Explicit Estimate on Primes between Consecutive Cubes

0.24 MB·English

by  Yuanyou Furui Cheng

EXPLICIT ESTIMATE ON PRIMES
BETWEEN CONSECUTIVE CUBES
8 Cheng, Yuan-You Fu-Rui
0
0
2 Abstract. We give an explicit form of Ingham’s Theorem on primes in the short
intervals, and show that there is at least one prime between every two consecutive
ct cubes x3 and(x+1)3 ifloglogx≥15.
O

2
1
1. Introduction

] Studies about certain problems in number theory are often connected to those
T
about the distribution of the prime numbers; problems about the distribution of
N
primes are among the central ones in number theory. One problem concerning the
.
h distributionofprimesisthedistributionofprimesincertainintervals. Forexample,
t Bertrand’s postulate asserts that there is a number B such that, for every x > 1,
a
m there is at least one prime number between x and Bx. If the interval [x,Bx] is
replaced by a “short interval” [x,x+xθ], then the problem is more difficult.
[
In1930,Hoheiselshowedthatthereisatleastoneprimeintheabovementioned

1 “shortinterval”withθ =1 1 forsufficientlylargex’s,see[13]. Ingham[15],in
v −33,000
3 1941,provedthatthereisatleastoneprimein[x,x+x3/5+ǫ],whereǫisanarbitrary
1 positive number tending to zero whenever x is tending to infinity, for “sufficiently
1 large” x’s. This implies that there is at least one prime between two consecutive
2
cubes if the numbers involvedare “largeenough.” One of the better results in this
.
0 direction, conjectured by using the Riemann Hypothesis, is that there is at least
1 one prime between [x,x+x1/2+ǫ] for “sufficiently large” x’s. The latter has not
8
been proved or disproved; though better results than Hoheisel’s and Ingham’s are
0
available. For example, one may see [2, 3, 12, 15, 17, 18, 19, 26, 28].
:
v
Thesekindsofresultswouldhavemanyusefulapplicationsiftheywere“explicit”
i
X (with all constants being determined explicitly). For references in other directions
withexplicitresults,onecansee[4,8,22,23,24,25]. Tofigureoutthe“sufficiently
r
a large” x’s related to θ as mentioned above, one needs to investigate the proof
in a “slightly different” way. As a starting step in this direction, we study the
distributionofprimesbetweenconsecutivecubes. Inthisarticle,wegiveanexplicit
form of Ingham’s Theorem; specifically, we show that there is at least one prime
between consecutive cubes if the numbers involved are larger than the cubes of x
0
2000Mathematics Subject Classification. 11Y35,11N05.
Keywords and phrases. Ingham’sTheorem,Primesinshortinterval,explicitestimates,Den-
sityTheorem,Divisors.
I wish to thank the referee for many helpful comments including an improved version for
Lemma3.2.
TypesetbyAMS-TEX
1
2 CHENG, YUAN-YOU FU-RUI
where x =exp(exp(15)) and we also set T =exp(exp(18)) throughout this paper
0 0
accordingly.
Our main task is to prove the Density Theorem or to estimate the number of
zeros in the strip σ > 1 for the Riemann zeta function, see Theorem 1 in the
2
follows. We let β = (ρ) and I (u) be the unit step function at the point u = β;
β
ℜ
that is, I (u) = 1 for 0 u β and I (u) = 0 for β < u 1. One defines
β β
≤ ≤ ≤
N(u,T):= I (u) and N(T):=N(0,T).
0≤ℑ(ρ)<T β
Theorem 1P. Let 5 σ <1 and T T . One has
8 ≤ ≥ 0
N(σ,T) CDT8(13−σ) log5T,
≤
where C :=453472.54.
D
Theorem 2. Let x x , h 3x2/3 and C be defined in Theorem 1. Then
0 D
≥ ≥
ψ(x+h) ψ(x) h(1 ǫ(x)),
− ≥ −
where
1
1 log x 3
ǫ(x) :=3192.34exp .
| | −273.79(cid:18)loglog x(cid:19) !
