explicit mertens’ theorems for number fields and …

EXPLICIT MERTENS’ THEOREMS FOR NUMBER FIELDS AND

DEDEKIND ZETA RESIDUE BOUNDS WITH GRH

STEPHAN RAMON GARCIA AND ETHAN SIMPSON LEE

Abstract. Assuming the Generalized Riemann Hypothesis we obtain uni-

form, effective number-field analogues of Mertens’ theorems, each with explicit

error bounds given in terms of the degree and discriminant of the number field.

To this end, we establish explicit bounds for the residue of the Dedekind zeta-

function at s = 1 by obtaining an explicit version of Duke’s short-sum theorem

for entire Artin L-functions.

1. Introduction

Assuming the Generalized Riemann Hypothesis, our aim is to establish uniform,

explicit number-field analogues of Mertens’ classical theorems:∑p≤x

log p

p= log x+O(1), (1)

∑p≤x

1

p= log log x+M +O

(1

log x

), (2)

∏p≤x

(1− 1

p

)=

e−γ

log x

(1 + o(1)

). (3)

Here p denotes a prime number, M = 0.2614 . . . is the Meissel–Mertens constant,

and γ = 0.5772 . . . is the Euler–Mascheroni constant. Mertens obtained these

results in 1874 [21], well before the prime number theorem was proved indepen-

dently by Hadamard [7] and de la Vallee Poussin [2]. See Ingham [9, Thm. 7] or

Montgomery–Vaughan [22, Thm. 2.7] for modern proofs of Mertens’ theorems, and

Rosen [28, (3.17) - (3.30)] for explicit unconditional error bounds.

Rosen [27, Lem. 2.3, Lem. 2.4, Thm. 2] generalized (1), (2) and (3) to the number-

field setting without explicit error terms; see also [15]. Assuming GRH, we provide

generalizations of (1), (2) and (3) in the number-field setting with explicit error

bounds. Our results are uniform (valid for x ≥ 2) and the constants involved depend

only on the degree and discriminant of the number field. This complements [4], in

which the authors obtained similar results in the unconditional context.

2010 Mathematics Subject Classification. 11N32, 11N05, 11N13.Key words and phrases. Number field, discriminant, Generalized Riemann Hypothesis, GRH,

Mertens’ theorem, prime ideal, prime counting function, Chebotarev density theorem, Artin L-

function, character.

SRG supported by NSF Grant DMS-1800123.

1

2 STEPHAN RAMON GARCIA AND ETHAN SIMPSON LEE

Definitions. Let K be a number field with ring of algebraic integers OK. Let nKdenote the degree of K and ∆K the absolute value of the discriminant of K. In what

follows, p ⊂ OK denotes a prime ideal, a ⊂ OK an ideal, and N(a) the norm of a.

The Dedekind zeta-function

ζK(s) =∑a⊆OK

1

N(a)s=∏p

(1− 1

N(p)s

)−1corresponding to K is analytic on C with exception of a simple pole at s = 1. The

analytic class number formula asserts that the residue of ζK(s) at s = 1 is

κK =2r1(2π)r2hKRK

wK√

∆K, (4)

where r1 is the number of real places of K, r2 is the number of complex places of K,

wK is the number of roots of unity in K, hK is the class number of K, and RK is the

regulator of K [14]. The nontrivial zeros of ζK lie in the critical strip, 0 < Re s < 1,

where there might exist an exceptional zero β, which is real and not too close to

Re s = 1 [31, p. 148]. The Generalized Riemann Hypothesis (GRH) claims that the

nontrivial zeros of ζK(s) satisfy Re s = 12 ; in particular, β does not exist.

Let K ⊆ L be a Galois extension of number fields with Galois group G =

Gal(L/K). Let nL denote the degree of L over K and let ∆L denote the abso-

lute value of the discriminant of L over Q. Suppose that P is a prime ideal of Llying above an unramified prime p of K. The Artin symbol [L/p] denotes the con-

jugacy class of Frobenius automorphisms corresponding to prime ideals P|p. For

each conjugacy class C ⊂ G, the prime ideal counting function is

πC(x,L/K) = #{p : p is unramified in L,

[L/Kp

]= C, NK(p) ≤ x

},

in which NK(·) denotes the norm in K. If K = L, then we may specialize our

notation so that the prime ideal counting function is πK(x) =∑N(p)≤x 1.

Statement of results. Assuming GRH, we obtain the following number-field ana-

logues of Mertens’ theorems (1), (2) and (3), each valid for x ≥ 2, with explicit

constants that depend only upon the degree and discriminant of the number field.

The proof involves a recent estimate of Grenie–Molteni [6] for πK(x).

Theorem A. Assume GRH. For a number field K and x ≥ 2,∑N(p)≤x

logN(p)

N(p)= log x+AK(x), (A1)

∑N(p)≤x

1

N(p)= log log x+MK +BK(x), (B1)

∏N(p)≤x

(1− 1

N(p)

)=

e−γ

κK log x

(1 + CK(x)

), (C1)

in which

MK = γ + log κK +∑p

[1

N(p)+ log

(1− 1

N(p)

)],

EXPLICIT MERTENS’ THEOREMS FOR NUMBER FIELDS 3

|AK(x)| ≤ 4.73 log ∆K + 9.27nK + log 2, (A2)

|BK(x)| ≤ 13.47 log ∆K + (26.37 + 0.12 log x)nK√x

, and (B2)

|CK(x)| ≤ |EK(x)|e|EK(x)| with |EK(x)| ≤ nKx− 1

+ |BK(x)|. (C2)

In particular, EK(x) = o(1) (and hence CK(x) = o(1) as x→∞),

γ + log κK − nK ≤ MK ≤ γ + log κK, (M)

and

1

e134.5nK−26.76 log log ∆K≤ κK ≤ (log log ∆K)nK−1e152.74nK−63.23 (K)

For nK = 1 (K = Q), Schoenfeld [29, Cor. 2-3] provides smaller constants. In

particular, κQ = 1 so we can largely restrict our attention to nK ≥ 2. Although

unconditional bounds for κK are available (see [17–20, 31] and the discussion in

[4]), stronger bounds are possible under GRH. To establish (K), we first prove the

following explicit version of Duke’s short-sum theorem [3, Prop. 5].

