21.11.04 · number-theory / dirichlet-series-arithmetic-functions

Perron's Formula and Mellin Inversion

shipped3 tiersLean: none

Anchor (Master): Perron 1908 *J. reine angew. Math.* 134, 95 (the original formula); Riemann 1859 *Monatsber. Berliner Akad.* (the contour-integral programme); Davenport 2000 *Multiplicative Number Theory* §11-12; Montgomery-Vaughan 2007 *Multiplicative Number Theory I* §5; Titchmarsh 1986 *The Theory of the Riemann Zeta-Function* (2nd ed., rev. Heath-Brown, Oxford) §3; Tenenbaum 2015 *Introduction to Analytic and Probabilistic Number Theory* (3rd ed., AMS) §II.2

Intuition Beginner

Suppose you have packaged a sequence of numbers $a_{1}, a_{2}, a_{3}, \dots$ into a single function by forming the sum $A (s) = a_{1} + a_{2} / 2^{s} + a_{3} / 3^{s} + \dots$ . This packaging is called a Dirichlet series, and the Riemann zeta function is the case where every $a_{n}$ equals one. The packaged function $A (s)$ is smooth and analytic, and the tools of calculus apply to it. The original numbers are hidden inside.

Perron's formula is the recipe for getting the original numbers back out, and in particular for recovering the running total $a_{1} + a_{2} + \dots$ up to some cutoff $x$ . The recipe is a contour integral: you integrate $A (s)$ multiplied by a simple steering factor $x^{s} / s$ along a vertical line in the complex plane, and the answer is exactly the partial sum you wanted. The steering factor acts like a switch. For each integer $n$ , it turns the term $a_{n}$ fully on when $n$ is below the cutoff and fully off when $n$ is above it.

Why does this matter? Questions about the original numbers — how the primes pile up, how divisors are distributed — are hard to attack directly. Once the running total is written as a contour integral, you can slide the contour around, pick up contributions from the poles of $A (s)$ , and read off the answer. This is the engine behind the proof that the primes thin out like $x / ln x$ .

Visual Beginner

A vertical line in the complex plane at horizontal position $c$ (to the right of where the series stops converging). Along this line the integration runs from bottom to top. The integrand carries the factor $x^{s} / s$ , which behaves like an on-off switch keyed to the cutoff $x$ .

  imaginary
    axis
     ^
     |        contour runs upward
     |        along the vertical line Re(s) = c
     |              |
     |              |   integrand: A(s) * x^s / s
-----+--------------|----------------> real axis
     |              |
     |             Re(s) = c
     |
   switch value of (1/2 pi i) * integral of y^s/s :
        y = n/x < 1  (n below cutoff x)  ->  1   (term kept)
        y = n/x > 1  (n above cutoff x)  ->  0   (term dropped)
        y = 1        (n exactly at x)    ->  1/2 (half weight)

The table inside the diagram is the heart of the matter: the contour integral of $y^{s} / s$ outputs $1$ , $\frac{1}{2}$ , or $0$ depending on whether $y$ is below, at, or above $1$ . Setting $y = n / x$ turns this into the switch that keeps exactly the terms with $n \leq x$ .

Worked example Beginner

Take the simplest Dirichlet series, where $a_{1} = a_{2} = a_{3} = \dots = 1$ , so $A (s)$ is the Riemann zeta function. The running total up to a cutoff $x$ is just the count of integers from $1$ to $x$ . Let us check the switch directly with small numbers instead of integrating.

Step 1. The switching rule says: for a number $y$ bigger than $1$ the switch outputs $0$ , for $y$ smaller than $1$ it outputs $1$ , and at $y = 1$ it outputs $1/2$ . Put the cutoff at $x = 3.5$ .

Step 2. Test each integer $n$ by computing $y = n / x$ and reading the switch. For $n = 1$ : $y = 1/3.5 = 0.286 < 1$ , switch outputs $1$ . For $n = 2$ : $y = 0.571 < 1$ , outputs $1$ . For $n = 3$ : $y = 0.857 < 1$ , outputs $1$ . For $n = 4$ : $y = 1.143 > 1$ , outputs $0$ . For $n = 5$ and beyond: $y > 1$ , outputs $0$ .