Theorem 3. Let x exp(exp(45)) and h 3x23. Then
≥ ≥
1
1 log x 3
π(x+h) π(x) h 1 3192.34exp .
− ≥ − −283.79 loglog x !!
(cid:18) (cid:19)
Corollary. Let x exp(exp(15)). Then there is at least one prime between each
≥
pair of consecutive cubes x3 and (x+1)3.
The proof of Theorem 1 is delayed until Section 5. We shall prove Theorem 2
and 3 in Section 2. The proof of Theorem 2 is based on Theorem 1 and Laudau’s
approximate formula, which is in Section 6. Then, it is not difficult to prove
Theorem 3 from Theorem 2, as shown in Section 2.
2. Proof of Theorem 2 and 3
From [25], one has
T log T
N(T) +?? ??.
≤ 2π −
The following proposition follows straightforward.
Proposition 2.1. For T 6, one has N(T) TlogT.
≥ ≤ 2π
Proposition 2.2. Let C be defined in Theorem 1. Assume that the Riemann
D
zeta-function does not vanish for σ > 1 z(t). Suppose that T T < x3/8. For
0
− ≤
any h>0, one has
(x+h)ρ xρ 2CDT38z(t)logxlog5T
− h.
(cid:12) ρ (cid:12)≤ xz(t)(log x 8log T)
(cid:12)(cid:12)|ℑ(Xρ)|≤T (cid:12)(cid:12) − 3
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
EXPLICIT ESTIMATE ON PRIMES BETWEEN CONSECUTIVE CUBES 3
Proof. Notes that
(x+h)ρ xρ x+h
− = uρ−1du hxβ−1,
where β = (ρ) is t(cid:12)(cid:12)(cid:12)(cid:12)he realρpart of(cid:12)(cid:12)(cid:12)(cid:12)ρ; (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Zx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)≤
ℜ
β
xβ =1+log x xudu;
Z0
and
β 1
xudu= xuI (u)du,
β
Z0 Z0
where I (u) is the unit step function or I (u)=1 for 0 u β and I (u)=0 for
β β β
≤ ≤
β <u 1. After interchanging the summation and integration, one has
≤
(x+h)ρ xρ h
− xβ
(cid:12) ρ (cid:12)≤ x
(2.1) (cid:12)(cid:12)|ℑ(Xρ)|≤T (cid:12)(cid:12) |ℑ(Xρ)|≤T
(cid:12) (cid:12)
(cid:12) h (cid:12) 1
(cid:12) (cid:12) 1+logx xu I (u) du .
β
≤ x   
|ℑ(Xρ)|≤T Z0 |ℑ(Xρ)|≤T
   
If the Riemann zeta-function does not vanish in the region σ > 1 z(t), then the
−
expression in the outmost parenthesis in (2.1) is bounded by
5 1−z(t)
8
(2.2) 2N(0,T)+2N(0,T)logx xudu+2logx xuN(u,T)du.
Z0 Z85
SinceT 6,onecanapplyProposition2.1. Thesumofthefirsttwotermsin(2.2)
≥
5/8 x5/8T logT
(2.3) 2 1+logx xudu N(0,T)=2x5/8N(0,T) .
Z0 ! ≤ π
From Theorem 1, one sees that the last term in (2.2) is bounded by
1−z(t)
(2.4) 2CDlog xlog5T xuT38(1−σ)du
Z5/8
1−z(t) x u
=2CDT38 logxlog5T T8/3 du
Z5/8 (cid:16) (cid:17)
2CDT38 logxlog5T x 1−z(t) x 85
=
log x− 83log T (cid:18)T83(cid:19) −(cid:18)T83(cid:19) !