Theorem B (Explicit Duke’s Theorem). Let L(s, χ) be an entire Artin L-function

that satisfies GRH, in which χ has degree d and conductor N . Then∣∣∣∣ logL(1, χ)−∑

p≤(logN)12

χ(p)

p

∣∣∣∣ ≤ 89.5 + 134.5d. (5)

The constants in Theorems A and B are rounded here for aesthetic purposes; the

proofs exhibit these constants with greater accuracy. Without explicit constants,

one can improve upon (K) somewhat. Assuming GRH, Cho–Kim obtained(1

2+ o(1)

)ζ(nK)

eγ log log ∆K≤ κK ≤ (2 + o(1))

nK−1 (eγ log log ∆K)nK−1 (6)

for nK ≥ 2 [1]. We have chosen here to pursue an explicit version of Duke’s short-

sum theorem since that result could be of independent interest. For example,

Theorem B provides an explicit estimate for the class number hK of a totally real

number field K whose normal closure has the symmetric group SnK as its Galois

group. Indeed, the class number formula implies

hK =

√∆K

2nK−1RKL(1, χ),

in which L(s, χ) = ζK(s)/ζ(s) [3, p. 103].

Slight changes to the proof of Theorem A (see Remark 1) yield the following.

Theorem C. Assume GRH and let K ⊆ L be a Galois extension of number fields.

For x ≥ 2, ∑N(p)≤x

logN(p)

N(p)=

#C#G

log x+AL(x),

∑N(p)≤x

1

N(p)=

#C#G

log log x+ML +BL(x),


∏N(p)≤x

(1− 1

N(p)

)=

e−γ

κL(log x)#C/#G(1 + CL(x)

),

in which p is a prime ideal in K that does not ramify in L, κL denotes the residue of

ζL(s) at s = 1, and AL(x), BL(x), CL(x), ML are defined as AK(x), BK(x), CK(x),

MK are in Theorem A, but with nL, ∆L, κL in place of nK, ∆K, κK. Our explicit

bounds for MK and κK also carry forward for ML and κL.

Outline of the paper. Section 2 contains the proof of Theorems A and Theorem

C, except for (K). The proof of Theorem B is in Section 3, which finally yields (K).

Acknowledgements. We thank Karim Belabas, Eduardo Friedman, Lenny Fuk-

shansky, Peter Kim, Stephane Louboutin, and Tim Trudgian for helpful feedback.

2. Proof of Theorem A

We prove (A1), (B1), and (C1), along with the associated error bounds (A2),

(B2), and (C2), respectively, in separate subsections. We then prove (M), which is

a consequence of the computations in the proof of (C1). We defer the proof of (K)

until Section 3 since it uses the explicit version of Duke’s theorem (Theorem B).

2.1. Preliminaries. We require an explicit version of the prime ideal theorem,

which was originally established by Landau in 1903 (without explicit constants) [13].

Recall that πK(x) =∑N(p)≤x 1 is the prime-ideal counting function and let

Li(x) =

∫ x

2

dt

log t

denote the offset logarithmic integral. Under GRH,

πK(x) = Li(x) +RK(x), (7)

in whichRK(x) = O(x1/2 log x). To boundRK(x) explicitly, one can follow Lagarias–

Odlyzko [12]. Since the quality of the error term one obtains depends upon the

zero-free region of ζK one uses, GRH naturally enters the conversation. In 2019,

Grenie–Molteni [6, Cor. 1] proved that GRH implies that (7) holds for x ≥ 2 with

|RK(x)| ≤√x

[(1

2π+

3

log x

)log ∆K +

(log x

8π+

1

4π+

6

log x

)nK

]. (8)

This strengthens a result announced by Oesterle [24] and improves the constants

Winckler obtains [32, Thm. 1.2] by using methods from Schoenfeld [29], Lagarias–

Odlyzko [12], and previous work of Grenie–Molteni [5].

The final important ingredient we need is the following formula for a continuously-

differentiable function f . Integration by parts and (7) provide∑N(p)≤x

f(N(p)) = f(x)πK(x)−∫ x

2

f ′(t)πK(t) dt

= f(x) Li(x)−∫ x

2

f ′(t) Li(t) dt+ f(x)RK(x)−∫ x

2

f ′(t)RK(t) dt

=

∫ x

2

f(t)

log tdt+ f(x)RK(x)−

∫ x

2

f ′(t)RK(t)dt. (9)


Remark 1. We briefly remark how to adjust the proof below to obtain Theorem

C. For x ≥ 2, Grenie–Molteni [6, Cor. 1] proved

πC(x) =#C#G

Li(x) +RL/K(x),

in which

|RL/K(x)| ≤ #C#G√x

[(1

2π+

3

log x

)log ∆L +

(log x

8π+

1

4π+

6

log x

)nL

].

Since #C/#G ≤ 1, we can bound |RL/K(x)| above by (8). The proof below goes

through mutatis mutandis. For example, any occurrence of Li(x) should be replaced

by (#C/#G) Li(x).

2.2. Proof of (A1) and (A2). Let f(t) = log t/t in (9) and obtain∑N(p)≤x

logN(p)

N(p)= log x+ f(x)RK(x)−

∫ x

2

f ′(t)RK(t)dt− log 2︸︷︷︸AK(x)

. (10)

For x ≥ 2, log x/√x achieves its maximum value 2/e at x = e2 and (log x)2/

√x

achieves its maximum value 16/e2 at x = e4. Thus, (8) yields

|f(x)RK(x)| ≤ log x√x

[(1

2π+

3

log x

)log ∆K +

(log x

8π+

1

4π+

6

log x

)nK

].

