21.12.04 · number-theory / prime-number-theorem

The Riemann-von Mangoldt Explicit Formula

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Anchor (Master): Davenport 2000 *Multiplicative Number Theory* 3e §17 (the exact and truncated explicit formulas, the residue sum over the critical zeros); Montgomery-Vaughan 2007 *Multiplicative Number Theory I* (Cambridge UP) §12.1, §13.1-13.2 (the explicit formula and the zero-counting $N(T)$); Edwards 1974 *Riemann's Zeta Function* (Academic Press) Ch. 3 (Riemann's $J(x)$ and $\pi(x)$ via $\operatorname{Li}$); Riemann 1859; von Mangoldt 1895; Backlund 1918 (the remainder in $N(T)$)

Intuition Beginner

The previous units handed us a weighted prime count $ψ (x)$ , the running total of $lo g p$ over every prime power up to $x$ , and showed that it grows like $x$ . The explicit formula is the exact bookkeeping behind that approximate statement. It says: take the simple straight-line growth $x$ , then subtract one small correction wave for every place the zeta function vanishes. Add the main growth to all the waves and you do not get an estimate of the prime count — you get the prime count exactly, on the nose.

Think of a struck bell. The pure tone is the main term $x$ . The overtones, the faint ringing on top, are the waves coming from the zeros. Each zero of the zeta function sits at a complex location, and that location fixes both the pitch and the loudness of its wave. The whole subtlety of the primes is encoded as the spectrum of one sound. This is the precise sense in which "the primes are the music of the zeros."

There is a second formula of the same kind that counts something else: how many zeros the zeta function has up to a given height. This zero-counting formula grows in a clean, predictable way, roughly a height times the logarithm of that height. The two formulas are two halves of one dictionary. One says the zeros govern the primes; the other says exactly how many zeros there are to do the governing.

Visual Beginner

Picture the staircase of $ψ (x)$ , jumping up by $lo g p$ at each prime power, overlaid on the straight diagonal line $y = x$ . The staircase wiggles around the diagonal. The explicit formula explains every wiggle: each zero of the zeta function adds one gentle wave, and the waves sum to the exact gap between the staircase and the line.

   psi(x)
     ^                                   staircase psi(x)
     |                              _____/
     |                        _____/   <- jumps by log p
     |                  _____/
     |            _____/    .......... diagonal y = x
     |      _____/    wiggle = sum of zero-waves
     |_____/
     +-------------------------------------> x

   adding the waves:   x  +  wave(rho_1)  +  wave(rho_2) + ...  =  psi(x) exactly

$T$ (height)	$N (T)$ , zeros up to height $T$	$(T /2 π) lo g (T /2 π) - T /2 π$
$25$	$0$	$0.0$
$50$	$10$	$9.7$
$100$	$29$	$28.1$

The right column is the smooth Riemann-von Mangoldt estimate; the middle column is the true zero count. They track each other closely, the numerical shadow of the zero-counting formula.

Worked example Beginner

Let us see, with the smallest pieces, how the zeros build the wiggle in $ψ (x)$ . The explicit formula reads $$ \psi(x) = x - (\text{a wave for each zero}) - (\text{tiny fixed leftovers}). $$

Step 1. Locate the lowest zeros. The zeta function's first vanishing in the upper half-plane sits at height about $14.13$ ; it comes paired with its mirror image at height $- 14.13$ . Together this pair makes the first and loudest wave.

Step 2. Size the wave. A zero on the critical line contributes a wave whose height is the square root of $x$ divided by the height of the zero. At $x = 100$ the square root is $10$ , and the zero height is about $14.13$ , so the first wave has size about $10/14.13 \approx 0.71$ — a small ripple on top of the main term $100$ .

Step 3. Add the next pair. The second pair of zeros sits at height about $21.0$ , giving a wave of size about $10/21.0 \approx 0.48$ . The waves shrink as the zeros climb, so the first few already capture most of the wiggle.

What this tells us: the main term $x = 100$ does the heavy lifting, and the zeros add ever-smaller ripples, $0.71$ , then $0.48$ , then less. The prime count is the diagonal plus a quickly fading chorus of waves, one per zero.

Check your understanding Beginner

Exercise (easy, multiple choice).

In the explicit formula $ψ (x) = x - (waves) - (leftovers)$ , what does the leading term $x$ represent?

A. the contribution of the first zero

B. the smooth main growth, coming from the single pole of zeta at the point $1$

C. the number of primes up to $x$

D. the height of the first zero

Hint

The straight diagonal line $y = x$ is what the staircase hugs; it comes from the one place where the zeta function blows up, not from a zero.

Answer

B. The term $x$ is the smooth main growth, produced by the single pole (blow-up point) of the zeta function at $1$ ; the zeros only add the correction waves on top. Feedback-correct: pole gives the main term, zeros give the ripples — that split is the whole content of the formula. Feedback-wrong: the first zero and the leftovers are corrections, not the main growth, and the raw prime count $π (x)$ is a different (smaller) function.

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 21.12.01 with its simple pole at $s = 1$ , $ψ (x) = \sum_{n \leq x} Λ (n)$ is the Chebyshev function with $Λ$ the von Mangoldt function, and $ρ = β + iγ$ denotes a critical zero of $ζ$ (a zero with $0 < Re (ρ) < 1$ ). The partial-fraction expansion of $ζ^{'} / ζ$ and the Hadamard product for the completed zeta $ξ$ are those established in 21.12.01 and 06.01.17. We follow Davenport ^{[Davenport §17]}.

