21.12.02 · number-theory / prime-number-theorem

The Prime Number Theorem via Contour Integration

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Anchor (Master): Davenport 2000 *Multiplicative Number Theory* 3e §17-18 (the full PNT contour proof with error term); Montgomery-Vaughan 2007 *Multiplicative Number Theory I* (Cambridge UP) §6.2-6.3 (zero-free region, $\psi(x)=x+O(x\exp(-c\sqrt{\log x}))$); Newman 1980 *Amer. Math. Monthly* 87 (the short Tauberian proof); Korevaar 1982 *Math. Intelligencer* 4 (Newman's method); de la Vallée Poussin 1896; Hadamard 1896

Intuition Beginner

The prime number theorem is the statement that primes thin out at a predictable rate: up to a large number $x$ , the count of primes is about $x$ divided by the natural logarithm of $x$ . The cleaner way to say it weights each prime power by the logarithm of its base prime and adds the weights up to $x$ , giving the running total $ψ (x)$ of the previous unit. The theorem then reads $ψ (x)$ is about $x$ . The whole task of this unit is to prove that one short sentence.

The proof is a piece of detective work in the complex plane. The previous unit handed us a magic function, the logarithmic derivative of the zeta function, whose hidden numbers are exactly the prime-power weights. Perron's formula from the other prerequisite turned the running total $ψ (x)$ into an integral of that magic function along a vertical line. So the entire question becomes: what is the value of one specific integral? We answer it by sliding the line of integration to the left and watching what we sweep past.

The one thing we sweep past is a single spike, a pole, sitting at the point $1$ on the real axis. That spike contributes exactly $x$ . Everything else — the wandering of the line after the slide — turns out to be small, provided the zeta function never equals zero right on the vertical line through $1$ . That no-zeros-on-the-line fact is the heart of the matter, and there is a beautiful one-line trigonometric reason for it.

Visual Beginner

Picture the complex plane with a tall vertical line drawn just to the right of the point $1$ . The integral that computes $ψ (x)$ runs up this line. Now imagine grabbing the line and dragging it leftward, past the point $1$ . As it crosses $1$ it catches a single spike — the pole — and that catch is worth exactly $x$ . The dragged line, now sitting to the left of $1$ , contributes only a small leftover.

   imaginary
     axis
      ^
      |   original line          shifted line
      |   Re(s) = 1 + small      Re(s) = 1 - c/log T
      |        :                      |
      |        :   <-- drag left -->  |
  ----+--------X----------------------|----> real axis
      |     pole at                   |
      |      s = 1   (residue = x)    |
      |        :                      |
      |   shaded band: no zeros of zeta here

$x$	$ψ (x)$ (approx.)	$x / lo g x$	$π (x)$ (true count)
$1 0^{2}$	$94$	$21.7$	$25$
$1 0^{4}$	$10013$	$1086$	$1229$
$1 0^{6}$	$\approx 1 0^{6}$	$72382$	$78498$

The middle column tracks the raw prime count $x / lo g x$ ; the right column is the true count $π (x)$ . Their ratio creeps toward one, the numerical shadow of the theorem.

Worked example Beginner

Let us see, with arithmetic instead of integration, why one single point on the line $Re (s) = 1$ being zero-free is what the whole proof needs. The explicit formula from the previous unit says the running total splits as a main piece plus a wave for each zero of zeta: $$ \psi(x) = x - (\text{a wave from each zero } \rho) - (\text{small constants}). $$

Step 1. Measure the size of one wave. A zero sitting at the complex point $ρ$ produces a wave whose height is $x$ raised to the real part of $ρ$ , divided by the size of $ρ$ . So the real part of $ρ$ is the exponent that controls everything.

Step 2. Suppose a zero sat exactly on the line, with real part equal to $1$ . Then its wave has height $x$ raised to the power $1$ , that is, height proportional to $x$ itself — the same size as the main term. The running total would wobble by a fixed fraction of $x$ forever, and $ψ (x) / x$ would never settle to $1$ .

Step 3. Now suppose every zero has real part strictly below $1$ , say at most $1 - δ$ for some gap $δ$ . Then each wave has height at most $x$ raised to $1 - δ$ , which is $x$ divided by $x^{δ}$ . Since $x^{δ}$ grows without bound, each wave is tiny compared to the main term $x$ .

What this tells us: pushing the zeros even a hair off the line $Re (s) = 1$ shrinks every wave below the main term, so $ψ (x)$ is forced to track $x$ . The proof is the job of producing that hair-width gap.

Check your understanding Beginner

Exercise (easy, multiple choice).

When the vertical line of integration is dragged leftward past the point $1$ , the single spike it catches contributes which value to $ψ (x)$ ?

A. $lo g x$

B. $x$

C. $1$

D. $x^{2}$

Hint

The pole sits at the point where the zeta function itself blows up, and the magic function has a spike of strength one there. Combined with the steering factor, the catch is the main term of the prime number theorem.

Answer

B. The pole of the logarithmic derivative at the point $1$ has strength $+ 1$ , and the steering factor of Perron's formula evaluates to $x$ there, so the residue caught is exactly $x$ . Feedback-correct: this single catch is the entire main term of the prime number theorem; everything else is error. Feedback-wrong: $1$ is the strength of the spike, but the steering factor multiplies it by $x$ ; $lo g x$ and $x^{2}$ are the wrong growth rates entirely.

Formal definition Intermediate+

Throughout, $s = σ + i t$ with $σ, t$ real, $lo g$ is the natural logarithm, $ζ$ is the Riemann zeta function of 06.01.16 with its Euler product for $σ > 1$ , and $ψ (x) = \sum_{n \leq x} Λ (n)$ is the Chebyshev function of 21.12.01, with $Λ$ the von Mangoldt function. We follow Davenport ^{[Davenport §17]}.

