Editing Prime number theorem (section)

== Proof sketch ==
Here is a sketch of the proof referred to in one of [[Terence Tao]]'s lectures.<ref>{{cite web |last1=Tao |first1=Terence |author-link = Terence Tao|title=254A, Notes 2: Complex-analytic multiplicative number theory |url=https://terrytao.wordpress.com/2014/12/09/254a-notes-2-complex-analytic-multiplicative-number-theory/ |website=Terence Tao's blog|date=10 December 2014 }}</ref> Like most proofs of the PNT, it starts out by reformulating the problem in terms of a less intuitive, but better-behaved, prime-counting function. The idea is to count the primes (or a related set such as the set of prime powers) with ''weights'' to arrive at a function with smoother asymptotic behavior. The most common such generalized counting function is the [[Chebyshev function]] {{math|''ψ''(''x'')}}, defined by
: <math>\psi(x) = \sum_{k \geq 1} \sum_\overset{p^k \le x,}{\!\!\!\!p \text{ is prime}\!\!\!\!} \log p \; .</math> <!-- The various instance of "\!\!\!" are to avoid the space added by wide limits like "p is prime" which are low enough that they can overlap preceding and following operators.  In real TeX, I might use something like \makebox[0pt]{p\text{ is prime}} (pretend it's zero-width), but <math> is easily confused.  After some experimentation, it's more robust to take the space out of the limit (symmetrically so it remains centred), so the spacing determined by the other symbols is not affected.-->

This is sometimes written as
: <math>\psi(x) = \sum_{n\le x} \Lambda(n) \; ,</math>
where {{math|''Λ''(''n'')}} is the [[von Mangoldt function]], namely
: <math>\Lambda(n) = \begin{cases} \log p & \text{ if } n = p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}</math>

It is now relatively easy to check that the PNT is equivalent to the claim that
: <math>\lim_{x\to\infty} \frac{\psi(x)}{x} = 1 \; .</math>
Indeed, this follows from the easy estimates
: <math>\psi(x) = \sum_\overset{p\le x}{\!\!\!\! p \text{ is prime}\!\!\!\!} \log p \left\lfloor \frac{\log x}{\log p} \right\rfloor \le \sum_\overset{p\le x}{\!\!\!\! p \text{ is prime}\!\!\!\!} \log x = \pi(x)\log x</math>
and (using [[big O notation|big {{mvar|O}} notation]]) for any {{math|''ε'' > 0}},
: <math>\psi(x) \ge \sum_{\!\!\!\!\overset{x^{1-\varepsilon}\le p\le x}{p \text{ is prime}}\!\!\!\!} \log p \ge \sum_{\!\!\!\!\overset{x^{1-\varepsilon}\le p\le x}{p \text{ is prime}}\!\!\!\!} (1-\varepsilon)\log x=(1-\varepsilon)\left(\pi(x)+O\left(x^{1-\varepsilon}\right)\right)\log x \; .</math>

The next step is to find a useful representation for {{math|''ψ''(''x'')}}. Let {{math|''ζ''(''s'')}} be the Riemann zeta function. It can be shown that {{math|''ζ''(''s'')}} is related to the [[von Mangoldt function]] {{math|''Λ''(''n'')}}, and hence to {{math|''ψ''(''x'')}}, via the relation
: <math>-\frac{\zeta'(s)}{\zeta(s)} = \sum_{n = 1}^\infty \Lambda(n) \, n^{-s} \; .</math>

A delicate analysis of this equation and related properties of the zeta function, using the [[Mellin transform]] and [[Perron's formula]], shows that for non-integer {{mvar|x}} the equation
: <math>\psi(x) = x \; - \; \log(2\pi) \; - \!\!\!\! \sum\limits_{\rho :\, \zeta(\rho) = 0} \frac{x^\rho}{\rho}</math><!--This is a special case.  We don't want a wide space between the minus and the sum, because the lower limit can overlap on the left, but we do need extra space on the right, because the fraction is tall enough to collide with the lower limit.-->
holds, where the sum is over all zeros (trivial and nontrivial) of the zeta function. This striking formula is one of the so-called [[Explicit formulae (L-function)|explicit formulas of number theory]], and is already suggestive of the result we wish to prove, since the term {{mvar|x}} (claimed to be the correct asymptotic order of {{math|''ψ''(''x'')}}) appears on the right-hand side, followed by (presumably) lower-order asymptotic terms.