Step 3. Add the kept terms: $1 + 1 + 1 = 3$ . The count of integers up to $3.5$ is indeed $3$ (namely $1, 2, 3$ ).

What this tells us: the steering factor $x^{s} / s$ , integrated along the vertical line, is exactly a device that keeps the terms with $n \leq x$ and discards the rest. The contour integral does the bookkeeping of a partial sum automatically.

Check your understanding Beginner

Formal definition Intermediate+

Let $A (s) = \sum_{n = 1}^{\infty} a_{n} n^{- s}$ be a Dirichlet series with coefficients $a_{n} \in C$ , absolutely convergent for $Re (s) > σ_{a}$ (the abscissa of absolute convergence). Write $σ = Re (s)$ throughout, and fix a real number $c > σ_{a}$ . The notation $\sum_{n \leq x}^{'}$ denotes the partial sum in which the term $n = x$ (when $x$ is an integer) is counted with weight $\frac{1}{2}$ . ^{[Montgomery-Vaughan §5.1]}

Definition (Mellin transform). For a function $f : R_{> 0} \to C$ that is locally integrable and suitably decaying, the Mellin transform is

f (s) := \int_{0}^{\infty} f (x) x^{s - 1} d x,

defined on the vertical strip where the integral converges absolutely. The Mellin transform is the multiplicative analogue of the Fourier transform: the substitution $x = e^{- u}$ converts it into the bilateral Laplace transform of $f (e^{- u})$ , and hence into a Fourier transform after a further rotation. Its inversion theorem reads

f (x) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} f (s) x^{- s} d s

at points of continuity of $f$ , for any $c$ inside the strip of convergence.

Definition (discontinuous integral). For $y > 0$ , $c > 0$ , define the principal-value vertical integral

I (y, c) := \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} \frac{y ^{s}}{s} d s = T \to \infty lim \frac{1}{2 π i} \int_{c - i T}^{c + i T} \frac{y ^{s}}{s} d s .

The discontinuous-integral lemma states

I (y, c) = ⎩ ⎨ ⎧ 1 \frac{1}{2} 0 y > 1, y = 1, 0 < y < 1.

The discontinuity across $y = 1$ is the feature that makes the integral a switch.

Definition (Perron's formula). With $A (s)$ and $c > σ_{a}$ as above, and $x > 0$ not necessarily an integer,

n \leq x \sum^{'} a_{n} = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} A (s) \frac{x ^{s}}{s} d s .

This is the (full) Perron formula: substituting $A (s) = \sum_{n} a_{n} n^{- s}$ and integrating term by term applies $I (x / n, c)$ to each coefficient, selecting precisely those $n \leq x$ .

Counterexamples to common slips

The contour line must lie strictly to the right of $σ_{a}$ . Choosing $c \leq σ_{a}$ destroys the term-by-term interchange because the Dirichlet series no longer converges absolutely on the line of integration. The standard practice is $c = σ_{a} + 1/ ln x$ , which keeps $x^{c} = x^{σ_{a}} \cdot e$ controlled while staying in the convergence region.
The principal value is essential. The integrand $y^{s} / s$ is not absolutely integrable on the vertical line — $∣ y^{s} / s ∣ = y^{c} /∣ s ∣$ decays only like $1/∣ t ∣$ , and $\int d t /∣ t ∣$ diverges. Convergence of $I (y, c)$ holds only as a symmetric limit $T \to \infty$ ; dropping the symmetry breaks it.
The boundary weight is $\frac{1}{2}$ , not $1$ or $0$ . When $x$ is an integer, the term $n = x$ contributes $I (1, c) = \frac{1}{2}$ . Writing $\sum_{n \leq x}$ (full weight) instead of $\sum_{n \leq x}^{'}$ introduces an error of $\frac{1}{2} a_{x}$ that matters in sharp asymptotic work.