Definition (the exact explicit formula). The Riemann-von Mangoldt explicit formula is the identity, valid for $x > 1$ not a prime power, $$ \psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \log(2\pi) - \frac{1}{2}\log!\big(1 - x^{-2}\big), $$ the sum taken over the critical zeros $ρ$ in order of increasing $∣ Im ρ ∣$ (equivalently, paired as $ρ, \overset{ρ}{ˉ}$ ). The term $x$ is the residue of the Perron integrand $- \frac{ζ ^{'} ( s )}{ζ ( s )} \frac{x ^{s}}{s}$ at the pole $s = 1$ ; each $- x^{ρ} / ρ$ is its residue at a zero $ρ$ ; $- lo g (2 π) = - ζ^{'} (0) / ζ (0)$ is the residue at $s = 0$ ; and $- \frac{1}{2} lo g (1 - x^{- 2}) = \sum_{k \geq 1} x^{- 2 k} / (2 k)$ collects the residues at the elementary zeros $s = - 2, - 4, \dots$ .

Definition (the truncated explicit formula). For a truncation height $T \geq 2$ , the truncated explicit formula restricts the zero-sum to $∣ Im ρ ∣ \leq T$ at the cost of an error term: $$ \psi(x) = x - \sum_{|\operatorname{Im}\rho| \le T} \frac{x^\rho}{\rho} + O!\left(\frac{x \log^2(xT)}{T} + \log x\right). $$ This is the form used for estimates: choosing $T$ as a function of $x$ trades the length of the zero-sum against the size of the remainder.

Definition (the Riemann-von Mangoldt zero-counting formula). Let $N (T)$ count the zeros $ρ = β + iγ$ of $ζ$ with $0 < β < 1$ and $0 < γ \leq T$ . Then $$ N(T) = \frac{T}{2\pi}\log\frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T). $$ The main term measures the density of zeros: the average gap between consecutive heights $γ$ near $T$ is about $2 π / lo g T$ , so zeros become denser as one climbs.

The symbols $ψ (x)$ , $Λ (n)$ , $s = σ + i t$ , the critical zeros $ρ = β + iγ$ , the completed zeta $ξ (s)$ , the counting function $N (T)$ , the logarithmic integral $Li (x)$ , and Riemann's prime-counting $J (x)$ are recorded in _meta/NOTATION.md. The elementary zeros are the points $s = - 2, - 4, \dots$ from the functional equation of 21.12.01.

Counterexamples to common slips

"The zero-sum $\sum_{ρ} x^{ρ} / ρ$ converges absolutely." It does not. The series is only conditionally convergent, and only when the zeros are paired $ρ, \overset{ρ}{ˉ}$ (or summed in order of increasing $∣ γ ∣$ ). Pairing $ρ = β + iγ$ with $\overset{ρ}{ˉ} = β - iγ$ makes $x^{ρ} / ρ + x^{\overset{ρ}{ˉ}} / \overset{ρ}{ˉ}$ real and of size $O (x^{β} lo g γ / γ)$ , summable against the zero density $d N \sim lo g γ d γ$ only because of the cancellation.
" $N (T)$ counts zeros on the critical line." $N (T)$ counts all critical-strip zeros up to height $T$ , with no assumption that they lie on $Re (s) = \frac{1}{2}$ . The Riemann hypothesis is the separate claim $β = \frac{1}{2}$ for every such zero; the formula for $N (T)$ holds unconditionally.
"The explicit formula proves the prime number theorem by itself." The identity is exact but unconditional, and a zero with $β = 1$ would make its wave as large as $x$ . The PNT needs the extra input $ζ (1 + i t) \neq = 0$ of 21.12.02, which forces every $β < 1$ and so every wave to be $o (x)$ .

Key theorem with proof Intermediate+

The signature result is the Riemann-von Mangoldt zero-counting formula. It is the companion to the explicit formula: the explicit formula expresses primes through the zeros, and $N (T)$ counts those zeros, supplying the density input that bounds the conditionally convergent zero-sum. The proof is the argument principle applied to the completed zeta $ξ$ , whose zeros are exactly the critical zeros of $ζ$ 06.01.17. ^{[Davenport §17]}

Theorem (Riemann-von Mangoldt zero-counting formula). Let $N (T)$ count, with multiplicity, the zeros $ρ = β + iγ$ of $ζ$ with $0 < γ \leq T$ (and $0 < β < 1$ ). Then for $T \geq 2$ , $$ N(T) = \frac{T}{2\pi}\log\frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T). $$

Proof. Recall the completed zeta $ξ (s) = \frac{1}{2} s (s - 1) π^{- s /2} Γ (s /2) ζ (s)$ of 06.01.17, entire of order one, with $ξ (s) = ξ (1 - s)$ , whose zeros are precisely the critical zeros of $ζ$ . Apply the argument principle to $ξ$ on the rectangle $R$ with vertices $2 - i T$ , $2 + i T$ , $- 1 + i T$ , $- 1 - i T$ (taking $T$ to avoid the ordinate of any zero). The number of zeros inside is $$ N(T) = \frac{1}{2\pi},\Delta_R \arg\xi(s), $$ the total change of argument around $\partial R$ , divided by $2 π$ . By the symmetry $ξ (s) = ξ (1 - s)$ and the reflection $ξ (\overset{s}{ˉ}) = \overline{ξ (s)}$ , the contour splits into two halves contributing equally, so $Δ_{R} ar g ξ = 2 Δ_{L} ar g ξ$ where $L$ runs from $2$ up to $2 + i T$ , then to $\frac{1}{2} + i T$ .