=1√x

(3 log ∆K + 6nK) +log x√x

(log ∆K

2π+nK4π

)+

(log x)2√x

nK8π

≤ 1√2

(3 log ∆K + 6nK) +2

e

(log ∆K

2π+nK4π

)+

2nKe2π

< 2.23843 log ∆K + 4.3874nK. (11)

Observe that

f ′(t) =1− log t

t2

undergoes a sign change from positive to negative at x = e and that

log x

8π+

1

4π+

6

log x, whose derivative is

1

8πx− 6

x(log x)2,

decreases until e4√3π ≈ 215,328.6 (in particular, it is decreasing on [2, e]). Conse-

quently, we split the integral in (10) at x = e. First note that∣∣∣∣∫ e

2

f ′(t)RK(t) dt

∣∣∣∣≤∫ e

2

1− log t

t3/2

[(1

2π+

3

log t

)log ∆K +

(log t

8π+

1

4π+

6

log t

)nK

]dt

≤(

4√e−√

2(1 + log 2)

)[(1

2π+

3

log 2

)log ∆K +

(log 2

8π+

1

4π+

6

log 2

)nK

]< 0.14203 log ∆K + 0.27737nK. (12)

For x ≥ e, symbolic integration provides∣∣∣∣∫ x

e

f ′(t)RK(t) dt

∣∣∣∣


≤∫ ∞e

log t− 1

t3/2

[(1

2π+

3

log t

)log ∆K +

(log t

8π+

1

4π+

6

log t

)nK

]dt

=(3π√eEi(− 1

2 ) + 6π + 2) log ∆K

π√e

+(12π√eEi(− 1

2 ) + 24π + 7)nK

2π√e

< 2.34600 log ∆K + 4.59546nK, (13)

in which Ei(x) is the exponential integral.

In light of (10), we add (11), (12), and (13), and obtain the bound

|AK(x)| < 4.72646 log ∆K + 9.26023nK + log 2.

This concludes the proof of (A2). �

2.3. Proof of (B1) and (B2). Let f(t) = 1/t in (9) and obtain∑N(p)≤x

1

N(p)=

∫ x

2

dt

t log t+RK(x)

x+

∫ x

2

RK(t) dt

t2

= log log x− log log 2 +RK(x)

x+

∫ ∞2

RK(t) dt

t2−∫ ∞x

RK(t) dt

t2

= log log x+

∫ ∞2

RK(t)

t2dt− log log 2︸︷︷︸MK

+RK(x)

x−∫ ∞x

RK(t)

t2dt︸︷︷︸

BK(x)

,

in which the convergence of the improper integral is guaranteed by (8).

For x ≥ 2, (8) provides

|RK(x)|x

≤ 1√x

[(1

2π+

3

log x

)log ∆K +

(log x

8π+

1

4π+

6

log x

)nK

]and∫ ∞

x

|RK(t)| dtt2

≤∫ ∞x

1

t3/2

[(1

2π+

3

log t

)log ∆K +

(log t

8π+

1

4π+

6

log t

)nK

]dt

= log ∆K

∫ ∞x

1

t3/2

(1

2π+

3

log t

)dt+ nK

∫ ∞x

1

t3/2

(log t

8π+

1

4π+

6

log t

)dt

≤ log ∆K

∫ ∞x

1

t3/2

(1

2π+

3

log x

)dt+ nK

∫ ∞x

1

t3/2

(log t

8π+

1

4π+

6

log x

)dt

=1√x

[(1

π+

6

log x

)log ∆K +

(log x

4π+

1

π+

12

log x

)nK

].

For x ≥ 2, the triangle inequality provides

|BK(x)| ≤ 1√x

[(3

2π+

9

log 2

)log ∆K +

(3 log x

8π+

5

4π+

18

log 2

)nK

](14)

<1√x

[13.47 log ∆K + (26.37 + 0.120 log x)nK] .

Now we find the constant MK, following Ingham [9]. Since this part of the proof

is identical to that in [4], we present an abbreviated version here.


For Re s > 1, partial summation and (B1) provide∑p

1

N(p)s= limx→∞

( ∑N(p)≤x

1

N(p)s

)= limx→∞

( ∑N(p)≤x

1

N(p)s−1N(p)

)

= limx→∞

(1

xs−1

∑N(p)≤x

1

N(p)

)+ (s− 1)

∫ ∞2

( ∑N(p)≤t

1

N(p)

)dt

ts

= (s− 1)

∫ ∞2

( ∑N(p)≤t

1

N(p)

)dt

ts

= (s− 1)

∫ ∞2

MK

tsdt+ (s− 1)

∫ ∞2

BK(t)

tsdt+ (s− 1)

∫ ∞2

log log t

tsdt. (15)

As s → 1+, the first integral tends to MK, the second tends to zero by (B2), and

the substitution ts−1 = ey converts the third integral into

(s− 1)

∫ ∞2

log log t

tsdt =

∫ ∞log(2s−1)

e−y log y dy − 21−s log(s− 1),

which tends to −γ − log(s− 1). As s→ 1+, the Euler product formula yields

MK − γ + o(1)

= log(s− 1) +∑p

[1

N(p)s+ log

(1− 1

N(p)s

)]−∑p

log

(1− 1

N(p)s

)= log

((s− 1)ζK(s)

)+∑p

[1

N(p)s+ log

(1− 1

N(p)s

)]= log κK +

∑p

[1

N(p)+ log

(1− 1

N(p)

)]+ o(1), (16)

in which the last sum converges, by comparison with∑

pN(p)−2. �

2.4. Proofs of (C1) and (C2). The proof is identical to that in [4] so we just

sketch it here for the sake of completeness. From (16), we may write

− γ − log κK +MK =∑

N(p)≤x

[1

N(p)+ log

(1− 1

N(p)

)]+ FK(x). (17)

Since

0 ≤ −y − log(1− y) ≤ y2

1− y, (18)

it follows that

|FK(x)| = −∑

N(p)>x

[1

N(p)+ log

(1− 1

N(p)

)]≤

∑N(p)>x

1

N(p)(N(p)− 1)

≤∑p>x

∑fi

1

pfi(pfi − 1)< nK

∑m>x

1

m(m− 1)≤ nKx− 1

, (19)


in which∑fi

denotes the sum over the inertia degrees fi of the prime ideals lying

over p. Then (B1) and (17) imply

−γ − log κK +MK =∑

N(p)≤x

log

(1− 1

N(p)

)+ log log x+MK +BK(x) + FK(x)︸︷︷︸

EK(x)

.