Key theorem with proof Intermediate+

The signature result is the truncated Perron formula — the effective version that replaces the infinite contour by a finite segment of length $2 T$ and bounds the resulting error. This is the form used in every contour proof of the prime number theorem, because an infinite contour cannot be estimated, while a finite one can. ^{[Davenport §11]}

Theorem (truncated Perron formula). Let $A (s) = \sum_{n = 1}^{\infty} a_{n} n^{- s}$ be absolutely convergent for $Re (s) > σ_{a}$ . Let $c > max (0, σ_{a})$ , $x \geq 1$ , and $T \geq 1$ . Then

n \leq x \sum^{'} a_{n} = \frac{1}{2 π i} \int_{c - i T}^{c + i T} A (s) \frac{x ^{s}}{s} d s + R (x, T),

where the error term satisfies

∣ R (x, T) ∣ \leq n = 1 \sum \infty ∣ a_{n} ∣ (\frac{x}{n})^{c} min (1, \frac{1}{T ∣ ln ( x / n ) ∣}) .

Proof. The proof rests on the truncated discontinuous-integral lemma: for $y > 0$ , $y \neq = 1$ , $c > 0$ , $T \geq 1$ ,

\frac{1}{2 π i} \int_{c - i T}^{c + i T} \frac{y ^{s}}{s} d s - 1_{{y > 1}} \leq y^{c} min (1, \frac{1}{T ∣ ln y ∣}),

where $1_{{y > 1}}$ is $1$ for $y > 1$ and $0$ for $y < 1$ .

Establish this lemma first. Suppose $y > 1$ . Consider the rectangle with vertical side $[c - i T, c + i T]$ closed to the left by a far vertical line $Re (s) = - U$ with $U \to \infty$ . The integrand $y^{s} / s$ has a single simple pole inside, at $s = 0$ , with residue $y^{0} = 1$ . By the residue theorem 21.03.01 the closed contour integral equals $1$ . The contributions of the two horizontal sides $Im (s) = \pm T$ are bounded by integrating $∣ y^{s} / s ∣ = y^{σ} /∣ s ∣$ across $σ \in [- U, c]$ ; since $∣ s ∣ \geq T$ on these sides and $\int_{- \infty}^{c} y^{σ} d σ = y^{c} / ln y$ converges for $y > 1$ , each horizontal side contributes at most $y^{c} / (2 π T ln y)$ . The left side vanishes as $U \to \infty$ because $y^{- U} \to 0$ . Hence the vertical segment differs from $1$ by at most $y^{c} / (π T ln y)$ , and the cruder uniform bound $y^{c}$ holds by estimating the segment directly. Taking the smaller of the two bounds gives the stated $min$ .

For $0 < y < 1$ the same argument closes the rectangle to the right, where the integrand has no pole, so the residue is $0$ and the same horizontal-side estimates apply with $∣ ln y ∣$ in place of $ln y$ . This proves the truncated lemma.

Now apply it term by term. Because $A (s)$ converges absolutely on the line $Re (s) = c$ , the series and the finite-length integral may be interchanged:

\frac{1}{2 π i} \int_{c - i T}^{c + i T} A (s) \frac{x ^{s}}{s} d s = n = 1 \sum \infty a_{n} \cdot \frac{1}{2 π i} \int_{c - i T}^{c + i T} \frac{( x / n ) ^{s}}{s} d s .

Replace each inner integral by $1_{{x / n > 1}} = 1_{{n < x}}$ plus its error, controlled by the truncated lemma with $y = x / n$ . The main terms sum to $\sum_{n < x} a_{n}$ , which agrees with $\sum_{n \leq x}^{'} a_{n}$ up to the half-weight boundary term absorbed into the convention. The accumulated error is bounded by $\sum_{n} ∣ a_{n} ∣ (x / n)^{c} min (1, 1/ (T ∣ ln (x / n) ∣))$ , exactly $R (x, T)$ . $□$

Bridge. The truncated formula builds toward the contour proof of the prime number theorem, where this is exactly the step that converts $ψ (x) = \sum_{n \leq x} Λ (n)$ into a contour integral of $- ζ^{'} (s) / ζ (s) \cdot x^{s} / s$ , and the asymptotic $ψ (x) \sim x$ appears again in 21.03.01 as the residue of the pole at $s = 1$ . The foundational reason the method works is that the discontinuous integral is a switch, and the central insight is that sliding the contour past a pole reads off a main term while the shifted line gives a small remainder — putting these together, the location of the zeros of $ζ (s)$ controls the size of the error, which generalises to every $L$ -function whose logarithmic derivative one wishes to sum. The bridge is the Mellin transform: Perron's formula is Mellin inversion applied to the summatory function, and this is dual to the Dirichlet-series packaging that produced $A (s)$ in the first place.

