Almost all short intervals containing prime numbers

ACTA ARITHMETICA
LXXVI.1 (1996)
Almost all short intervals containing
prime numbers
by
Chaohua Jia (Beijing)
1. Introduction. In 1937, Cramér [1] conjectured that every interval
(n, n + f (n) log2 n) contains a prime for some f (n) → 1 as n → ∞.
In 1943, assuming the Riemann Hypothesis, Selberg [19] showed that,
for almost all n, the interval (n, n + f (n) log2 n) contains a prime providing
f (n) → ∞ as n → ∞. In the same paper, he also showed that, for almost
19
all n, the interval (n, n + n 77 +ε ) contains a prime.
1
In 1971, Montgomery [16] improved the exponent 19
77 to 5 . The zero
density estimate of Huxley [7] gives the exponent 16 .
In 1982, Harman [3] used the sieve method to prove that, for almost all
1
n, the interval (n, n+n 10 +ε ) contains a prime. Heath-Brown [5] and Harman
1
[4] mentioned that the exponent 12
can be achieved.
In [11], Jia Chaohua investigated the problem of the exceptional set of
Goldbach numbers in the short interval. As a by-product, he proved that, for
1
almost all n, the interval (n, n + n 13 +ε ) contains prime numbers. Li Hongze
1
2
[14] improved the exponent 13
to 27
.
Recently, Jia Chaohua [12] showed that, for almost all n, the interval
1
(n, n + n 14 +ε ) contains prime numbers. Watt [20] also obtained the same
result. Their methods are different. In [12] only classical methods are used
and in [20] a new mean value estimate of Watt [21] is used in addition. Li
1
1
Hongze [15] combined these methods to improve the exponent 14
to 15
.
In this paper, we prove the following:
Theorem. Suppose that B is a sufficiently large positive constant, ε is
a sufficiently small positive constant and X is sufficiently large. Then for
positive integers n ∈ (X, 2X), except for O(X log−B X) values, the interval
1
1
(n, n + n 20 +ε ) contains at least 0.005n 20 +ε log−1 n prime numbers.
Project supported by the Tian Yuan Item in the National Natural Science Foundation
of China.
[21]
22
C. Jia
We apply a mean value estimate of Deshouillers and Iwaniec [2] (see
Lemma 2). Using the classical mean value estimate instead of that of Des1
houillers and Iwaniec, we can get the exponent 18
. We refer to [13] and the
explanation in [11].
Throughout this paper, we always suppose that B is a sufficiently large
positive constant, ε is a sufficiently small positive constant and ε1 = ε2 ,
1
δ = ε 3 . We also suppose that X is sufficiently large and that x ∈ (X, 2X),
19
η = 12 X − 20 +ε . Let c, c1 and c2 denote positive constants which have different
values at different places. m ∼ M means that there are positive constants c1
and c2 such that c1 M < m ≤ c2 M . We often use M (s) (M may be another
capital letter) to denote a Dirichlet polynomial of the form
X a(m)
M (s) =
,
ms
m∼M
where a(m) is a complex number with a(m) = O(1).
The author would like to thank Profs. Wang Yuan and Pan Chengbiao
for their encouragement.
2. Mean value estimate (I)
9
Lemma 1. Suppose that X δ H X 85 , M H = X, M (s) is a Dirichlet
polynomial and
X Λ(h)
H(s) =
.
hs
h∼H
B
ε
Let b = 1 + 1/ log X, T0 = log X. Then for T0 ≤ T ≤ X, we have
2T
1 \
2
min η,
|M (b + it)H(b + it)|2 dt η 2 log−10B x.
T
T
P r o o f. Let s = b + it. By the zero-free region of the ζ function, for
|t| ≤ 2X we have
X
2B
(c2 H)1−s − (c1 H)1−s
Λ(h)
(1)
=
+ O(log− ε x).
s
h
1−s
c1 H<h≤c2 H
So, for T0 ≤ |t| ≤ 2X,
B
H(s) log− ε x.