2CDxT83z(t)logxlog5T 2CDx85T logxlog5T
= .
xz(t)(log x 8log T) − log x 8log T
− 3 − 3
4 CHENG, YUAN-YOU FU-RUI
Oneseesthatthesumoftheupperboundin(2.3)andthesecondtermontheright
side in (2.4) is negative. Finally, one combines (2.1) and the first term in the last
expression in (2.4) to finish the proof of Lemma 2.1. (cid:3)
Proof of Theorem 2. From Lemma 9.1 and Proposition 2.2, one sees that
ψ(x+h) ψ(x)=h+hǫ(x),
−
with
1 (x+h)ρ xρ
ǫ(x) − + E(x+h) + E(x)
| |≤ h(cid:12) ρ (cid:12) | | | |
(cid:12)(cid:12)|ℑ(ρX)|≤Tu (cid:12)(cid:12)
(2.5) 2CD(cid:12)(cid:12)T83z(t)log xlog5T (cid:12)(cid:12) (x+h)log2(x+h)
(cid:12) +10(cid:12).52 +
≤ xz(t)(log x 8log T) hT
− 3
(x+h)log2T log2T
+66.976 +6 .
hTlogx hx
Let 3x32 h. Also, let
≤
1
1 1 log x 3
T =T(x):=x3 exp ,
256.59 loglog x
(cid:18) (cid:19) !
with some undetermined constant u>1. Then,
1
1 1 log x 3
log T = log x+ 0.34log x,
3 256.59 loglog x ≤
(cid:18) (cid:19)
loglog T loglog x,
≤
and
1
8 8 8 log x 3
T3 =x9 exp .
3 256.59 loglog x
!
× (cid:18) (cid:19)
From [10], it is known that the Riemann zeta function does not vanish for T
≥
1 z(T) with
− 1
z(T)= .
58.51log2/3T(loglogT)1/3
Let Z(x):=z(T(x)). Then
1
Z(x) ,
≥ 28.51log2/3x(loglog x)2/3
Z(x)
x z(T) x19
 
(cid:18)T38(cid:19) ≥ exp 8 logx 13
 3×256.59 loglogx 
 (cid:18) (cid:16) (cid:17) (cid:19)
1
1 log x 3 8
=exp .
256.59(cid:18)loglog x(cid:19) − 3 256.59 28.51log31 x(loglog x)1o3r2!
× ×
EXPLICIT ESTIMATE ON PRIMES BETWEEN CONSECUTIVE CUBES 5
It follows that for x exp(exp(45)) the right side in (2.5) is bounded from above
≥
by
1 1
1 log x 3 1 log x 3
3192.34exp +1.76exp +
−273.79(cid:18)loglog x(cid:19) ! −256.6(cid:18)loglog x(cid:19) !
1 log x 13 0.24log2+4.65log x+260.48
+2.59exp + .
−256.6(cid:18)loglog x(cid:19) ! x
Conclude that one has proved Theorem 2. (cid:3)
Proof of Theorem 3. By definition of π(x) and ψ(x), one has
logp
π(x+h) π(x)= 1
− ≥ logx
x<p≤x+h x<p≤x+h
X X
ψ(x+h) ψ(x) h
= − (1 ǫ(x)).
logx ≥ logx −
This finishes the proof of Theorem 3. (cid:3)
Proof of Corollary. Let X =x3 and h=(x+1)3 x3. Then h 3x2 =3X32. By
− ≥
Theorem 3,
3x2
π (x+1)3 π x3 1 ǫ(x3) >1.
− ≥ 3log x −
This proves the cor(cid:0)ollary. (cid:3)(cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
3. Three auxiliary functions
Three auxiliary functions U , V , and W are introduced in this section. For ref-
A A A
erences, one may see [4], [17], [26].
Definitions of Three Auxiliary Functions. LetAbe apositiveinteger. Define
A
µ(n)
U (s)= .
A ns
n=1
X
Here µ is the Mo¨bius µ-function. Then,
V (s)=ζ(s)U (s) 1, W (s)=1 V2(s).
A A − A − A
Lemma 3.1. Let ν(n)= µ(m). Then ν(n) d(n) and
m≤A:m|n | |≤
P
ν(n)
V (s)= .
A ns
n>A
X
Every non-trivial zero of ζ(s) is a zero of W (s).
