ESTIMATES OF θ(x; k, l) FOR LARGE VALUES OF x 1. Introduction Let

MATHEMATICS OF COMPUTATION Volume 71, Number 239, Pages 1137–1168 S 0025-5718(01)01351-5 Article electronically published on November 21, 2001

ESTIMATES OF θ(x; k, l) FOR LARGE VALUES OF x PIERRE DUSART

Abstract. We extend a result of Ramaré and Rumely, 1996, about the Chebyshev function θ in arithmetic progressions. We find a map ε(x) such that (∀a > 0), whereas ε(x) is | θ(x; k, l) − x/ϕ(k) |< xε(x) and ε(x) = O ln1a x a constant. Now we are able to show that, for x > 1531, x | θ(x; 3, l) − x/2 |< 0.262 ln x and, for x > 151, x π(x; 3, l) > . 2 ln x

1. Introduction q ln x Let R = 9.645908801 and X = R . Rosser [6] and Schoenfeld [7, Th. 11 p. 342] showed that, for x > 101, | θ(x) − x |, | ψ(x) − x |< xε(x), where

r

8 X 1/2 exp(−X). 17π We adapt their work to the case of arithmetic progressions. Let us recall the usual notations for nonnegative real x: X ln p, where p is a prime number, θ(x; k, l) = ε(x) =

p≡lmodk p6x

ψ(x; k, l) =

X

Λ(n),

where Λ is Von Mangold’s function,

n≡lmodk n6x

and ϕ is Euler’s function. We show, for x > x0 (k) where x0 (k) can be easily computed, that | θ(x; k, l) − x/ϕ(k) |, | ψ(x; k, l) − x/ϕ(k) |< xε(x), where

s ε(x) = 3

k X 1/2 exp(−X) ϕ(k)C1 (k)

Received by the editor February 23, 1998 and, in revised form, December 17, 1998 and August 21, 2000. 2000 Mathematics Subject Classification. Primary 11N13, 11N56; Secondary 11Y35, 11Y40. Key words and phrases. Bounds for basic functions, arithmetic progression. c

2001 American Mathematical Society

1137

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

1138

PIERRE DUSART

for an explicit constant C1 (k). We apply the above results for k = 3. For small values, we use Ramaré and Rumely’s results [3]. We show that for x > 1531, x (1) . | θ(x; 3, l) − x/2 |< 0.262 ln x If we assume that the Generalized Riemann Hypothesis is true, then we can show that, for x > 1 and k 6 432, | ψ(x; k, l) − x/ϕ(k) |

x ln x

Letting π(x; k, l) =

for x > 17. X

1,

p6x, p≡lmodk

we show an analogous result in the case of arithmetic progression with k = 3 and l = 1 or 2, x for x > 151. π(x; 3, l) > 2 ln x This result, inferred from (1), implies (2) and cannot be proved with Ramaré and Rumely’s results. The method used for k = 3 can also be applied for other fixed integers k. 2. Preliminary lemmas Notations. We will always denote by ρ a nontrivial zero of Dirichlet’s function L, that is to say a zero such that 0 < 1000, and ρ = β+iγ is a zero of L(s, χ) with |γ| > H, then there exists a computable constant C1 (χ, H) such that 1−β >

1 Note

1 . R ln(k|γ|/C1 (χ, H))

that our GRH is an acronym for the usual Generalized Riemann Hypothesis.


ESTIMATES OF θ(x; k, l) FOR LARGE VALUES OF x

1139

Examples. Some examples, extracted from [3, p. 409], appear in the following table. k 1 3 420

Hk 545000000 10000 2500

C1 (χ, Hk ) 38.31 20.92 56.59

Proof. See Theorem 3.6.3 of Ramaré and Rumely [3, p. 409]. Remark. For k > 1 and Hk > 1000, C1 (χ, H) > C1 (χ0 , 1000) > 9.14. As C1 (χ, H) could be large, we limit C1 (χ, H) up to 32π to make some computations. So we have in our hypothesis 9.14 6 C1 (χ, H) 6 32π. From now on, (3)

C1 (k) = min( min C1 (χ, Hχ ), 32π). χmodk

2.2. GRH(k, H) and N (T, χ). Lemma 1 (McCurley [1]). Let C2 = 0.9185 and C3 = 5.512. Write F (y, χ) = y ky and R(y, χ) = C2 ln(ky) + C3 . If χ is a character of Dirichlet with π ln 2πe conductor k, if T > 1 is a real number, and if N (T, χ) denotes the number of zeros β + iγ of L(s, χ) in the rectangle 0 < β < 1, |γ| 6 T , then |N (T, χ) − F (T, χ)| 6 R(T, χ). Lemma 2 (deduced from [3, Theorem 2.1.1, p. 399] and [9]). • GRH(1, H) is true for H = 5.45 × 108 . • GRH(k, H) is true for H = 10000 and k 6 13. • GRH(k, 2500) is true for sets E1 = {k 6 72}, E2 = {k 6 112, k not prime}, E3 = {116, 117, 120, 121, 124, 125, 128, 132, 140, 143, 144, 156, 163, 169, 180, 216, 243, 256, 360, 420, 432}. 2.3. Estimates of |ψ(x; k, l) − x/ϕ(k)| using properties of zeros of L(s, χ). As in Ramaré and Rumely, we remove the zeros with β = 0 and we consider only primitive L-series by adding small terms. Here we take the version stated in [3, Theorem 4.3.1] which is deduced from [1]. Theorem 2 (McCurley [1]). Let x > 2 be a real number, m and k two positive integers, δ a real number such that 0 < δ < x−2 mx , and T a positive real. Let m 1 X m (4) (1 + jδ)m+1 . A(m, δ) = m δ j=0 j


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PIERRE DUSART

Assume GRH(k, 1). Then y ϕ(k) max |ψ(y; k, l) − | < x 16y6x ϕ(k)

A(m, δ)

X X χ

ρ∈℘(χ) |γ|>T

xβ−1 |ρ(ρ + 1) · · · (ρ + m)|

mδ mδ X X xβ−1 ˜ + + R/x, + 1+ 2 |ρ| 2 χ ρ∈℘(χ) |γ|6T

P

˜ where χ denotes the summation over all Pcharacters modulo k, R 1 ϕ(k)[(f (k) + 0.5) ln x + 4 ln k + 13.4] and f (k) = p|k p−1 .

=

2.4. One more explicit form of estimates. The next lemma can be found in [3] with the difference that the authors assumed GRH(k,H) but in fact they used only GRH(k,1). Since we must apply it with T > H, we repeat the proof. Lemma 3. Let χ be a character modulo k. Assume GRH(k, 1). Then, for any T > 1, we have X 1 ˜ ) 6 E(T |ρ| |γ|6T ρ∈℘(χ)

˜ )= with E(T

1 2π

ln2 (T ) +

ln( π

k 2π

)

ln(T ) + C2 + 2

1 π

ln

k 2πe

+ C2 ln k + C3 .