Exercises Intermediate+

Exercise 6 (medium, multiple choice).

In the contour proof of the prime number theorem, which Dirichlet series is fed into Perron's formula to produce the Chebyshev function $ψ (x) = \sum_{n \leq x} Λ (n)$ ?

A. $ζ (s)$

B. $1/ ζ (s)$

C. $- ζ^{'} (s) / ζ (s)$

D. $ζ (s)^{2}$

Hint

The von Mangoldt function $Λ (n)$ is the coefficient of the Dirichlet series whose value is the logarithmic derivative of $ζ$ .

Answer

C. The identity $- ζ^{'} (s) / ζ (s) = \sum_{n = 1}^{\infty} Λ (n) n^{- s}$ for $Re (s) > 1$ packages the von Mangoldt function as a Dirichlet series. Feeding $A (s) = - ζ^{'} (s) / ζ (s)$ into Perron's formula gives $ψ (x)$ as a contour integral; the pole of $A (s)$ at $s = 1$ (from the zero of $ζ$ 's reciprocal vanishing, i.e. the simple pole of $ζ$ ) contributes the main term $x$ , yielding $ψ (x) \sim x$ . Option A produces a counting function, not the weighted prime sum; B is $\sum μ (n) n^{- s}$ ; D is $\sum d (n) n^{- s}$ .

Exercise 7 (hard, symbolic).

Using the truncated Perron formula for $A (s) = ζ (s)$ with $a_{n} = 1$ , derive the classical estimate $\sum_{n \leq x} 1 = x + O (x^{1 + ϵ} / T) + (contour integral)$ , identifying which terms of the error sum dominate.

Hint

Split the error sum $\sum_{n} (x / n)^{c} min (1, 1/ (T ∣ ln (x / n) ∣))$ into the ranges $n \leq x /2$ , $x /2 < n < 2 x$ , and $n \geq 2 x$ . The middle range, where $∣ ln (x / n) ∣$ is small, is the delicate one.

Answer

Take $c = 1 + 1/ ln x$ , so $x^{c} ≍ x$ . The error sum is $E = \sum_{n = 1}^{\infty} (x / n)^{c} min (1, 1/ (T ∣ ln (x / n) ∣))$ .

For $n \leq x /2$ or $n \geq 2 x$ , $∣ ln (x / n) ∣ \geq ln 2$ , so each term is $\leq (x / n)^{c} / (T ln 2)$ . Summing $(x / n)^{c}$ over $n \geq 1$ gives $x^{c} ζ (c) ≍ x \cdot ln x$ (since $ζ (c) \sim 1/ (c - 1) = ln x$ near $c = 1$ ). This range contributes $O (x ln x / T)$ .

For the near-diagonal range $x /2 < n < 2 x$ with $n \neq = x$ , write $n = x + r$ with $∣ r ∣ < x$ . Then $∣ ln (x / n) ∣ = ∣ ln (1 + r / x) ∣ ≍ ∣ r ∣/ x$ , so the term is $≍ min (1, x / (T ∣ r ∣))$ . Summing over integer $r$ with $1 \leq ∣ r ∣ < x$ : the terms with $∣ r ∣ \leq x / T$ contribute $O (x / T \cdot 1) = O (x / T)$ each capped at $1$ , totalling $O (x / T)$ ; the terms with $∣ r ∣ > x / T$ contribute $\sum x / (T ∣ r ∣) = O ((x / T) ln T)$ . The integer nearest $x$ (if $x$ is close to an integer) gives an extra $O (1)$ from the $min (1, \cdot)$ cap.

Collecting, $E = O (x ln x / T)$ , which for any fixed $ϵ > 0$ is $O (x^{1 + ϵ} / T)$ once $T$ is a fixed power of $x$ . Choosing $T = x$ gives error $O (x^{ϵ})$ , recovering the elementary $\sum_{n \leq x} 1 = ⌊ x ⌋ = x + O (1)$ up to the contour main term $x$ (the residue at $s = 1$ ). $□$

Exercise 8 (hard, symbolic).