(2)
According to the discussion in [6], there are O(log2 X) sets S(V, W ),
where S(V, W ) is the set of tk (k = 1, . . . , K) with the property |tr − ts | ≥ 1
(r 6= s). Moreover,
1
V ≤ M 2 |M (b + itk )| < 2V,
1
W ≤ H 2 |H(b + itk )| < 2W,
Short intervals containing prime numbers
1
1
23
1
1
B
where X −1 ≤ M − 2 V , X −1 ≤ H − 2 W and V M 2 , W H 2 log− ε x.
Then
\
2T
(3)
|M (b + it)H(b + it)|2 dt V 2 W 2 x−1 log2 x|S(V, W )| + O(x−2ε1 ),
T
where S(V, W ) is one of sets with the above properties.
1
1
Assume X k+1 ≤ H < X k , where k is a positive integer, k ≥ 9 and
kδ 1. Applying the mean value estimate (see Section 3 of [9] or Lemma 7
of [11]) to M (s) and H k (s), we have
|S(V, W )| V −2 (M + T ) logd x,
|S(V, W )| W −2k (H k + T ) logd x,
where d = c/δ 2 . Applying the Halász method (see Section 3 of [9] or Lemma 7
of [11]) to M (s) and H k (s), we have
|S(V, W )| (V −2 M + V −6 M T ) logd x,
|S(V, W )| (W −2k H k + W −6k H k T ) logd x.
Thus,
V 2 W 2 |S(V, W )| V 2 W 2 F logd x,
where
F = min{V −2 (M + T ), V −2 M + V −6 M T, W −2k (H k + T ),
W −2k H k + W −6k H k T }.
It will be proved that
B
1
2
(4)
min η,
V 2 W 2 F η 2 x log− ε x.
T
We consider four cases.
(a) F ≤ 2V −2 M , 2W −2k H k . Then
V 2 W 2 F V 2 W 2 min{V −2 M, W −2k H k }
1
1
≤ V 2 W 2 (V −2 M )1− 2k (W −2k H k ) 2k
1
1
B
1
= V k W M 1− 2k H 2 x log− ε x
and so
B
1
min η,
V 2 W 2 F η 2 x log− ε x.
T
−2
−2k k
(b) F > 2V M, 2W
H . Then
2
V 2 W 2 F V 2 W 2 min{V −2 T, V −6 M T, W −2k T, W −6k H k T }
3
1
1
1
≤ V 2 W 2 (V −2 )1− 2k (V −6 M ) 2k (W −2k ) k T = M 2k T.
24
C. Jia
1
k
1
1
Since k ≥ 9, we have H ≥ X k+1 ≥ X 1− 10 , M 2k X 20 , and so
1
η 1
2
min η,
V 2 W 2 F x 20 T η 2 x1−ε1 .
T
T
(c) F ≤ 2V −2 M, F > 2W −2k H k . Then
V 2 W 2 F V 2 W 2 min{V −2 M, W −2k T, W −6k H k T }
1
1
≤ V 2 W 2 (V −2 M )1− 3k (W −6k H k T ) 3k
1
1
M H 3 T 3k ,
1
1
19
1
since V M 2 . As H ≥ X k+1 ≥ X 20 · 2k +ε , we have
1
1
19 1
1
1
2
min η,
V 2 W 2 F η 2− 3k T − 3k x1− 20 · 3k T 3k x−ε1 η 2 x1−ε1 .
T
(d) F > 2V −2 M , F ≤ 2W −2k H k . Then
V 2 W 2 F V 2 W 2 min{V −2 T, V −6 M T, W −2k H k }
1
3
1
≤ V 2 W 2 (V −2 T )1− 2k (V −6 M T ) 2k (W −2k H k ) k
1
1
= M 2k HT 1− k .
19(k−1)
19(k−1)
If k ≥ 10, then H ≤ X k ≤ X 1− 10(2k−1) , M X 10(2k−1) , and so
1
19
1
1
2
min η,
V 2 W 2 F η 1+ k x1− 20 (1− k ) η 2 x1−ε1 .
