\input fremtex
\input mt
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\def\perim{\mathop{\text{perim}}}

\frfilename{n26221.tex}
\versiondate{14.4.26}

\Centerline{\bf The Prime Number Theorem}

\Centerline{\smc D.H.Fremlin}

\bigskip

\noindent The Prime Number Theorem, estimating the number of primes below a
large number $n$, was one of the glories of nineteenth-century mathematics.
Generations of young mathematicians have been taught that there is no easy
proof.   Slowly, however, ways have been found to proofs which are not exactly
easy, but are short enough to be accessible.   Here I will present a version of
the proof in {\smc Zagier 97}, itself derived from {\smc Newman 80}, expanded 
to include those fragments of the theory of contour integration which are 
required.

\bigskip

\noindent {\bf 1. Line integrals}  The first step is to get hold of some 
especially easy examples of integrals of complex functions along lines in the
complex plane.

\medskip

{\bf 1A Vertical and horizontal line segments (a)} Suppose that $f$ is a
bounded Borel measurable
complex-valued function defined (at least) on a horizontal line segment from
$z_0$ to $z_1$ where $z_0$, $z_1$ are complex numbers with
$\Imag(z_0)=\Imag(z_1)$  and $\Real(z_0)<\Real(z_1)$.   Then I define
$\int_{[z_0,z_1]}f(z)dz$ to be
$\int_{\Real(z_0)}^{\Real(z_1)}f(x+i\Imag(z_0))dx$, where this
integral is with respect to ordinary one-dimensional Lebesgue measure on 
$\Bbb R$.   

Similarly, for a vertical line segment from
$z_0$ to $z_1$ where $z_0$, $z_1$ are complex numbers with
$\Real(z_0)=\Real(z_1)$  and $\Imag(z_0)<\Imag(z_1)$, 
$\int_{[z_0,z_1]}f(z)dz$ will be 
$i\int_{\Imag(z_0)}^{\Imag(z_1)}f(\Real(z_0)+iy)dy$.

\medskip

{\bf (b)(i)}
Note that if $z_0$, $z_1$, $z_2$ are on the same vertical or horizontal line,
with $z_1$ strictly between the other two, then
$\int_{[z_0,z_1]}f(z)dz+\int_{[z_1,z_2]}f(z)dz=\int_{[z_0,z_2]}f(z)dz$ for any
function $f$ for which the third integral is defined.

\medskip

\quad{\bf (ii)} We have a simple form for an integral $\int_{[z_0,z_1]}z\,dz$, 
which in both 
the cases considered in (a) above is $\bover12(z_1^2-z_0^2)$, as the
formalism leads us to expect.   \Prf\ For a horizontal segment, we have

$$\eqalign{\int_{[z_0,z_1]}z\,dz&
=\int_{\Real(z_0)}^{\Real(z_1)}x+i\Imag(z_0)dx\cr&
=\Bover12((\Real(z_1)^2-(\Real(z_0))^2)+i(\Real(z_1)-\Real(z_0))\Imag(z_0)\cr&
=\Bover12((\Real(z_1)^2+2i\Real(z_1)\Imag(z_0)-\Imag(z_0)^2)\cr&
\mskip30mu
   -\Bover12(\Real(z_0)^2+2i\Real(z_0)\Imag(z_0)-\Imag(z_0)^2))\cr&
=\Bover12((\Real(z_1)+i\Imag(z_1))^2-(\Real(z_0)+i\Imag(z_0))^2)
=\Bover12(z_1^2-z_0^2),\cr}$$

\noindent and the calculation for vertical sections is similar.\ \Qed

\medskip

{\bf (c)} More generally, if $f$ is any bounded Borel measurable complex-valued
function and $[z_0,z_1]$ is either a vertical or horizontal line segment
included in $\dom f$, then
$|\int_{[z_0,z_1]}f(z)dz|\le\sup_{z\in[z_0,z_1]}|f(z)|\cdot|z_1-z_0|$.
\Prf\ Translating this into one-dimensional integrals as in (a)-(b), this is
the inequality 
$|\int_{x_0}^{x_1}f(x)dx|\le\int_{x_0}^{x_1}|f(x)|dx\le M(x_1-x_0)$ 
where $M=\sup_{x\in[x_0,x_1]}|f(x)|$.\ \Qed

\medskip

{\bf (d)(i)} The formula in (c) echoes a similar formula when we have a function
which is differentiable, as a complex function, along a line.   Suppose that
$z_0$ and $z_1$ are two points in $\Bbb C$, and write $[z_0,z_1]$ for the
straight line segment from $z_0$ to $z_1$, that is, the set
$\{\alpha z_0+(1-\alpha)z_1:\alpha\in[0,1]\}$.   Now suppose that $\phi$ is a
complex-valued function defined on and $[z_0,z_1]$ and differentiable along
$[z_0,z_1]$ at each point of $[z_0,z_1]$, that is, 
$\phi'(w)=\lim_{z\in[z_0,z_1],z\to w}\Bover{f(z)-f(w)}{z-w}$ in
$\Bbb C$ for every $w\in[z_0,z_1]$.   Then
$|\phi(z_1)-\phi(z_0|\le\sup_{w\in[z_0,z_1]}|\phi'(w)|\cdot|z_1-z_0|$.   
\Prf\ Apply the one-dimensional
real-variable version of the result to the function
$\alpha\mapsto\Real(v\phi(\alpha z_0+(1-\alpha)z_1)):[0,1]\to\Bbb R$ where
$v\in\Bbb C$ and $|v|=1$.\ \Qed

\medskip

\quad{\bf (ii)} Next, suppose that $z_0$, $z_1$ are such that 
$\Imag(z_0)=\Imag(z_1)$ and
$\Real(z_0)<\Real(z_1)$, and that $\phi$ is a bounded Borel measurable 
complex-valued function with domain including the line segment $[z_0,z_1]$.
For $w\in\Bbb C\setminus[z_0,z_1]$ set 
$f(w)=\int_{[z_0,z_1]}\Bover1{z-w}\phi(z)dz$.   Then $f$ is differentiable and
$f'(w)=-\int_{[z_0,z_1]}\Bover1{(z-w)^2}\phi(z)dz$ for every 
$w\in\Bbb C\setminus[z_0,z_1]$.   \Prf\ Note that as $w$ does not belong to the
compact set $[z_0,z_1]$, there is a $\delta>0$ such that $|z-w|\ge 2\delta$ for
every $z\in[z_0,z_1]$, so $z\mapsto\Bover1{z-w}\phi(z)$ and
$z\mapsto\Bover1{(z-w)^2}\phi(z)$ are both bounded on $[z_0,z_1]$;  as 
they are Borel measurable, the integrals are defined, for any particular $w$.
Moreover, taking $U$ to be the open disc $\{v:|v-w|<\delta\}$,
$\sup\{|\Bover1{(z-v)^2}\phi(z)|:v\in U$, $z\in[z_0,z_1]\}
\le\Bover1{\delta^2}\sup_{z\in[z_0,z_1]}|\phi(z)|$ is finite, so the argument
for Corollary 123D in {\smc Fremlin 11}, adapted to complex-valued integrands,
applies to the function
$(z,v)\mapsto\Bover1{z-v}\phi(z):[z_0,z_1]\times U\to\Bbb C$, and
$f'(w)=-\int_{[z_0,z_1]}\Bover1{(z-w)^2}\phi(z)dz$.\ \Qed

Similarly, this formula will be valid if $[z_0,z_1]$ is a vertical line
segment.   Furthermore, we can use the same argument to show that the second
derivative $f''(w)$ is defined and equal to 

\Centerline{$-\int_{[z_0,z_1]}
      \Bover{\partial}{\partial w}\Bover1{(z-w)^2}\phi(z)dz
   =2\int_{[z_0,z_1]}\Bover1{(z-w)^3}\phi(z)dz$}

\noindent for every $w\in\Bbb C\setminus[z_0,z_1]$.

\medskip

{\bf (e)} Suppose that $[z_0,z_1]$ is a horizontal or vertical
line segment as in (a)-(d), and
$\sequencen{\phi_n}$ a uniformly bounded sequence of Borel measurable
complex-valued functions, all with domain including $[z_0,z_1]$ and such that
$\phi(z)=\lim_{n\to\infty}\phi_n(z)$ is defined in $\Bbb C$ for each
$z\in[z_0,z_1]$.   Setting $f(w)=\int_{[z_0,z_1]}\Bover1{z-w}\phi(z)dz$ and
$f_n(w)=\int_{[z_0,z_1]}\Bover1{z-w}\phi_n(z)dz$ for $n\in\Bbb N$ and
$w\in\Bbb C\setminus[z_0,z_1]$, $f(w)=\lim_{n\to\infty}f_n(w)$ and
$f'(w)=\lim_{n\to\infty}f'_n(w)$ for every $w\in\Bbb C\setminus[z_0,z_1]$, by
Lebesgue's Dominated Convergence Theorem.