Exponentiate and simplify to get∏N(p)≤x

(1− 1

N(p)

)=

e−γ

κK log xe−EK(x) =

e−γ

κK log x

(1 + CK(x)

),

where |CK(x)| ≤ |EK(x)|e|EK(x)| since |et − 1| ≤ |t|e|t| for t ∈ R. �

2.5. Proof of (M). Since there are no prime ideals of norm less than 2, (16) yields

MK = γ + log κK + FK(2− δ) for δ ∈ (0, 1),

in which FK(x) is defined in (17) In particular, (19) reveals that

|FK(2− δ)| ≤ nK as δ → 0+.

In light of (18), each summand in FK(2 − δ) is non-positive, so FK(2 − δ) ≤ 0 as

δ → 0+. Consequently, −nK ≤ FK(2− δ) ≤ 0 and hence

−nK ≤MK − γ − log κK ≤ 0,

which is equivalent to (M). �

To truly complete the proof of Theorem A, we must prove (K). This requires

Theorem B, which is our next task.

3. Proof of Theorem B

3.1. Prerequisites. Most of the relevant background for what follows can be found

in [10, Ch. 5]. Let K be a number field of degree nK ≥ 2 over Q and let L/Q denote

the Galois closure of the extension K/Q. Then we may write

L(s, χ) =ζK(s)

ζ(s)=

∞∑k=1

χ(k)

ks, for Re s > 1, (20)

in which χ is the character of a certain (n − 1)-dimensional representation of

Gal(L/Q). This representation is a direct sum of irreducible representations, none

of which are trivial. Under these circumstances, the conductor N of χ satisfies

N = ∆K [30, p. 159], [23, Cor. 11.8]. In particular, N ≥ 3 by Minkowski’s bound.

It is known that L(s, χ) is meromorphic on C; Artin’s conjecture asserts that it

is entire. The Euler product formula

ζK(s) =∏p

(1−N(p)−s

)−1, for Re s > 1,

ensures that any zeros of ζK(s) in the right half plane occur in the critical strip

0 ≤ Re s ≤ 1. GRH implies that these zeros lie on the critical line Re s = 12

(we adhere to the wide interpretation of GRH, in which one extends the Riemann

Hypothesis to all global L-functions and not merely the Dirichlet L-functions; some

call this the “Extended Riemann Hypothesis”). Unconditionally, L(s, χ) has no


zeros or poles on the vertical line Re s = 1 [10, Cor. 5.47]. Since ζ(s) has a simple

pole at s = 1 with residue 1, (20) implies that

κK = Ress=1 ζK(s) = lims→1

(s− 1)ζ(s)L(s, χ) = lims→1

L(s, χ) = L(1, χ). (21)

In particular, this ensures that L(1, χ) > 0 since ζK(s)/ζ(s) > 0 for s > 1 and

because ζK(s) has a simple pole at s = 1.

Dirichlet convolution shows that the coefficients in (20) satisfy

χ(k) =∑i|k

akµ

(k

i

),

in which ak denotes the number of ideals in K of norm k and µ is the Mobius

function. Therefore,

χ(p) = apµ(1) + a1µ(p) = ap − 1 ∈ [−1, nK − 1]. (22)

Indeed, if p is inert in K then N(p) = pnK 6= p and hence χ(p) = −1. The other

extreme λ(p) = nK − 1 occurs if p splits totally in K.

Let d denote the degree of χ and N its conductor. The function L(s, χ) satisfies

the functional equation

Λ(1− s, χ) = εχNs− 1

2 Λ(s, χ), (23)

in which

Λ(s, χ) = πds2 Γ(s

2

) d+`2

Γ(s+ 1

2

) d−`2

L(s, χ), (24)

|εχ| = 1, and ` is the value of χ on complex conjugation. Moreover, L(s, χ) is of

finite order with a canonical product of genus 1 [3, p. 112]; this is needed below for

a Phragmen–Lindelof argument.

3.2. Preparatory lemmas. Let L(s, χ) be an entire Artin L-function, in which χ

has degree d and conductor N . We assume that L(s, χ) satisfies the GRH, so that

logL(s, χ) is analytic on Re s > 1/2. First, we derive an upper bound for L(s, χ).

Lemma 2. For Re s ≥ δ > 1,

|L(s, χ)| ≤ ζ(δ)d. (25)

Proof. For Re s > 1, (20) implies

logL(s, χ) =∑p

∞∑m=1

1

mχ(pm)p−ms, (26)

in which |χ(pm)| ≤ d [3, (23), p. 111]. Therefore, (26) converges absolutely since

| logL(s, χ)| ≤∑p

∞∑m=1

1

m|χ(pm)|p−mRe s ≤ d

∑p

∞∑m=1

p−mδ

m= d log ζ(δ). (27)

It follows that for δ > 1,

|L(s, χ)| = |elogL(s,χ)| ≤ e| logL(s,χ)| ≤ ed log ζ(δ) = ζ(δ)d. �

Next, we derive an upper bound for L(s, χ) on a slightly larger half plane. As is

customary, we write s = σ + it for a complex variable, in which σ, t ∈ R.