Definition (the two equivalent forms of the PNT). The prime number theorem is either of the asymptotic statements $$ \psi(x) \sim x \qquad (x \to \infty), \qquad\qquad \pi(x) \sim \frac{x}{\log x} \qquad (x \to \infty), $$ where $π (x) = \sum_{p \leq x} 1$ counts primes. The two are equivalent (proved below by partial summation), and a third equivalent form replaces $x / lo g x$ by the logarithmic integral $Li (x) = \int_{2}^{x} d t / lo g t$ , which is asymptotically larger and gives a better error term.

Definition (zero-free region). A zero-free region for $ζ$ is an open set $R \subset C$ on which $ζ (s) \neq = 0$ . The classical (de la Vallée Poussin) region is $$ \mathcal{R}_c = \Big{ s = \sigma + it : \sigma > 1 - \frac{c}{\log(|t| + 2)} \Big} $$ for an absolute constant $c > 0$ ; its left boundary hugs the line $σ = 1$ , narrowing as $∣ t ∣ \to \infty$ . The boundary case $σ = 1$ — the non-vanishing $ζ (1 + i t) \neq = 0$ for all real $t$ — is established in 06.01.16 and is the qualitative core; the quantitative width $c / lo g (∣ t ∣ + 2)$ is what converts $ψ (x) \sim x$ into a power-saving error term.

Definition (the basic trigonometric inequality). The non-vanishing on the line rests on the elementary inequality $$ 3 + 4\cos\theta + \cos 2\theta = 2(1 + \cos\theta)^2 \ge 0 \qquad (\theta \in \mathbb{R}). $$ Applied with $θ = t lo g p$ across the Euler product, it forces $ζ (s)$ away from zero near $σ = 1$ .

The symbols $ψ (x)$ , $π (x)$ , $Λ (n)$ , $s = σ + i t$ , the critical zeros $ρ$ , $Li (x)$ , and the region $R_{c}$ are recorded in _meta/NOTATION.md. A critical zero of $ζ$ is a zero with $0 < Re (ρ) < 1$ ; the elementary zeros are the points $s = - 2, - 4, \dots$ from the functional equation of 06.01.16.

Counterexamples to common slips

"The pole at $s = 1$ being caught is the whole proof." Catching the residue $x$ is only the main term. The proof is not finished until the four contour segments — the shifted vertical line and the two horizontal connectors and the truncation error — are each bounded below $x$ . Without the zero-free region those bounds are unavailable.
" $ζ (1 + i t) \neq = 0$ alone gives the error term." Mere non-vanishing on the line gives the qualitative $ψ (x) \sim x$ (the Newman-Wiener-Ikehara route). The power-saving error $O (x exp (- c lo g x))$ needs the quantitative width $σ > 1 - c / lo g (∣ t ∣ + 2)$ , because the contour can be shifted left only as far as the region permits.
" $π (x) \sim x / lo g x$ and $π (x) \sim Li (x)$ say the same thing with the same accuracy." They are equivalent as leading-order asymptotics, but $Li (x)$ is the true secondary approximation: $Li (x) = x / lo g x + x / (lo g x)^{2} + \dots$ , and the contour proof naturally produces $Li (x)$ , not $x / lo g x$ , in the error term.

Key theorem with proof Intermediate+

The signature result is the non-vanishing of $ζ$ on the line $Re (s) = 1$ , sharpened to the classical zero-free region. This is the analytic fact that, fed into the Perron contour, yields the prime number theorem. The argument is the $3 + 4 cos θ + cos 2 θ$ device of 06.01.16, here carried into a quantitative width. ^{[Davenport §18]}

Theorem (de la Vallée Poussin zero-free region). There is an absolute constant $c > 0$ such that $ζ (σ + i t) \neq = 0$ whenever $σ \geq 1 - c / lo g (∣ t ∣ + 2)$ . In particular $ζ (1 + i t) \neq = 0$ for all real $t$ .

Proof. For $σ > 1$ , the logarithm of the Euler product gives $$ \log|\zeta(\sigma + it)| = \operatorname{Re}\log\zeta(\sigma + it) = \sum_p \sum_{m=1}^{\infty} \frac{\cos(t m \log p)}{m, p^{m\sigma}}. $$ Form the combination $3 lo g ζ (σ) + 4 lo g ∣ ζ (σ + i t) ∣ + lo g ∣ ζ (σ + 2 i t) ∣$ . Term by term its summand is $$ \frac{1}{m, p^{m\sigma}}\big(3 + 4\cos(t m \log p) + \cos(2 t m \log p)\big) = \frac{2(1 + \cos(t m \log p))^2}{m, p^{m\sigma}} \ge 0, $$ using the identity $3 + 4 cos θ + cos 2 θ = 2 (1 + cos θ)^{2}$ . Summing over $p, m$ gives the key inequality $$ 3\log\zeta(\sigma) + 4\log|\zeta(\sigma + it)| + \log|\zeta(\sigma + 2it)| \ge 0, \quad\text{i.e.}\quad \zeta(\sigma)^3, |\zeta(\sigma + it)|^4, |\zeta(\sigma + 2it)| \ge 1. \tag{ $*$ } $$