\bigskip

{\bf 1B Notation} The theory here will involve integrals around closed curves;
and the point of this note is that we can restrict attention to one very simple
type.   

\medskip

{\bf (a)} I will say that an {\bf upright rectangle} is a set of the form
$R=\{z:\alpha_0\le\Real(z)\le\alpha_1$, $\beta_0\le\Imag(z)\le\beta_1\}$ where
$\alpha_0$, $\alpha_1$, $\beta_0$, $\beta_1$ are real numbers,
$\alpha_0<\alpha_1$ and $\beta_0<\beta_1$.   Its boundary $\partial R$ is the
union of the line segments $[z_0,z_1]$, $[z_1,z_2]$, $[z_3,z_2]$ and
$[z_0,z_3]$ where $z_0=\alpha_0+\beta_0i$, $z_1=\alpha_1+\beta_0i$,
$z_2=\alpha_1+\beta_1i$ and $z_3=\alpha_0+\beta_1i$.   If $f$ is a
bounded Borel measurable complex-valued function and $\partial R\subseteq\dom f$
then we shall need to look at the oriented sum
$\int_{[z_0,z_1]}f(z)dz+\int_{[z_1,z_2]}f(z)dz
-\int_{[z_3,z_2]}f(z)dz-\int_{[z_0,z_3]}f(z)dz$;  
I will call it $\int_{\partial R}f(z)dz$.

We can think of this as the integral along the oriented path $z_0\to z_1\to
z_2\to z_3\to z_0$, where the leftwards and downwards segments are counted as
negatives of rightwards and upwards segments $[z_3,z_2]$ and $[z_0,z_3]$.
If you interpret the integral $\int_{[z_3,z_2]}f(z)dz$ as
$\int_{z_3}^{z_2}f(z)dz$ this makes sense.   But the whole point of the ideas to
follow is that $\int_{[z_0,z_1]}+\int_{[z_1,z_2]}$ may be different from
$\int_{[z_0,z_3]}+\int_{[z_3,z_2]}$.   (See 1Ce below.)   So we cannot safely
think of either as being $\int_{z_0}^{z_2}$.

\medskip

{\bf (b)} If $R$, $R'$ are upright rectangles which intersect in a shared edge,
so that their union $R\cup R'$ is again an upright rectangle, then
$\int_{\partial(R\cup R'}f(z)dz)
=\int_{\partial R}f(z)dz+\int_{\partial R'}f(z)dz$ 
for every bounded Borel measurable complex-valued $f$
defined on $\partial R\cup\partial R'$.   \Prf\ If $R$ and $R'$ share a
vertical edge, so that they can be expressed as 
$\{z:\alpha_0\le\Real(z)\le\alpha_1$, $\beta_0\le\Imag(z)\le\beta_1\}$ and
$\{z:\alpha_1\le\Real(z)\le\alpha_2$, $\beta_0\le\Imag(z)\le\beta_1\}$ where 
$\alpha_0<\alpha_1<\alpha_2$ and $\beta_0<\beta_1$, then

$$\eqalignno{&\int_{\partial R}f(z)dz+\int_{\partial R'}f(z)dz\cr&
\mskip 20mu =\int_{[z_0,z_1]}f(z)dz+\int_{[z_1,z_2]}f(z)dz
   -\int_{[z_3,z_2]}f(z)dz-\int_{[z_0,z_3]}f(z)dz\cr&
\mskip 50mu   +\int_{[z_1,z_4]}f(z)dz+\int_{[z_4,z_5]}f(z)dz
   -\int_{[z_2,z_5]}f(z)dz-\int_{[z_1,z_2]}f(z)dz
\Displaycause{where $z_0,\ldots,z_3$ are as in (a), 
$z_4=\alpha_2+i\beta_0$ and $z_5=\alpha_2+i\beta_1$}
\mskip 20mu =\int_{[z_0,z_4]}f(z)dz+\int_{[z_4,z_5]}f(z)dz
   -\int_{[z_3,z_5]}f(z)dz-\int_{[z_0,z_3]}f(z)dz\cr&
\mskip 20mu =\int_{\partial(R\cup R'}f(z)dz\cr}$$

\noindent since
$\int_{[z_0,z_1]}f(z)dz+\int_{[z_1,z_4]}f(z)dz=\int_{[z_0,z_4]}f(z)dz$ and
$\int_{[z_3,z_2]}f(z)dz+\int_{[z_2,z_5]}f(z)dz=\int_{[z_3,z_5]}f(z)dz$ 
by 1A(b-i) above.  \Qed

The same idea works if one of $R$, $R'$ is stacked on top of the
other.

\bigskip

{\bf 1C Some special cases} Take an upright rectangle $R$ with vertices
$z_0,\ldots,z_3$ starting at bottom left and going anticlockwise, as in 1Bb.

\medskip

{\bf (a)} Corresponding to the special cases treated in 1Ab we find that if
$f(z)=c_0+c_1z$ for $z\in\Bbb C$, where $c_0$, $c_1\in\Bbb C$, than
$\int_{\partial R}f(z)dz=0$.  \Prf\ 

$$\eqalignno{\int_{\partial R}f(z)dz&=\int_{[z_0,z_1]}f(z)dz+\int_{[z_1,z_2]}f(z)dz
        -\int_{[z_3,z_2]}f(z)dz-\int_{[z_0,z_3]}f(z)dz\cr&
=c_0\int_{[z_0,z_1]}1\,dz+c_1\int_{[z_0,z_1]}z\,dz
   +c_0\int_{[z_1,z_2]}1\,dz+c_1\int_{[z_1,z_2]}z\,dz\cr&
\mskip50mu   -c_0\int_{[z_3,z_2]}1\,dz-c_1\int_{[z_0,z_3]}z\,dz
   -c_0\int_{[z_0,z_3]}1\,dz-c_1\int_{[z_0,z_3]}z\,dz
\Displaycause{the formulae in 1Aa make it plain that integration is a linear
operator}
=c_0(z_1-z_0)+\Bover12c_1(z_1^2-z_0^2)+c_0(z_2-z_1)
     +\Bover12c_1(z_2^2-z_1^2)\cr&
\mskip50mu  -c_0(z_2-z_3)-\Bover12c_1(z_2^2-z_3^2)
     -c_0(z_3-z_0)-\Bover12c_1(z_3^2-z_0^2)
\Displaycause{1Ab}
=0.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} If $f$ is a bounded Borel measurable
function with domain including $\partial R$, then
$|\int_{\partial R}f(z)dz)|\le\perim R\cdot\sup_{z\in\partial R}|f(z)|$.
\Prf\ Apply 1Ac to each of the four sides of $R$.\ \Qed

\medskip

{\bf (c)} If $\phi$ is a bounded Borel
measurable complex-valued function with domain including $\partial R$,
then $f(w)=\int_{\partial R}\Bover1{z-w}\phi(z)dz$ is defined
for $w\in\Bbb C\setminus\partial R$, and
$f$ is differentiable, with 
$f'(w)=-\int_{\partial R}\Bover1{(z-w)^2}\phi(z)dz$ for
$w\in\Bbb C\setminus\partial R$.   \Prf\ Apply 1Ad to each side of $R$.\ \QeD\
In the same way, the second
derivative $f''(w)$ is defined and equal to 

\Centerline{$-\int_{\partial R}
      \Bover{\partial}{\partial w}\Bover1{(z-w)^2}\phi(z)dz
   =2\int_{\partial R}\Bover1{(z-w)^3}\phi(z)dz$}

\noindent for every $w\in\Bbb C\setminus\partial R$.

\medskip

{\bf (d)} Suppose that $R$ is an upright rectangle and
$\sequencen{\phi_n}$ a uniformly bounded sequence of Borel measurable
complex-valued functions with domain including $\partial R$, and such that
$\phi(z)=\lim_{n\to\infty}\phi_n(z)$ is defined in $\Bbb C$ for each
$z\in\partial R$.   Setting $f(w)=\int_{\partial R}\Bover1{z-w}\phi(z)dz$ and
$f_n(w)=\int_{\partial R}\Bover1{z-w}\phi_n(z)dz$ for $n\in\Bbb N$ and
$w\in\Bbb C\setminus\partial R$, $f(w)=\lim_{n\to\infty}f_n(w)$ and
$f'(w)=\lim_{n\to\infty}f'_n(w)$ for every $w\in\Bbb C\setminus\partial R$, as
in 1Ae.