The next step in the proof involves a study of the zeros of the zeta function. The trivial zeros −2, −4, −6, −8, ... can be handled separately:
: <math>\sum_{n=1}^\infty \frac{1}{2n\,x^{2n}} = -\frac{1}{2}\log\left(1-\frac{1}{x^2}\right),</math>
which vanishes for large {{mvar|x}}. The nontrivial zeros, namely those on the critical strip {{math|0 ≤ Re(''s'') ≤ 1}}, can potentially be of an asymptotic order comparable to the main term {{mvar|x}} if {{math|Re(''ρ'') {{=}} 1}}, so we need to show that all zeros have real part strictly less than 1.

=== Non-vanishing on Re(''s'') = 1 ===

To do this, we take for granted that {{math|''ζ''(''s'')}} is [[Meromorphic function|meromorphic]] in the half-plane {{math|Re(''s'') > 0}}, and is analytic there except for a simple pole at {{math|''s'' {{=}} 1}}, and that there is a product formula
: <math>\zeta(s)=\prod_p\frac{1}{1-p^{-s}} </math>
for {{math|Re(''s'') > 1}}. This product formula follows from the existence of unique prime factorization of integers, and shows that {{math|''ζ''(''s'')}} is never zero in this region, so that its logarithm is defined there and
: <math>\log\zeta(s)=-\sum_p\log \left(1-p^{-s} \right)=\sum_{p,n}\frac{p^{-ns}}{n} \; .</math>

Write {{math|''s'' {{=}} ''x'' + ''iy''}} ; then
: <math>\big| \zeta(x+iy) \big| = \exp\left( \sum_{n,p} \frac{\cos ny\log p}{np^{nx}} \right) \; .</math>

Now observe the identity
: <math> 3 + 4 \cos \phi+ \cos 2 \phi = 2 ( 1 + \cos \phi )^2\ge 0 \; ,</math>
so that
: <math>\left| \zeta(x)^3 \zeta(x+iy)^4 \zeta(x+2iy) \right| = \exp\left( \sum_{n,p} \frac{3 + 4 \cos(ny\log p) + \cos( 2 n y \log p )}{np^{nx}} \right) \ge 1</math>
for all {{math|''x'' > 1}}. Suppose now that {{math|''ζ''(1 + ''iy'') {{=}} 0}}. Certainly {{mvar|y}} is not zero, since {{math|''ζ''(''s'')}} has a simple pole at {{math|''s'' {{=}} 1}}. Suppose that {{math|''x'' > 1}} and let {{mvar|x}} tend to 1 from above. Since <math>\zeta(s)</math> has a simple pole at {{math|''s'' {{=}} 1}} and {{math|''ζ''(''x'' + 2''iy'')}} stays analytic, the left hand side in the previous inequality tends to 0, a contradiction.

Finally, we can conclude that the PNT is heuristically true. To rigorously complete the proof there are still serious technicalities to overcome, due to the fact that the summation over zeta zeros in the explicit formula for {{math|''ψ''(''x'')}} does not converge absolutely but only conditionally and in a "principal value" sense. There are several ways around this problem but many of them require rather delicate complex-analytic estimates. Edwards's book<ref>{{cite book |last = Edwards |first = Harold M. |author-link = Harold Edwards (mathematician) |title = Riemann's zeta function |publisher = Courier Dover Publications |year = 2001 |isbn = 978-0-486-41740-0}}</ref> provides the details. Another method is to use [[Ikehara's Tauberian theorem]], though this theorem is itself quite hard to prove. D.J. Newman observed that the full strength of Ikehara's theorem is not needed for the prime number theorem, and one can get away with a special case that is much easier to prove.