Proof. For |γ| 6 1, we have GRH(k, 1) and so X X 1 1 6 6 2N (1, χ). |ρ| |1/2 + iγ| |γ|61 |γ|61 ρ∈℘(χ)

For |γ| > 1, X 1T ρ∈℘(χ)

X X √ X xβ xβ 1 + 6 x φ (γ) + x . m m+1 m+1 m+1 |γ| |γ| |γ| |γ|>T |γ|>T |γ|>T ρ∈℘(χ)

ρ∈℘(χ)

ρ∈℘(χ)

Let us rewrite Lemma 7 of [6] to adapt it to the new functions F (y, χ) and R(y, χ) which we use. Lemma 5. Write N (y) = N (y, χ), F (y) = F (y, χ), and R(y) = R(y, χ). Let 1 < U 6 V and φ(y) be a positive and differentiable function for U 6 y 6 V . Let (W − y)φ0 (y) > 0 for U < y < V , where W does not necessarily belong to [U, V ]. Let Y be that one of the numbers U, V, W which is not numerically the least or greatest (or is the repeated one, if two among U, V, W are equal). Take j = 0 or 1, accordingly as W < V or W > V . Then Z Z V X ky 1 V φ(y) dy + Bj (Y, U, V ), φ(|γ|) 6 φ(y) ln dy + (−1)j C2 π U 2π y U U F (y) − R(y). Z X ky 1 V φ(|γ|) 6 [(N (y) − F (y) + R(y))φ(y)]VU + ln φ(y)dy π U 2π U W . Take Y = max(U, W ). Split the integral at Y . Then −φ0 (y) 6 0 for y ∈ [U, Y ] and −φ0 (y) > 0 for y ∈ [Y, V ]. Replacing N (y) by F (y) − R(y) in the first part and by F (y) + R(y) in the second part, we obtain Z Z Y Z V X ky 1 V φ(|γ|) 6 ln R0 (y)φ(y)dy − R0 (y)φ(y)dy φ(y)dy + π U 2π Y U

U 0, √ C1 (k) exp(X/ m + 1). (7) Wm = k Corollary 1 (Corollary from Lemma 5). Under the hypothesis of Lemma 5, if moreover 2π ke 6 U , then Z X V φ(|γ|) 6 1/π + (−1)j q(Y ) φ(y) ln(ky/2π)dy + Bj (Y, U, V ), U

U 2π/(ke). • Case (j = 0), then Y = max(U, W ). Z Z V X 1 V ky dy + φ(|γ|) < B0 (Y, U, V ) + φ(y) ln R0 (y)φ(y)dy. π U 2π Y U 1 an integer, H > 1000 a real number. Assume GRH(k, H). Let x0 > 2 be a real number, m a positive integer, and δ a real number such that 0 < δ < (x0 − 2)/(mx0 ) and let Y be defined as in Lemma 5. We write Z Z ∞ ky 1 ∞ φm (y) (8) dy, φm (y) ln dy + C2 A˜H = π H 2π y H ˜H = B0 (Y, H, ∞), (9) B kH 1 (10) + 1/m , C˜H = ln mπH m 2π C2 ˜H = (11) 2C2 ln(kH) + 2C3 + /H m+1 . D m+1 Then for all x > x0 , we have y ϕ(k) max |ψ(y; k, l) − | 16y6x x ϕ(k)

√ ϕ(k) ˜ ˜H + (C˜H + D ˜ H )/ x 6 A(m, δ) AH + B 2 √ mδ mδ ˜ ˜ + R/x. x+ ϕ(k)E(H)/ + 1+ 2 2

Remark. We find a version of Theorem 4.3.2 of [3] where x0 is replaced by x in A˜ ˜ and B. Proof. According to Theorem 2, y ϕ(k) max |ψ(y; k, l) − | < x 16y6x ϕ(k)

A(m, δ)

X X χ

ρ∈℘(χ) |γ|>H

xβ−1 |ρ(ρ + 1) · · · (ρ + m)|

mδ mδ X X xβ−1 ˜ + + R/x. + 1+ 2 |ρ| 2 χ ρ∈℘(χ) |γ|6H


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PIERRE DUSART

We separately examine the different parts: • We have X X xβ−1 X X xβ−1 6 . |ρ(ρ + 1) · · · (ρ + m)| |γ|m+1 χ ρ∈℘(χ) χ ρ∈℘(χ) |γ|>H

|γ|>H

By Lemma 4, X X χ

ρ∈℘(χ) |γ|>H

xβ−1 |γ|m+1

 =

X 1  X xβ−1 X xβ−1  +   m+1 2 ρ∈℘(χ) |γ| |γ|m+1 ρ∈℘(χ) χ 

6



|γ|>H

|γ|>H

 X X X 1 1  1  φm (γ) + √  . m+1 2 χ x |γ|>H |γ| |γ|>H ρ∈℘(χ)

ρ∈℘(χ)

Using Lemma 5 with U = H, V = ∞, φ = φm , and W = Wm , X ˜H . φm (γ) 6 A˜H + B |γ|>H ρ∈℘(χ)

Integration by parts gives X |γ|>H ρ∈℘(χ)

1 |γ|m+1

˜H. 6 C˜H + D

• By GRH(k, H) we have β = 1/2 for all |γ| 6 H, and by Lemma 3, X xβ−1 √ ˜ 6 E(H)/ x. |ρ| ρ∈℘(χ) |γ|6H

2.5. The leading term (A˜H ). To obtain an upper bound for the leading term, we proceed like Rosser and Schoenfeld with upper bounds on the integrals. The next three lemmas are issued directly from [6, p. 251-255]. Lemma 6 (Functions of incomplete Bessel type). Let Z 1 ∞ ν−1 z t H (t)dt, Kν (z, u) = 2 u where z > 0, u > 0, and z H z (t) = {H(t)}z = exp{− (t + 1/t)}. 2 Further, write Kν (z, 0) = Kν (z). Then r 3 π exp(−z) 1 + (12) , K1 (z) 6 2z 8z r 15 105 π exp(−z) 1 + + (13) . K2 (z) 6 2z 8z 128z 2 Lemma 7. Kν (z, x) + K−ν (z, x) = Kν (z). Hence, Kν (z, x) 6 Kν (z) (ν > 0).



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Lemma 8. Let Qν (z, x) =

xν+1 exp{−z(x + 1/x)/2}. z(x2 − 1)

If z > 0 and x > 1, then K1 (z, x) < Q1 (z, x) and K2 (z, x) < (x + 2/z)Q1 (z, x). The term A˜H can be expressed using incomplete Bessel functions. q √ ln x and Um = Lemma 9. Let X be defined by (6). Let zm = 2X m = 2 m R q 2m kH Rm kH zm ln C1 (k) = ln x ln C1 (k) . m k 2 ln x K2 (zm , Um ) A˜H = π Rm C1 (k) r m C1 (k) k ln x 2 K1 (zm , Um ) + ln π 2π Rm C1 (k) s m+1 k ln x K1 (zm+1 , Um+1 ). +2C2 R(m + 1) C1 (k) Proof. This is by straightforward algebraic manipulation; for example, we write Z ∞ − ln x dy C2 . exp I= m+1 y R ln(ky/C (k)) y 1 H Changing variables: r ky R(m + 1) ln , t = ln x C1 (k) r R(m + 1) dy . dt = ln x y Now   − ln x − ln x  q = exp  exp R ln(ky/C1 (k)) Rt/ R(m+1) ln x ! r −zm+1 1 (m + 1) ln x 1 = exp = exp R t 2 t and 1 y m+1

=

k C1 (k)

Consequently, Z I=

m+1

C2



(m + 1)t  exp − q = R(m+1) ln x

s

∞

Um+1



ln x R(m + 1)

k C1 (k)

k C1 (k)

m+1 exp

m+1

z m+1 . exp −t 2

−zm+1 (t + 1/t) . 2


1146

PIERRE DUSART

˜ 2.6. Study P of1 f (k) which appears in the expression of R. Remember that . f (k) = p|k p−1 Lemma 10. For an integer k > 1, f (k) 6

ln k . ln 2

f (k) 6

ln k . ln 2

Proof. We prove by recursion that

2 ln k For k = 1, it is obvious. For k = 2, f (k) = 1 6 ln ln 2 . Assume f (k) 6 ln 2 holds for k 6 n. Find an upper bound for f (n + 1). If (n + 1) is prime, then f (n + 1) = 1/n 6 ln n/ ln 2. If (n + 1) is not prime, then there exists p 6 n, which divides n. If p = 2 and 2α k n + 1, n+1 n+1 α ·2 =f + f (2) f (n + 1) = f 2α 2α ln 2 ln(n + 1) n+1 +1− 6 = 1+f α 2 ln 2 ln 2 ln(n + 1) . 6 ln 2