\medskip

{\bf (e)} If $R\subseteq\Bbb C$ is a square with centre $w$, and
$f(z)=\bover1{z-w}$ for $z\in\Bbb C\setminus\{w\}$, then 
$\int_{\partial R}f(z)dz=2\pi i$.
\Prf\ If $w=\gamma_0+i\gamma_1$ and $R$ has side length $2\delta$, then the
integral $\int_{[z_0,z_1]}\bover1{z-w}dz$ becomes

\Centerline{$\int_{\gamma_0-\delta}^{\gamma_0+\delta}
  \Bover1{(x+i(\gamma_1-\delta))-(\gamma_0+i\gamma_1)}dx
=\int_{-\delta}^{\delta}\Bover1{x-i\delta}dx$,}

\noindent and similarly we have

\Centerline{$\int_{[z_1,z_2]}\Bover1{z-w}dz
=i\int_{-\delta}^{\delta}\Bover1{\delta+iy}dy$,\qquad
$\int_{[z_3,z_2]}\Bover1{z-w}dz
=\int_{-\delta}^{\delta}\Bover1{x+i\delta}dx$,}

\Centerline{$\int_{[z_0,z_3]}\Bover1{z-w}dz
=i\int_{-\delta}^{\delta}\Bover1{-\delta+iy}dy$},

\noindent so that

$$\eqalign{\int_{[z_0,z_1]}\Bover1{z-w}dz-\int_{[z_3,z_2]}\Bover1{z-w}dz&
=\int_{-\delta}^{\delta}\Bover1{x-i\delta}-\Bover1{x+i\delta}dx
=\int_{-\delta}^{\delta}\Bover{2i\delta}{x^2+\delta^2}dx\cr&
=2i[\arctan\Bover{x}{\delta}]_{-\delta}^{\delta}
=2i(\Bover{\pi}4-(-\Bover{\pi}4))
=\pi i,\cr}$$

$$\eqalign{\int_{[z_1,z_2]}\Bover1{z-w}dz-\int_{[z_0,z_3]}\Bover1{z-w}dz&
=i\int_{-\delta}^{\delta}\Bover1{\delta+iy}-\Bover1{-\delta+iy}dy\cr&
=i\int_{-\delta}^{\delta}\Bover{2\delta}{\delta^2+y^2}dy
=\pi i\cr}$$

\noindent and $\int_{\partial R}f(z)dz=2\pi i$.\ \Qed
\bigskip

{\bf 2. Holomorphic functions}

\medskip

{\bf 2A}  I will use the phrase 
{\bf holomorphic function} to
mean a complex-valued function with domain an open subset of $\Bbb C$ which is
differentiable, as a complex function, at every point of its domain.
     
If $f$ and $g$ are holomorphic functions then the composition $fg$ is
holomorphic, with domain $\{z:z\in\dom g$, $g(z)\in\dom f\}$;  
$f+g$ and $f\times g$ are
holomorphic, with domain $\dom f\cap\dom g$;  and $f/g$ is holomorphic, with
domain $\{z:z\in\dom f\cap\dom g$, $g(z)\ne 0\}$. 

\bigskip

{\bf 2B Cauchy's theorem, primitive form} Let $f$ be a holomorphic function and
$R\subseteq\dom f$ an upright rectangle.   Then $\int_{\partial R}f(z)dz=0$.

\medskip

\noindent{\bf proof} \Quer\ Suppose otherwise.   Let $z_0,\ldots,z_3$ be
the vertices of $R$, taken
anticlockwise from bottom left as in 1B and 1C, so that

$$\eqalign{\int_{\partial R}f(z)dz&
=\int_{[z_0,z_1]}f(z)\,dz+\int_{[z_1,z_2]}f(z)\,dz\cr&
\mskip 50mu
  -\int_{[z_0,z_3]}f(z)\,dz-\int_{[z_3,z_2]}f(z)\,dz.\cr}$$

\noindent Dissecting $R=R_0$ into four similar rectangles, with corners

$$\eqalign{(R_{00}):&\quad
z_0,\quad\Bover{z_0+z_1}2,\quad\Bover{z_0+z_1+z_2+z_3}4,\quad\Bover{z_0+z_3}2,\cr
(R_{01}):&\quad
\Bover{z_0+z_1}2,\quad z_1,\quad\Bover{z_1+z_2}2,\quad\Bover{z_0+z_1+z_2+z_3}4,\cr
(R_{02}):&\quad
\Bover{z_0+z_1+z_2+z_3}4,\quad\Bover{z_1+z_2}2,\quad
z_2,\quad\Bover{z_2+z_3}2,\cr
(R_{03}):&\quad
\Bover{z_0+z_3}2,\quad\Bover{z_0+z_1+z_2+z_3}4,\quad\Bover{z_2+z_3}2,\quad
z_3\cr}$$

\noindent we see that

$$\eqalign{\int_{\partial R_0}f(z)dz&
=\int_{\partial(R_{00}\cup R_{01})}f(z)dz
   +\int_{\partial(R_{02}\cup R_{03})}f(z)dz\cr&
=\int_{\partial(R_{00})}f(z)dz+\int_{\partial(R_{01})}f(z)dz
   +\int_{\partial(R_{02})}f(z)dz+\int_{\partial(R_{03})}f(z)dz.\cr}$$

\noindent So there must be some $k\le 3$ such that
$|\int_{\partial R_{0k}}f(z)dz|\ge\bover14|\int_{\partial R_0}f(z)dz|$;  setting
$R_1=R_{0k}$, $R_1$ is an upright rectangle, included in $R_0$, such that
$|\int_{\partial R_1}f(z)dz|\ge\bover14|\int_{\partial R_0}f(z)dz|$, 
while its perimeter $\perim R_1$ is just $\bover12\perim R_0$.

Repeating this argument with $R_1$, we can find an upright rectangle 
$R_2\subseteq R_1$ such that 
$|\int_{\partial R_1}f(z)dz|\ge\bover14|\int_{\partial R_1}f(z)dz|
\ge\bover1{16}|\int_{\partial R_0}f(z)dz|$ and $\perim R_2=\bover14\perim R_0$;  
continuing, we have a decreasing sequence 
$R_0\supseteq R_1\supseteq R_2\supseteq\ldots$ of rectangles with
$|\int_{\partial R_n}f(z)dz|\ge 4^{-n}|\int_{\partial R}f(z)dz|$ and 
$\perim R_n=2^{-n}\perim R$ for each $n$.
Because $R$ is compact, the intersection $\bigcap_{n\in\Bbb N}R_n$ will 
have a point $w\in R$, and $f$ will be differentiable at $w$.   Consider the
function $g$ defined by saying that $g(z)=f(z)-f(w)-(z-w)f'(w)$ for 
$z\in\dom g$.  Then $g$ is holomorphic and $g(w)=g'(w)=0$, while 
$\int_{\partial R_n}g(z)dz=\int_{\partial R_n}f(z)dz$ for every $n$, by 1Ca.

Take $\epsilon>0$.
Let $\delta>0$ be such that $w+z\in\dom g$ and
$|g(w+z)|\le\epsilon|z|$ whenever $|z|\le\delta$, and take
$n\ge 0$ such that $\perim R_n\le\delta$, so that 
$|g(z)|\le\epsilon|z-w|\le\epsilon\perim R_n$ 
for every $z\in R_n$.   
Now $|\int_{\partial R_n}g(z)dz|\le\epsilon(\perim R_n)^2$ by 1Cb, so

$$\eqalign{4^{-n}|\int_{\partial R}f(z)dz|&
\le|\int_{\partial R_n}f(z)dz|=|\int_{\partial R_n}g(z)dz|\cr&
\le\epsilon(\perim R_n)^2=4^{-n}\epsilon(\perim R)^2\cr}$$

\noindent and $|\int_{\partial R}f(z)dz|\le\epsilon(\perim R)^2$.   
As $\epsilon$ is
arbitrary, $\int_{\partial R}f(z)dz=0$, which is what we wanted to know.

\bigskip

{\bf 2C Cauchy's Integral Formula} Let $f$ be a holomorphic function and
$w$ a point in $\dom f$.  If
$R\subseteq\dom f$ is an upright rectangle and 
$w\in\interior R=R\setminus\partial R$,
then $f(w)=\Bover1{2\pi i}\int_{\partial R}\Bover1{z-w}f(z)dz$.

\medskip

\noindent{\bf proof} Set $g(z)=\Bover1{z-w}f(z)$ for $z\in\dom f\setminus\{w\}$.
Then $g$ is holomorphic.   Take $\epsilon>0$.   Let 
$\delta\in\ocint{0,\bover12\epsilon}$ be such that
the square $R'$ with side length $2\delta$ and centre $w$ 
is included in $\interior R$, 
and $|\Bover{f(z)-f(w)}{z-w}-f'(w)|\le 1$ for $z\in R'\setminus\{w\}$.
Projecting the left and right sides of $R'$ to reach $\partial R$,
$R$ is divided into non-overlapping upright
rectangles $R'$, $R_1$, $R_2$, $R_3$, $R_4$, and for $1\le k\le 4$ we have 
$R_k\subseteq\dom g$ so $\int_{\partial R_k}g(z)dz=0$, by 2B.   It follows that
$\int_{\partial R}g(z)dz=\int_{\partial R'}g(z)dz$.