Write $ξ (s) = \frac{1}{2} s (s - 1) π^{- s /2} Γ (s /2) ζ (s)$ and track the argument of each factor along $L$ . The algebraic factor and $π^{- s /2}$ contribute $O (1)$ . Stirling's formula for $Γ$ gives $$ \Delta_L \arg!\Big(\pi^{-s/2}\Gamma!\big(\tfrac{s}{2}\big)\Big) = \frac{T}{2}\log\frac{T}{2\pi} - \frac{T}{2} + O(1), $$ the main contribution. The remaining factor contributes $Δ_{L} ar g ζ (s) = S (T) := \frac{1}{π} ar g ζ (\frac{1}{2} + i T)$ , and a contour bound on the number of zeros of $Re ζ$ on horizontal segments gives $S (T) = O (lo g T)$ ^{[Montgomery-Vaughan §13]}. Doubling and dividing by $2 π$ , $$ N(T) = \frac{1}{\pi}\Big(\frac{T}{2}\log\frac{T}{2\pi} - \frac{T}{2}\Big) + S(T) + O(1) = \frac{T}{2\pi}\log\frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T). \qquad \square $$

Bridge. This counting formula builds toward the convergence of the explicit formula's zero-sum and appears again in 21.12.03 as the input to effective error terms. The foundational reason it matters is that the average gap between zero heights near $T$ is $2 π / lo g T$ , so the density $d N / d T \sim (lo g T) /2 π$ is exactly what is needed to bound the conditionally convergent $\sum_{ρ} x^{ρ} / ρ$ : this is exactly the place where the analytic count of zeros controls the arithmetic fluctuation of the primes. Putting these together, $N (T)$ generalises the qualitative statement "infinitely many zeros" into a quantitative law, and the central insight is that the same completed function $ξ$ whose Hadamard product of 06.01.17 gave the partial-fraction expansion of $ζ^{'} / ζ$ now, through its argument, counts the very zeros that expansion sums over — the bridge is the argument principle reading off from $ξ$ both the density of zeros and, via residues, their arithmetic effect.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Show that pairing a zero $ρ = β + iγ$ with its conjugate $\overset{ρ}{ˉ} = β - iγ$ makes the combined term $x^{ρ} / ρ + x^{\overset{ρ}{ˉ}} / \overset{ρ}{ˉ}$ real, and find its size in terms of $x^{β}$ , $γ$ , and $lo g x$ .

Hint

Write $x^{ρ} = x^{β} e^{iγ l o g x}$ and $1/ ρ = \overset{ρ}{ˉ} /∣ ρ ∣^{2}$ . The sum of a complex number and its conjugate is twice its real part.

Answer

Since $\overset{ρ}{ˉ} = \overline{ρ}$ and $x^{\overset{ρ}{ˉ}} = \overline{x^{ρ}}$ (as $x > 0$ is real), the two terms are complex conjugates, so their sum equals $2 Re (x^{ρ} / ρ)$ , which is real. Writing $x^{ρ} = x^{β} (cos (γ lo g x) + i sin (γ lo g x))$ and $1/ ρ = (β - iγ) / (β^{2} + γ^{2})$ , $$ \frac{x^\rho}{\rho} + \frac{x^{\bar\rho}}{\bar\rho} = 2,\frac{x^\beta}{\sqrt{\beta^2 + \gamma^2}}\cos!\big(\gamma\log x - \arg\rho\big). $$ For large $γ$ , $β^{2} + γ^{2} \sim γ$ , so the size is $O (x^{β} / γ)$ in amplitude; summed against the zero density $d N \sim (lo g γ /2 π) d γ$ the tail is controlled. Rubric: full credit for the conjugate-pairing reality argument, the explicit cosine form, and the $O (x^{β} / γ)$ amplitude.

Exercise 4 (medium, symbolic).

From the truncated explicit formula $ψ (x) = x - \sum_{∣ γ ∣ \leq T} x^{ρ} / ρ + O ((x lo g^{2} (x T)) / T + lo g x)$ and the bound $∣ x^{ρ} / ρ ∣ \leq x /∣ γ ∣$ on the line $β = 1$ (worst case), together with $N (T) = O (T lo g T)$ , estimate the size of the truncated zero-sum and identify the choice $T ≍ x$ that makes the remainder manageable.

Hint

Bound $\sum_{∣ γ ∣ \leq T} 1/∣ γ ∣$ by an Abel-summation against the density $d N \sim lo g γ d γ$ , giving $O (lo g^{2} T)$ . Then balance against the remainder $x lo g^{2} (x T) / T$ .

Answer

By Abel summation against $N (t) \sim (t /2 π) lo g t$ , $\sum_{0 < γ \leq T} 1/ γ = \int_{1}^{T} t^{- 1} d N (t) = O (lo g^{2} T)$ , since $d N \sim (lo g t /2 π) d t$ gives $\int^{T} (lo g t) / t d t = O (lo g^{2} T)$ . Hence the truncated zero-sum is at most $x^{β} \cdot O (lo g^{2} T)$ ; in the worst case $β$ near $1$ this is $O (x lo g^{2} T)$ , but the zero-free region of 21.12.02 pushes $β$ below $1$ , shrinking it. The remainder $x lo g^{2} (x T) / T$ is made $o (x)$ by taking $T \to \infty$ with $x$ ; the standard choice $T ≍ x$ gives remainder $O (lo g^{2} x)$ , leaving the zero-sum dominant. Rubric: full credit for the Abel summation producing $lo g^{2} T$ , the worst-case $x^{β}$ amplitude, and the role of $T ≍ x$ .