Lemma 3. Let c1 = 12πζ( 3

2 ) < 4.1035. Then

|L(s, χ)| ≤ cd1N(t2 + 25

4

)d/2for σ ≥ 1

2 .

Proof. We require two gamma-function estimates from Rademacher [26, Lem. 1 &

2]. The first estimate states that for a ≥ 0 and − 12 ≤ σ ≤

12 ,∣∣∣∣Γ(a2 + 1−s

2 )

Γ(a2 + s2 )

∣∣∣∣ ≤ ( 12 |a+ 1 + s|) 1

2−σ. (28)

The second estimate states that for a− 12 ≥ 0 and − 1

2 ≤ σ ≤12 ,∣∣∣∣Γ(a2 + 1−s

2 )

Γ(a2 + s2 )

∣∣∣∣ ≤ ( 12 |a+ s|) 1

2−σ. (29)

From (23) and (24),

Λ(1− s, χ) = εχNs− 1

2πds2 Γ(s

2

) d+`2

Γ(s+ 1

2

) d−`2

L(s, χ).

Substituting 1− s and χ in (24) yields

Λ(1− s, χ) = πd(1−s)

2 Γ(1− s

2

) d+`2

Γ(

1− s

2

) d−`2

L(1− s, χ).

Equate the previous two formulas, simplify, and obtain

L(s, χ) =εχπ

d( 12−s)

Ns− 12

(Γ( 1−s

2 )

Γ( s2 )

) d+`2(

Γ(1− s2 )

Γ( s+12 )

) d−`2

L(1− s, χ).

From (28) with a = 0 and (29) with a = 1, for − 12 ≤ σ ≤

12 we have

|L(s, χ)| ≤∣∣(Nπd) 1

2−s∣∣ ∣∣∣∣Γ( 1−s

2 )

Γ( s2 )

∣∣∣∣d+`2∣∣∣∣Γ( 1

2 + 1−s2 )

Γ( 12 + s

2 )

∣∣∣∣d−`2

|L(1− s, χ)|

≤ (Nπd)12−σ

[(12 |1 + s|

) 12−σ

] d+`2[(

12 |1 + s|

) 12−σ

] d−`2

|L(1− s, χ)|

= (Nπd)12−σ

[(12 |1 + s|

) 12−σ

]d|L(1− s, χ)|.

Let σ = − 12 , use (26) to pass from χ to χ, and (25) to bound |L( 3

2 + it, χ)|. Then

|L(− 12 + it, χ)| ≤ Nπd

2d|1 + s|d|L( 3

2 + it, χ)|

≤ N(π2 ζ( 3

2 ))d|1 + s|d = cd1N |1 + s|d.

Applying the Phragmen–Lindelof theorem to the vertical strip − 12 ≤ Re s ≤ 3

2

yields |L(s, χ)| ≤ Ncd1|1 + s|d in the strip.

For − 12 ≤ σ ≤

32 , we also obtain

|1 + s|2 = |(1 + σ) + it|2 = (1 + σ)2 + t2 ≤ t2 + 254 .

It follows that

|L(s, χ)| ≤ cd1N(t2 + 25

4

) d2 for 1

2 ≤ Re s ≤ 32 .

This bound also holds for Re s > 32 by (25) since

|L(s, χ)| ≤ ζ(σ)d ≤ ζ( 32 )d ≤ cd1 ≤ cd1N

(t2 + 25

4

)d/2. �


Our next lemma gives estimates L′(s, χ)/L(s, χ) in a certain vertical strip.

Lemma 4. Let c2 = 38423π < 5.3144. Then∣∣∣∣L′(s, χ)

L(s, χ)

∣∣∣∣ ≤ c2d log(N

1d (|t|+ 4)

)for 5

8 ≤ Re s ≤ 278 .

Proof. GRH implies that logL(s, χ) is analytic on Re s > 12 . Lemma 3 gives

Re logL(s, χ) = log |L(s, χ)| ≤ log(cd1N

(254 + t2

)d/2)= d log c1 + logN + d

2 log(254 + (Im s)2

)for Re s ≥ 1

2 . If f is analytic on a neighbourhood of |z| ≤ R, [11, (3.7.12)] implies

|f ′(z)| ≤ 4R

π(R2 − |z|2)sup|ξ|=R

|Re f(ξ)− α|

for any α ∈ R. Fix t ∈ R and apply the previous inequality to

f(z) = logL(z + 2 + it, χ) = logL(s, χ)

with

R = 32 , | s− (2 + it)︸︷︷︸

z

| ≤ 118 , and α = d log c1.

For |s− (2 + it)| ≤ 118 , we obtain∣∣∣∣L′(s, χ)

L(s, χ)

∣∣∣∣ ≤ 4( 32 )

π(( 32 )2 − ( 11

8 )2) sup|ξ|= 3

2

(Re logL(s+ ξ, χ)− α

)≤ 384

23πsup|ξ|= 3

2

(logN + d

2 log(254 + Im(s+ ξ)2

) )≤ c2

(logN + d

2 log(254 + (|t|+ Im ξ)2

)≤ c2

(logN + d

2 log(254 + (|t|+ 3

2 )2)

≤ c2(

logN + d2 log((|t|+ 4)2)

)= c2d

(logN

1d + log(|t|+ 4)

)= c2d log(N

1d (|t|+ 4)).

Since t ∈ R is arbitrary, we obtain the desired bound for 58 ≤ Re s ≤ 27

8 . �

Duke’s approach to the previous lemma uses a result of Littlewood [16, Lem. 4],

which is [3, Lem. 2] in Duke’s paper, in place of the sharper [11, (3.7.12)] used

above. More information about “sharp real part theorems” for the derivative can

be found in [11, Ch. 5], along with a host of historical references.

Our next result is useful for an important step in the proof of Lemma 6.