Set $h(z)=g(z)-\Bover{f(w)}{z-w}-f'(w)$ for 
$z\in\dom f\setminus\{w\}$.   Then $h$ is holomorphic, and 
$\int_{\partial R'}h(z)dz=\int_{\partial R'}g(z)dz-2\pi if(w)$ by 1Ce and 1Ca.   
We have $|h(z)|\le 1$ for $z\in\partial R'$, so 
$|\int_{\partial R'}h(z)|\le\perim R'=8\delta$.   Accordingly

$$\eqalign{|f(w)-\Bover1{2\pi i}\int_{\partial R}\Bover1{z-w}f(z)dz|&
=\Bover1{2\pi}|2\pi if(w)-\int_{\partial R'}g(z)dz|\cr&
=\Bover1{2\pi}|\int_{\partial R'}h(z)dz|
\le\Bover{8\delta}{2\pi}\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$f(w)=\Bover1{2\pi i}\int_{\partial R}\Bover1{z-w}f(z)dz$.

\bigskip

{\bf 2D Corollary} If $f$ is a holomorphic function, its derivative is
holomorphic.

\medskip

\noindent{\bf proof} Take any $w\in\dom f$.   Then there is an upright
rectangle $R$ such that $w\in\interior R$ and $R\subseteq\dom f$.   Set
$g(v)=\Bover1{2\pi i}\int_{\partial R}\Bover1{z-v}f(z)$ for 
$v\in\Bbb C\setminus\partial R'$.
By 1Cc, $g$ is twice differentiable.   But 2C tells us that $f$ and $g$ agree on
the open set $\interior R$.   So $f''(w)=g''(w)$ is defined.   As $w$ is
arbitrary, $f'$ is holomorphic.

\bigskip

{\bf 2E Corollary} Let $\sequencen{f_n}$ be a sequence of holomorphic functions,
and $G\subseteq\bigcap_{n\in\Bbb N}\dom f_n$ an open set such that (i)
$\{f_n\restr G:n\in\Bbb N\}$ is locally uniformly bounded (ii)
$f(z)=\lim_{n\to\infty}f_n(z)$ is defined in $\Bbb C$ for every $z\in G$.   Then
$f$ is holomorphic and $f'(z)=\lim_{n\to\infty}f'_n(z)$ for every $z\in G$.

\medskip

\noindent{\bf proof (a)} Take any $w\in G$.   Then there is an upright rectangle
$R\subseteq G$ such that $w\in\interior R$.   Because $R$ is compact,
$\{f_n\restr R:n\in\Bbb N\}$ is uniformly bounded.   By 2C, 
$f_n(w)=\int_{\partial R}\Bover1{z-w}f_n(z)dz$ for $n\in\Bbb N$, so that

$$\eqalign{f(w)&
=\lim_{n\to\infty}f_n(w)
=\lim_{n\to\infty}\int_{\partial R}\Bover1{z-w}f_n(z)dz\cr&
=\int_{\partial R}\lim_{n\to\infty}\Bover1{z-w}f_n(z)dz
=\int_{\partial R}\Bover1{z-w}f(z)dz\cr}$$

\noindent as in 1Cd.   Moreover, 1Cd tells us that $f'(w)$ is defined and equal
to $\lim_{n\to\infty}f'_n(w)$.

\bigskip

{\bf 2F Taylor's theorem, primitive form} Let $f$ be a holomorphic function,
$R\subseteq\dom f$ an upright rectangle and $v$ a point in the interior of $R$.
Set $\rho=\inf_{z\in\partial R}|z-v|$ and for $k\in\Bbb N$ set
$a_k=\Bover1{2\pi i}\int_{\partial R}\Bover1{(z-v)^{k+1}}f(z)dz$.   Then
$f(w)=\sum_{k=0}^{\infty}a_k(w-v)^k$ whenever $|w-v|<\rho$.

\medskip

\noindent{\bf proof} For every $n\in\Bbb N$ and $z\in\partial R$, 

$$\sum_{k=0}^{n-1}\Bover{(w-v)^k}{(z-v)^k}
={{1-(\Bover{w-v}{z-v})^n}\over{1-(\Bover{w-v}{z-v})}}
=(1-(\Bover{w-v}{z-v})^n)\Bover{z-v}{z-w},$$

\noindent

\Centerline{$\sum_{k=0}^{n-1}\Bover{(w-v)^k}{(z-v)^{k+1}}
=(1-(\Bover{w-v}{z-v})^n)\Bover1{z-w}$,}

\Centerline{$|\sum_{k=0}^{n-1}\Bover{(w-v)^k}{(z-v)^{k+1}}-\Bover1{z-w}|
=(\Bover{|w-v|}{|z-v|})^n\Bover1{|z-w|}
\le(\Bover{|w-v|}{\rho})^n\Bover1{\rho-|w-v|}$.}

\noindent So 

\Centerline{$|\sum_{k=0}^{n-1}\Bover{(w-v)^k}{(z-v)^{k+1}}f(z)-\Bover1{z-w}f(z)|
\le M\Bover{|w-v|^n}{\rho^n}\Bover1{\rho-|w-v|}$}

\noindent where $M=\sup_{z\in\partial R}|f(z)|$, and
$\sum_{k=0}^{n-1}\Bover{(w-v)^k}{(z-v)^{k+1}}f(z)\to\Bover1{z-w}f(z)$, uniformly
for $z\in\partial R$, as $n\to\infty$.   Integrating along $\partial R$,

$$\eqalign{f(w)&=\Bover1{2\pi i}\int_{\partial R}\Bover1{z-w}f(z)dz
=\Bover1{2\pi i}\lim_{n\to\infty}\sum_{k=0}^{n-1}
   \int_{\partial R}\Bover{(w-v)^k}{(z-v)^{k+1}}f(z)dz
=\sum_{k=0}^{\infty}a_k(w-v)^k\cr}$$

\noindent as claimed.

\bigskip

{\bf 2G Corollary} Let $f$ be a holomorphic function, and $w\in\dom f$ a zero of
$f$.   Then either there is a neigbourhood of $w$ where $f$ is constant and
zero, or $\lim_{z\to w}\Bover{(z-w)f'(z)}{f(z)}$ is defined and a strictly
positive integer.

\medskip

\noindent{\bf proof} By 2F, there are a  
$\rho>0$ and a sequence $\sequencen{a_n}$ of complex numbers such that
$f(z)=\sum_{n=0}^{\infty}a_n(z-w)^n$ whenever $|z-w|<3\rho$.   
If every $a_n$ is
zero, then we have the first alternative.   Otherwise, there is a first $m$ such
that $a_m\ne 0$.   As $a_0=f(w)=0$, $m$ is greater than $0$, while
$f(z)=(z-w)^m\sum_{k=0}^{\infty}a_{k+m}(z-w)^k$ if $|z-w|<3\rho$.
Since there is a $v$ such that $2\rho<|v-w|<3\rho$, so that 
$\sum_{k=0}^{\infty}a_k(v-w)^k$ is defined
and $M=\sup_{k\in\Bbb N}|a_k(v-w)^k|<\infty$, we see that 
$|a_{k+m}(z-w)^k|\le M\Bover{|z-w|^k}{|v-w|^{k+m}}\le\Bover{2^{-k}M}{|v-w|^m}$
for $k\ge 0$ and 
$\sum_{k=l}^{\infty}|a_{k+m}(z-w)^k|\le\Bover{2^{-l+1}M}{|v-w|^m}$ whenever
$l\ge 0$ and $|z-w|<\rho$.   

Accordingly $g(z)=\sum_{k=0}^{\infty}a_{k+m}(z-w)^k$ is defined and 
the limit of the
uniformly bounded sequence of its partial sums on $\{z:|z-w|<\rho\}$,
and is holomorphic on $\{z:|z-w|<\rho\}$, while $g(w)=a_m$ is non-zero, and
$f(z)=(z-w)^mg(z)$.   So
$f'(z)=m(z-w)^{m-1}g(z)+(z-w)^mg'(z)$ and
$\Bover{(z-w)f'(z)}{f(z)}=m+(z-w)\Bover{g'(z)}{g(z)}\to m$ as $z\to w$.

\bigskip

{\bf 2H Remarks (a)} At a different level, we need to know that the exponential
function is defined everywhere in $\Bbb C$ by the Maclaurin series 
$\exp(z)=\sum_{n=0}^{\infty}\Bover{z^n}{n!}$
and that $\exp$ is holomorphic, with $\exp'=\exp$.   Some of the proof can be
shortened using the results above;  but the basic theory of Maclaurin and
Taylor series can be proved for complex numbers by the same methods as for real
numbers.
The point of 2B-2G above is that holomorphic complex functions from $\Bbb C$ to
$\Bbb C$  are far
more tightly circumscribed than differentiable real functions from $\Bbb R^2$ to
$\Bbb R^2$.

\medskip

{\bf (b)} If you have taken a first course in contour integration, everything
here is likely to be familiar, though you may have been allowed to be bolder in
some of the proofs.   I have taken the trouble to spell this material out 
in order to make it plain that there is nothing genuinely difficult for anyone
familiar with real $\epsilon$-$\delta$ analysis.   Of course I have omitted some
of the most important ideas relating to integration along curves rather than
simple polygons.

