Our first proof of the prime number theorem will run through a complex analytic argument due to Newman. The particular paper I'm referencing here is a very nice exposition of Newman's paper by D. Zagier. There is a wonderful stroll through the Zagier paper in Helgason's notes for a complex analysis class on MIT's OCW. Tao's notes follow a similar line, but works through a different set of functions.
We aim to prove the following description of the asymptotic density of the primes in the integers. Let \(\pi(x)\) denote the number of primes less than or equal to \(x\text{.}\) We will use the notation \(f \sim g\) (that is, \(f\) and \(g\) are asymptotically equal) to mean \(\lim_{x \to \infty} f(x)/g(x) = 1\text{.}\) We will also use the order notation \(g = \mathrm{O}(f)\) as \(x \to \infty\) to mean that there exists some \(C>0\) so that \(\abs{g(x)} \leq Cf(x)\) for all sufficiently large \(x\text{.}\) (Every order argument we make in this section will be at infinity, so we may sometimes just write \(g = \mathrm{O}(x)\text{.}\))
Finally, our asymptotic prime checking will be done by the function
\begin{equation*}
\vtheta(s) = \sum_{p \leq x} \log p.
\end{equation*}
Essentially, the argument goes by showing that there can be no zeros of \(\zeta\) on the line \(\RE(s) = 1\text{,}\) and then using that fact to force an asymptotic estimate on \(\vtheta\text{.}\) The steps to the proof are as follows:
Product formula.
Prove the product formula for \(\zeta\) for \(\RE s > 1\text{.}\) That is,
The idea is to write the function as a power series and show that it converges absolutely by comparison. Both functions are analytic on \(\RE s > 1\text{,}\) so we'll use that as a basis for our series. In order to leverage bounding in integral, note that we can write \(1/(s-1)\) in a sort of continuous \(p\)-series form via
\begin{align*}
\abs{ \int_n^{n+1} (\frac{1}{n^s} - \frac{1}{x^s})\, dx} \amp= \abs{s \int_n^{n+1} \int_n^x \frac{1}{u^{s+1}}\, du dx}\\
\amp\leq \max_{n \leq u \leq n+1} \abs{\frac{s}{u^{s+1}}}\\
\amp= \frac{\abs{s}}{n^{\RE s + 1}}.
\end{align*}
Thus, while we defined the series for \(\RE s > 1\text{,}\) the terms of the series will converge absolutely for \(\RE s > 0\text{,}\) (for example by successive application of the \(M\)-test on an appropriate family of half-disks).
\begin{equation*}
\vtheta(x) - \vtheta(\frac{x}{2}) \leq C x
\end{equation*}
for any \(C > \log 2\) for all \(x > x_0\) for some \(x_0\) whose choice depends on \(C\text{.}\) Now, given some large value of \(x\text{,}\) consider the sequence
Notice that the series on the right hand side converges for \(\RE s > \frac{1}{2}\text{.}\) From Step II above, we know that \(\zeta - \frac{1}{s-1}\) is holomorphic on \(\RE s > 0\text{,}\) and so
is meromorphic on \(\RE s \geq 0\) with a simple pole at \(s = 1\text{.}\) Likewise, \(\zeta\dd\) will be meromorphic with a pole at \(s = 1\) of order 2. Then \(\zeta\dd/\zeta\) is also a meromorphic function on \(\RE s > 0\) with poles at \(s = 1\) and at the zeros of \(\zeta\text{,}\) which implies for \(\Phi\text{,}\) which is
that we have a meromorphic extension of \(\Phi\) to \(\RE s > \frac{1}{2}\) with poles at \(s = 1\) and the zeros of \(\zeta\text{.}\)
Now let's analyze the zeros. First, because \(\zeta(s) = \cc{\zeta(\cc s)}\text{,}\) we see that if \(s\) is zero of \(\zeta\) then so is \(\cc{s}\text{.}\) Now let \(a\in \R\text{.}\) If \(s_0 = 1 + ia\) is a zero of \(\zeta\) of order \(\mu \geq 0\text{,}\) then \(\zeta = (s - s_0)^\mu h(s)\) for some holomorphic function \(h\) that is non-zero near \(s_0\text{.}\) Computing the logarithmic derivative gets us
for some function \(H = \log h\) holomorphic near \(s_0\text{.}\) The function \(\Phi(s_0 + \eps)\) then has a simple pole at \(\eps = 0\text{,}\) and by the residue theorem
and so \(0 \leq \mu \leq 1\text{.}\) As we intend to show that \(\mu = 0\text{,}\) we need more. So assume as well that \(\zeta\) has a zero at \(s_1 = 1 \pm i2a\) with order \(\nu \geq 0\text{.}\) (That is, we are admitting the possibility that there is no zero at those points.) Again, note that