Suppose, for contradiction, that $ζ (β + iγ) = 0$ with $β$ close to $1$ and $γ$ fixed, $γ \neq = 0$ . As $σ ↓ 1$ : the pole of $ζ$ at $s = 1$ gives $ζ (σ) \leq C / (σ - 1)$ ; the assumed zero gives $∣ ζ (σ + iγ) ∣ \leq C^{'} (σ - β)$ (the function vanishes at $β + iγ$ , so it is $O (σ - β)$ as $σ \to β$ ); and $ζ$ is bounded near $1 + 2 iγ$ , so $∣ ζ (σ + 2 iγ) ∣ \leq C^{''} lo g (∣ γ ∣ + 2)$ . Insert these into $(*)$ with $t = γ$ : $$ 1 \le \zeta(\sigma)^3 |\zeta(\sigma + i\gamma)|^4 |\zeta(\sigma + 2i\gamma)| \le \frac{C^3}{(\sigma - 1)^3},(C')^4(\sigma - \beta)^4, C''\log(|\gamma| + 2). $$ Write $σ = 1 + δ$ with $δ ↓ 0$ ; then $σ - β = δ + (1 - β)$ . The right side behaves like a constant times $lo g (∣ γ ∣ + 2) (δ + (1 - β))^{4} / δ^{3}$ . Minimising over $δ > 0$ (the minimum is near $δ ≍ 1 - β$ ) forces $$ 1 \lesssim \log(|\gamma| + 2),(1 - \beta), \qquad\text{hence}\qquad 1 - \beta \gtrsim \frac{1}{\log(|\gamma| + 2)}. $$ Thus any zero satisfies $β \leq 1 - c / lo g (∣ γ ∣ + 2)$ for an absolute $c > 0$ : no zero lies in $R_{c}$ . Taking $γ$ arbitrary and $δ \to 0$ also rules out $β = 1$ , giving $ζ (1 + i t) \neq = 0$ . $□$

Bridge. This zero-free region builds toward the contour proof of the prime number theorem: feeding $- ζ^{'} / ζ$ into the truncated Perron formula of 21.11.04 and shifting the line left to the boundary of $R_{c}$ is exactly the step where the residue at $s = 1$ supplies the main term $x$ , and the asymptotic $ψ (x) \sim x$ appears again in 21.12.01 as the leading term of the explicit formula. The foundational reason the inequality $(*)$ matters is that the pole of $ζ$ at $s = 1$ is so strong it cannot tolerate a nearby zero — this is exactly the mechanism by which an analytic feature of one function controls the distribution of primes, and putting these together with the contour shift generalises Chebyshev's $ψ (x) ≍ x$ of the prerequisite chain into the sharp $ψ (x) = x + O (x exp (- c lo g x))$ . The central insight is that the width $c / lo g (∣ t ∣ + 2)$ of the region is dual to the size $exp (- c lo g x)$ of the error: the bridge is the optimisation of the contour position against the truncation height $T$ .

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Prove the equivalence $ψ (x) \sim x ⟺ π (x) \sim x / lo g x$ via the auxiliary function $ϑ (x) = \sum_{p \leq x} lo g p$ , using $ψ (x) = ϑ (x) + O (x lo g^{2} x)$ and partial summation.

Hint

First show $ψ - ϑ$ collects only prime powers $p^{m}$ with $m \geq 2$ , bounded by $O (x lo g^{2} x)$ . Then relate $ϑ (x) = \sum_{p \leq x} lo g p$ to $π (x) = \sum_{p \leq x} 1$ by Abel summation against the weight $lo g p$ .

Answer

The difference $ψ (x) - ϑ (x) = \sum_{m \geq 2} \sum_{p^{m} \leq x} lo g p = \sum_{p \leq x} lo g p ⌊ lo g_{p} x ⌋ + \dots$ is dominated by $ϑ (x) lo g x + ϑ (x^{1/3}) lo g x + \dots = O (x lo g^{2} x)$ , since $ϑ (y) = O (y)$ by Chebyshev. Hence $ψ (x) \sim x ⟺ ϑ (x) \sim x$ . Now apply Abel summation to $ϑ (x) = \sum_{p \leq x} lo g p = \int_{2^{-}}^{x} lo g t d π (t) = π (x) lo g x - \int_{2}^{x} \frac{π ( t )}{t} d t$ . If $ϑ (x) \sim x$ , then $π (x) lo g x = ϑ (x) + \int_{2}^{x} π (t) / t d t$ ; the integral is $O (x / lo g x) = o (x)$ (using the crude $π (t) = O (t / lo g t)$ ), so $π (x) lo g x \sim x$ , i.e. $π (x) \sim x / lo g x$ . Conversely, partial summation of $ϑ (x) = π (x) lo g x - \int_{2}^{x} π (t) / t d t$ with $π (x) \sim x / lo g x$ gives $ϑ (x) \sim x$ . Rubric: full credit for bounding $ψ - ϑ$ , the Abel-summation identity, and the $o (x)$ control of the integral.

Exercise 4 (medium, symbolic).

From the key inequality $ζ (σ)^{3} ∣ ζ (σ + i t) ∣^{4} ∣ ζ (σ + 2 i t) ∣ \geq 1$ , deduce that $ζ (1 + i t) \neq = 0$ for $t \neq = 0$ , assuming only the simple pole of $ζ$ at $s = 1$ and continuity.

Hint

If $ζ (1 + i t_{0}) = 0$ , expand each factor as $σ ↓ 1$ : $ζ (σ) \sim 1/ (σ - 1)$ , $∣ ζ (σ + i t_{0}) ∣ = O (σ - 1)$ , and $ζ (σ + 2 i t_{0})$ stays bounded. Track the power of $(σ - 1)$ .

Answer

Suppose $ζ (1 + i t_{0}) = 0$ with $t_{0} \neq = 0$ . As $σ ↓ 1$ : the pole gives $ζ (σ) = (σ - 1)^{- 1} (1 + o (1))$ , so $ζ (σ)^{3} ≍ (σ - 1)^{- 3}$ ; the assumed simple zero gives $ζ (σ + i t_{0}) = (σ - 1) ζ^{'} (1 + i t_{0}) (1 + o (1))$ , so $∣ ζ (σ + i t_{0}) ∣^{4} ≍ (σ - 1)^{4}$ ; and $ζ (σ + 2 i t_{0}) \to ζ (1 + 2 i t_{0})$ , finite (the point $1 + 2 i t_{0} \neq = 1$ ). The product is then $≍ (σ - 1)^{- 3} (σ - 1)^{4} \cdot O (1) = O (σ - 1) \to 0$ . But the inequality demands the product be $\geq 1$ . Contradiction. Hence $ζ (1 + i t_{0}) \neq = 0$ . Rubric: full credit for the three asymptotic factors, the net power $(σ - 1)^{+ 1} \to 0$ , and the contradiction with the lower bound $1$ .