If p > 2 and pα k n + 1,

n+1 n+1 α · p = f + f (p) pα pα n+1 1 ln p ln(n + 1) 1 +f + − 6 α p−1 p ln 2 p − 1 ln 2 1 ln p ln(n + 1) because − < 0 for p > 2. ln 2 p − 1 ln 2

f (n + 1) = f = 6

3. The method with m = 1 Theorem 4. Let k be an integer, H > 1250, q and H > k. Assume GRH(k, H). Let C1 (k) defined by (3). Let x > 1. Write X =

ln x R

and

s

1 (15/16 + ln(C1 (k)/(2π))) X 3/4 exp(−X). 1+ ε(x) = 2 2X √ , then If ε(x) 6 0.2 and X > 2 ln CkH 1 (k) kϕ(k) √ C1 (k) π

max | ψ(y; k, l) − y/ϕ(k) |6 xε(x)/ϕ(k).

16y6x

√ , then W1 > H. Proof. Take m = 1 in Theorem 3. Assuming X > 2 ln CkH 1 (k) ˜ In this situation, Y = W1 and BH < 2R(W1 )φ1 (W1 ). For y > 1, R(y)/ ln y is



1147

decreasing; hence, ˜H B

R(H) φ1 (W1 ) ln W1 < 2R(W1 )φ1 (W1 ) < 2 ln H C1 (k) R(H) X √ + ln φ1 (W1 ) = 2 ln H k 2 √ C1 (k) R(H) X √ + ln (k/C1 (k))2 exp(−2 2X). = 2 ln H k 2

Inserting the upper bounds (12) and (13) into the bound for A˜H in Lemma 9, r k 15 105 π ˜ exp(−2X) 1 + + AH < 2 X 2 /π C1 (k) 4X 16X 512X 2 r 3 π 1 C1 (k) X exp(−2X) 1 + + ln π 2π 4X 16X r √ 3 kX π √ √ exp(−2 2X) 1 + √ . + C2 C1 (k) 2 4 2X 16 2X Put 2 15 k C1 (k) 1 1 3/2 X + ln exp(−2X) 1 + . F1 := √ π C1 (k) 16 2π 2X In Lemma 11 below it is shown that √ ˜H + (C˜H + D ˜ H + 3E(H))/ ˜ ˜ x+R A˜H + B

2 < F1 . xϕ(k)

We must choose δ to minimize A(1, δ) ϕ(k)F1 + δ/2. 2 As A1 (δ) = (δ 2 + 2δ + 2)/δ, we must q minimize g(δ) = 2f , and the value (δ/2 + 1 + 1/δ)f + δ/2. The minimum value here is at δ = 1+f q p 2f ) = f + 2f (1 + f ). there is g( 1+f It is a simple matter to prove that for 0 6 f 6 0.202, p p f + 2f (1 + f ) < 2 f . √ , then x0 > exp(122.5), and it is obvious that δ As X > X0 := 2 ln CkH (k) 1 meets the hypothesis 0 < δ < (x0 − 2)/x0 in Theorem 3 since √ p x0 . 0 < δ < 2 f < 0.6357 < x0 − 2 Write f = ϕ(k)F1 .

Lemma 11. √ ˜H + (C˜H + D ˜ H + 3E(H))/ ˜ ˜ x+R A˜H + B

2 < F1 . xϕ(k)


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PIERRE DUSART

˜H < F1 : Proof. First we prove that A˜H + B k √ X 3/2 e−2X 1 + (15/16 + ln(C1 (k)/2π))/X F1 = C1 (k) π 1 15 ln(C1 (k)/2π) + ln2 (C1 (k)/2π))/X 2 , + (225/1024 + 32 4 C1 (k) 105 1 3 k 15 √ X 3/2 e−2X 1 + + + + ln A˜H < C1 (k) π 16X 512X 2 2π X 16X 2 ! √ √ kπ p √ exp(−2( 2 − 1)X)(1/X + 3/(16 2X 2 )) , + C2 C1 (k) 2 2 √ ˜H < √ k X 3/2 exp(−2X) exp(−2( 2 − 1)X) B πC1 (k) √ 1 1 2k π + √ ln(C1 (k)/k) . (C2 ln(kH) + C3 ) √ × C1 (k) ln H 2X X X ˜H > 0 if This yields F1 − A˜H − B 9 C1 (k) 1 1 2 C1 (k) 15 + ln + ln F2 := 2 X 1024 32 2π 4 2π √ √ 1 C2 πk exp(−2( 2 − 1)X) √ > C1 (k) 2X ! "r r 3 π 2 √ × 3/2 X 16X 2 2 !# √ 2 C1 (k) ln k + C3 /C2 ln . 1+ + 2 1+ ln H X k This holds if we can show that √ √ C2 k π 1 exp(−2( 2 − 1)X) √ · 16.9, F2 > C1 (k) 2X √ since C1 (k) 6 32π, H > 1250, X > 2 ln(1250/32π), and k 6 H. It remains to be proved that √ √ 2C (k) √ 1 (15/1024 + · · · ) > X 3/2 exp(−2( 2 − 1)X). kC2 π · 16.9 √ , But for X > X0 := 2 ln CkH 1 (k) −(1+a) kH C1 (k) ! 1+a 1 3/4 C1 (k) ln3/2 (kH/C1 (k)) ·2 , k H ka

√ 3/2 X 3/2 exp(−2( 2 − 1)X) < X0 =

√ √ 3/2 1 (k)) reaches its where a = 2 2( 2 − 1) − 1 ≈ 0.17157. The map k 7→ ln (kH/C ka 3 C1 (k) maximum for k = e 2a H . Hence 3/2 √ C1 (k) 3/4 3 3/2 2 exp(−2( 2 − 1)X) < /e3/2 . X kH 2a



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We must compare

√ 3 3/2 23/4 ( 2a ) 2 √ (15/1024 + · · · ) with . C2 π · 16.9 He3/2

Since C1 (k) > 9.14 (see the remark above (3)) and C2 = 0.9185, it remains to be proved that 0.007976 >

3 3/2 ) 23/4 ( 2a (≈ 0.00776), 3/2 He

which is true since H > 1250. √ ˜ H + 3E(H))/ ˜ ˜ 2 are x+R We show below that the remaining terms (C˜H + D xϕ(k) negligible. ˜ ˜ H ) + 3 ϕ(k) E(H) ˜ √ (C˜H + D + R/x. • We will find an upper bound for A(1, δ) ϕ(k) 2 2 x √ √ ; hence, X > X0 := 2 ln 1250 ≈ 3.5644. It We assume that X > 2 ln CkH 32π 1 (k) is straightforward but tedious to check that ˜ 2R 1250(ln H ln k)2 if k 6= 1, ˜ H + 3E(H) ˜ √ 6 + Rest := C˜H + D if k = 1. 1250(ln H)2 ϕ(k) x √ , Let us consider the case k 6= 1. As X > 2 ln CkH 1 (k) X kH . > exp √ C1 (k) 2 This yields Rest 6 1250(ln H ln k)2

6

√ (ln H ln k)2 1250 2 exp(X 2) kH C1 (k)

6 6

2 √ 1 ln 1250 exp(X 2) 1250C12(k) 2 e 1250 √ K exp(X 2) because C1 (k) 6 32π,

where K := 55.65. Now compare √ √ K exp(X 2) √ = K exp(X 2 − RX 2 /2) x with the term involving 1/X 2 in F1 k 1 √ X 3/2 exp(−2X). × X2 C1 (k) π We may compute c such that √ 1 k √ X 3/2 exp(−2X) K exp(X 2 − RX 2 /2) 6 c × 2 × X C1 (k) π q √ √ X2 ⇔ c > K 32π π exp(X 2 − RX 2 /2 + 2X) 3/2 X ⇔ c > 0.7 · 10−18 for X > X0 . Thus, the rest is negligible and absorbed by rounding up the constants.