This requires that \(\mu = 0\) since \(\mu, \nu \geq 0\text{.}\) We have shown that if \(\zeta\) has a zero of the form \(s_0 = 1 + ia\text{,}\) then it must have order 0, and hence not be a zero. That is, \(\zeta\) has no zeros on the line \(\RE s > 1\text{.}\)
Finally, recall that the poles of \(\Phi\) are \(s = 1\) and the zeros of \(\zeta\text{,}\) none of which can lie on \(\RE s = 1\text{.}\) If we trace through the form of \(\Phi\) from (3.3.1) and the implications of Step II, we conclude that \(\Phi - \frac{1}{s-1}\) is holomorphic on \(\RE s \geq 1\text{.}\)
In the next step, we hone our understanding of the behavior of \(\vtheta\text{.}\)
We'll use Stieltjes integrals and their properties. Assume that \(\RE s > 1\text{.}\) First, note that \(\vtheta\) is a step function increasing with a step at each new prime number \(p\text{.}\) Then we can take the view that
We're setting up to invoke a sort of dominated integral theorem from Newman's original paper, so we now introduce functions \(f\) and \(g\text{.}\) Define the functions
By Step III, \(\vtheta\) is \(O(x)\text{,}\) and so \(f\) is bounded. Proceeding formally for a moment, (that is, ignoring issues of convergence, which will be pinned down shortly) the same change of variables from above will give
We'll proceed by cases on the assumption that \(\vtheta(x) \nsim x\text{.}\) First, assume that there exists \(\la > 1\) so that \(\vtheta(x) > \la x\) for sufficiently large \(x\text{.}\) Given such an \(x\text{,}\) as \(\vtheta\) is an increasing function,
where this constant does not depend on \(x\text{.}\) However, the convergence in Step V means that for all \(\eps > 0\text{,}\) there exists some \(M\) for which \(M_1, M_2 \geq M\) we have
As is typical in mathematics, the technical lemma that drove our proof above, Newman's so-called Analytic Theorem, is the key to the entire enterprise. Indeed, according to J. Korevaar's On Newman's quick way to the prime number theorem 1 , the simplicity of Newman's approach boils down to his proof of that theorem. Between the 1930s and the 1980s, the standard proof of the prime number theorem went by way of Fourier analysis, and a method due to Weiner called Tauberian integrals 2 . In fact, the Analytic Theorem appears in a paper by Ingham from the 1930s with a different proof and more general statement. We restate Newman's result here.
Theorem3.3.10.Analytic Theorem.
Let \(f(t)\) be bounded on \((0, \infty)\) and integrable on any finite subinterval, and with the Laplace transform of \(f\)
well-defined and analytic for \(\RE s > 0\text{.}\) If \(g(s)\) extends to a holomorphic function on \(\RE s \geq 0\) (that is, \(g\) is holomorphic in a neighborhood of every point on the imaginary axis, then
as an improper integral and equals \(g(0)\text{.}\)
Korevaar points out that this is related to a very useful theorem called the Ikehara-Weiner theorem 3 . (The structure of this should remind you of the functions we used in the proof of the prime number theorem.)
Theorem3.3.11.Ikehara-Weiner.
Let \(f(x)\) be nonnegative and nondecreasing on \([1,\infty)\) and such that the Mellin transform
can be continued analytically to a neighborhood of every point on the line \(\RE s = 1\text{.}\) Then
\begin{equation*}
\lim_{x \to \infty} \frac{f(x)}{x} = c.
\end{equation*}
Notice that the content of the above theorem is essentially Steps IV-V of Newman's argument. The ideas that we're tossing around here are really ideas in analytic number theory, where complex analysis is used to analyze objects like Dirichlet series 4 via the theory of generating functions. Whereas the Ikehara-Weiner theorem is quite difficult, the proof of the Analytic Theorem requires little more than Cauchy's theorem and careful choice of contour.
extends holomorphically to \(\RE s \geq 0\text{.}\) Choose some radius \(R\) and then choose a small \(\delta\) so that \(g\) is analytic on and inside the contour \(C_1 \cup C_2\text{.}\) Set the function