A
6 CHENG, YUAN-YOU FU-RUI
Lemma 3.2. One has
7.9
V (2+it)2 .
A
| | ≤ A
If A 8, then both (W (2+it)) and W (2+it) does not vanish; if A 16, then
A A
≥ ℜ ≥
V (2+it)2 < 1 and W (2+it) > 1.
| A | 2 | A | 2
Lemma 3.3. Let b =5.134. For σ 1 and t 3.297, one has
1 ≥ 4 ≥
V (s) t3/2,
A
| |≪
and
16 16
W (s) A3/4t3/2+b A3/4t1/2 A3/4t3/2+b A3/4t1/2+2 .
A 1 1
| |≤ 9 9
(cid:18) (cid:19)(cid:18) (cid:19)
Proof of Lemma 3.1. Easy. (cid:3)
Proof of Lemma 3.2. We may assume A 5. Observe
≥
d(n) 1 1 1
=ζ(2) +
n2 m2 m2 n2
nX>A mX>A mX≤A nX>mA
ζ(2) 1 1
+
≤ A m2 A
m≤A m
X
ζ(2) (cid:4) (cid:5)1 1
+ +
≤ A m(A m) m2
m≤XA2+1 − A2+1X<m≤A
ζ(2) 1 4 log(A 1) 1
+ + + − +
≤ A A 1 A2 4 A A
− −
2.8
<
A
for A 5. (cid:3)
≥
One needs the following proposition.
Proposition 3.1. Let σ 1 and t 3.297. Then
≥ 4 ≥
⌊t2⌋
1
ζ(s)= +B(s),
ns
n=1
X
where B(s) b t1/2 with b :=5.134.
1 1
| |≤
Proof. Note that in [4], [17], [26]
N 1 ∞ u [u] 1
ζ(s)= s − du+ (σ >0,s=1).
ns − us+1 (s 1)Ns−1 6
n=1 ZN −
X
EXPLICIT ESTIMATE ON PRIMES BETWEEN CONSECUTIVE CUBES 7
From this, we have
1 N1−σ 1 1 t
ζ(s) + + (σ >0,s=1).
(cid:12)(cid:12) −n≤N ns(cid:12)(cid:12)≤ t sσ02 t20Nσ 6
(cid:12) X (cid:12)
(cid:12) (cid:12)
Using this id(cid:12)(cid:12)entity, Proposi(cid:12)(cid:12)tion 3.1 follows. (cid:3)
Proof of Lemma 3.3. Using its definition, one sees that
A
1
U (s) .
| A |≤ nσ
n=1
X
If 0<σ <1, one gets
A 1 A1−σ
U (s) du= ;
| A |≤ uσ 1 σ
Z0 −
if σ 1, one has
≥
A
1
U (s) log A+1.
A
| |≤ n ≤
n=1
X
For σ 1, one obtains
≥ 4
4 4
(3.2) U (s) max A3/4,log A+1 A3/4.
A
| |≤ 3 ≤ 3
(cid:26) (cid:27)
Similarly, one gets
⌊t2⌋
1 4
t3/2.
nσ ≤ 3
n=1
X
Combining with the result in Proposition 3.2, one has
4
(3.3) ζ(s) t3/2+b t1/2,
1
| |≤ 3
for σ 1 and t 3.297.
≥ 4 ≥
Recalling the definition of V (s) one has
A
V (s) ζ(s) U (s) +1;
A A
| |≤| || |
from (3.1), one gets
W (s) ζ(s) U (s)(2+ ζ(s) U (s)).
A A A
| |≤| || | | || |
Conclude with (3.2) and (3.3) that one proves Lemma 3.3. (cid:3)
8 CHENG, YUAN-YOU FU-RUI
4. Representing the number of zeros by an integral
Notation N (σ,T). Let F(s) be a complex function and T > 0. The notation
F
N (σ,T)expressesthe numberofzerosinthe formβ+iγ forF(s)withσ β and
F
≤
0 γ <T.