Derive Perron's formula from Mellin inversion. Specifically, identify the summatory function $\sum_{n \leq x}^{'} a_{n}$ as the inverse Mellin transform of $A (s) / s$ and explain the role of the $1/ s$ factor.

Hint

The Mellin transform of the indicator $1_{[0, 1]} (t)$ , suitably scaled, is $1/ s$ . Convolution in the multiplicative group $R_{> 0}$ becomes multiplication of Mellin transforms.

Answer

Let $S (x) = \sum_{n \leq x}^{'} a_{n}$ , the right-continuous step function jumping by $a_{n}$ at each integer $n$ . Its Mellin transform, computed via integration by parts (or directly on the jumps), is

\int_{0}^{\infty} S (x) x^{- s - 1} d x \cdot s = n = 1 \sum \infty a_{n} \int_{n}^{\infty} x^{- s - 1} d x \cdot s = n = 1 \sum \infty a_{n} n^{- s} = A (s),

so the Mellin transform of $S$ is $A (s) / s$ , valid for $Re (s) > σ_{a}$ . The factor $1/ s$ arises because $\int_{n}^{\infty} x^{- s - 1} d x = n^{- s} / s$ : it is the Mellin transform of the tail-indicator, i.e. integrating the coefficient mass from $n$ to infinity. Applying the Mellin inversion theorem,

S (x) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} \frac{A ( s )}{s} x^{s} d s,

which is exactly Perron's formula. The $1/ s$ is thus the signature of the summatory (rather than pointwise-coefficient) inversion: it converts the Dirichlet series of the coefficients into the Mellin transform of their running total. $□$

Advanced results Master

Effective truncation with smoothing. The sharp error in the truncated formula can be reduced by integrating against a smooth weight rather than the hard cutoff. Replacing $x^{s} / s$ by $x^{s} \cdot w (s)$ for a Mellin pair $(w, w)$ with $w$ smooth and compactly supported produces a smoothed summatory function $\sum_{n} a_{n} w (n / x)$ whose contour integral converges absolutely, so no principal value is needed and the error decays faster than any power of $T$ . The price is that one recovers $\sum_{n} a_{n} w (n / x)$ rather than the sharp partial sum; the difference is controlled by the support width of $w$ . This smoothed Perron formula is the modern workhorse in sieve theory and in the study of moments of $L$ -functions ^{[Tenenbaum §II.2]}.

The explicit formula as a Perron computation. Feeding $A (s) = - ζ^{'} (s) / ζ (s)$ into the full Perron formula and moving the contour leftward past the pole at $s = 1$ and past every nontrivial zero $ρ$ of $ζ$ produces the Riemann-von Mangoldt explicit formula

ψ (x) = x - ρ \sum \frac{x ^{ρ}}{ρ} - \frac{ζ ^{'} ( 0 )}{ζ ( 0 )} - \frac{1}{2} ln (1 - x^{- 2}),

valid for non-integer $x > 1$ . Each term is a residue: $x$ is the residue at the pole $s = 1$ of $ζ$ , the sum over $ρ$ collects residues at the zeros, and the final terms come from the pole at $s = 0$ and the negative-even zeros ^{[Titchmarsh §3]}. The convergence of $\sum_{ρ} x^{ρ} / ρ$ is conditional and must be read symmetrically in $∣ Im ρ ∣ \leq T$ , the same principal-value structure inherited from the discontinuous integral.

Abscissa shifting and the zero-free region. The size of $ψ (x) - x$ is governed by how far left the contour can be pushed without crossing a zero. If $ζ (s) \neq = 0$ for $σ \geq 1 - c_{0} / ln (∣ t ∣ + 2)$ , the contour can be moved to $σ = 1 - c_{0} / ln T$ , on which $∣ x^{s} / s ∣ \leq x^{1 - c_{0} / l n T} /∣ s ∣$ . Balancing the truncation error $x ln x / T$ against the shifted-line contribution $x exp (- c_{0} ln x / ln T)$ and optimising $T$ yields the classical prime number theorem error $ψ (x) = x + O (x exp (- c ln x))$ . The entire mechanism is Perron's formula plus the residue theorem 21.03.01; the analytic input is the zero-free region.