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PIERRE DUSART

4. The method with m = 2 Lemma 12. Let A(m, δ) be defined as in formula (4). Write m Rm (δ) = 1 + (1 + δ)m+1 . Then A(m, δ) 6

Rm (δ) . δm

Proof. The proof appears in [4, p. 222]. Theorem 5. Let an integer k > 1. Remember that R = 9.645908801. Let H > 1000. Assume GRH(k, H). Let C1 (k) be defined by (3). Let X0 , X1 , X2 , and X3 be such that s eX1 kϕ(k) eX0 √ , =H = 10ϕ(k), 2πC1 (k) X1 X0 X2 = kC1 (k)/(2πϕ(k)),

X3 =

2kπe . C1 (k)ϕ(k)

Let X4 := max(10, X0 , X1 , X2 , X3 ). Write s k X 1/2 exp(−X). ε(X) = 3 ϕ(k)C1 (k) q Then for all real x such that X = lnRx > X4 , we have ! r ln x , max |ψ(y; k, l) − y/ϕ(k)| < xε 16y6x R ! r ln x . max |θ(y; k, l) − y/ϕ(k)| < xε 16y6x R Corollary 2. With the notations and the hypothesis of Theorem 5, let X5 > X4 and c := ε(X5 ). For x > exp(RX52 ), we have |ψ(x; k, l) − x/ϕ(k)|,

|θ(x; k, l) − x/ϕ(k)| < cx.

Proof. The idea is to judiciously split the integral into two parts, and bound each part optimally, using an m = 0 estimate in the first part and an m = 2 estimate in the second part. We want to split the integral at T , where T will optimally be chosen later. We take T in the same form as Wm (formula (7)): (14)

T :=

C1 (k) exp(νX), k

where ν is a parameter. √ Assume that T > H and 1/ m + 1 6 ν 6 1. Hence Wm 6 T 6 W0 . This hypothesis is needed to apply Corollary 1. We use Theorem 2 and split the sums at T : X X mδ X X xβ−1 mδ xβ−1 + 1+ + + A(m, δ) |ρ(ρ + 1) · · · (ρ + m)| 2 |ρ| 2 χ ρ∈℘(χ) χ ρ∈℘(χ) |γ|>T


|γ|6T

last

˜ R . x


1151

Define A˜1

:=

X X xβ−1 , |ρ| χ ρ∈℘(χ) |γ|6T

A˜2

:=

X X χ

ρ∈℘(χ) |γ|>T

Bounding the term A˜1 , we get  X  X xβ−1 + A˜1 =  |ρ| ρ∈℘(χ) χ |γ|6H

=

 √ X x 1  X +  x χ |ρ| ρ∈℘(χ) |γ|6H

xβ−1 . |ρ(ρ + 1) · · · (ρ + m)| 

X

β−1

ρ∈℘(χ) HT



X xβ xβ  1 X X +   m+1 2x χ |γ| |γ|m+1 ρ∈℘(χ) ρ∈℘(χ)

6



|γ|>T

|γ|>T

 X X X √ 1  1  φm (γ) + x x m+1  2x χ |γ| ρ∈℘(χ) ρ∈℘(χ)

=

|γ|>T

|γ|>T

by Lemma 4. By using Corollary 1 (j = 0) on [U, V ] = [T, ∞), Z ∞ X ky φm (γ) 6 {1/π + q(T )} φm (y) ln( )dy + B0 (T, T, ∞). 2π T ρ∈℘(χ) |γ|>T

We have B0 (T, T, ∞) < 2R(T )φm (T ). Moreover, X ρ∈℘(χ) |γ|>T

1 ˜T . 6 C˜T + D |γ|m+1



1153

Let us study more precisely Z ∞ ky φm (y) ln dy I2 := 2π T m 2 k 2md zm K1 (zm , Um ) , K2 (zm , Um ) + = 2m2 C1 (k) zm √ kT ln = ν m. Now, by writing where d = ln C12π(k) and U 0 := Um = 2m zm C1 (k) z = zm and using Lemma 8, 2dm K1 (z, U 0 ) < (U 0 + 2/z + 2dm/z)Q1(z, U 0 ) K2 (z, U 0 ) + z 2 √ 0 0 z U0 1 + dm e− 2 (U +1/U ) . m ν+ 6 mX z(U 0 2 − 1) √ √ √ But z2 (U 0 + 1/U 0 ) = X m(ν m + 1/(ν m)) = mνX + X/ν = mνX + (Y 00 + 2X − νX), where Y 00 = X(1 − ν)2 /ν. Hence −(m−1) kT 2dm m 0 0 −Y 00 −1 −2X K1 (z, U ) < G1 e X e , K2 (z, U ) + z 2(m − 1) C1 (k) 1+dm U02 because where G1 := m−1 m U 0 2 −1 ν + mX m−1 kT eνX(m−1) = C1 (k) √

and

m z

= 12 X −1 . This yields Z ∞ 00 k G1 e−Y Xe−2X T −(m−1) φm (y) ln(ky/2π)dy < I2 = m − 1 C1 (k) T

Let G2 :=

Rm (δ) 2m (1

+ πq(T )). So, by using Lemma 12, Z ∞ ϕ(k) (1/π + q(T )) φm (y) ln(ky/2π)dy A(m, δ) 2 T ( ) m 00 2 ϕ(k) G2 kG1 e−Y −2X −(m−1) Xe T . < δ 2 π (m − 1)C1 (k)

The results above yield (1 + mδ/2)A˜1 + A(m, δ)A˜2 00

(15)

1) ˜ R 1 ε1 := (1 + mδ/2)A˜1 + A(m, δ)A˜2 + mδ + 2 x ˜ 00 R 2kϕ(k) 1 1/m 1−1/m G1 G0 +δ +r+ . < mG2 Xe−2X e−Y 2 δ(m − 1)πC1 (k) x The expression between braces can be minimized by choosing 1/2 2kϕ(k) 1−1/m 1/m −Y 00 (17) G1 e X 1/2 e−X . δ = G0 (m − 1)πC1 (k) Hence, we write (by replacing the above value of δ in (16)) 1/2 1/2m 2C1 (k) G1 Y 00 (m − 1)πe /G0 (18) X −1/2 eX T = G0 kϕ(k) and (19)

1/2 ˜ R m 1−1/m 1/m −Y 00 2kϕ(k) √ G1 e X 1/2 e−X + r + . ε1 < G2 G0 πC1 (k) x m−1

m . For the remainder of the The value m = 2 minimizes the expression √m−1 argument, we fix m = 2. We now have two definitions for T . On the one hand (equation (18)), s 1/4 2πC1 (k) −1/2 X G1 Y 00 /2 X e e T = G30 kϕ(k)

with Y 00 = X(1 − ν)2 /ν, and on the other hand (equation (14)) T =

C1 (k) exp(νX). k

These two equations are compatible if and only if there exists ν such that f (ν) = 1, where 1/2 2 C1 (k)ϕ(k) G30 Xe−X(1−ν) /ν e−2X(1−ν) . f (ν) = 2πk G1 √ √ Here we have m = 2 and our assumption 1/ m + 1 6 ν 6 1 gives 1/ 3 6 ν 6 1. Note that G0

= ν 2 (ν + d/X),

ν2 1 + 2d 1 + dm = 2 ν+ . ν+ G1 = mX 2ν − 1 2X √ It is easy to check that on the interval 1/ 2 6 ν 6 1, G30 /G1 is increasing, and hence, f (ν) is strictly increasing. Moreover, limν→(1/√2)+ f (ν) = 0 and f (1) > 1 m − 1 U 02 m U 02 − 1



1155

√ 2πk (for all X > C1 (k)ϕ(k) ). So there exists a unique ν ∈ ]1/ 2, 1[ such that f (ν) = 1. √ For 1/ 2 < ν < 1, we have (m = 2) H(ν) := Write, for X > X3 :=

[ν 2 (ν + d/X)]3 G30 < (ν + d/X)2 . = ν2 1+2d G1 (ν + ) 2 2ν −1 2X

2πke C1 (k)ϕ(k) ,

1 ln ν0 = 1 − 2X

(20)

C1 (k)ϕ(k)X 2kπ

.