\bigskip

{\bf 3. The Riemann zeta function}

\medskip

{\bf 3A Definition} For complex $z$ such that $\Real(z)>1$, set

\Centerline{$\zeta_0(z)=\sum_{n=1}^{\infty}\Bover1{n^z}$.}

\noindent For each $n\ge 1$, $z\mapsto\Bover1{n^z}=\exp(-z\ln n)$ is
holomorphic and $|\Bover1{n^z}|=\Bover1{n^{\Real z}}$.   So 
$\sum_{n=1}^{\infty}|\Bover1{n^z}|$ is finite whenever $\Real(z)>1$, and the
series is uniformly 
absolutely summable on $\{z:\Real(z)>1+\delta\}$ for every $\delta>0$.   By
2E, $\zeta_0$ is holomorphic.
\leaveitout{with $\zeta'_0(z)=\sum_{n=1}^{\infty}\Bover1{\ln n}{n^z}$}

\medskip

\noindent{\bf Remark} I write $\zeta_0$ here, and $\zeta_1$ in 3E below, 
to emphasize
that these functions are restrictions of the true zeta function $\zeta$ to
$\{z:\Real(z)>1\}$ and $\{z:\Real(z)>0\}$ respectively.

\bigskip

{\bf 3B Theorem} If $\Real z>1$, 
$\zeta_0(z)=\prod_{p\text{ prime}}\Bover1{1-p^{-z}}$.

\medskip

\noindent{\bf proof} Enumerate the primes in increasing order as 
$\langle p_k\rangle_{k\ge 1}$.   For any $n\ge 1$,

$$\eqalign{\prod_{k=1}^n\Bover1{1-p_k^{-z}}&
  =\prod_{k=1}^n\sum_{r=0}^{\infty}p_k^{-rz}
  =\sum_{r_1,\ldots,r_n=0}^{\infty}p_1^{-r_1z}\ldots p_n^{-r_nz}\cr&
  =\sum_{r_1,\ldots,r_n=0}^{\infty}(p_1^{r_1}\ldots p_n^{r_n})^{-z}
  =\sum_{m\in K_n}\Bover1{m^z},\cr}$$

\noindent where $K_n$ is the set of those $m\ge 1$ not divisible by any
prime greater than $p_n$.   
Because the series $\langle\Bover1{m^z}\rangle_{m\ge 1}$ is absolutely summable,
and $\langle K_n\rangle_{n\ge 1}$ is non-decreasing with union 
$\Bbb N\setminus\{0\}$, 

\Centerline{$\zeta_0(z)
=\lim_{n\to\infty}\sum_{m\in K_n}\Bover1{m^z}
=\lim_{n\to\infty}\prod_{k=1}^n\Bover1{1-p_k^{-z}}
%=\prod_{k=1}^{\infty}\Bover1{1-p_k^{-z}
=\prod_{p\text{ prime}}\Bover1{1-p^{-z}}$.}

%\leaveitout{the product is uniformly convergent
%on $\{z:\Real(z)>1+\delta\}$}

\bigskip

{\bf 3C Corollary} If $\Real(z)>1$, $\zeta_0(z)\ne 0$ and
$\Bover{\zeta'_0(z)}{\zeta_0(z)}=\sum_{p\text{ prime}}\Bover{\ln p}{1-p^z}$.

\medskip

\noindent{\bf proof} For any particular $z$ with $\Real(z)>1$,
$|p^{-z}|<1$ and $1-p^{-z}\ne 0$ for every prime $p$, 
while $\sum_{p\text{ prime}}|p^{-z}|\le\sum_{n=1}^{\infty}n^{-\Real z}$ is
finite.   So $\prod_{p\text{ prime}}(1-p^{-z})$ is non-zero and its reciprocal
$\prod_{p\text{ prime}}\Bover1{1-p^{-z}}$ is non-zero.   Moreover, if 
$\delta>0$, then in the region $\{z:\Real(z)\ge 1+\delta\}$ the terms 
$1-p^{-z}$ are bounded away from $0$ and the sums
$\sum_{p\text{ prime}}|p^{-z}|$ are bounded, so the products
$\prod_{p\text{ prime}}(1-p^{-z})$ are uniformly convergent.

Enumerating the primes as $\langle p_k\rangle_{k\ge 1}$ 
as in 3B, we see that

$$\eqalign{\Bover{d}{dz}\prod_{k=1}^m\Bover1{1-p_k^{-z}}&
=-\sum_{k=1}^m\Bover{\bover{d}{dz}(1-p_k^{-z})}{1-p_k^{-z}}
   \prod_{k=1}^m\Bover1{1-p_k^{-z}}\cr&
=-\sum_{k=1}^m\Bover{p_k^{-z}\ln p_k}{1-p_k^{-z}}\prod_{k=1}^m\Bover1{1-p_k^{-z}}
=-\sum_{k=1}^m\Bover{\ln p_k}{p_k^z-1}\prod_{k=1}^m\Bover1{1-p_k^{-z}}\cr}$$

\noindent whenever $m\ge 1$ and $\Real(z)>1$.   
Now 2E and 3B tell us not just that 
$z\mapsto\zeta_0(z)=\lim_{m\to\infty}\prod_{k=1}^m\Bover1{1-p_k^{-z}}$ 
is holomorphic on $\{z:\Real(z)>1\}$ but also that $\zeta'_0(z)
=\lim_{m\to\infty}\Bover{d}{dz}\prod_{k=1}^m\Bover1{1-p_k^{-z}}$.
So

\Centerline{$\Bover{\zeta'_0(z)}{\zeta_0(z)}
=-\lim_{m\to\infty}\sum_{k=1}^m\Bover{\ln p}{p_k^z-1}
=\sum_{p\text{ prime}}\Bover{\ln p}{1-p^z}$}

\noindent whenever $\Real(z)>1$.
   
\bigskip

{\bf 3D Theorem} There is a holomorphic function $\varrho$ with domain
$\{z:\Real(z)>0\}$ such that $\zeta_0(z)=\varrho(z)+\Bover1{z-1}$ when
$\Real(z)>1$.

\medskip

\noindent{\bf proof} If $n\ge 1$ and $\Real(z)>0$,

$$\eqalignno{|\int_n^{n+1}(\Bover1{n^z}-\Bover1{x^z})dx|&
=|\int_n^{n+1}\int_n^x\Bover{z}{t^{z+1}}dt\,dx|
\Displaycause{the integrals here are with respect to Lebesgue measure}
\le\sup_{t\in[n,n+1]}|\Bover{z}{t^{z+1}}|
=\Bover{|z|}{n^{\Real(z)+1}}.\cr}$$

\noindent For each $n\ge 1$, 
$z\mapsto\int_n^{n+1}(\Bover1{n^z}-\Bover1{x^z})dx:\Bbb C\to\Bbb C$ 
is holomorphic, just as in 1Ad above, while 
$\langle\int_n^{n+1}(\Bover1{n^z}-\Bover1{x^z})dx\rangle_{n\ge 1}$ is uniformly
summable on $\{z:\Real(z)\ge\delta$, $|z|\le\bover1{\delta}\}$ for every 
$\delta>0$.   So if we set
$\varrho(z)=\sum_{n=1}^{\infty}\int_n^{n+1}(\Bover1{n^z}-\Bover1{x^z})dx$ for
$\Real(z)>0$, $\varrho$ will be holomorphic, by 2E again.

Now observe that if $\Real(z)>1$,

\Centerline{$\varrho(z)
=\sum_{n=1}^{\infty}\Bover1{n^z}-\int_1^{\infty}\Bover1{x^z}dx
=\zeta_0(z)+[\Bover1{(z-1)x^{z-1}}]_1^{\infty}
=\zeta_0(z)-\Bover1{z-1}$,}

\noindent so $\zeta_0(z)=\varrho(z)+\Bover1{z-1}$.

\bigskip

{\bf 3E Corollary} Define $\zeta_1$ by saying that $\dom\zeta_1=\{z:\Real(z)>0$,
$z\ne 1\}$ and $\zeta_1(z)=\varrho(z)+\Bover1{z-1}$.   Then $\zeta_1$ is
holomorphic and extends $\zeta_0$.

\bigskip

{\bf 3F Proposition} Set 
$\varphi_0(z)=\sum_{p\text{ prime}}\Bover{\ln p}{p^z}$ when $\Real(z)>1$, 
$\varkappa(z)=\sum_{p\text{ prime}}\Bover{\ln p}{p^z(p^z-1)}$  
when $\Real(z)>\bover12$, and
$\varphi(z)=-\Bover{\zeta_1'(z)}{\zeta_1(z)}-\varkappa(z)$ when
$\Real(z)>\bover12$, $z\ne 1$ and $\zeta_1(z)\ne 0$.
Then $\varphi_0$, $\varkappa$ and $\varphi$ are
holomorphic, and $\varphi$ extends $\varphi_0$.