Mellin-Barnes and the analytic continuation of integrals. The discontinuous integral is the simplest Mellin-Barnes integral. The general Mellin-Barnes representation writes a product of Gamma factors $\frac{1}{2 π i} \int \prod_{j} Γ (a_{j} + s) \prod_{k} Γ (b_{k} - s) x^{- s} d s$ and evaluates it by residue summation, producing hypergeometric functions. Perron's formula is the degenerate case where the only Gamma factor is the pole $1/ s = Γ (s) /Γ (s + 1)$ at $s = 0$ . This places Perron's formula inside the broader theory of Mellin-Barnes integrals that compute Feynman amplitudes and special-function identities.

Synthesis. The truncated Perron formula is the foundational reason analytic number theory can pass from a Dirichlet series back to its coefficients, and the central insight is that the discontinuous integral is a switch whose only singularity is the simple pole at $s = 0$ . This is exactly the structure that generalises: each pole one crosses while sliding the contour contributes a residue, so the main term $x$ for $ψ (x)$ is the residue at $ζ$ 's pole, the oscillating corrections are residues at the zeros, and putting these together yields the explicit formula. The bridge is Mellin inversion — Perron's formula is the inverse Mellin transform of $A (s) / s$ — and the method is dual to the Dirichlet-series packaging that built $A (s)$ . The same mechanism appears again in 21.03.01 through the zero-free region that bounds the error, and it generalises to every $L$ -function whose summatory coefficients one wishes to extract, making Perron's formula the universal engine that converts analytic facts about $A (s)$ into arithmetic facts about $\sum_{n \leq x} a_{n}$ .

Full proof set Master

Proposition (full discontinuous-integral lemma). For $c > 0$ and $y > 0$ ,

\frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} \frac{y ^{s}}{s} d s = ⎩ ⎨ ⎧ 1 \frac{1}{2} 0 y > 1, y = 1, 0 < y < 1,

the integral understood as a symmetric principal value.

Proof. For $y > 1$ write $y^{s} = e^{s l n y}$ with $ln y > 0$ . On a left-closing rectangle with corners $c \pm i T$ and $- U \pm i T$ , the integrand $y^{s} / s$ has its unique pole at $s = 0$ inside, with residue $lim_{s \to 0} s \cdot y^{s} / s = y^{0} = 1$ . By the residue theorem 06.01.03 the closed contour integral equals $1$ . On the horizontal sides $Im (s) = \pm T$ , $∣ y^{s} / s ∣ \leq y^{σ} / T$ and $\int_{- U}^{c} y^{σ} d σ \leq y^{c} / ln y$ , so each contributes $O (y^{c} / (T ln y)) \to 0$ as $T \to \infty$ . On the left side $Re (s) = - U$ , $∣ y^{s} ∣ = y^{- U} \to 0$ as $U \to \infty$ , killing that contribution. Hence the vertical integral equals the residue, namely $1$ .

For $0 < y < 1$ , $ln y < 0$ , so $y^{σ}$ decays as $σ \to + \infty$ ; close the rectangle to the right with corners $c \pm i T$ , $U \pm i T$ . No pole lies inside (the pole $s = 0$ is to the left of $Re (s) = c > 0$ ), so the closed integral is $0$ , and the same horizontal-side estimates with $∣ ln y ∣$ give the vertical integral the value $0$ .

For $y = 1$ the integrand is $1/ s$ . On the line $s = c + i t$ , $\frac{1}{2 π i} \int_{- T}^{T} \frac{1}{c + i t} i d t = \frac{1}{2 π} \int_{- T}^{T} \frac{d t}{c + i t}$ . Splitting the real and imaginary parts, the imaginary part is odd in $t$ and cancels, while the real part is $\frac{1}{2 π} \int_{- T}^{T} \frac{c}{c ^{2} + t ^{2}} d t = \frac{1}{2 π} \cdot 2 arctan (T / c) \to \frac{1}{2 π} \cdot π = \frac{1}{2}$ as $T \to \infty$ . $□$

Proposition (term-by-term interchange is valid). If $A (s) = \sum_{n} a_{n} n^{- s}$ converges absolutely on $Re (s) = c$ , then for each fixed $T$ ,

\frac{1}{2 π i} \int_{c - i T}^{c + i T} A (s) \frac{x ^{s}}{s} d s = n = 1 \sum \infty a_{n} \cdot \frac{1}{2 π i} \int_{c - i T}^{c + i T} \frac{( x / n ) ^{s}}{s} d s .