Let us study H(ν0 ): H(ν0 ) < 1

if

equivalently which holds if As f (ν) =

C1 (k)ϕ(k) 2kπ

ν0 + d/X 6 1, C1 (k)ϕ(k)X ln(C1 (k)/2π) 1 ln 6 1, + 1− 2X 2πk X kC1 (k) . X > X2 := 2πϕ(k) G30 G1

1/2 X exp(−X(1 − ν)2 /ν) exp(−2X(1 − ν)),

replacing ν0 by (20), we obtain 3 1/2 G0 C1 (k)ϕ(k)X 2 exp − ln /(4ν0 X) . f (ν0 ) = G1 2kπ Assume that ν0 > 0, then, for X > X2 , f (ν0 ) < 1 = f (ν) and hence ν0 < ν. We will require X > X2 . The assumption T > H holds if T > C1k(k) exp(ν0 X) > H. Using (20), rewrite q C1 (k) 2πC1 (k) X− 1 ln X 2 exp(ν X) = . Let X0 satisfy 0 k kϕ(k) e s kϕ(k) X0 − 12 ln X0 . =H e 2πC1 (k) We have T > H provided that X > X0 . We will require X > X0 . For X > X3 = C1 2kπe (k)ϕ(k) , ν0 is an increasing function of X. We will require that X > max(X3 , 10). Then since C1 (k) 6 32π and X > 10, we have ν0 > 0.7462413 and ν0 < ν < 1. √ The assumption ν > 1/ 2 is satisfied. We want to evaluate p 00 (21) K := G2 ( G0 G1 e−Y )1/2 , which appears in (19). Again using C1 (k) 6 32π and X > 10, we find ν0 1 + 2d ν0 + G0 G1 < (1 + d/X) 2 2ν0 − 1 2X < 8.995. The following results will be needed in later computations.


1156

PIERRE DUSART

√ 1. Since X > X0 and exp(X)/ X is increasing for X > 1/2, s 1 kϕ(k) X 1/2 exp(−X) 6 . 2πC1 (k) H 2. Since G0 G1 < 9,

s p kϕ(k) 4 00 X 1/2 e−X = 2 G0 G1 exp(−Y /2) 2πC1 (k) √ 6 2 3/H.

δ

In particular, for H > 1000, we have δ 6 0.00347. 3. G2 =

R2 (δ) (1 + πq(T )) < (1 + 3.012 · δ/2)2 (1 + πq(T )), 22

because

R2 (δ) 22

= =

(1 + δ)3 + 1 2

2

2 2 1 3.012 δ 1 + δ(3 + 3δ + δ 2 ) < 1 + 2 2

since 1 + δ + δ 2 /3 < 1.0035. 4. Since T > H, q(T ) = 6

C2 T ln(kT /2π) C2 . H ln(kH/2π)

But exp(−Y 00 /2) 6 1 and H > 1000, so this yields K

< (8.995)1/4 G2 < (8.995)1/4 1 +

πC2 1000 ln(1000/(2π))

×

√ !2 3.012 2 3 1+ 2 1000

< 1.751. Inserting this upper bound of K (see formula (21) in (19), we obtain s r ˜ R 2 kϕ(k) 1/2 K X exp(−X) + r + ε1 < 2 π C1 (k) x s ˜ R kϕ(k) 1/2 (22) X exp(−X) + r + . < 2.7941 C1 (k) x ˜

Now we want to bound r and R x. • An upper bound for ϕ(k)(1 + δ)R(T )φ0 (T ) and ϕ(k)A(2, δ)R(T )φ2 (T ). Recall that R(T ) = φ0 (T ) = φm (T ) =

C2 ln(kT ) + C3 , 1 exp −X 2 / ln(kT /C1 (k)) , T φ0 (T )T −m.



1157

Now φ0 (T ) = and

1 1 1 1 exp(−X 2 /(νX)) = exp(− X) 6 exp(−X) T T ν T

s 1/2 1/4 kϕ(k) G0 G0 1 1/2 = X exp(−X) , T C1 (k) 2πeY 00 G1

hence R(T )φ0 (T ) 6 6

C2 ln(kT ) + C3 exp(−X) Ts √ −X kϕ(k) G0 1/2 G0 1/4 −X Xe e (C2 ln(kT ) + C3 ) . C1 (k) 2πeY 00 G1

But G0

6

G0 6 G1 exp(Y 00 ) > ln(kT ) =

1+

ln(C1 (k)/2π) , X

2ν 2 − 1 < 1

(m = 2),

1, νX + ln(C1 (k)) 6 X + ln(C1 (k)) 6 X + ln(32π).

So, since X > 10 and C1 (k) 6 32π, # " 1/2 1/4 G0 G0 exp(−X) (1 + δ)ϕ(k) (C2 ln(kT ) + C3 ) 2πeY 00 G1 r √ ! 1 + ln 16/10 2 3 [C2 (X + ln 32π) + C3 ] exp(−X) 6 ϕ(k) 1 + 1000 2π 6

0.857ϕ(k)X exp(−X).

Furthermore, if X1 is defined by exp(X1 )/X1 = 10ϕ(k), and if we require that X > X1 , then this term is bounded by 0.0857. Hence, under the hypotheses on X in Theorem 5, an upper bound for ϕ(k)(1 + δ)R(T )φ0 (T ) is s kϕ(k) 1/2 X exp(−X). 0.09 · C1 (k) Next, by (16)

r δT = 2

G1 . G0

Hence, by Lemma 12 A(2, δ)/T 2 6

R2 (δ) R2 (δ) G0 R2 (δ) 6 6 2 2 (δT ) 2 G1 22

and ϕ(k)A(2, δ)R(T )φ2 (T ) 6 ϕ(k)

R2 (δ) R(T )φ0 (T ). 22


1158

PIERRE DUSART

√ √ Using δ 6 2 3/H 6 2 3/1000, we get R2 (δ)/22 6 1.0147. Under the hypotheses on X in Theorem 5, an upper bound for ϕ(k)A(2, δ)R(T )φ2 (T ) is therefore s kϕ(k) 1/2 X exp(−X). 0.087 · C1 (k) The sum of the two terms can be bounded by s kϕ(k) 1/2 X exp(−X). (23) 0.2 · C1 (k) ϕ(k) ˜ T ) + R/x. ˜ ˜ ) ϕ(k) √ + A(2, δ) √ (C˜T + D • An upper bound for (1 + δ)E(T x 2 x P 1 observe that (Lemma 10) For f (k) = p|k p−1 ln k . ln 2 We can explicitly rewrite for m = 2, H > 1000, and C1 (k) 6 32π the following expressions: k 1 1 2 ˜ ln T + ln ln T + C2 3E(T ) = 3 2π π 2π k 1 ln +2 + C2 ln k + C3 , π 2πe kT 1 + 1/2 , ln C˜T = 2πT 2 2π ˜ T = (2C2 ln(kT ) + 2C3 + C2 /3)/T 3, D f (k) 6

˜ R √ ϕ(k) x

6

√ [(f (k) + 0.5) ln x + 4 ln k + 13.4] / x.