Exercise 5 (medium, symbolic).

The classical zero-free region permits shifting the contour to $σ = 1 - c / lo g T$ for the truncation height $T$ . On this line, $∣ x^{s} / s ∣ \leq x^{1 - c / l o g T} /∣ s ∣$ . Show that the size factor $x^{1 - c / l o g T}$ equals $x \cdot exp (- c lo g x / lo g T)$ , and identify the choice of $T$ that minimises the resulting error against a truncation term of order $x / T$ .

Hint

Write $x^{1 - c / l o g T} = x \cdot x^{- c / l o g T}$ and convert $x^{- c / l o g T}$ to an exponential. Balance $x exp (- c lo g x / lo g T)$ against $x lo g x / T$ by setting the two exponents comparable; try $lo g T = lo g x$ .

Answer

Since $x^{- c / l o g T} = exp (- \frac{c}{l o g T} lo g x)$ , the size factor is $x^{1 - c / l o g T} = x exp (- c lo g x / lo g T)$ . The shifted-line contribution is then of order $x exp (- c lo g x / lo g T)$ , while the truncated-Perron error is of order $x lo g x / T = x exp (lo g lo g x - lo g T)$ . Set $lo g T = λ lo g x$ . The shifted line gives exponent $- c lo g x / (λ lo g x) = - (c / λ) lo g x$ ; the truncation gives exponent $- λ lo g x + lo g lo g x$ . The two power-saving exponents match when $c / λ = λ$ , i.e. $λ = c$ , giving $T = exp (c lo g x)$ and total error $x exp (- c^{'} lo g x)$ for $c^{'} = c$ . Rubric: full credit for the exponential rewrite, the two competing error terms, and the optimisation $lo g T ≍ lo g x$ producing $exp (- c^{'} lo g x)$ .

Exercise 7 (hard, symbolic).

Sketch the full contour argument: starting from the truncated Perron formula $$ \psi(x) = \frac{1}{2\pi i}\int_{c - iT}^{c + iT} \Big(-\frac{\zeta'(s)}{\zeta(s)}\Big)\frac{x^s}{s},ds + R(x, T), $$ with $c = 1 + 1/ lo g x$ , replace the segment by the three sides of a rectangle reaching left to $σ = 1 - c_{0} / lo g T$ , and account for the residue at $s = 1$ and the four error contributions. State the final bound.

Hint

The rectangle has corners $c \pm i T$ and $1 - c_{0} / lo g T \pm i T$ . Inside sits the pole $s = 1$ (residue $x$ ). Bound: (i) the truncation error $R$ , (ii) the two horizontal sides at $Im (s) = \pm T$ , (iii) the shifted left vertical side, each using $∣ ζ^{'} / ζ ∣ ≪ lo g^{2} (∣ t ∣ + 2)$ in the region.

Answer

By the residue theorem the closed rectangular contour integral equals the residue at $s = 1$ , which is $x$ (the pole of $- ζ^{'} / ζ$ has strength $+ 1$ and $x^{s} / s ∣_{s = 1} = x$ ). Hence $$ \psi(x) = x - \frac{1}{2\pi i}\Big(\int_{\text{left}} + \int_{\text{top}} + \int_{\text{bottom}}\Big)\Big(-\frac{\zeta'}{\zeta}\Big)\frac{x^s}{s},ds + R(x, T). $$ In the zero-free region $σ \geq 1 - c_{0} / lo g (∣ t ∣ + 2)$ one has the standard bound $∣ ζ^{'} (s) / ζ (s) ∣ ≪ lo g^{2} (∣ t ∣ + 2)$ . Then: (i) the truncated Perron error is $R ≪ x lo g^{2} x / T$ ; (ii) each horizontal side at height $T$ has length $O (1)$ in $σ$ and integrand $≪ x^{c} lo g^{2} T / T ≪ x lo g^{2} T / T$ ; (iii) the left vertical side at $σ_{0} = 1 - c_{0} / lo g T$ has integrand $≪ x^{σ_{0}} lo g^{2} T \int_{- T}^{T} d t /∣ s ∣ ≪ x^{σ_{0}} lo g^{3} T = x exp (- c_{0} lo g x / lo g T) lo g^{3} T$ . Summing, $ψ (x) = x + O (x lo g^{2} x / T) + O (x exp (- c_{0} lo g x / lo g T) lo g^{3} T)$ . Choosing $lo g T = lo g x$ (Exercise 5) balances the two and yields $$ \psi(x) = x + O!\big(x\exp(-c'\sqrt{\log x})\big). $$ Rubric: full credit for the residue $x$ , the four error pieces, the $∣ ζ^{'} / ζ ∣ ≪ lo g^{2}$ input, and the optimisation producing the de la Vallée Poussin error.

Exercise 8 (hard, symbolic).

Newman's Tauberian route. Set $Φ (s) = \sum_{p} (lo g p) p^{- s} = - ζ^{'} (s) / ζ (s) + (holomorphic on σ \geq 1)$ . Granting $ζ (1 + i t) \neq = 0$ , explain why $Φ (s) - 1/ (s - 1)$ extends holomorphically to a neighbourhood of $σ \geq 1$ , and state the Tauberian step that upgrades this to $\int_{1}^{\infty} (ϑ (x) - x) x^{- 2} d x$ convergent, hence $ϑ (x) \sim x$ .