\medskip

\noindent{\bf proof} $\varphi_0$ is holomorphic because 
$\sum_{p\text{ prime}}|\Bover{\ln p}{p^z}|<\infty$, just as for $\zeta_0$ in
3A, while $\varkappa$ is well-defined and holomorphic because
if $n\ge 4$ and $\gamma\ge\bover12$ then
$\bover12n^{\gamma}\ge 1$ and $n^{2\gamma}-n^{\gamma}\ge\bover12n^{2\gamma}$.
So if $\delta>0$ and $\Real z\ge\bover12+\delta$,
then $|n^z(n^z-1)|\ge|n^{2z}|-|n^z|\ge\bover12n^{1+\delta}$ whenever
$n\ge 4$, and 

\Centerline{$\sum_{n=4}^{\infty}|\Bover{\ln n}{n^z(n^z-1)}|
\le 2\sum_{n=4}^{\infty}\Bover{\ln n}{n^{1+\delta}}<\infty$.}

Now 

$$\eqalign{\varphi_0(z)+\varkappa(z)&
=\sum_{p\text{ prime}}\ln p(\Bover1{p^z}+\Bover1{p^z(p^z-1)})\cr&
=\sum_{p\text{ prime}}\Bover{\ln p}{p^z-1}
=-\Bover{\zeta'_0(z)}{\zeta_0(z)}\cr}$$

\noindent by 3C, so $\varphi$ extends $\varphi_0$.

\bigskip

{\bf 3G Remark} Note that all the functions we have been examining here,
from $\exp$ and $z\mapsto n^z$ and $\zeta_0$ to $\varphi_0$, commute with
complex conjugation, that is, $f(\bar z)$ is defined and equal to
$\overline{f(z)}$ whenever $z\in\dom f$.

\bigskip

{\bf 3H Theorem} $\zeta_1(z)\ne 0$ whenever $\Real(z)\ge 1$ and $z\ne 1$.

\medskip

\noindent{\bf proof (a)} If $\Real(z)>1$ then $\zeta_1(z)=\zeta_0(z)\ne 0$, as
observed in 3C.   

\medskip

{\bf (b)} If $\Real(w)=1$ and $w\ne 1$, 
$\zeta_1(w+\epsilon)=\zeta_0(w+\epsilon)\ne 0$ for every $\epsilon>0$, and
$\zeta_1$ is not constant zero on any neighbourhood of $w$.   So
$\lim_{z\to w}(z-w)\Bover{\zeta_1'(z)}{\zeta_1(z)}$ is
zero if $\zeta_1(w)\ne 0$, and otherwise is a strictly positive integer,
by 2G.  Also $\lim_{z\to w}(z-w)\varkappa(z)= 0\cdot\varkappa(w)=0$, so
$\lim_{z\to w}(z-w)\varphi(z)$ is
zero if $\zeta_1(w)\ne 0$, and otherwise is a strictly negative integer.

As for the case $w=1$, we have $\zeta_0(z)=\varrho(z)+\Bover1{z-1}$ and
$\zeta'_0(z)=\varrho'(z)-\Bover1{(z-1)^2}$ when $\Real(z)>1$, by 3D, so

\Centerline{$\Bover{\zeta'_0(z)}{\zeta_0(z)}
  =\Bover{(z-1)^2\varrho'(z)-1}{(z-1)^2\varrho(z)+z-1}$}

\noindent when $\Real z>1$. and

$$\eqalign{\lim_{\epsilon\downarrow 0}\epsilon\varphi(1+\epsilon)&
=-\lim_{\epsilon\downarrow 0}
\Bover{\epsilon\zeta'_0(\epsilon)}{\zeta_0(\epsilon)}-\lim_{\epsilon\downarrow
0}\epsilon\varkappa(1+\epsilon)\cr&
=-\lim_{\epsilon\downarrow 0}
   \Bover{\epsilon^2\varrho'(1+\epsilon)-1}{\epsilon\varrho(1+\epsilon)+1}
=1.\cr}$$

\medskip

{\bf (c)} \Quer\ If $\Real z=1$ and $z\ne 1$ and $\zeta_1(z)=0$, then 
$z=1+e^{i\alpha}$ for some $\alpha\ne 0$, and 
$\lim_{\epsilon\downarrow 0}\epsilon\varrho_0(1+e^{i\alpha}+\epsilon)=-k$ 
for some integer $k>0$.   Also 
$\lim_{\epsilon\downarrow 0}\epsilon\varrho_0(1+e^{2i\alpha}+\epsilon)=-l$ 
for some
integer $l\ge 0$.   Since $\varphi_0$ commutes with conjugation,
$\lim_{\epsilon\downarrow 0}\varphi_0(e^{-i\alpha}+\epsilon)=-k$ 
and $\lim_{\epsilon\downarrow 0}\epsilon\varphi_0(e^{-2i\alpha}+\epsilon)=-l$.
And we have just seen that 
$\lim_{\epsilon\downarrow 0}\epsilon\varphi(1+\epsilon)=1$.   But now consider

$$\eqalign{h(\epsilon)&
=\varphi_0(1+e^{-2i\alpha}+\epsilon)
 +4\varphi_0(1+e^{-i\alpha}+\epsilon)\cr&
\mskip50mu +6\varphi_0(1+\epsilon)
 +4\varphi_0(1+e^{i\alpha}+\epsilon)
 +\varphi_0(1+e^{2i\alpha}+\epsilon)\cr}$$

\noindent for $\epsilon>0$.   Then 
$\lim_{\epsilon\downarrow 0}\epsilon h(\epsilon)=-l-4k+6-4k-l<0$.   
On the other hand,

$$\eqalign{h(\epsilon)&
=\sum_{p\text{ prime}}\Bover{\ln p}{p^{1+\epsilon}}
    (p^{-2i\alpha}+4p^{-i\alpha}+6+4p^{i\alpha}+p^{2i\alpha})\cr&
=\sum_{p\text{ prime}}\Bover{\ln p}{p^{1+\epsilon}}
    (p^{-i\alpha/2}+p^{i\alpha/2})^4
>0\cr}$$

\noindent for $\epsilon>0$.\ \Bang

\medskip

{\bf (d)} Together with (a). this shows that
$\zeta_1(z)\ne 0$ whenever $\Real(z)\ge 1$ and $z\ne 1$.

\bigskip

{\bf 4. The Prime Number Theorem}

\medskip

{\bf 4A Proposition} Setting $\vartheta(x)=\sum_{p\text{ prime},p\le x}\ln p$,
$\vartheta(x)\le 4x\ln 2$ for every real $x>0$.

\medskip

\noindent{\bf proof} For any integer $n\ge 1$,

$$\eqalign{\vartheta(2n)-\vartheta(n)&
=\sum_{p\text{ prime},n<p\le 2n}\ln p
=\ln\bigl(\prod_{p\text{ prime},n<p\le 2n}p\bigr)\cr&
\le\ln(\Bover{(2n)!}{(n!)^2})
\le\ln(2^{2n})
=2n\ln 2.\cr}$$

\noindent In particular, $\vartheta(2^{k+1})\le\vartheta(2^k) + 2^{k+1}\ln 2$ for
$k\ge 0$;  inducing on $k$, $\vartheta(2^k)\le 2\ln 2\cdot 2^k$ for all $k$.
Now for any $x>\bover12$ there is a $k\in\Bbb N$ such that $x\le 2^k<2x$, so that

\Centerline{$\vartheta(x)\le\vartheta(2^k)\le 2\ln 2\cdot 2^k\le 4x\ln 2$.}

\bigskip

{\bf 4B Lemma} If $\beta>0$ and $\max(|\Real(z)|,|\Imag(z)|)=\beta$ then
$|1+\Bover{z^2}{\beta^2}|\le\Bover{2|z\Real(z)|}{\beta^2}$.

\medskip

\noindent{\bf proof} If $z$ is of the form $x\pm i\beta$, then 

$$\eqalign{|1+\Bover{z^2}{\beta^2}|&
=\Bover1{\beta^2}|\beta^2+x^2\pm 2i\beta x-\beta^2|
=\Bover1{\beta^2}|x||x\pm 2i\beta|\cr&
\le\Bover1{\beta^2}|x||2x\pm 2i\beta|
=\Bover{2|x||z|}{\beta^2}
=\Bover{2|z\Real(z)|}{\beta^2}.\cr}$$

\noindent If $z$ is of the form $iy\pm\beta$ where $|y|\le\beta$, then 

$$\eqalign{|1+\Bover{z^2}{\beta^2}|&
=\Bover1{\beta^2}|2\beta^2-y^2\pm 2i\beta y|
\le\Bover1{\beta^2}|2\beta^2\pm 2i\beta y|\cr&
=\Bover1{\beta^2}|\Real(z)||2(\beta\pm iy)|
=\Bover{2|z\Real(z)|}{\beta^2}.\cr}$$

\bigskip

{\bf 4C Lemma} Suppose that $f:\coint{0,\infty}\to\Bbb C$ is bounded and
Borel measurable, and that $g$ is a holomorphic function with domain
including $\{z:\Real z\ge 0\}$ and such that $g(z)=\int_0^{\infty}f(t)e^{-zt}dt$
whenever $\Real(z)>0$.   Then 
$\lim_{\alpha\to\infty}\int_0^{\alpha}f(t)dt=g(0)$.