It is tedious but easy to check that the sum of the above quantities is less than √ 1000(ln T ln k)2 for k 6= 1, for k = 1. 1000 ln2 T Now we want to find a number c such that √ 1/2 kϕ(k) 1000(ln T ln k)2 √ 6c X 1/2 exp(−X) A(2, δ)ϕ(k) x C1 (k) q 1/2 2 1 and by (16), T = G with X = lnRx . But A(2, δ) 6 R2δ(δ) 2 G0 δ , so A(2, δ) 6 Moreover,

√1 x

c > 1000 G0 G1

R2 (δ) 2 G0 T . 22 G1

= exp(−RX 2 /2), hence

√ R2(δ) G0 2 T ϕ(k)(ln T ln k)2 2 2 G1

< 1, T 2 =

C12 (k) k2

exp(2νX) 6

C1 (k) kϕ(k)

1/2

X −1/2 exp(X − RX 2 /2).

C12 (k) k2

exp(2X), hence it suffices to take 1/2 C1 (k)ϕ(k) R2(δ) 2 ln k 2 exp(3X − RX 2 /2). c > 1000 2 C1 (k) 2 (ln(C1 (k)/k) + X) 2 k kX

As



1159 √

√

2 3 But ϕ(k) 6 1, lnk2k 6 1, and R2 (δ) 6 (1 + 3.012δ/2)2 with δ 6 2H3 6 1000 . So, k finally, it suffices to take √ !2 p 3.012 3 1000 1+ C12 (k)(ln C1 (k) + X)2 C1 (k)X −1/2 exp(3X − RX 2 /2). c> 4 1000

Since C1 (k) 6 32π and X > 10, we can take c = 0.643 · 10−187 .

(24)

In the case k = 1, we can replace the upper bound lnk2k 6 1 by 1, and obtain the same result. Combining (22), (23), and (24), we obtain the result in Theorem 5; more precisely, for all X satisfying the conditions of the theorem, s k X 1/2 exp(−X). | ψ(x; k, l) − x/ϕ(k) | /x 6 2.9941 ϕ(k)C1 (k) We also wish to allow θ instead of ψ, which can be done by recalling Theorem 13 of [5]: √ 0 6 ψ(x; k, l) − θ(x; k, l) 6 ψ(x) − θ(x) 6 1.43 x for x > 0. √ Using X > 10, we find 1.43 x/x 6 d · 3(kϕ(k))/C1 (k) X 1/2 exp(−X), where d = 1.17 · 10−204 . This difference is absorbed by rounding up the constants. 5. Application for k = 3 Now we are able to compute x0 and c such that, for x > x0 , | θ(x; 3, l) − x/2 |< cx/ ln x. This would not have been possible if we had used only the results of [3]. According to Theorem 5, r 6 3 X 1/2 exp(−X) ε(X) = 2 20.92 for k = 3. To determine for which x this bound is valid, let us solve for the constants X0 , X1 , X2 , X3 in Theorem 5. Noting that H3 = 10000 by the table in Theorem 1, we need X0 to satisfy r 1 6 ≈ 2136.51. exp(X0 − ln X0 ) > 10000 2 2π · 20.92 X0 ≈ 8.76 works. Find X1 such that exp(X1 − ln X1 ) > 20. X1 ≈ 4.5 works. Compute the two other bounds: X2 ≈ 4.99, X3 ≈ 1.22. Thus we can take X = max(10, X0 , X1 , X2 , X3 ) = 10 in Theorem 5. q q • For lnRx > 10, write X = lnRx , then ε(X) ln x = RX 2 ε(X). Find the value c such that ε(X) < c/ ln(x).


1160

PIERRE DUSART

q For any x such that x > exp(964.59 · · · ),

ln x R

> 10, c 6 R · 102 ε(10) 6 0.12. Hence we have for

|θ(x; 3, l) − x/2| 6 0.12

x . ln x

We want to extend the above result for x 6 exp(964.59 · · · ). Olivier Ramaré has kindly computed some additional values supplementing Table 1 in [3]. We have |θ(x; 3, l) − x/2| < c˜ · x/2 with c˜ = c˜ =

0.0008464421 for ln x >= 400 (m = 3, δ = 0.00042325), 0.0006048271 for ln x >= 500 (m = 3, δ = 0.00030250),

c˜ =

0.0004190635 for ln x >= 600 (m = 2, δ = 0.00027950).

Hence, • For e600 6 x 6 e964.59... c 6 0.0004190635 · 964.6/ϕ(3) 6 0.203. • For e400 6 x 6 e600 c 6 0.0008464421 · 600/ϕ(3) 6 0.254. Using the computations of [3], • For 10100 6 x 6 e400 c 6 0.001310 · 400/ϕ(3) 6 0.262. • For 1030 6 x 6 10100 c 6 0.001813 · 100 ln 10/ϕ(3) 6 0.42/2 6 0.21. • For 1013 6 x 6 1030 c 6 0.001951 · 30 ln 10/ϕ(3) 6 0.14/2 6 0.07. • For 1010 6 x 6 1013 c 6 0.002238 · 13 ln 10/ϕ(3) 6 0.067/2 6 0.00335. • For 4403 6 x 6 1010 √ | θ(x; 3, l) − x/2 |< 2.072 x (Theorem 5.2.1 of Ramaré and Rumely [3]) We choose c = 0.262. We check that this bound is also valid for 1531 6 x 6 4403. Theorem 6. For x > 1531, | θ(x; 3, l) − x/2 |6 0.262

x . ln x



1161

6. Results assuming GRH(k,∞) Assuming GRH(k,∞), we obtain more precise results. Under this hypothesis, one can show that function ψ has the following asymptotic behaviour: Proposition 1 ([8, p. 294]). Assume GRH(k, ∞). Then √ x + O( x ln2 x). ψ(x; k, l) = ϕ(k) Theorem 7. Let x > 1010 . Let k be a positive integer. Assume GRH(k, ∞). 1) If k 6

4 5

ln x, then |ψ(x; k, l) −

√ x | 6 0.085 x ln2 x. ϕ(k)

|ψ(x; k, l) −

√ x | 6 0.061 x ln2 x. ϕ(k)

2) If k 6 432, then

Proof. Let x0 = 1010 . Applying Theorem 2 in the same way as Theorem 3 (assume that T > 1), x ϕ(k) |ψ(x; k, l) − | x ϕ(k) X X 6 A(m, δ) χ |γ|>T

+(1 + mδ/2)

x−1/2 |ρ(ρ + 1) · · · (ρ + m)|

X X x−1/2 ˜ + mδ/2 + R/x |ρ| χ |γ|6T

6

˜ mδ R mδ 1 X X 1 1 1 X X √ ) + + + (1 + A(m, δ) √ x χ |γ|m+1 2 x χ |ρ| 2 x

6

ϕ(k) ˜ ) + mδ + R/x. ˜ ˜ T ) + (1 + mδ ) ϕ(k) √ E(T A(m, δ) √ (C˜T + D x 2 x 2

|γ|>T

|γ|6T

Take m = 1 and let (25)