≤
It is well known that ζ(s) does not vanish for σ 1; so one may restrict our
≥
discussion to σ <1.
Lemma 4.1. Let T =14 and A 16. Then for σ <σ <1 and T T , one has
1 0 1
≥ ≥
1 1 T (594?)16T
N (σ;T) V (σ +it)2 dt+ +1+c (T) ,
ζ A 0 A
≤ σ−σ0 2π ZT1 | | 2πA !
with
3/2 1/2
log 16A3/4 T + 7 +b A3/4 T + 7
9 4 1 4
cA(T):= (cid:18) (cid:16) (cid:17) (cid:16) (cid:17) (cid:19)
log(7/6)
3/2 1/2
log 16A3/4 T + 7 +b A3/4 T + 7 +2
9 4 1 4 log 2
+ (cid:18) (cid:16) (cid:17) (cid:16) (cid:17) (cid:19) + .
log(7/6) log(7/6)
Corollary. Let A 595T and T exp(exp(18)). Then
≤ 594 ≥
c (T) 29.193log T +11.978.
A
≤
Notation N (σ;T,T ). LetF(s), σ, andT as in the lastdefinition. The notation
F 1
N (σ;T,T ) expresses the number of zeros in the form β+iγ for F(s) with σ β
F 1
≤
and T γ <T.
1
≤
Bedefinition,oneseesthatN (σ;T,T )=N (σ,T) N (σ,T )foranycomplex
F 1 F F 1
−
function F. Note here, see [9] or [17], that there is no zero for the Riemann zeta
functionζ(σ+it)for0 t 14. IfonetakesT =14,thenN (σ;T,T )=N (σ,T).
1 ζ 1 ζ
≤ ≤
For an analytic function, a zero is isolated and the number of zeros in any
compact region is finite. Fix σ and T. Let ǫ , ǫ , and ǫ be sufficiently small
1 2 3
positive numbers and λ=σ ǫ and T =T +ǫ . One may assume that λ is not
1 2 2
−
therealpartandT isnottheimaginarypartsofanyzerosforthefunctionW (s).
2 A
Recalling the second part of Lemma 3.1, one gets the following proposition.
Proposition 4.1. Let T =14 and ǫ and ǫ be small positive numbers such that
1 1 2
λ = σ ǫ is not the real part and T = T +ǫ is not the imaginary part of any
1 2 2
−
zero for the function W (s). Then
A
N (σ,T) N (λ;T ,T ).
ζ ≤ WA 2 1
EXPLICIT ESTIMATE ON PRIMES BETWEEN CONSECUTIVE CUBES 9
Let σ < λ. Since N (λ;T ,T ) is a non-increasing function of λ by the
0 WA 2 1
definition, one sees that
1 λ
N (λ;T ,T ) N (ρ;T ,T )dρ.
WA 2 1 ≤ λ σ WA 2 1
− 0 Zσ0
Noting that
λ 2
N (ρ;T ,T )dρ N (ρ;T ,T )dρ,
WA 2 1 ≤ WA 2 1
Zσ0 Zσ0
one has the next proposition.
Proposition 4.2. Let σ <λ<1 and T >T . Assume that λ is not the real part
0 2 1
and T is not the imaginary part of any zero for W (s). Then
2 A
1 2
N (λ;T ,T ) N (ρ;T ,T )dρ.
WA 2 1 ≤ λ σ WA 2 1
− 0 Zσ0
Using the arguments in [26, p.213 and p.220], one gets the following result.
Proposition 4.3. Let 1 < σ < 2. Assume that T is not the imaginary part
2 0 2
of any zero for W (s). Also, let be the number of zeros for (W (s)) on the
A k A
N ℜ
segment between σ +it and 2+it on the line t=T for k =1 and 2 respectively.
0 k
Then
2 1 T2 W (σ +it) +
A 0 1 2
(4.5) N (ρ;T ,T )dρ log | | dt+ N N +1.
WA 2 1 ≤ 2π W (2+it) 2
Zσ0 ZT1 (cid:18) | A | (cid:19)
The following result can be found in [4].