\medskip

\noindent{\bf proof} Take any $\epsilon>0$.   

\medskip

{\bf (a)} Set $M=\sup_{t\ge 0}|f(t)|$.   Let $\beta>0$ be such that
$\Bover{8M}{\beta}\le\epsilon$, and let $\delta\in\ooint{0,\beta}$ be such
that the upright rectangle 
$R=\{z:-\delta\le\Real(z)\le\beta$, $|\Imag(z)|\le\beta\}$ is
included in $\dom g$.   Set $K=\sup_{z\in\partial R}|g(z)|$
and let $m\ge 0$ be such that 

\Centerline{$\int_{[-\delta+\beta i,\beta i]}e^{m\Real(z)}dz
    +\int_{[-\delta-\beta i,-\delta+\beta i]}e^{m\Real(z)}dz
    +\int_{[-\delta-\beta i,-\beta i]}e^{m\Real(z)}dz
\le\Bover{\epsilon\delta}{3K}$;}

\noindent the point here is that if $\Real(z)<0$ then 
$0\le e^{m\Real(z)}\le 1$ for all $m$ and
$\lim_{m\to\infty}e^{m\Real(z)}\to 0$ as $m\to\infty$, so the integrals tend to
$0$ by 1Ae.

\medskip

{\bf (b)} Take any $\alpha\ge m$.   Set 
$g_{\alpha}(z)=\int_0^{\alpha}f(t)e^{-zt}dt$ for $z\in\Bbb C$;  then
$g_{\alpha}$ is holomorphic, as in 1Ad above.   Now set

\Centerline{$h(z)=(g(z)-g_{\alpha}(z))(1+\Bover{z^2}{\beta^2})e^{\alpha z}$}

\noindent for $z\in\dom g$, so that $h$ is holomorphic and defined on $R$
and $h(0)=g(0)-g_{\alpha}(0)$.   Then

$$\eqalignno{g(0)-g_{\alpha}(0)&
=h(0)=\Bover1{2\pi i}\int_{\partial R}\Bover{h(z)}{z}
\Displaycause{by 2C}
=\Bover1{2\pi i}\int_{\partial R}(g(z)-g_{\alpha}(z))
(1+\Bover{z^2}{\beta^2})\Bover{e^{\alpha z}}zdz.\cr}$$
   
%in the formulae of Zagier 97, p 707, T represents \alpha and R represents \beta

\medskip

{\bf (c)} To estimate the integral, I break it into three parts.   

\medskip

\quad{\bf (i)} To the right, we have 

\Centerline{$\partial R\cap\{z:\Real(z)\ge 0\}
=[-\beta i,\beta(1-i)]\cup[\beta(1-i),\beta(i+1)]\cup[\beta i,\beta(i+1)]$}

\noindent giving a contribution

\Centerline{$J_1=\int_{[-\beta i,\beta(1-i)]}\Bover{h(z)}{z}
+\int_{[\beta(1-i),\beta(i+1)]}\Bover{h(z)}{z}
-\int_{[\beta i,\beta(i+1)]}\Bover{h(z)}z$.}

\noindent In these integrals, we have (except at the endpoints $-\beta i$, 
$\beta i$)

$$\eqalignno{|g(z)-g_{\alpha}(z)|&
=|\int_{\alpha}^{\infty}f(t)e^{-zt}dt|
\le M\int_{\alpha}^{\infty}|e^{-zt}|dt\cr&
=M\int_{\alpha}^{\infty}e^{-t\Real(z)}dt
=\Bover{Me^{-\alpha\Real(z)}}{\Real(z)}\cr}$$

\noindent and

\Centerline{$|\Bover1ze^{\alpha z}(1+\Bover{z^2}{\beta^2})|
\le\Bover1{|z|}e^{\alpha\Real(z)}\cdot\Bover{2|z\Real(z)|}{\beta^2}
=e^{\alpha\Real(z)}\cdot\Bover{2|\Real(z)|}{\beta^2}$}

\noindent by 4B, so 

\Centerline{$|h(z)|\le\Bover{Me^{-\alpha\Real(z)}}{\Real(z)}e^{\alpha\Real(z)}
   \Bover{2|\Real(z)|}{\beta^2}|
=\Bover{2M}{\beta^2}$;}

\noindent as the sum of the lengths of the intervals is $4\beta$, 
$|J_1|\le\Bover{8M}{\beta}\le\epsilon$. 

\medskip

\quad{\bf (ii)} To the left, we have 

\Centerline{$\partial R\cap\{z:\Real(z)\le 0\}
=[-\delta+\beta i,\beta i]\cup[-\delta-\beta i,-\delta+\beta i]
    \cup[-\delta-\beta i,-\beta i]$.}

\noindent Setting $h_2(z)=g(z)(1+\Bover{z^2}{\beta^2})e^{\alpha z}$ for
$z\in\dom g$ and $h_3(z)=g_{\alpha}(z)(1+\Bover{z^2}{\beta^2})e^{\alpha z}$
for all $z\in\Bbb C$, we have a contribution $J_2-J_3$ to the integral, where

\Centerline{$J_2=-\int_{[-\delta+\beta i,\beta i]}\Bover{h_2(z)}zdz
    -\int_{[-\delta-\beta i,-\delta+\beta i]}\Bover{h_2(z)}zdz
      +\int_{[-\delta-\beta i,-\beta i]}\Bover{h_2(z)}zdz$,}

\Centerline{$J_3=-\int_{[-\delta+\beta i,\beta i]}\Bover{h_3(z)}zdz
    -\int_{[-\delta-\beta i,-\delta+\beta i]}\Bover{h_3(z)}zdz
      +\int_{[-\delta-\beta i,-\beta i]}\Bover{h_3(z)}zdz$.}

\noindent Now 

\Centerline{$|\Bover{h_2(z)}z|\le\Bover3{\delta}|g(z)|e^{\alpha\Real z}
\le\Bover{3K}{\delta}e^{m\Real(z)}$}

\noindent whenever $z\in\partial R$ and $\Real z\le 0$, so $|J_2|\le\epsilon$,
by the choice of $m$.   

\medskip

\quad{\bf (iii)} As for $J_3$, 
$h_3:\Bbb C\to\Bbb C$ is holomorphic and
$z\mapsto\Bover{h_3(z)}z:\Bbb C\setminus\{0\}\to\Bbb C$ is holomorphic.
So if $R'$ is the upright
rectangle $\{z:-\beta\le\Real(z)\le-\delta$, $|\Imag(z)|\le\beta\}$ then
$\int_{\partial R'}\Bover{h_3(z)}zdz=0$ and

$$\eqalign{J_3&=J_3+\int_{\partial R'}\Bover{h_3(z)}zdz\cr&
=-\int_{[-\beta+\beta i,\beta i]}\Bover{h_3(z)}zdz
    -\int_{[-\beta-\beta i,-\beta+\beta i]}\Bover{h_3(z)}zdz
      +\int_{[-\beta-\beta i,-\beta i]}\Bover{h_3(z)}zdz.\cr}$$

\noindent Now if $\Real(z)<0$, 

$$\eqalign{|g_{\alpha}(z)|&
=|\int_0^{\alpha}f(t)e^{-zt}dt|
\le M\int_0^{\alpha}|e^{-zt}|dt\cr&
=M\int_0^{\alpha}e^{-t\Real(z)}dt
=\Bover{M(e^{-\alpha\Real(z)}-1)}{|\Real{z}|}
\le\Bover{Me^{-\alpha\Real(z)}}{|\Real{z}|}\cr}$$

\noindent while $|\Bover1ze^{\alpha z}(1+\Bover{z^2}{\beta^2})|
=e^{\alpha\Real(z)}\cdot\Bover{2|\Real(z)|}{\beta^2}$ as in (i), so
$|h_3(z)|\le\Bover{2M}{\beta^2}$ and 
$|J_3|\le\Bover{8M}{\beta}\le\epsilon$, as before.

\medskip

{\bf (d)} Collecting these, we see that

\Centerline{$|g(0)-g_{\alpha}(0)|=|\Bover1{2\pi i}(J_1+J_2-J_3)|
\le\Bover{3\epsilon}{2\pi}\le\epsilon$}

\noindent for every $\alpha\ge\alpha_0$;  as $\epsilon$ is arbitrary,

\Centerline{$g(0)=\lim_{\alpha\to\infty}g_{\alpha}(0)
=\lim_{\alpha\to\infty}\int_0^{\alpha}f(t)dt$}

\noindent as claimed.

\bigskip
 
{\bf 4D Proposition} $\lim_{\beta\to\infty}
  \int_1^{\beta}\Bover{\vartheta(t)-t}{t^2}dt$ exists in $\Bbb R$.
 