δ R1 (δ) ϕ(k) ˜ δ ϕ(k) ˜ ˜ ˜ √ √ E(T ) + + R/x, (CT + DT ) + 1 + εk (x, T, δ) := δ 2 2 x x

where C˜T

=

˜T D

=

˜ ) = E(T

kT 1 ln +1 , πT 2π 1 (2C2 ln(kT ) + 2C3 + C2 /2) , T2 k 1 1 1 2 ln T + ln(k/(2π)) ln T + C2 + 2 ln + C2 ln k + C3 . 2π π π 2πe

Choose (26)

T =

2R1 (δ) δ(2 + δ)


1162

PIERRE DUSART

to minimize in (25) the preponderant terms involving T . So kR1 (δ) 2(2 + δ) R1 (δ) ˜ ˜ (CT + DT ) = ln +1 δ 4π πδ(2 + δ) 2kR1 (δ) πδ(2 + δ) 2C2 ln + 2C3 + C2 /2 , + 2R1 (δ) δ(2 + δ) 2R1 (δ) 2R1 (δ) 2+δ 2 ˜ ln + 2 ln(k/(2π)) ln (1 + δ/2)E(T ) = 4π δ(2 + δ) δ(2 + δ) 1 ln(k/(2πe)) + C2 ln k + C3 . +2πC2 + 4π π With the choice of T , the main terms of εk are δ ϕ(k) 1 2 2R1 (δ) √ ln + . δ(δ + 2) 2 x 2π These terms are minimized by choosing (27)

δ=

ϕ(k) ln x √ . π x

Now, replacing (26) and (27) in (25), we only have a function of x for fixed k: εk (x) := εk (x, T, δ). We simplify expression (25): εk (x, T, δ) 6 ε˜k (x, T, δ) ϕ(k) := By choosing T = Hence, √ ε˜k (x) x

˜ √ √ R1 (δ) ˜ ˜ )/ x + δ + R . ˜ T )/ x + (1 + δ )E(T (CT + D δ 2 2 xϕ(k)

2R1 (δ) δ(2+δ)

and δ =

A = 2C2 ln

x > x0 and

ε˜k (x, T, δ) became ε˜k (x).

√ √ k 2π x R1 (δ) 2π x R1 (δ) 2+δ 2 · · = ln + 2 ln ln 4π ln x 2+δ 2π ln x 2+δ √ ˜ R ln x k x R1 (δ) ln x 2 + δ √ · (A) + + + √ +2 ln ln x 2 + δ x R1 (δ) 2πϕ(k) ϕ(k) x 1 2+δ (2 + 2πC2 + 4π( ln(k/(2πe)) + C2 ln k + C3 ) + 4π π

with

Let δ1 =

ln√x , π x

ln√x0 π x0 . 2+δ R1 (δ)
x0 , of find an upper bound 0.06012.


because

√ εk (x) x , ϕ(k) ln2 x

we


1163

To obtain 1) in Theorem 7, we will study the sum in brackets for 1 6 k 6 45 ln x: 1 2 2πd1 4 ln x 2πd1 2πd1 ln x + ln2 [· · · ] = + ln x ln + 2 ln ln 4 ln x ln x 10π ln x ln x 4 ln x 1 ln x + ln x + ln(4d1 /5) + √ (A) + ln 10π 2 x 2πd1 1 2 ln x + ln x(ln = + 1/2 + ln(4 ln x/(10π))) 4 ln x 4 ln x 2πd1 2πd1 ln x + 2 ln ln + ln(4d1 /5) + √ (A) . + ln2 ln x 10π ln x x We conclude that

√ 1 εk (x) x , = lim 2 x→+∞ 8π ln x which is the same asymptotic bound as Schoenfeld’s [7] for ψ. √ The bound εk (x) x is an increasing function of k. Choose k = 45 ln x. Now √ εk (x) x/ ln2 x is a decreasing function of x bounded by 0.0849229 for x > x0 . Remark. If we take k = 1 in Theorem 7, our upper bound is twice as bad as the result of Schoenfeld [7, p. 337]: for x > 73.2, 1 √ x ln2 x. |ψ(x) − x| 6 8π These differences are explained by: • an exact P computation of zeros with γ 6 D ≈ 158 (the preponderant ones!) in 1 the sum |ρ| ; • a better knowledge of R(T ) (k fixed, k = 1). Corollary 3. Assume GRH(k, ∞). For all k used in Lemma 2 and x > 224, √ ψ(x; k, l) − x 6 1 x ln2 x. ϕ(k) 4π Proof. We use Theorem 5.2.1 of [3]: for all k noted in Lemma 2 and 224 6 x 6 1010 , √ x |6 x |ψ(x; k, l) − ϕ(k) √ 2 1 √ and x < 4π x ln x for x > 35. We conclude by Theorem 7. 7. Estimates for π(x; 3, l) Definition 1. Let π(x; k, l) =

X

1

p6x p≡lmodk

be the number of primes smaller than x which are congruent to l modulo k. Our aim is to have bounds for π(x; 3, l). We show that Theorem 8. For l = 1 or 2, x (i) 2 ln x < π(x; 3, l) for x > 151, (ii) π(x; 3, l) < 0.55 lnxx for x > 229869.


1164

PIERRE DUSART

From this, we can deduce that for all x > 151, x < π(x) ln x because π(x) = π(x; 3, 1) + π(x; 3, 2) + 1. 7.1. The upper bound. First we give the proof of Theorem 8 (ii). Rx Lemma 13. Let In = a lndtn t . Then In = lnxn x − lnan a + nIn+1 . Furthermore, (x − a)/ lnn (x) 6 In 6 (x − a)/ lnn (a).

for a > e,

Theorem 9 (Ramaré and Rumely [3]). For 1 6 x 6 1010 , for all k 6 72, for all l relatively prime with k, √ y |6 2.072 x. max | θ(y; k, l) − 16y6x ϕ(k) Furthermore, for x > 1010 and k = 3 or 4, | θ(x; k, l) −

x x |6 0.002238 . ϕ(k) ϕ(k)

Write first π(x; k, l) − π(x0 ; k, l) =

θ(x; k, l) θ(x0 ; k, l) − + ln(x) ln(x0 )

Z

x

x0

θ(t; k, l) dt. t ln2 t

Put x0 := 105 . Preliminary computations : π(105 , 3, 1) = 4784. θ(105 , 3, 1) = 49753.417198 · · · 5 π(105 , 3, 2) = 4807. θ(10 , 3, 2) = 49930.873458 · · · 1.002238 and K = maxl (π(105 , 3, l) − θ(105 , 3, l)/ ln(105 )) ≈ 470. Put c0 := 2 20 • For 10 6 x, Z x θ(x; k, l) θ(105 ; k, l) θ(t; k, l) 5 − + dt. π(x; k, l) − π(10 ; k, l) = 2 ln(x) ln(105 ) 105 t ln t But Z