Lemma 5. For N ≥ 1 and − 78 ≤ σ ≤ −

38 ,∫ ∞

−∞|Γ(σ + it)| log

(N

1d (|t|+ 4)

)dt ≤ c3 logN

d+ c4,

in which

c3 =2

538 e

43

7√π

< 30.1794 and c4 =2

538 e

43

7√π

(log 4− e2π Ei(−2π)

)< 46.0461.


Here Ei(x) denotes the exponential integral function.

Proof. For s = σ + it with σ ≥ 0, we have the inequality [25, 5.6E9]:

|Γ(s)| ≤√

2π|s|σ− 12 e−π|t|/2 exp

(1

6|s|

).

Since Γ(s+ 1) = sΓ(s) for s ∈ C, the previous inequality yields

|Γ(s)| = |Γ(s+ 1)||s|

≤√

2π|s+ 1|σ+ 12 e−π|t|/2

|s|exp

(1

6|s+ 1|

). (30)

For − 78 ≤ σ ≤ −

38 ,

√2π|s+ 1|σ+ 1

2

|s|exp

(1

6|s+ 1|

)︸︷︷︸

f(s)

≤√

2πe( 58 + |t|) 1

8

|t|→ 0

uniformly in σ as |t| → ∞, so we may use numerical methods to maximize f(s) in

the strip − 78 ≤ σ ≤ −

38 . This reveals that f(s) attains its maximum value

α =

√π 2

378 e

43

7< 23.7029

at s = − 78 . Therefore, (30) infers that

|Γ(σ + it)| ≤ αe−π|t|/2, for − 78 ≤ σ ≤ −

38 .

Because, |Γ(z)| = |Γ(z)| for z ∈ C,∫ ∞−∞|Γ(σ + it)| log

(N

1d (|t|+ 4)

)dt = 2

∫ ∞0

|Γ(σ + it)| log(N

1d (|t|+ 4)

)dt

≤ 2α

∫ ∞0

e−π|t|/2 log(N

1d (|t|+ 4)

)dt

=4α logN

πd+

4α(

log 4− e2π Ei(−2π))

π

= c3logN

d+ c4. �

3.3. Proof of Theorem B. The first step in our proof of Theorem B is Lemma

6, which is an explicit version of [3, Lem. 4].

Lemma 6. Let L(s, χ) be an entire Artin L-function that satisfies GRH, in which

χ has degree d and conductor N . For 1 ≤ u ≤ 32 and x > 1,∣∣∣∣∑

p

(log p)χ(p)p−ue−p/x +L′(u, χ)

L(u, χ)

∣∣∣∣ ≤ c2dx58−u

2π

(c3 logN

d+ c4

)+ d.

Proof. For Re s > 1, the derivative of (26) yields∑p

log p

∞∑m=1

χ(pm)p−ms = −L′(s, χ)

L(s, χ). (31)

Substitute y = pm/x in the classical Cahen–Mellin integral (see [8])

e−y =1

2πi

∫ 2+i∞

2−i∞y−sΓ(s) ds, for y > 0,


and obtain

e−pm/x =

1

2πi

∫ 2+i∞

2−i∞p−msxsΓ(s) ds for x > 0.

For 1 ≤ u ≤ 32 and x > 0, it follows from (31) that

∑p

log p

∞∑n=1

χ(pm)p−mue−pm/x

=∑p

log p

∞∑m=1

χ(pm)p−mu(

1

2πi

∫ 2+i∞

2−i∞p−msxsΓ(s) ds

)

=1

2πi

∫ 2+i∞

2−i∞

(∑p

log p

∞∑m=1

χ(pm)p−m(s+u)

)xsΓ(s) ds

= − 1

2πi

∫ 2+i∞

2−i∞

L′(s+ u, χ)

L(s+ u, χ)xsΓ(s) ds.

Shift the integration contour of integration to the vertical line Re s = 58 −u . Since

Γ has a simple pole with residue 1 at s = 0 and 58 − u < 0, we pick up the residue

Ress=0

(L′(s+ u, χ)

L(s+ u, χ)xsΓ(s)

)=L′(u, χ)

L(u, χ),

and obtain∑p

log p

∞∑m=1

χ(pm)

pmuepm/x+L′(u, χ)

L(u, χ)= − 1

2πi

∫ 58−u+i∞

58−u−i∞

L′(s+ u, χ)

L(s+ u, χ)xsΓ(s) ds. (32)

Since 58 − u ∈ [− 7

8 ,−38 ], we estimate integral on the right-hand side of (32) with

Lemmas 4 and 5:∣∣∣∣∣∫ 5

8−u+i∞

58−u−i∞

L′(s+ u, χ)

L(s+ u, χ)xsΓ(s) ds

∣∣∣∣∣ ≤∫ ∞−∞

∣∣∣∣L′( 58 + it, χ)

L( 58 + it, χ)

∣∣∣∣x 58−u|Γ( 5

8 − u+ it)|dt

≤ c2dx58−u

∫ ∞−∞|Γ( 5

8 − u+ it)| log(N

1d (|t|+ 4)

)dt

≤ c2dx58−u

(c3 logN

d+ c4

).

It follows from the triangle inequality and (32) that∣∣∣∣∣∑p

log p

∞∑m=1

χ(pm)

pmuepm/x+L′(u, χ)

L(u, χ)

∣∣∣∣∣≤

∣∣∣∣∣∑p

log pχ(p)

puep/x+L′(u, χ)

L(u, χ)

∣∣∣∣∣+

∣∣∣∣∣∑p

log p

∞∑m=2

χ(pm)

pmuepm/x

∣∣∣∣∣≤ c2dx

58−u

2π

(c3 logN

d+ c4

)+

∣∣∣∣∣∑p

log p

∞∑m=2

χ(pm)

pmuepm/x

∣∣∣∣∣ . (33)


Finally, we bound the sum over m ≥ 2 in (33) using |χ(pm)| ≤ d, the inequality

1 ≤ u ≤ 32 , and numerical summation:1∣∣∣∣∑

p

log p

∞∑m=2

χ(pm)

pmuepm/x

∣∣∣∣ ≤ d∑p

log p

∞∑m=2

p−mu

≤ d∑p

log p

pu(pu − 1)

≤ d∑p

log p

p(p− 1)

< d. �

The next step in the proof of Theorem B is the following lemma, which is an ex-

plicit version of the argument at the top of [3, p. 113]. In what follows, the notation

f(t) = O?(g(t)) means that |f(t)| ≤ |g(t)| for all values of t under consideration.