Exercise 5 (medium, symbolic).

Riemann's prime-counting function is $J (x) = \sum_{p^{m} \leq x} \frac{1}{m}$ , related to $π (x)$ by Möbius inversion $π (x) = \sum_{m \geq 1} \frac{μ ( m )}{m} J (x^{1/ m})$ . Granting Riemann's explicit formula $J (x) = Li (x) - \sum_{ρ} Li (x^{ρ}) - lo g 2 + \int_{x}^{\infty} \frac{d t}{t ( t ^{2} - 1 ) l o g t}$ , identify the main term of $π (x)$ and the leading correction.

Hint

The $m = 1$ term of the Möbius sum is $J (x)$ itself; the $m \geq 2$ terms are $O (Li (x))$ , smaller. The main term of $J (x)$ is $Li (x)$ .

Answer

The Möbius sum is $π (x) = J (x) - \frac{1}{2} J (x^{1/2}) - \frac{1}{3} J (x^{1/3}) + \dots$ , and since $J (x^{1/ m}) = O (Li (x^{1/ m})) = O (x^{1/ m} / lo g x)$ , only $m = 1$ contributes to the leading order: the main term is $Li (x)$ . The leading correction inside $J (x)$ is the zero-sum $- \sum_{ρ} Li (x^{ρ})$ , whose terms have size $∣ Li (x^{ρ}) ∣ ≍ x^{β} / (∣ γ ∣ lo g x)$ . Thus $$ \pi(x) = \operatorname{Li}(x) - \tfrac12\operatorname{Li}(x^{1/2}) - \sum_\rho\operatorname{Li}(x^\rho) + (\text{smaller}), $$ the famous secondary term $- \frac{1}{2} Li (x)$ being why $Li (x)$ overestimates $π (x)$ on average. Rubric: full credit for isolating $Li (x)$ as the main term via the Möbius sum, the $- \frac{1}{2} Li (x)$ correction, and the $Li (x^{ρ})$ zero-sum.

Exercise 7 (hard, symbolic).

Derive the exact explicit formula by summing residues. Starting from the truncated Perron representation $ψ (x) = \frac{1}{2 π i} \int_{c - i T}^{c + i T} (- \frac{ζ ^{'} ( s )}{ζ ( s )}) \frac{x ^{s}}{s} d s + R$ with $c = 1 + 1/ lo g x$ , push the contour to $Re (s) \to - \infty$ and collect the residue at each pole of the integrand.

Hint

The poles are: $s = 1$ (pole of $- ζ^{'} / ζ$ , residue $+ 1$ , kernel $x$ ); each critical zero $s = ρ$ (residue $- 1$ , kernel $x^{ρ} / ρ$ ); $s = 0$ (kernel pole, $- ζ^{'} / ζ$ holomorphic); the elementary zeros $s = - 2 k$ . Use the partial-fraction expansion of 21.12.01 to justify uniform contour bounds.

Answer

The integrand $F (s) = - \frac{ζ ^{'} ( s )}{ζ ( s )} \frac{x ^{s}}{s}$ has poles at:

(i) $s = 1$ : $- ζ^{'} / ζ$ has a simple pole of residue $+ 1$ (21.12.01) and $x^{s} / s ∣_{s = 1} = x$ , giving residue $x$ .

(ii) each critical zero $s = ρ$ (simple): residue of $- ζ^{'} / ζ$ is $- 1$ and the kernel is $x^{ρ} / ρ$ , giving $- x^{ρ} / ρ$ .

(iii) $s = 0$ : the kernel $1/ s$ has residue $1$ and $- ζ^{'} (0) / ζ (0) = - lo g (2 π)$ , giving $- lo g (2 π)$ .

(iv) each elementary zero $s = - 2 k$ ( $k \geq 1$ ): residue $- 1$ of $- ζ^{'} / ζ$ and kernel $x^{- 2 k} / (- 2 k)$ , giving $\sum_{k \geq 1} x^{- 2 k} / (2 k) = - \frac{1}{2} lo g (1 - x^{- 2})$ .

The partial-fraction expansion $ζ^{'} / ζ = B - 1/ (s - 1) + \sum_{ρ} (1/ (s - ρ) + 1/ ρ) - \frac{1}{2} Γ^{'} /Γ (s /2 + 1) + \dots$ of 21.12.01 bounds $∣ ζ^{'} / ζ ∣$ on the shifting horizontal segments so the rectangular contours' horizontal and far-left pieces vanish in the limit (for $x > 1$ ). Summing all residues, $$ \psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \log(2\pi) - \tfrac12\log(1 - x^{-2}). $$ Rubric: full credit for the four pole types with correct residues, the use of the partial-fraction bound to kill the contour, and the conditionally convergent $ρ, \overset{ρ}{ˉ}$ pairing of the zero-sum.

Exercise 8 (hard, symbolic).

Prove the truncated explicit formula's error term: show that replacing the exact zero-sum by its truncation $\sum_{∣ γ ∣ \leq T}$ and using the truncated Perron remainder produces an error $O ((x lo g^{2} (x T)) / T + lo g x)$ . Sketch the two error sources.