Hint

$Φ$ inherits the pole of $- ζ^{'} / ζ$ at $s = 1$ (residue $1$ ) and the poles at zeros of $ζ$ ; non-vanishing on $σ = 1$ removes boundary poles. Newman's analytic lemma converts holomorphy of $\int_{0}^{\infty} (ϑ (e^{t}) e^{- t} - 1) e^{- z t} d t$ up to $Re z = 0$ into convergence at $z = 0$ .

Answer

For $σ > 1$ , $- ζ^{'} (s) / ζ (s) = \sum_{n} Λ (n) n^{- s}$ , and $Φ (s) = \sum_{p} (lo g p) p^{- s}$ differs from it only by the prime-power tail $\sum_{m \geq 2}$ , which converges for $σ > 1/2$ and is holomorphic there. So $Φ$ has the same poles as $- ζ^{'} / ζ$ on $σ \geq 1$ : a simple pole at $s = 1$ with residue $1$ (the pole of $ζ$ ), and a pole at each zero $ρ$ of $ζ$ on the line $σ = 1$ . The hypothesis $ζ (1 + i t) \neq = 0$ removes the latter, so $Φ (s) - 1/ (s - 1)$ is holomorphic on an open set containing ${σ \geq 1}$ . Newman's analytic theorem: if $f (t) = ϑ (e^{t}) e^{- t} - 1$ is bounded (Chebyshev: $ϑ (x) = O (x)$ ) and $g (z) = \int_{0}^{\infty} f (t) e^{- z t} d t$ , holomorphic for $Re z > 0$ , extends holomorphically to $Re z \geq 0$ , then $\int_{0}^{\infty} f (t) d t$ converges and equals $g (0)$ . Here $g (z) = Φ (z + 1) / (z + 1) - 1/ z$ is exactly that extension, so $\int_{1}^{\infty} (ϑ (x) - x) x^{- 2} d x$ converges. Convergence of this integral with $ϑ$ nondecreasing forces $ϑ (x) \sim x$ (else a fixed-sign chunk would make the tail diverge), hence $ψ (x) \sim x$ . Rubric: full credit for the prime-power-tail holomorphy, the boundary-pole removal by non-vanishing, the statement of Newman's lemma, and the monotonicity Tauberian step.

Advanced results Master

Theorem 1 (PNT with the de la Vallée Poussin error). Combining the truncated Perron formula of 21.11.04 for $- ζ^{'} / ζ$ with the zero-free region $R_{c}$ and the bound $∣ ζ^{'} / ζ ∣ ≪ lo g^{2} (∣ t ∣ + 2)$ on its boundary, the contour shift past $s = 1$ yields ^{[Davenport §18]} $$ \psi(x) = x + O!\big(x\exp(-c'\sqrt{\log x})\big), \qquad \pi(x) = \operatorname{Li}(x) + O!\big(x\exp(-c'\sqrt{\log x})\big), $$ the second following by partial summation against $lo g p$ . The error is super-polynomially small in $lo g x$ but weaker than any power of $x$ , a direct image of the logarithmic narrowing of $R_{c}$ .

Theorem 2 (Newman's Tauberian proof). Granting only $ζ (1 + i t) \neq = 0$ on the line ^{[Newman 1980]}, the function $Φ (s) = \sum_{p} (lo g p) p^{- s}$ satisfies that $Φ (s) - 1/ (s - 1)$ extends holomorphically across $σ = 1$ , and Newman's analytic lemma — a single contour integral over a finite circular arc with a convergence-forcing kernel $e^{z T} (1/ z + z / R^{2})$ — converts this into the convergence of $\int_{1}^{\infty} (ϑ (x) - x) x^{- 2} d x$ . Monotonicity of $ϑ$ then forces $ϑ (x) \sim x$ , hence the PNT. This proof uses no zero-free region and no explicit formula; it produces no error term, trading quantitative strength for brevity ^{[Korevaar 1982]}.

Theorem 3 (equivalence of the PNT forms). The statements $ψ (x) \sim x$ , $ϑ (x) \sim x$ , $π (x) \sim x / lo g x$ , and $π (x) \sim Li (x)$ are mutually equivalent; $ψ - ϑ = O (x lo g^{2} x)$ links the first two, and Abel summation against $lo g p$ links $ϑ$ to $π$ . The error terms are not equivalent: the Perron contour produces $Li (x)$ , and only $Li (x)$ — not $x / lo g x$ — carries the genuine de la Vallée Poussin error without losing a factor $x / (lo g x)^{2}$ .

Theorem 4 (the Wiener-Ikehara theorem). The Tauberian content of Theorem 2 is the Wiener-Ikehara theorem: if $A (s) = \sum_{n} a_{n} n^{- s}$ has non-negative coefficients, converges for $σ > 1$ , and $A (s) - C / (s - 1)$ extends continuously to $σ \geq 1$ , then $\sum_{n \leq x} a_{n} \sim C x$ ^{[Korevaar 1982]}. Applied to $A = - ζ^{'} / ζ$ with $C = 1$ it gives $ψ (x) \sim x$ . Wiener-Ikehara is the abstract statement that boundary non-vanishing alone — without any region of width — already pins the leading asymptotic.

Theorem 5 (error term and the Riemann hypothesis). The exponent $lo g x$ is an artefact of the logarithmic region. A zero-free region of the Vinogradov-Korobov shape $σ > 1 - c / (lo g ∣ t ∣)^{2/3} (lo g lo g ∣ t ∣)^{1/3}$ improves the error to $x exp (- c (lo g x)^{3/5} (lo g lo g x)^{- 1/5})$ ^{[Montgomery-Vaughan §6.3]}; the Riemann hypothesis $Re (ρ) = \frac{1}{2}$ collapses it to $ψ (x) = x + O (x^{1/2} lo g^{2} x)$ , the best possible, equivalent to $π (x) = Li (x) + O (x^{1/2} lo g x)$ .