χ has degree d and conductor N . For x > 1,∣∣∣∣∣ logL(1, χ)−∑p

χ(p)p−1e−p/x

∣∣∣∣∣ ≤ 5

2d+

c5 logN + c6d

x3/8 log x,

in which

c5 =c2c32π

< 25.5261 and c6 =c2c42π

< 38.9464.

Proof. Fix x > 1. We begin by integrating the expressions in the inequality from

Lemma 6 over u ∈ [1, 32 ] and obtain∫ 32

1

∑p

(log p)χ(p)p−ue−p/x du =∑p

(log p)χ(p)e−p/x∫ 3

2

1

p−u du

=∑p

χ(p)e−p/x√p− 1

p3/2

=∑p

χ(p)p−1e−p/x −∑p

χ(p)p−3/2e−p/x

=∑p

χ(p)p−1e−p/x +O?(d∑p

1

p3/2

)=∑p

χ(p)p−1e−p/x +O?(d)

by numerical summation.2 The initial interchange of integral and summation is

permissible by uniform convergence. Using the estimate (27), we obtain∫ 32

1

L′(u, χ)

L(u, χ)du = logL( 3

2 , χ)− logL(1, χ) = − logL(1, χ) +O?(d log ζ( 32 ))

= − logL(1, χ) +O?(d),

1Mathematica gives∑

plog p

p(p−1)≈ 0.755. A crude upper bound is log 2

2+∑∞

n=1log(2n+1)(2n+1)(2n)

< 0.86.2Mathematica’s built-in PrimeZetaP command or numerical summation give the upper bound 0.85.


since log ζ( 32 ) ≈ 0.9603 < 1. Integrate the upper bound in Lemma 6 and obtain∫ 3/2

1

[c2dx

58−u

2π

(c3 logN

d+ c4

)+ d

]du =

c2d

2π

(c3 logN

d+ c4

)( √x− 1

x7/8 log x

)+d

2

≤ c2d

2π

(c3 logN

d+ c4

)x4/8

x7/8 log x+d

2

=

(c2c32π

logN +c2c4d

2π

)1

x3/8 log x+d

2

=c5 logN + c6d

x3/8 log x+d

2.

Lemma 6 says that∑p

(log p)χ(p)p−ue−p/x = −L′(u, χ)

L(u, χ)+O∗

(c2dx

58−u

2π

(c3 logN

d+ c4

)+ d

).

Integrate this over u ∈ [1, 32 ], use the preceding computations, and obtain∑p

χ(p)p−1e−p/x +O?(d) = L(1, χ) +O∗ (d) +O∗(c5 logN + c6d

x3/8 log x+d

2

),

which is equivalent to the desired result. �

The sum that appears in the previous lemma involves the expression e−p/x. Since

e−p/x = 1 + O(1/x), one might seek to replace e−p/x with 1 if p/x is sufficiently

small (with all constants explicit, of course). This is the content of the next lemma.


χ has degree d and conductor N . For x = (logN)3 and y = x1/6 = (logN)1/2,∣∣∣∣∣∑p

χ(p)p−1e−p/x −∑p≤y

χ(p)p−1

∣∣∣∣∣ < 7.81469 d.

Proof. Let x = (logN)3 and y = x1/6 = (logN)1/2. Then∑p

χ(p)p−1e−p/x =∑p≤y

χ(p)p−1 +∑p≤y

χ(p)p−1(e−p/x − 1)︸︷︷︸I1

+∑

y<p≤x2

χ(p)p−1e−p/x

︸︷︷︸I2

+∑x2<p

χ(p)p−1e−p/x

︸︷︷︸I3

.

Bounding I1. Since p ≤ y < x, we may use t = p/x in the inequality

|1− et| ≤ (e− 1)|t|, for |t| < 1, (34)

which follows from the power-series representation of the exponential function and

the triangle inequality. Observe that I1 = 0 if√

logN = y < 2; that is, if N < e4 ≈54.6. For N ≥ 55, we have

|I1| =∣∣∣∑p≤y

χ(p)p−1(e−p/x − 1)∣∣∣ ≤ ∑

p≤y

|χ(p)| |e−p/x − 1|

p


= d∑p≤y

1− e−p/x

p≤ (e− 1)d

∑p≤y

1

p· px

(by (34))

= (e− 1)d∑p≤y

1

(logN)3≤ (e− 1)d

(logN)52

≤(

e− 1

(log 55)52

)d < 0.05346 d.

Bounding I2. If N = 3, then (y, x2] ≈ (1.048, 1.326] and hence I2 = 0. Conse-

quently, we assume that N ≥ 4. Apply (14) with K = Q and obtain∣∣∣∣∑p≤t

1

p− log log t−M

∣∣∣∣ ≤ 1√t

(3 log t

8π+

18

log t+

5

4π

)︸︷︷︸

m(t)

, for t ≥ 2, (35)

in which M ≈ 0.2615 is the Meissel–Mertens constant. To see that m(t) is decreas-

ing for t ≥ 2, observe that m(2) > 18.7, m′(2) > −17.9, and

m′(t) =−3(log t)3 − 4(log t)2 − 144π log t− 288π

16πt3/2(log t)2,

has only a single real zero (at t ≈ 0.137648).