Hint

Two sources: (a) the Perron truncation remainder $R ≪ \sum_{n} Λ (n) min (1, x / (T ∣ lo g (x / n) ∣))$ , bounded by $O (x lo g^{2} x / T)$ away from $x$ plus $O (lo g x)$ near $x$ ; (b) the tail $\sum_{∣ γ ∣ > T} x^{ρ} / ρ$ dropped from the sum, bounded using $N (t + 1) - N (t) = O (lo g t)$ .

Answer

(a) Truncated Perron remainder. The discontinuous-integral estimate of 21.12.02 gives $R = \frac{1}{2 π i} \int_{(c)} - (truncated) ≪ \sum_{n} Λ (n) min (1, \frac{x}{T ∣ l o g ( x / n ) ∣})$ . The terms with $n$ at distance $≫ x$ from the cutoff contribute $≪ (x lo g x) / T \cdot lo g x = O (x lo g^{2} x / T)$ (summing $Λ (n) / n^{c}$ against the kernel), and the few terms with $n$ within $O (1)$ of $x$ contribute $O (lo g x)$ from the nearest prime power.

(b) Dropped zero-tail. The zeros with $∣ γ ∣ > T$ removed from the exact sum contribute $\sum_{∣ γ ∣ > T} x^{ρ} / ρ$ . Grouping by unit intervals and using $N (t + 1) - N (t) = O (lo g t)$ , the tail is $≪ x \sum_{t > T} (lo g t) / t^{2} ≍ x lo g T / T$ , absorbed into $O (x lo g^{2} (x T) / T)$ .

Adding the two sources gives the truncated formula $$ \psi(x) = x - \sum_{|\gamma| \le T}\frac{x^\rho}{\rho} + O!\Big(\frac{x\log^2(xT)}{T} + \log x\Big). $$ Rubric: full credit for the Perron remainder bound $O (x lo g^{2} x / T + lo g x)$ , the dropped-tail bound $O (x lo g T / T)$ via zero density, and assembling the stated error.

Advanced results Master

Theorem 1 (the exact explicit formula). For $x > 1$ not a prime power ^{[von Mangoldt 1895]}, $$ \psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \log(2\pi) - \frac{1}{2}\log!\big(1 - x^{-2}\big), $$ the zero-sum conditionally convergent when paired $ρ, \overset{ρ}{ˉ}$ . The identity is exact and unconditional: it holds whatever the location of the zeros, and reduces the entire arithmetic of $ψ$ to the geometry of the zero set. At a prime power $x = p^{m}$ the left side is replaced by $\frac{1}{2} (ψ (x^{+}) + ψ (x^{-}))$ , the midpoint of the jump.

Theorem 2 (the truncated explicit formula). For $2 \leq T \leq x$ ^{[Davenport §17]}, $$ \psi(x) = x - \sum_{|\operatorname{Im}\rho| \le T} \frac{x^\rho}{\rho} + O!\left(\frac{x\log^2(xT)}{T} + \log x\right). $$ This is the workhorse: combined with the de la Vallée Poussin zero-free region of 21.12.02 (every $β \leq 1 - c / lo g T$ ) and the choice $lo g T = lo g x$ , it reproduces $ψ (x) = x + O (x exp (- c^{'} lo g x))$ , and under the Riemann hypothesis $β = \frac{1}{2}$ it yields $ψ (x) = x + O (x^{1/2} lo g^{2} x)$ .

Theorem 3 (Riemann-von Mangoldt zero count). With $S (T) = π^{- 1} ar g ζ (\frac{1}{2} + i T)$ ^{[Montgomery-Vaughan §13]}, $$ N(T) = \frac{T}{2\pi}\log\frac{T}{2\pi} - \frac{T}{2\pi} + \frac{7}{8} + S(T) + O!\Big(\frac1T\Big), \qquad S(T) = O(\log T). $$ The smooth part comes from Stirling applied to the $Γ$ -factor in $ξ$ ; the fluctuating $S (T)$ is the argument of $ζ$ on the critical line, whose typical size is $lo g lo g T$ (Selberg) though only $O (lo g T)$ is provable unconditionally.

Theorem 4 (Riemann's prime-counting formula). Riemann's weighted count $J (x) = \sum_{p^{m} \leq x} 1/ m = \sum_{n \leq x} Λ (n) / lo g n$ satisfies ^{[Riemann 1859]} $$ J(x) = \operatorname{Li}(x) - \sum_\rho \operatorname{Li}(x^\rho) - \log 2 + \int_x^\infty \frac{dt}{t(t^2 - 1)\log t}, $$ and $π (x) = \sum_{m \geq 1} \frac{μ ( m )}{m} J (x^{1/ m}) = Li (x) - \frac{1}{2} Li (x^{1/2}) - \dots$ . The term $Li (x^{ρ})$ is the integral-logarithm image of the wave $x^{ρ} / ρ$ ; the secondary $- \frac{1}{2} Li (x)$ explains the persistent bias $Li (x) > π (x)$ for accessible $x$ , reversed infinitely often (Littlewood).

Theorem 5 (the duality primes $\leftrightarrow$ zeros). The explicit formula is a self-dual pairing: the von Mangoldt weights $Λ (n)$ at the primes and the residues at the zeros $ρ$ are the two sides of one Fourier-type identity, generalised by Weil to any test function $g$ with Mellin transform $\overset{g}{^}$ , $$ \sum_n \Lambda(n),g(n) = \int_0^\infty g(t),dt - \sum_\rho \hat g(\rho) + (\text{archimedean term}), $$ the von Mangoldt formula being the case $g = 1_{[1, x]}$ . The primes and zeros are dual data of the same distribution ^{[Davenport §17]}.