Proof. On the compact segment $s = c + i t$ , $t \in [- T, T]$ , the partial sums $\sum_{n \leq N} a_{n} n^{- s}$ converge uniformly to $A (s)$ because $\sum_{n > N} a_{n} n^{- s} \leq \sum_{n > N} ∣ a_{n} ∣ n^{- c} \to 0$ independently of $t$ . The factor $x^{s} / s$ is bounded on the segment (it is continuous on a compact set avoiding $s = 0$ when $c > 0$ ). Uniform convergence of the integrand on a finite-length contour permits interchange of the sum and the integral by the uniform-limit theorem for integrals. $□$

Connections Master

Residue theorem 06.01.03. Perron's formula is evaluated by exactly the contour-closing argument the residue theorem provides: every value of the discontinuous integral is a residue at $s = 0$ , and the prime number theorem main term $x$ is the residue at the pole $s = 1$ of $ζ$ . The residue theorem is the computational engine that turns a vertical contour integral into a sum over poles.
Riemann zeta function 06.01.16. The zeta function is the model Dirichlet series to which Perron's formula is applied, and its Euler product, pole at $s = 1$ , and zeros are exactly the analytic data that the contour shift converts into the asymptotics of $\sum_{n \leq x} a_{n}$ . Perron's formula is the bridge that carries the analytic structure of $ζ$ over to the counting of integers and primes.
Riemann zeta function, number-theoretic anchor 21.03.01. The contour proof of the prime number theorem feeds $- ζ^{'} (s) / ζ (s)$ into the truncated Perron formula; the zero-free region of $ζ$ established there is precisely the input that bounds the Perron error term and yields $ψ (x) = x + O (x e^{- c l n x})$ . The explicit formula is a Perron computation read as a sum of residues at the zeros.

Historical & philosophical context Master

Oskar Perron published the formula bearing his name in 1908 in Crelle's Journal ^{[Perron 1908]}, in a paper on the general theory of Dirichlet series. The contour-integral idea was older: Riemann's 1859 memoir had already represented prime-counting functions by contour integrals of $ζ$ , and the formula is in this sense a precise and general statement of a method Riemann had used in a special case. Perron's contribution was to state the inversion as a theorem about an arbitrary Dirichlet series and to isolate the discontinuous integral as its analytic core. The systematic theory of the Mellin transform on which the inversion rests was developed by Hjalmar Mellin in the same period, culminating in his 1910 Mathematische Annalen synthesis ^{[Mellin 1910]}, which placed the Gamma function, hypergeometric functions, and the inversion integral in a single framework.

The formula became the standard engine of multiplicative number theory through its treatment in Landau's 1909 Handbuch and, in the modern canon, in Davenport's Multiplicative Number Theory and Montgomery-Vaughan's text, where the truncated form with explicit error term is the version used to prove the prime number theorem with error. The discontinuous integral itself dates to Cahen's 1894 thesis on Dirichlet series; Cahen's evaluation contained a gap that Perron and later Landau repaired. The truncated form, with its $min (1, 1/ (T ∣ ln (x / n) ∣))$ error, is the practical instrument: an exact infinite contour integral admits no asymptotic estimate, while a finite segment of length $2 T$ can be balanced against the analytic continuation of the series past its abscissa of convergence.

Bibliography Master

@article{Perron1908,
  author = {Perron, Oskar},
  title = {Zur Theorie der Dirichletschen Reihen},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume = {134},
  year = {1908},
  pages = {95--143},
  note = {Original statement of Perron's formula for Dirichlet series}
}

@article{Mellin1910,
  author = {Mellin, Hjalmar},
  title = {Abriss einer einheitlichen Theorie der Gamma- und der hypergeometrischen Funktionen},
  journal = {Mathematische Annalen},
  volume = {68},
  year = {1910},
  pages = {305--337},
  note = {Systematic theory of the Mellin transform and its inversion}
}

@book{Davenport2000,
  author = {Davenport, Harold},
  title = {Multiplicative Number Theory},
  edition = {3},
  publisher = {Springer},
  series = {Graduate Texts in Mathematics},
  volume = {74},
  year = {2000},
  note = {Revised by H. L. Montgomery. \S11-12: Perron's formula and the contour proof of PNT}
}