Synthesis. The prime number theorem is the foundational reason analytic number theory exists as a discipline: it is the first deep arithmetic fact extracted from a single contour integral, and the central insight is that one pole of $ζ$ at $s = 1$ supplies the main term $x$ while the zeros supply only correction waves. This is exactly the bridge from the elementary $ψ (x) ≍ x$ of the Chebyshev chain to the analytic $ψ (x) \sim x$ : where the elementary theory squeezed $ψ$ between constant multiples of $x$ , the contour reads the residue at $s = 1$ and pins the constant to $1$ , and putting these together with the $3 + 4 cos θ + cos 2 θ$ inequality — which forbids a zero on $Re (s) = 1$ — converts the qualitative squeeze into the quantitative error $O (x exp (- c lo g x))$ . The central insight recurs in the duality between region and error: the width $c / lo g (∣ t ∣ + 2)$ of the zero-free region is dual to the size $exp (- c lo g x)$ of the remainder, and this is exactly the same duality the Riemann hypothesis sharpens to $x^{1/2}$ . The Newman-Tauberian proof generalises the same boundary non-vanishing into the Wiener-Ikehara theorem, showing that the leading asymptotic needs only the line $Re (s) = 1$ , while the contour proof needs the region; the bridge between them is that both read the pole at $s = 1$ as the source of $x$ , and the modern theory of $L$ -functions is the systematic repetition of this single contour computation for every Dirichlet series whose primes one wishes to count.

Full proof set Master

Proposition 1 (the key inequality $(*)$ ). For $σ > 1$ and real $t$ , $ζ (σ)^{3} ∣ ζ (σ + i t) ∣^{4} ∣ ζ (σ + 2 i t) ∣ \geq 1$ .

Proof. For $σ > 1$ , $lo g ∣ ζ (σ + i t) ∣ = \sum_{p} \sum_{m \geq 1} (m p^{mσ})^{- 1} cos (t m lo g p)$ , the real part of the absolutely convergent $lo g ζ$ . Then $$ 3\log\zeta(\sigma) + 4\log|\zeta(\sigma + it)| + \log|\zeta(\sigma + 2it)| = \sum_{p}\sum_{m \ge 1}\frac{3 + 4\cos(\theta) + \cos(2\theta)}{m p^{m\sigma}}, \quad \theta = t m\log p. $$ Each numerator is $2 (1 + cos θ)^{2} \geq 0$ , so the whole sum is $\geq 0$ . Exponentiating gives $ζ (σ)^{3} ∣ ζ (σ + i t) ∣^{4} ∣ ζ (σ + 2 i t) ∣ \geq 1$ . $□$

Proposition 2 (non-vanishing on the line). $ζ (1 + i t) \neq = 0$ for every real $t \neq = 0$ .

Proof. Suppose $ζ (1 + i t_{0}) = 0$ , $t_{0} \neq = 0$ . As $σ ↓ 1$ : $ζ (σ) = (σ - 1)^{- 1} + O (1)$ , so $ζ (σ)^{3} = (σ - 1)^{- 3} (1 + o (1))$ ; the zero gives $ζ (σ + i t_{0}) = O (σ - 1)$ , so $∣ ζ (σ + i t_{0}) ∣^{4} = O ((σ - 1)^{4})$ ; and $∣ ζ (σ + 2 i t_{0}) ∣ = O (1)$ since $1 + 2 i t_{0} \neq = 1$ . The product is $O ((σ - 1)^{- 3} (σ - 1)^{4}) = O (σ - 1) \to 0$ , contradicting Proposition 1's lower bound $1$ . Hence no zero on the line away from $s = 1$ , and the pole at $s = 1$ is not a zero. $□$

Proposition 3 (quantitative width). There is $c > 0$ with $ζ (σ + i t) \neq = 0$ for $σ \geq 1 - c / lo g (∣ t ∣ + 2)$ .

Proof. Let $β + iγ$ be a zero with $γ$ fixed, $∣ γ ∣ \geq 1$ . Using $ζ^{'} / ζ (σ) \leq (σ - 1)^{- 1} + A$ , $Re ζ^{'} / ζ (σ + iγ) \leq A lo g (∣ γ ∣ + 2) - (σ - β)^{- 1}$ (the zero contributes $- (σ - β)^{- 1}$ to the real part of the logarithmic derivative), and $Re ζ^{'} / ζ (σ + 2 iγ) \leq A lo g (∣ γ ∣ + 2)$ , the logarithmic-derivative form of $(*)$ , $$ 3\operatorname{Re}\frac{\zeta'}{\zeta}(\sigma) + 4\operatorname{Re}\frac{\zeta'}{\zeta}(\sigma + i\gamma) + \operatorname{Re}\frac{\zeta'}{\zeta}(\sigma + 2i\gamma) \le 0 $$ (the sign-flipped $(*)$ for $- ζ^{'} / ζ$ , whose coefficients are $\geq 0$ ) gives $$ \frac{4}{\sigma - \beta} \le \frac{3}{\sigma - 1} + A'\log(|\gamma| + 2). $$ Put $σ = 1 + δ$ , $δ = λ / lo g (∣ γ ∣ + 2)$ for a small constant $λ$ . Then $4/ (σ - β) \leq 3/ δ + A^{'} lo g (∣ γ ∣ + 2)$ , and solving for $σ - β$ shows $σ - β \geq$ a positive multiple of $1/ lo g (∣ γ ∣ + 2)$ , whence $1 - β \geq c / lo g (∣ γ ∣ + 2)$ for absolute $c > 0$ . $□$

Proposition 4 (equivalence $ψ \sim x ⟺ π \sim x / lo g x$ ). The two asymptotics are equivalent.