Synthesis. The explicit formula is the foundational reason analytic number theory is a spectral subject: it states that $ψ (x)$ is the diagonal $x$ corrected by one wave per zero, and the central insight is that the residue calculus of 21.12.02 and the partial-fraction expansion of 21.12.01 are two readings of the same contour integral — one reads the pole at $s = 1$ as the main term, the other reads the zeros as the spectrum. This is exactly the bridge from the asymptotic $ψ (x) \sim x$ to an exact identity: where the contour proof produced the leading $x$ plus an error, the explicit formula names every term of that error as $- x^{ρ} / ρ$ , and putting these together with the zero-counting $N (T)$ — whose density $lo g T /2 π$ governs the convergence of the zero-sum — gives the quantitative law that a zero at height $γ$ with real part $β$ contributes a wave of amplitude $x^{β} / γ$ . The central insight recurs as duality: the explicit formula is dual to the Hadamard product of 06.01.17, the prime-side $Λ (n)$ is dual to the zero-side residues, and Weil's generalisation makes this Fourier duality the structural template for every $L$ -function. The bridge is everywhere the completed $ξ$ : its argument counts the zeros, its Hadamard product expands $ζ^{'} / ζ$ , and its functional equation symmetrises the explicit formula — the single object from which both halves of the dictionary, primes and zeros, are read.

Full proof set Master

Proposition 1 (exact explicit formula by residues). For $x > 1$ not a prime power, $ψ (x) = x - \sum_{ρ} x^{ρ} / ρ - lo g (2 π) - \frac{1}{2} lo g (1 - x^{- 2})$ .

Proof. By Perron's formula of 21.12.02, $ψ (x) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} (- \frac{ζ ^{'} ( s )}{ζ ( s )}) \frac{x ^{s}}{s} d s$ for $c > 1$ , interpreted as a symmetric limit. Apply the residue theorem on rectangles with right edge $Re (s) = c$ and left edge $Re (s) = - U$ for $U = 2 K + 1 \to \infty$ through odd integers (avoiding the elementary zeros). The integrand $F (s) = - \frac{ζ ^{'} ( s )}{ζ ( s )} \frac{x ^{s}}{s}$ has simple poles at $s = 1$ (residue $x$ , since $- ζ^{'} / ζ$ has residue $+ 1$ there and $x^{s} / s \to x$ ), at each critical zero $ρ$ (residue $- x^{ρ} / ρ$ ), at $s = 0$ (residue $- ζ^{'} (0) / ζ (0) = - lo g 2 π$ ), and at each elementary zero $s = - 2 k$ (residue $- x^{- 2 k} / (2 k)$ ). The partial-fraction expansion of 21.12.01 gives $∣ ζ^{'} (s) / ζ (s) ∣ ≪ lo g (∣ s ∣ + 2)$ on horizontal lines $Im (s) = T_{j}$ chosen between consecutive zero ordinates, so the top and bottom edges contribute $o (1)$ , and on the far-left edge $Re (s) = - U$ the factor $x^{s} = x^{- U} \to 0$ for $x > 1$ . Summing residues, $$ \psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \log(2\pi) - \sum_{k \ge 1}\frac{x^{-2k}}{2k}, \qquad \sum_{k\ge1}\frac{x^{-2k}}{2k} = \tfrac12\log\frac{1}{1 - x^{-2}} = -\tfrac12\log(1 - x^{-2}), $$ which is the claim. $□$

Proposition 2 (zero-counting via the argument principle). $N (T) = \frac{T}{2 π} lo g \frac{T}{2 π} - \frac{T}{2 π} + O (lo g T)$ .

Proof. As in the Key theorem: $N (T) = \frac{1}{2 π} Δ_{R} ar g ξ$ on the rectangle with vertices $2 \pm i T$ , $- 1 \pm i T$ . Symmetry $ξ (s) = ξ (1 - s)$ and $ξ (\overset{s}{ˉ}) = \overline{ξ (s)}$ reduce this to twice the variation along $L : 2 \to 2 + i T \to \frac{1}{2} + i T$ . The factor $\frac{1}{2} s (s - 1) π^{- s /2}$ contributes $O (1)$ ; Stirling gives $ar g Γ (\frac{s}{2})$ along $L$ equal to $\frac{T}{2} lo g \frac{T}{2} - \frac{T}{2} + O (1)$ , and with $π^{- s /2}$ contributing $- \frac{T}{2} lo g π$ , the smooth part is $\frac{T}{2} lo g \frac{T}{2 π} - \frac{T}{2}$ . The $ζ$ -factor contributes $ar g ζ (\frac{1}{2} + i T) = π S (T)$ , and Jensen's inequality applied to $Re ζ$ on the segment bounds the number of sign changes, giving $S (T) = O (lo g T)$ . Doubling and dividing by $2 π$ yields the formula. $□$

Proposition 3 (conditional convergence of the zero-sum). The series $\sum_{ρ} x^{ρ} / ρ$ , summed in order of increasing $∣ Im ρ ∣$ , converges.