For 4 ≤ N ≤ 1012, numerical summation provides3

|I2| =∣∣∣∣ ∑y<p≤x2

χ(p)p−1e−p/x∣∣∣∣ ≤ d ∑

p≤(log(1012))6p−1e−p/x < 2.49045d. (36)

For N > 1012, (35) yields

|I2| =∣∣∣∣ ∑y<p≤x2

χ(p)p−1e−p/x∣∣∣∣ ≤ d ∑

y<p≤x2

p−1e−p/x ≤ d∑

y<p≤x2

p−1

≤ d( ∑p≤x2

p−1 −∑p≤y

p−1)

≤ d((

log log(x2) +M +O?(m(x2)))−(

log log y +M +O?(m(y))))

≤ d(

log

(log(x2)

log y

)+m(x2) +m(y)

)= d

(log

(6 log logN12 log logN

)+m(x2) +m(y)

)= d(

log 12 +m(x2) +m(y))

≤ d(

log 12 +m((log(1012))6

)+m

(√log(1012)

))< 7.47604d.

It follows that |I2| < 7.47604d for N ≥ 3.

3The cutoff 1012 is chosen so that (36) is amenable to computation. In Mathematica, run

Sum[Exp[-p/x]/p, {p, Prime@Range@PrimePi@r}] with x = Log[10^12]^3 and r = x^2.


Bounding I3. We use a geometric series argument. Observe that

|I3| =∣∣∣ ∑x2<p

χ(p)p−1e−p/x∣∣∣ ≤ d ∑

x2<p

p−1e−p/x

≤ d

x2

∑x2<p

e−p/x ≤ d

x2

∞∑k=bx2c

(e−1/x)k

=d

x2(e−1/x)bx

2c

1− e−1/x≤ d(e−1/x)x

2

x2(1− e−1/x)

≤ de−x

x2(1− e−1/x)≤ de−(log 3)3

(log 3)6(1− e−1/(log 3)3)

< 0.28519 d,

since the expression that multiplies d in the penultimate line is decreasing in x and

is thus maximized by x = (log 3)3 because N ≥ 3. Putting this all together we have∣∣∣∣∣∑p

χ(p)p−1e−p/x −∑p≤y

χ(p)p−1

∣∣∣∣∣ < 7.81469 d. �

The numerical constant that appears in the statement of the previous lemma

can be given in closed form by examining the constants that bound I1, I2, and I3in the proof. However, this does not seem worthwhile to pursue.

Remark 9. Under the Riemann Hypothesis, Schoenfeld proved [29, Cor. 2]:∣∣∣∣∣∑p≤t

1

p− log log t−M

∣∣∣∣∣ ≤ 3 log t+ 4

8π√t

, for t ≥ 13.5.

For large t, this is better than our estimate (35). However, the range of validity is

problematic since we require estimates on∑p≤y p

−1, in which y =√

logN . Since√logN ≥ 13.5 requires N ≥ 1.413× 1079, we do not use Schoenfeld’s estimate and

instead revert to our own (35) since it is valid for t ≥ 2.

We are now in a position to complete the proof of Theorem B. With x = (logN)3,

Lemmas 7 and 8 ensure that∣∣∣∣ logL(1, χ)−∑

p≤(logN)1/2

χ(p)

p

∣∣∣∣ ≤ 5

2d+

c5 logN + c6d

x3/8 log x+ 7.81469 d

≤ c5 logN

x3/8 log x+

c6d

x3/8 log x+ 10.3147 d

=c5

3(logN)1/8 log logN+

c6d

3(logN)9/8 log logN+ 10.3147 d

<c5

3(log 3)1/8 log log 3+

c6d

3(log 3)9/8 log log 3+ 10.3147 d

< 89.4146 + 134.494 d. �


3.4. Proof of (K). In this section, we establish a particular application of Theorem

B. First note that κQ = 1 since the Riemann zeta-function has a simple pole at

s = 1 with residue 1. Thus, we may assume that nK ≥ 2.

Let L(s, χ) denote the Artin L-function from (20) and recall that its conductor is

N = ∆K and χ is the character of a certain (nK−1)-dimensional representation. Let

Cd = 89.5+134.5 d denote the bound on the right-hand side of (5); here d = nK−1.

Since the sums below contain no summands unless N ≥ e4 ≈ 54.6, we may assume

that N ≥ 55. By (22), (35), and Theorem B it follows that

logL(1, χ) =∑

p≤(logN)12

χ(p)

p+O?(Cd)

≤ d∑

p≤(logN)12

1

p+ Cd

≤ d log(12 log logN

)+Md+ dm

(√log 55

)+ Cd

≤ d(

log log logN + log 12 +M +m

(√log 55

))+ Cd

≤ (nK − 1) log log logN + 152.74nK − 63.23.

Since L(1, χ) > 0, it follows that

L(1, χ) ≤ (log logN)nK−1e152.74nK−63.23.

For N ≥ 55, (22), (35), and Theorem B provide

logL(1, χ) ≥ −∑

p≤(logN)12

1

p+O?(Cd)

≥ − log log(√

logN)−M −m(√

logN)− Cd≥ − log log logN − log 1

2 −M −m(√

log 55)− Cd> − log log logN − 134.5nK + 26.76.

Thus,

L(1, χ) ≥ 1

e134.5nK−26.76 log logN.

Since (21) ensures that κK = L(1, χ), this completes the proof of (K). �

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Department of Mathematics, Pomona College, 610 N. College Ave., Claremont, CA

91711

Email address: [email protected]

URL: http://pages.pomona.edu/~sg064747

School of Science, UNSW Canberra at the Australian Defence Force Academy,

Northcott Drive, Campbell, ACT 2612

Email address: [email protected]

URL: https://www.unsw.adfa.edu.au/our-people/mr-ethan-lee

https://arxiv.org/abs/1311.5715

http://pages.pomona.edu/~sg064747

https://www.unsw.adfa.edu.au/our-people/mr-ethan-lee

explicit mertens’ theorems for number fields and …

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