Proof. As in Exercise 3: $ψ (x) - ϑ (x) = O (x lo g^{2} x) = o (x)$ , so $ψ (x) \sim x ⟺ ϑ (x) \sim x$ . Abel summation gives $ϑ (x) = π (x) lo g x - \int_{2}^{x} π (t) / t d t$ and $π (x) = ϑ (x) / lo g x + \int_{2}^{x} ϑ (t) / (t lo g^{2} t) d t$ . With the crude Chebyshev bounds the integrals are $o (x / lo g x)$ relative to the main terms, so $ϑ (x) \sim x ⟺ π (x) lo g x \sim x ⟺ π (x) \sim x / lo g x$ . $□$

Proposition 5 (Newman's analytic lemma). Let $f : [0, \infty) \to R$ be bounded and locally integrable, and let $g (z) = \int_{0}^{\infty} f (t) e^{- z t} d t$ for $Re z > 0$ . If $g$ extends holomorphically to an open set containing ${Re z \geq 0}$ , then $\int_{0}^{\infty} f (t) d t$ converges and equals $g (0)$ .

Proof. For $R > 0$ and a contour $Γ$ = the boundary of ${∣ z ∣ \leq R, Re z \geq - δ}$ (the $δ$ chosen so $g$ is holomorphic there), set $g_{T} (z) = \int_{0}^{T} f (t) e^{- z t} d t$ , entire. By Cauchy, $$ g(0) - g_T(0) = \frac{1}{2\pi i}\int_\Gamma (g(z) - g_T(z)),e^{zT}\Big(\frac1z + \frac{z}{R^2}\Big),dz. $$ On the right semicircle $Re z > 0$ , the integrand is $O (∥ f ∥_{\infty} / R)$ after using $∣ g (z) - g_{T} (z) ∣ \leq ∥ f ∥_{\infty} e^{- T Re z} / Re z$ and the kernel bound $∣ e^{z T} (1/ z + z / R^{2}) ∣ = e^{T Re z} 2 Re z / R^{2}$ ; this contributes $O (∥ f ∥_{\infty} / R)$ . On the left part, $g_{T}$ is entire so its integral is bounded by $O (∥ f ∥_{\infty} / R)$ by the same kernel, while the $g$ -piece $\to 0$ as $T \to \infty$ for fixed $R$ by holomorphy and dominated decay. Letting $T \to \infty$ then $R \to \infty$ , $g_{T} (0) = \int_{0}^{T} f \to g (0)$ . $□$

Connections Master

The truncated Perron formula and the discontinuous-integral switch are imported wholesale from 21.11.04; this unit is the application that the Perron machinery was built for, feeding $A (s) = - ζ^{'} / ζ$ into the contour and shifting the line past $s = 1$ , so the abstract inversion theorem of that unit becomes the concrete asymptotic $ψ (x) \sim x$ here.
The von Mangoldt identity $- ζ^{'} (s) / ζ (s) = \sum_{n} Λ (n) n^{- s}$ , the residue $+ 1$ at $s = 1$ , and the explicit formula $ψ (x) = x - \sum_{ρ} x^{ρ} / ρ - \dots$ are exactly those of 21.12.01; this unit supplies the analytic input — the zero-free region — that bounds the zero-sum in that explicit formula and turns it into a theorem with an error term.
The analytic continuation, functional equation, simple pole at $s = 1$ , and the non-vanishing $ζ (1 + i t) \neq = 0$ via $3 + 4 cos θ + cos 2 θ$ are the facts about $ζ$ established in 06.01.16; the contour proof reads the pole there as the main term $x$ and the line non-vanishing as the guarantee that the shifted contour stays in a zero-free strip.
The two-track strategy — a contour proof with error term and a Tauberian proof without — connects to the general Tauberian theory whose Riemann zeta instance is 21.03.01; the Wiener-Ikehara theorem is the abstract version of the boundary-non-vanishing argument and recurs for every $L$ -function in 21.03.02.

Historical & philosophical context Master

The prime number theorem was conjectured in the form $π (x) \sim x / lo g x$ by Legendre and, more precisely as $Li (x)$ , by Gauss in the 1790s from extensive tables. Riemann's 1859 memoir laid the analytic programme — the continuation of $ζ$ , the product over the zeros, the contour representation of the prime count — but did not prove the theorem; the missing analytic estimates, above all the non-vanishing of $ζ$ on the line $Re (s) = 1$ , were supplied independently in 1896 by Jacques Hadamard ^{[Hadamard 1896]} and Charles-Jean de la Vallée Poussin ^{[de la Vallée Poussin 1896]}. Hadamard's route went through his theory of entire functions of finite order applied to $ξ$ ; de la Vallée Poussin's went through the explicit region $σ > 1 - c / lo g (∣ t ∣ + 2)$ and produced the first error term, the $O (x exp (- c lo g x))$ that still bears his name. The $3 + 4 cos θ + cos 2 θ$ device that makes the non-vanishing elementary is due to Franz Mertens (1898), simplifying both original arguments.

For most of a century the proof was held to require complex analysis essentially; Atle Selberg and Paul Erdős found an elementary (real-variable) proof in 1949, which sharpened the sense of what the zeros contribute without displacing the analytic method as the source of error terms. The modern short proof is Donald Newman's 1980 contour argument ^{[Newman 1980]}, which replaces the zero-free region and the explicit formula by a single application of an analytic lemma to $\sum_{p} (lo g p) p^{- s}$ , exposited and extended by Jacob Korevaar ^{[Korevaar 1982]} into the Wiener-Ikehara circle. Davenport's Multiplicative Number Theory and Montgomery-Vaughan's text are the canonical modern treatments of the contour proof with error.