Proof. Pair $ρ = β + iγ$ with $\overset{ρ}{ˉ}$ ; the combined term is $2 Re (x^{ρ} / ρ) = 2 x^{β} ∣ ρ ∣^{- 1} cos (γ lo g x - ar g ρ)$ , of amplitude $≪ x^{β} / γ$ for $γ \geq 1$ . Grouping zeros in dyadic blocks $T < γ \leq 2 T$ , each block has $N (2 T) - N (T) ≪ T lo g T$ zeros, each term $≪ x^{β} / T$ , so the block contributes $≪ x^{β} lo g T$ . Partial summation over blocks against the oscillating cosine — whose phase $γ lo g x$ sweeps many periods per block, forcing cancellation — gives convergence of the ordered sum; without the pairing/ordering the series is not absolutely convergent, since $\sum_{γ} 1/ γ$ diverges like $\sum (lo g γ) / γ$ . $□$

Proposition 4 ( $ψ$ to $π$ via $J$ and Möbius inversion). $π (x) = Li (x) - \frac{1}{2} Li (x^{1/2}) - \sum_{ρ} Li (x^{ρ}) + O (1)$ .

Proof. Define $J (x) = \sum_{p^{m} \leq x} 1/ m$ . Then $J (x) = \sum_{m \geq 1} \frac{1}{m} π (x^{1/ m})$ , a finite sum (terms vanish once $x^{1/ m} < 2$ ), and Möbius inversion gives $π (x) = \sum_{m \geq 1} \frac{μ ( m )}{m} J (x^{1/ m})$ . Integrating the explicit formula for $ψ$ against $1/ lo g t$ (Abel summation, $dJ = d ψ / lo g t$ up to the prime-power correction) converts each term: $x \mapsto Li (x)$ , $- x^{ρ} / ρ \mapsto - Li (x^{ρ})$ , and the constants into $- lo g 2$ plus the convergent integral $\int_{x}^{\infty} d t / (t (t^{2} - 1) lo g t)$ . Hence $J (x) = Li (x) - \sum_{ρ} Li (x^{ρ}) - lo g 2 + \int_{x}^{\infty} \frac{d t}{t ( t ^{2} - 1 ) l o g t}$ . The Möbius sum keeps the $m = 1$ term $J (x)$ and the $m = 2$ term $- \frac{1}{2} J (x) = - \frac{1}{2} Li (x) + O (1)$ ; higher $m$ are $O (x^{1/3})$ . $□$

Connections Master

The exact and truncated explicit formulas are the residue-sum payoff of the contour machinery of 21.12.02: that unit shifts the Perron integral of $- ζ^{'} / ζ$ past the pole at $s = 1$ to extract the main term $x$ , and this unit completes the sum by collecting every remaining residue, turning the asymptotic $ψ (x) \sim x$ into the exact identity with the zero-sum as its error.
The partial-fraction expansion of $ζ^{'} / ζ$ and the residue $+ 1$ at $s = 1$ , residue $- 1$ at each simple zero, and the constant $- ζ^{'} (0) / ζ (0) = - lo g (2 π)$ are exactly those of 21.12.01; this unit assembles those residues into the explicit formula, so the analytic pole structure established there becomes the wave decomposition of $ψ (x)$ here.
The completed zeta $ξ (s)$ , its entire order-one Hadamard factorisation, and the count of its zeros via the argument principle rest on the Weierstrass-Hadamard factorisation theory of 06.01.17; the same $ξ$ whose product expanded $ζ^{'} / ζ$ now, through its argument along a contour, yields the zero-counting $N (T)$ that bounds the explicit formula's zero-sum.
The truncated explicit formula with its $O (x lo g^{2} (x T) / T)$ remainder is the precise input to the effective error terms of 21.12.03; pairing it with a quantitative zero-free region converts the conditionally convergent zero-sum into a power-saving bound on $ψ (x) - x$ , the step that makes the prime number theorem effective.

Historical & philosophical context Master

Bernhard Riemann's 1859 memoir ^{[Riemann 1859]} introduced the explicit formula in the form for the prime-counting function $J (x)$ , expressing it as $Li (x)$ minus a sum $\sum_{ρ} Li (x^{ρ})$ over the zeros plus lower-order terms, and stated the asymptotic for the zero count $N (T)$ . Riemann gave only sketches; the analytic estimates — the Hadamard product for the order-one entire function $ξ$ , the convergence of the zero-sum, the justification of the contour shift — were not in place. The first rigorous proof of the explicit formula, recast in the cleaner weighted form $ψ (x) = x - \sum_{ρ} x^{ρ} / ρ - lo g (2 π) - \frac{1}{2} lo g (1 - x^{- 2})$ , is due to Hans von Mangoldt in 1895 ^{[von Mangoldt 1895]}, who supplied the function $Λ (n)$ and the contour argument; the formula and its weighted variant now carry both names.

The zero-counting formula was stated by Riemann and proved by von Mangoldt and, with the sharpened remainder, by Ralf Backlund in 1918; the canonical modern derivation via the argument principle on $ξ$ is in Davenport's Multiplicative Number Theory and Montgomery-Vaughan's text ^{[Montgomery-Vaughan §13]}. Edwards's Riemann's Zeta Function reconstructs Riemann's own route to $J (x)$ and $π (x)$ through $Li (x^{ρ})$ and Möbius inversion ^{[Davenport §17]}. André Weil's 1952 reformulation recognised the explicit formula as a duality between a distribution on the primes and a distribution on the zeros, the form that governs the explicit formulas of all automorphic $L$ -functions.