Proposition 4.4. Suppose that s is a fixed complex number and f is a complex
0
function non-vanishing at s and regular for s s < R for positive number R.
0 0
| − |
Let 0 < r < R and M = max f(s). Then the number of zeros of f in
f |s−s0|=R| |
s s r, denoted by , multiple zeros being counted according to their order
0 f
| − | ≤ N
of multiplicity satisfies the following inequality.
logM log f(s )
f 0
− | |.
f
N ≤ logR logr
−
Proof of Lemma 4.1. From Proposition 4.1, 4.2, and 4.3, one has
1 1 T2 W (σ +it) +
A 0 1 2
(4.6) N(σ,T) log| |dt+ N N +1 ,
≤ λ−σ0 2π ZT1 |WA(2+it)| 2 !
where λ=σ ǫ as T =T +ǫ as in Proposition 4.1.
1 2 2
−
Clearly, we have
1
(W (σ+it))= W (σ+it)+W (σ it) .
A A A
ℜ 2 −
(cid:16) (cid:17)
10 CHENG, YUAN-YOU FU-RUI
Thenumberofzerosof (W (s))on arethesameforthefollowingregularfunc-
A k
ℜ S
tions
1
W(k)(s)= (W (s+iT )+W (s iT ))
0 2 A k A − k
on the real axis between σ and 2.
0
First,oneappliesProposition4.4toestimatethenumber ′ ofzerosforW(k)(s)
Nk 0
in s 2 3. It is obvious that ′ . One takes s = 2, R = 7, r = 3.
| − | ≤ 2 Nk ≤ Nk 0 4 2
Recalling (4.7) and Lemma 3.3, one acquires
1
max W(k)(s) ( max W (s+iT ) + max W (s iT ))
|s−2|=3 0 ≤ 2 |s−2|=3| A k | |s−2|=3| A − k |
2(cid:12) (cid:12) 2 2
(cid:12) (cid:12) 7 7
(cid:12) max(cid:12) W (s+iT ) W(1) T + W(1) T + ,
≤|s−2|=3| A k |≤ A k 4 ≤ A 2 4
2 (cid:18) (cid:19) (cid:18) (cid:19)
where W(1)(t) is the upper bound of W (s) in Lemma 3.3. Letting ǫ tend to
A | A | 2
zero, one sees that
7
max W(k)(s) W(1) T + .
|s−2|=3| 0 |≤ A 4
2 (cid:18) (cid:19)
Also, recall that W (2+it) > 1 from Lemma 3.2. This implies
| 0 | 2
log W(1) T + 7 log(1/2)
c (T):= A 4 − ,
k A
N ≤ log(7/6)
(cid:0) (cid:1)
for k =1 and 2. Hence,
+
1 2
(4.8) N N c (T).
A
2 ≤
Now, transform the integral in (4.6) into a one involving the function V (s)
A
instead of W (s).
A
Recalling the definition of W (s), using the triangular inequality in the form
A
x y x + y , and noting that log(1+x) x for x>0, one has
| − |≤| | | | ≤
log W (σ +it) =log 1 V2(σ +it)
A 0 0
(4.9) | | | − |
log(1+ V (σ +it)2) V (σ +it)2.
A 0 A 0
≤ | | ≤| |
Also, by the triangular inequality in the form x y x y , one gets 1
| − | ≥ | |−| | | −
V2(1+it) 1 V (1+it)2. Using the increasing property of the logarithmic
A | ≥ −| A |
function, one sees that log 1 V2(2+it) log(1 V (2+it)2). It follows that
| − A |≥ −| A |
log W (2+it) = log 1 V2(2+it) log(1 V (2+it)2).
− | A | − | − A |≤− −| A |
From the last part of Lemma 3.2, one sees that V (2+it)2 < 1 since A 16.
| A | 2 ≥
Applying log(1 x)<2x for 0<x< 1, one acquires
− − 2
7.9
(4.10) log W (2+it) 2V (2+it)2 < .
A A
− | |≤ | | A