://bit.ly/vF5EZP )



\beamertemplatearticlebibitems

\end{thebibliography}

\lyxframeend{}
\end{document}