Bibliography Master

@article{riemann1859,
  author  = {Riemann, Bernhard},
  title   = {Ueber die Anzahl der Primzahlen unter einer gegebenen Gr\"{o}sse},
  journal = {Monatsberichte der K\"{o}niglichen Preu\ss{}ischen Akademie der Wissenschaften zu Berlin},
  pages   = {671--680},
  year    = {1859}
}

@article{vonmangoldt1895,
  author  = {von Mangoldt, Hans},
  title   = {Zu Riemanns Abhandlung ``Ueber die Anzahl der Primzahlen unter einer gegebenen Gr\"{o}sse''},
  journal = {Journal f\"{u}r die reine und angewandte Mathematik (Crelle)},
  volume  = {114},
  pages   = {255--305},
  year    = {1895}
}

@article{backlund1918,
  author  = {Backlund, Ralf},
  title   = {\"{U}ber die Nullstellen der Riemannschen Zetafunktion},
  journal = {Acta Mathematica},
  volume  = {41},
  pages   = {345--375},
  year    = {1918}
}

@book{edwards1974,
  author    = {Edwards, Harold M.},
  title     = {Riemann's Zeta Function},
  publisher = {Academic Press},
  year      = {1974},
  note      = {Chapters 1, 3: Riemann's $J(x)$, $N(T)$, and the explicit formula}
}

@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: the explicit formula and the zero count}
}

@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      = {\S12.1: explicit formula; \S13.1--13.2: the Riemann-von Mangoldt $N(T)$}
}

Prerequisites

21.12.02
21.12.01
06.01.17

Tier anchors

beginner: Derbyshire 2003 *Prime Obsession* (Joseph Henry Press) Ch. 19-22 (the explicit formula as a main term plus a sum of waves, one per zeta zero); 3Blue1Brown 'Pi hiding in prime regularities' for the spectral picture of $\psi(x)$ as the diagonal $x$ corrected by zero-waves
intermediate: Davenport 2000 *Multiplicative Number Theory* 3e (Springer GTM 74) §17 (the truncated explicit formula for $\psi(x)$ with error term, the contour and residue calculus, the bound on the zero-sum tail); Apostol 1976 *Introduction to Analytic Number Theory* (Springer UTM) Ch. 13
master: Davenport 2000 *Multiplicative Number Theory* 3e §17 (the exact and truncated explicit formulas, the residue sum over the critical zeros); Montgomery-Vaughan 2007 *Multiplicative Number Theory I* (Cambridge UP) §12.1, §13.1-13.2 (the explicit formula and the zero-counting $N(T)$); Edwards 1974 *Riemann's Zeta Function* (Academic Press) Ch. 3 (Riemann's $J(x)$ and $\pi(x)$ via $\operatorname{Li}$); Riemann 1859; von Mangoldt 1895; Backlund 1918 (the remainder in $N(T)$)

References

Davenport, H. — Multiplicative Number Theory · Springer Graduate Texts in Mathematics 74, 3rd edition (2000), revised by H. L. Montgomery. §17. Derives the von Mangoldt explicit formula $\psi(x) = x - \sum_\rho x^\rho/\rho - \log(2\pi) - \tfrac12\log(1 - x^{-2})$ by applying the truncated Perron formula to $-\zeta'/\zeta$ and summing residues over the poles, using the partial-fraction expansion of $\zeta'/\zeta$ to control the contour as it is pushed left; states the truncated form $\psi(x) = x - \sum_{|\operatorname{Im}\rho| \le T} x^\rho/\rho + O((x\log^2(xT))/T + \log x)$ usable for estimates; and the Riemann-von Mangoldt zero-counting formula $N(T) = (T/2\pi)\log(T/2\pi) - T/2\pi + O(\log T)$ from the argument principle applied to $\xi$.
Riemann, B. — Ueber die Anzahl der Primzahlen unter einer gegebenen Grosse · Monatsberichte der Königlichen Preussischen Akademie der Wissenschaften zu Berlin (1859), pp. 671-680. States the explicit formula for the prime-counting function $J(x) = \sum_{p^m \le x} 1/m$ as $\operatorname{Li}(x)$ minus a sum $\sum_\rho \operatorname{Li}(x^\rho)$ over the zeros plus lower-order terms, and recovers $\pi(x)$ by Möbius inversion $\pi(x) = \sum_m \mu(m)/m \cdot J(x^{1/m})$; conjectures the form of the zero-counting function $N(T)$.
von Mangoldt, H. — Zu Riemanns Abhandlung 'Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse' · Journal für die reine und angewandte Mathematik (Crelle) 114 (1895), 255-305. Gives the first rigorous proof of Riemann's explicit formula in the weighted form $\psi(x) = x - \sum_\rho x^\rho/\rho - \log(2\pi) - \tfrac12\log(1 - x^{-2})$, justifying the contour shift and the convergence of the zero-sum.
Montgomery, H. L. & Vaughan, R. C. — Multiplicative Number Theory I: Classical Theory · Cambridge Studies in Advanced Mathematics 97 (2007), §12.1 (the explicit formula, exact and truncated) and §13.1-13.2 (the Riemann-von Mangoldt zero-counting formula $N(T) = (T/2\pi)\log(T/2\pi) - T/2\pi + 7/8 + S(T) + O(1/T)$, the function $S(T) = \pi^{-1}\arg\zeta(\tfrac12 + iT) = O(\log T)$).
Edwards, H. M. — Riemann's Zeta Function · Academic Press (1974), Ch. 1, 3. Translates and develops Riemann's 1859 memoir: the prime-counting $J(x)$, the Riemann-von Mangoldt $N(T)$, and the passage from $\psi$ to $\pi$ via $\operatorname{Li}(x^\rho)$ and Möbius inversion.

Estimated time

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