x

105

θ(t; k, l) dt = t ln2 t

Z

1010

105

θ(t; k, l) dt + t ln2 t

Z

√ x

1010

θ(t; k, l) dt + t ln2 t

Z

x

√ x

θ(t; k, l) dt t ln2 t

and, by Theorem 9 Z

1010

105 √ x

Z

1010

Z

x

√ x

θ(t; k, l) dt t ln2 t θ(t; 3, l) dt t ln2 t θ(t; 3, l) dt t ln2 t

Z

1010

< M := 1/ϕ(k) · √ x − 1010 < c0 ln2 1010 √ x− x < c0 2 √ . ln x

105

dt + 2.072 · ln2 t


Z

1010

105

dt √ 2 t ln t


1165

We compute M = 10381055.54 · · ·. Then

√ √ x− x x x − 1010 + K + M + c0 + 2√ π(x; 3, l) < c0 ln x ln2 1010 ln x 1020 − 1010 ln 1020 x c0 + K + M + c0 < ln x 1020 ln2 1010 x . < 0.545 ln x • For 1010 6 x 6 1020 , Z x Z 1010 θ(t; 3, l) θ(t; 3, l) x dt + dt + c0 π(x; 3, l) < K + 2 2 ln x 5 10 t ln t t ln t 10 10 ln x c0 c0 x c0 + K + M − 1010 2 10 + 2 10 ln x < ln x x ln 10 ln 10 x . < 0.5468 ln x • For 105 6 x 6 1010 , Z x Z Z x 1 x dt θ(t; k, l) dt √ 2 dt < + 2.072 2 2 2 105 ln t t ln t 105 t ln t 105 Z x Z x x 105 dt dt 1 √ 2 . − 2 +2 + 2.072 = 2 3 5 2 ln x ln 10 t ln t 105 ln t 105 h √ ib R R b dt b 2 t + 4 a √tdt . Now, a √t ln2 t = ln 2t ln3 t a Therefore √ x 1 x + 2.072 +K π(x; 3, l) < 2 ln x ln x Z x x 105 1 dt − + 2 + 3 2 ln2 x ln2 105 105 ln t ! √ √ Z x 2 105 dt 2 x √ 3 − +4 +2.072 ln2 x ln2 105 t ln t 105 x for x > 6 · 105 . < 0.55 ln x 7.2. The lower bound. Let KK = minl (π(105 , 3, l) − θ(105 , 3, l)/ ln(105 )) ≈ 462 . and c = 0.498881 = 1−0.002238 2 • For 1010 6 x, Z x θ(t; k, l) θ(x; 3, l) + dt π(x; 3, l) > KK + 2 ln x 105 t ln t cx > ln x because Z x θ(t; k, l) dt > 0. KK > 0 and 5 t ln2 t 10 • For 105 6 x 6 1010 . Lemma 14 (McCurley [2]). For x > 91807 and c2 = 0.49585, we have θ(x; 3, l) > c2 x.


1166

PIERRE DUSART

Remark. This bound is better than the one given in Theorem 9 for x 6 2.5 · 105 . Z x θ(t; k, l) θ(x; 3, l) + dt. π(x; 3, l) > KK + ln x 5 t ln2 t 10 Thus for any x0 , x1 with 105 6 x0 < x1 , Z x0 θ(t; k, l) θ(x; 3, l) + dt for x > x0 π(x; 3, l) > KK + ln x 5 t ln2 t 10 Z x0 ln x1 x θ(t) dt > c2 + KK + for x0 6 x 6 x1 . 2 ln x x1 105 t ln t Using the previous remark, we find Z x θ(t; k, l) dt 5 5 dt > c2 2 2 if 10 6 x 6 2.5 · 10 105 t ln t 105 ln t and √ Z x Z 2.5·105 dt t/2 − 2.072 t + dt if 2.5 · 105 6 x. > c2 ln2 t t ln2 t 105 2.5·105

Z

x

We use this to make step by step computations with Maple: x0

x1

105 2 · 106 3 · 107 3 · 108 3 · 109

2 · 106 3 · 107 3 · 108 3 · 109 1010

We conclude that π(x; 3, l) > 0.499 lnxx for 105 6 x 6 1010 . 7.3. Small values. We now check whether 0.49888 lnxx < π(x; 3, l) < 0.55 lnxx for x < 6 · 105 . It is sufficient to prove that π(p; 3, l) < 0.55

p for p ≡ l mod 3, ln p

and if 0.49888

p < π(p; 3, l) − 1 for p ≡ l mod 3. ln p

The highest value not satisfying the first inequality is p = 229849, and the highest value not satisfying the second is p = 151. Furthermore, π(229869; 3, l) 6 10241 < 151 0.55 ln229869 229869 ≈ 10241.0075 and π(151; 3, l) > 16 > 0.49888 ln 151 ≈ 15.01. The conclusion is x x < π(x; 3, l) < . 0.55 0.49888 x>229869 ln x x>151 ln x Remark. We cannot show that x/(2 ln x) < π(x; 3, l) by using the formula θ(x) < c · x. We have obtained other formulas (see Theorem 6) which we will use below.



1167

7.4. More precise lower bound of π(x; 3, l). Now we will give the proof of Theorem 8(i). Classically, Z x θ(x; 3, l) θ(105 ; 3, l) θ(t; 3, l) − + dt. π(x; 3, l) − π(105 ; 3, l) = 2 5 ln(x) ln(10 ) 105 t ln t x 1 − lnαx with α = ϕ(3) · 0.262 by use of Theorem 6. So we Now θ(t; 3, l) > ϕ(3) write θ(105 ; 3, l) , KK = min π(105 ; 3, l) − l ln(105 ) Z x α 1 1 − α/ ln t x 1− + dt. π(x; 3, l) > J(x, α) = KK + ϕ(k) ln x ln x ϕ(k) 105 ln2 t The derivative of J(x, α) with respect to x equals 1 − α/ ln x α 1 + 3 . ϕ(k) ln x ln x Moreover, the derivative of ϕ(k)xln x equals 1 1 1 − . ϕ(k) ln x ln2 x The inequality 1 1 1 − α/ ln x α 1 1 − 2 + 3 < ϕ(k) ln x ln x ϕ(k) ln x ln x holds if α − 1 < α/ ln x; this holds for all x > 1. The only thing to do is to find a value x1 such that x1 . J(x1 , α) > ϕ(k) ln x1 5

10 For x1 = 105 , J(105 , 0.524) ≈ 4607.75 and 2 ln 105 ≈ 4342.94. We verify by com5 puter that the inequality holds for x 6 10 and l = 1 or 2. We conclude that x < π(x; 3, l) for x > 151. 2 ln x

References [1] Kevin S. McCurley, “Explicit zero-free regions for Dirichlet L-functions”, J. Number Theory, 19, (1984) pp. 7-32. MR 85k:11041 [2] Kevin S. McCurley, “Explicit estimates for θ(x; 3, l) and ψ(x, 3, l)”, Math. Comp., 42, Number 165 (January 1984) pp. 287-296. MR 85g:11085 [3] O. Ramar´ e and R. Rumely, “Primes in arithmetic progressions”, Math. Comp., 65, Number 213 (January 1996) pp. 397-425. MR 97a:11144 [4] J. Barkley Rosser, “Explicit bounds for some functions of prime numbers”, Amer. J. Math., 63, (1941) pp. 211-232. MR 2:150e [5] J. Barkley Rosser and L. Schoenfeld, “Approximate formulas for some functions of prime numbers”, Illinois J. Math. 6 (1962) pp. 64-94. MR 25:1139 [6] J. Barkley Rosser and L. Schoenfeld, “Sharper bounds for the Chebyshev functions θ(x) and ψ(x)”, Math. Comp., 29, Number 129 (January 1975) pp. 243-269. MR 56:15581a [7] Lowell Schoenfeld, “Sharper bounds for the Chebyshev functions θ(x) and ψ(x). II”, Math. Comp., 30, Number 134 (April 1976) pp. 337-360. MR 56:15581b [8] G. Tenenbaum, “Introduction ` a la théorie analytique et probabiliste des nombres”, Institut Elie Cartan (1990) MR 97e:11005a


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[9] J. van de Lune, H. J. J. te Riele and D. T. Winter, “On the zeros of the Riemann zeta function in the critical strip. IV” Math. Comp., 46, Number 174 (April 1986) pp. 667-681. MR 87e:11102 D´ epartement de Math., LACO, 123 avenue Albert Thomas, 87060 Limoges cedex, France E-mail address: [email protected]


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ESTIMATES OF θ(x; k, l) FOR LARGE VALUES OF x 1. Introduction Let

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