Bibliography Master

@article{hadamard1896,
  author  = {Hadamard, Jacques},
  title   = {Sur la distribution des z\'{e}ros de la fonction $\zeta(s)$ et ses cons\'{e}quences arithm\'{e}tiques},
  journal = {Bulletin de la Soci\'{e}t\'{e} Math\'{e}matique de France},
  volume  = {24},
  pages   = {199--220},
  year    = {1896}
}

@article{valleepoussin1896,
  author  = {de la Vall\'{e}e Poussin, Charles-Jean},
  title   = {Recherches analytiques sur la th\'{e}orie des nombres premiers},
  journal = {Annales de la Soci\'{e}t\'{e} scientifique de Bruxelles},
  volume  = {20},
  pages   = {183--256, 281--397},
  year    = {1896}
}

@article{newman1980,
  author  = {Newman, Donald J.},
  title   = {Simple analytic proof of the prime number theorem},
  journal = {American Mathematical Monthly},
  volume  = {87},
  number  = {9},
  pages   = {693--696},
  year    = {1980}
}

@article{korevaar1982,
  author  = {Korevaar, Jacob},
  title   = {On Newman's quick way to the prime number theorem},
  journal = {The Mathematical Intelligencer},
  volume  = {4},
  number  = {3},
  pages   = {108--115},
  year    = {1982}
}

@book{davenport2000,
  author    = {Davenport, Harold},
  title     = {Multiplicative Number Theory},
  edition   = {3},
  series    = {Graduate Texts in Mathematics},
  volume    = {74},
  publisher = {Springer-Verlag},
  year      = {2000},
  note      = {Revised by H. L. Montgomery; \S17--18: the contour proof of the PNT with error}
}

@book{montgomeryvaughan2007,
  author    = {Montgomery, Hugh L. and Vaughan, Robert C.},
  title     = {Multiplicative Number Theory I: Classical Theory},
  series    = {Cambridge Studies in Advanced Mathematics},
  volume    = {97},
  publisher = {Cambridge University Press},
  year      = {2007},
  note      = {\S6.2--6.3: zero-free region and the de la Vall\'{e}e Poussin error}
}

Prerequisites

21.12.01
21.11.04
06.01.16

Tier anchors

beginner: Derbyshire 2003 *Prime Obsession* (Joseph Henry Press) Ch. 19-22 (the explicit formula and the role of the line $\operatorname{Re}(s)=1$ described informally); 3Blue1Brown 'Pi hiding in prime regularities' for the spectral picture of $\psi(x)$ as a main term plus zero-waves
intermediate: Davenport 2000 *Multiplicative Number Theory* 3e (Springer GTM 74) §17-18 (the truncated Perron integral for $\psi(x)$, the contour shift, the zero-free region, and the de la Vallée Poussin error term); Apostol 1976 *Introduction to Analytic Number Theory* (Springer UTM) Ch. 13
master: Davenport 2000 *Multiplicative Number Theory* 3e §17-18 (the full PNT contour proof with error term); Montgomery-Vaughan 2007 *Multiplicative Number Theory I* (Cambridge UP) §6.2-6.3 (zero-free region, $\psi(x)=x+O(x\exp(-c\sqrt{\log x}))$); Newman 1980 *Amer. Math. Monthly* 87 (the short Tauberian proof); Korevaar 1982 *Math. Intelligencer* 4 (Newman's method); de la Vallée Poussin 1896; Hadamard 1896

References

Davenport, H. — Multiplicative Number Theory · Springer Graduate Texts in Mathematics 74, 3rd edition (2000), revised by H. L. Montgomery. §17-18. Carries out the contour proof of the prime number theorem: applies the truncated Perron formula to $-\zeta'/\zeta$, shifts the line of integration to the left of $\operatorname{Re}(s)=1$ using the classical zero-free region $\sigma > 1 - c/\log(|t|+2)$, bounds the horizontal and shifted-vertical segments, and obtains $\psi(x)=x+O(x\exp(-c\sqrt{\log x}))$, equivalently $\pi(x)=\operatorname{Li}(x)+O(x\exp(-c\sqrt{\log x}))$.
de la Vallée Poussin, Ch.-J. — Recherches analytiques sur la théorie des nombres premiers · Annales de la Société scientifique de Bruxelles 20 (1896), 183-256, 281-397. Establishes the non-vanishing of $\zeta(1+it)$ and the quantitative zero-free region $\sigma > 1 - c/\log(|t|+2)$, then derives the prime number theorem with an explicit error term by contour integration of the logarithmic derivative.
Hadamard, J. — Sur la distribution des zéros de la fonction $\zeta(s)$ et ses conséquences arithmétiques · Bulletin de la Société Mathématique de France 24 (1896), 199-220. Independent proof of the prime number theorem from the non-vanishing of $\zeta$ on the line $\operatorname{Re}(s)=1$, using the Hadamard product theory for entire functions of finite order.
Newman, D. J. — Simple analytic proof of the prime number theorem · American Mathematical Monthly 87 (1980), no. 9, 693-696. A short Tauberian proof that bypasses the zero-free region and the explicit formula: it uses only $\zeta(1+it)\neq 0$ and an analytic-continuation contour lemma to deduce convergence of $\int_1^\infty (\psi(x)-x)x^{-2}\,dx$, whence $\psi(x)\sim x$.
Korevaar, J. — On Newman's quick way to the prime number theorem · Mathematical Intelligencer 4 (1982), no. 3, 108-115. Expands Newman's contour argument into the Wiener-Ikehara circle, giving the cleanest modern presentation of the analytic Tauberian route to $\psi(x)\sim x$.
Montgomery, H. L. & Vaughan, R. C. — Multiplicative Number Theory I: Classical Theory · Cambridge Studies in Advanced Mathematics 97 (2007), §6.2-6.3. The zero-free region via the $3+4\cos\theta+\cos2\theta$ inequality, the explicit formula with error term, and the contour-integral proof of the prime number theorem with the de la Vallée Poussin error $O(x\exp(-c\sqrt{\log x}))$.

Estimated time

beginner: 18m
intermediate: 48m
master: 90m