\medskip

\noindent{\bf proof (a)} Observe that if $\Real(z)>1$ then

$$\eqalign{z\int_0^{\infty}e^{-zt}\vartheta(e^t)dt&
=z\int_0^{\infty}e^{-zt}\sum_{p\text{ prime},p\le e^t}\ln p\,dt\cr&
=\sum_{p\text{ prime}}\int_{\ln p}^{\infty}ze^{-zt}\ln p\,dt
=\sum_{p\text{ prime}}\Bover{\ln p}{p^z}
=\varphi_0(z)
=\varphi(z)\cr}$$

\noindent by 3F.  If we set $f(t)=\vartheta(e^t)e^{-t}-1$ for $t\ge 0$, 
$-1\le f(t)\le 4\ln 2-1$ for every $t\ge 0$, by 4A, so $f$ is bounded and
Borel measurable, and if $\Real(z)>0$, then

\Centerline{$\int_0^{\infty}f(t)e^{-zt}dt
=\int_0^{\infty}\vartheta(e^t)e^{-(z+1)t}-e^{-zt}dt
=\Bover{\varphi(z+1)}{z+1}-\Bover1z$.}

\medskip

\query

{\bf (b)} We saw in 3E that $\zeta_1(z)=\varrho(z)+\Bover1{z-1}$ if
$\Real(z)>0$ and $z\ne 1$.   So $\zeta_1(z)\ne 0$ and

\Centerline{$\Bover{\zeta_1'(z)}{\zeta_1(z)}+\Bover1{z-1}
=(\rho'(z)-\Bover1{(z-1)^2})/(\rho(z)+\Bover1{z-1})+\Bover1{z-1}
=\Bover{(z-1)\rho'(z)+\rho(z)}{(z-1)\rho(z)+1}$}

\noindent if $z$ is close to $1$ and not equal to $1$.   But this means that if
we set $h(z)=\Bover{\zeta_1'(z)}{\zeta_1(z)}+\Bover1{z-1}+\varkappa(z)$ when
$\Real(z)>\bover12$ and $z\ne 1$ and $\zeta_1(z)\ne 0$, and
$h(1)=\rho(1)+\varkappa(1)$, then $h'(1)$ is defined and equal to 
$1+\kappa'(1)$, so $h$ will be holomorphic, with domain
$\{z:\Real(z)>\bover12$, $\zeta_1(z)\ne 0\}$, which is an open set including
$\{z:\Real(z)\ge 1\}$.   
3F tells us also that $h(z)=\Bover1{z-1}-\varphi(z)$ when $z\in\dom h$ and 
$z\ne 1$.

\medskip

{\bf (c)} Set

\Centerline{$g(z)=-\Bover{h(z+1)}{z+1}$ when 
$\Real(z)>-\bover12$ and $\zeta_1(z+1)\ne 0$}

\noindent so that $g$ is holomorphic and its domain includes
$\{z:\Real(z)\ge 0\}$, while 

\Centerline{$g(z)=\Bover{\varphi(z+1)}{z+1}-\Bover1z
=\int_0^{\infty}f(t)e^{-zt}dt$}

\noindent when $\Real(z)>0$.   

\medskip

{\bf (d)} By Lemma 4C, 
$\lim_{\beta\to\infty}\int_0^{\beta}f(t)dt$ is defined in $\Bbb R$, that is,

\Centerline{$\lim_{\beta\to\infty}\int_0^{\beta}\vartheta(e^t)e^{-t}-1\,dt$}

\noindent is defined, that is, 

\Centerline{$\lim_{\beta\to\infty}\int_0^{\ln\beta}
                  (\Bover{\vartheta(t)}{t}-1)\Bover1tdt$}

\noindent is defined, and this is the required limit 
$\lim_{\beta\to\infty}\int_0^{\beta}\Bover{\vartheta(t)-t}{t^2}dt$.

\bigskip

{\bf 4E Corollary} $\lim_{x\to\infty}\Bover{\vartheta(x)}x=1$.

\medskip

\noindent{\bf proof (a)} If $\lambda>1$, 
$\int_1^{\lambda}bover{\lambda-t}{t^2}dt>0$, so there is a $\beta_0>0$ such that

\Centerline{$\int_{\beta}^{\beta'}\Bover{\vartheta(t)-t}{t^2}dt
<\int_1^{\lambda}\Bover{\lambda-t}{t^2}dt$}

\noindent whenever $\beta_0\le\beta\le\beta'$.   Now if $x>0$ is such that 
$\vartheta(x)\ge\lambda x$, then

$$\int_x^{\lambda x}\Bover{\vartheta(t)-t}{t^2}dt
\ge\int_x^{\lambda x}\Bover{\vartheta(x)-t}{t^2}dt
\ge\int_x^{\lambda x}\Bover{\lambda x-t}{t^2}dt
=\int_1^{\lambda}\Bover{\lambda-t}{t^2}dt$$

\noindent so $x\le\beta_0$.   As $\lambda$ is arbitrary,
$\limsup_{x\to\infty}\Bover{\vartheta(x)}x\le 1$.

\medskip

{\bf (b)} If $\lambda<1$, then $\int_{\lambda}^1\Bover{\lambda-t}{t^2}dt<0$,
so there is a $\beta_0>0$ such that

\Centerline{$\int_{\beta}^{\beta'}\Bover{\vartheta(t)-t}{t^2}dt
>\int_{\lambda}^1\Bover{\lambda-t}{t^2}dt$}

\noindent whenever $\beta_0\le\beta\le\beta'$.   Now if $x>0$ is such that 
$\vartheta(x)\le\lambda x$, then

$$\int_{\lambda x}^x\Bover{\vartheta(t)-t}{t^2}dt
\le\int_{\lambda x}^x\Bover{\vartheta(x)-t}{t^2}dt
\le\int_{\lambda x}^x\Bover{\lambda x-t}{t^2}dt
=\int_{\lambda}^1\Bover{\lambda-t}{t^2}dt$$

\noindent and $\lambda x<\beta_0$.   So
$\liminf_{x\to\infty}\Bover{\vartheta(x)}x\ge 1$.   Together with (a), this
shows that $\lim_{x\to\infty}\Bover{\vartheta(x)}x=1$.

\bigskip

{\bf 4F The Prime Number Theorem} Writing $\pi(x)$ for the number of primes less
than or equal to $x$, $\lim_{x\to\infty}\Bover{\pi(x)\ln x}x=1$.

\medskip

\noindent{\bf proof} For any $x\ge 1$,

\Centerline{$\vartheta(x)=\sum_{p\text{ prime},p\le x}\ln p\le\pi(x)\ln x$}

\noindent so $\liminf_{x\to\infty}\Bover{\pi(x)\ln x}{x}
\ge\lim_{x\to\infty}\Bover{\vartheta(x)}x=1$.      

On the other side, given $\epsilon\in\ooint{0,1}$, we have
$\pi(x)=\pi(x^{1-\epsilon})+\#(\{p:p\text{ is prime}$, 
$x^{1-\epsilon}<p\le x\})$ for every $x>0$, so

$$\eqalignno{\Bover{\pi(x)\ln x}x&
\le\Bover{x^{1-\epsilon}\ln x}x
  +\sum_{p\text{ prime},x^{1-\epsilon}<p\le x}\Bover{\ln x}x\cr&
\le\Bover{x^{1-\epsilon}\ln x}x
  +\sum_{p\text{ prime},x^{1-\epsilon}<p\le x}\Bover{\ln p}{x(1-\epsilon)}
\Displaycause{because if $x^{1-\epsilon}\le p$ then 
$(1-\epsilon)\ln x\le\ln p$}
\le\Bover{\ln x}{x^{\epsilon}}
  +\Bover{\vartheta(x)}{x(1-\epsilon)}
\to 0+\Bover1{1-\epsilon}
\cr}$$

\noindent as $x\to\infty$.   So 
$\limsup_{x\to\infty}\Bover{\pi(x)\ln x}{x}\le\Bover1{1-\epsilon}$;  as
$\epsilon$ is arbitrary, $\limsup_{x\to\infty}\Bover{\pi(x)\ln x}{x}\le 1$.

Putting these together,

\Centerline{$\lim_{x\to\infty}\Bover{\pi(x)\ln x}{x}$ is defined and equal
to $1$.}

\bigskip


\bigskip

\noindent{\bf References}

Fremlin D.H.\ [11] {\it Measure Theory, Vol.\ 1:  The Irreducible Minimum.}   
Torres Fremlin, 2011\ifnum\stylenumber=2{

\noindent}\else{ }\fi
({\tt https://www1.essex.ac.uk/maths/people/fremlin/mtsales.htm}).

Newman D.J.\  [80] `Simple analytic proof of the Prime Number Theorem', 
Amer.\ Math.\ Monthly 87 (1980) 693-696.

Zagier D.\ [97] `Newman's short proof of the Prime Number Theorem',
Amer.\ Math.\ Monthly 104 (1997) 705-708.

\end