@book{MontgomeryVaughan2007,
  author = {Montgomery, Hugh L. and Vaughan, Robert C.},
  title = {Multiplicative Number Theory I: Classical Theory},
  publisher = {Cambridge University Press},
  series = {Cambridge Studies in Advanced Mathematics},
  volume = {97},
  year = {2007},
  note = {\S5.1: Perron's formula, Mellin inversion, the discontinuous integral}
}

@book{Titchmarsh1986,
  author = {Titchmarsh, E. C.},
  title = {The Theory of the Riemann Zeta-Function},
  edition = {2},
  publisher = {Oxford University Press},
  year = {1986},
  note = {Revised by D. R. Heath-Brown. \S3: explicit formula, truncated Perron}
}

@book{Tenenbaum2015,
  author = {Tenenbaum, G\'erald},
  title = {Introduction to Analytic and Probabilistic Number Theory},
  edition = {3},
  publisher = {American Mathematical Society},
  series = {Graduate Studies in Mathematics},
  volume = {163},
  year = {2015},
  note = {\S II.2: Mellin transform, Perron's formula, effective truncation}
}

Prerequisites

06.01.03
06.01.16
21.03.01

Tier anchors

beginner: Derbyshire 2003 *Prime Obsession* Ch. 19-21 (informal account of summing primes by contour integration); Stewart *Calculus* early-transcendentals on step functions and partial sums
intermediate: Davenport 2000 *Multiplicative Number Theory* (3rd ed., GTM 74) §11 (Perron's formula and the truncated form with error term); Montgomery-Vaughan 2007 *Multiplicative Number Theory I: Classical Theory* (Cambridge Studies 97) §5.1 (Perron, Mellin inversion, the discontinuous integral)
master: Perron 1908 *J. reine angew. Math.* 134, 95 (the original formula); Riemann 1859 *Monatsber. Berliner Akad.* (the contour-integral programme); Davenport 2000 *Multiplicative Number Theory* §11-12; Montgomery-Vaughan 2007 *Multiplicative Number Theory I* §5; Titchmarsh 1986 *The Theory of the Riemann Zeta-Function* (2nd ed., rev. Heath-Brown, Oxford) §3; Tenenbaum 2015 *Introduction to Analytic and Probabilistic Number Theory* (3rd ed., AMS) §II.2

References

Perron, O. — Zur Theorie der Dirichletschen Reihen · *Journal für die reine und angewandte Mathematik* 134 (1908), 95-143. The original derivation of the formula expressing the partial-sum function of a Dirichlet series as a vertical contour integral of the series against $x^s/s$.
Davenport, H. — Multiplicative Number Theory · Springer Graduate Texts in Mathematics 74, 3rd edition revised by H. L. Montgomery (2000). §11 (Perron's formula, the truncated form with explicit error term, application to the prime number theorem); §12 (the contour proof of PNT, moving the line of integration past the pole).
Montgomery, H. L. & Vaughan, R. C. — Multiplicative Number Theory I: Classical Theory · Cambridge Studies in Advanced Mathematics 97 (2007). §5.1 (Perron's formula via the discontinuous integral, Mellin inversion, the effective truncated form with the standard $O$-term).
Titchmarsh, E. C. — The Theory of the Riemann Zeta-Function · Oxford University Press, 2nd edition revised by D. R. Heath-Brown (1986). §3 (the function $\psi(x)$, the explicit formula, and the truncated Perron formula as the analytic engine of the prime number theorem with error term).
Tenenbaum, G. — Introduction to Analytic and Probabilistic Number Theory · American Mathematical Society Graduate Studies in Mathematics 163, 3rd edition (2015). §II.2 (the Mellin transform, Perron's formula, and the effective truncation, with sharp error bounds).
Mellin, H. — Abriss einer einheitlichen Theorie der Gamma- und der hypergeometrischen Funktionen · *Mathematische Annalen* 68 (1910), 305-337. The systematic theory of the Mellin transform and its inversion, the multiplicative-Fourier underpinning of Perron's formula.

Estimated time

beginner: 18m
intermediate: 45m
master: 80m