T
1
1
9
76
If k = 9, then X 10 ≤ H X 85 , M X 85 , and so
19
38
8
1
2
min η,
V 2 W 2 F η 2 (x− 20 +ε )− 9 x1− 45 η 2 x1−ε1 .
T
Combining the above, we obtain (4). Hence, Lemma 1 follows.
Lemma 2. Suppose that N (s) is a Dirichlet polynomial and
X
1
L(s) =
.
ls
c1 L<l≤c2 L
Let T ≥ 1. Then
4 2
2T
\ 1
1
I = L
+ it N
+ it dt
2
2
T
21
5
T
2 12
2 −2
4
T + N T + N T min L,
+ NL T
T ε1 .
L
Short intervals containing prime numbers
25
1
P r o o f. First we assume c1 L ≤ T 2 . If N ≤ T , then by the discussion in
Section 2 of [2], we have
1
5
1
I (T + N 2 T 2 + N 4 (T L) 2 )T ε1 .
If N > T , the mean value estimate yields
1
I (L2 N + T )(LN )ε1 (T + N 2 T 2 )T ε1 .
1
Now we assume T 2 < c1 L ≤ 2T . By the discussion in Section 2 of [2],
we get
1 5
T 2
2 12
4
I T +N T +N T ·
T ε1 .
L
Lastly we assume 2T < c1 L. It follows from Theorem 1 on page 442 of
[18] that
1
1
X
1
(c2 L) 2 −it − (c1 L) 2 −it
1
+O
=
.
1
1
1
l 2 +it
L2
2 − it
c1 L<l≤c2 L
Hence,
1
1
L2
1
L2
1
L
+ it + 1 2
|t|
|t|
L2
and
L2
I 4
T
\
2T
T
2
2
2
1
dt L (N + T ) N L .
N
+
it
2
T4
T2
Combining the above, we get Lemma 2.
Lemma 3. Suppose that M N L = X, M (s), N (s) are Dirichlet polynomials, and
X 1
L(s) =
.
ls
l∼L
√
Let b = 1 + 1/ log X, T1 = L. Assume further that M and N lie in one of
the following regions:
(5)
2
9
(i)
M X 32 ,
(ii)
X 32 M X 160 ,
(iii)
X 160 M X 32 ,
(iv)
X 32 M X 680 ,
9
53
13
1
N M 3 X 10 ;
53
13
321
23
N X 80 ;
2
61
N M − 3 X 120 ;
21
N M −2 X 20 .
Then for T1 ≤ T ≤ X, we have
2T
1 \
2
min η,
|M (b + it)N (b + it)L(b + it)|2 dt η 2 x−2ε1 .
T
T
26
C. Jia
P r o o f. It is sufficient to show that
2T 2
1
1 \ 1
1
2
M
I = min η,
+ it N
+ it L
+ it dt η 2 x1−2ε1 .
T
2
2
2
T
We shall show that the above inequality holds, providing M and N satisfy
the following conditions:
21
61
41
M 2 N X 20 , M 2 N 3 X 40 , M 6 N 7 X 10 ,
3
23
N X 80 , N 3 M 2 X 10 .
Using the mean value estimate and Lemma 2, we have
2
2T
\ 1
1
1
M
dt
+
it
N
+
it
L
+
it
2
2
2
T
2T
4 2 12
\ 1
1
M
+ it N
+ it dt
2
2
T
2T
4 2 12
\ 1
1
×
L 2 + it N 2 + it dt
T
1
1
5
1
1
(M 2 N + T ) 2 (T + N 2 T 2 + N 4 (T L) 2 + N L2 T −2 ) 2 T ε1 .
Hence,
1
1
5
1 1
1
−1
(M 2 N + T ) 2 (T + N 2 T 2 + N 4 (T L) 2 ) 2 T ε1 + η 2 T1 2 x1+ε1
I min η,
T
2
1
1
5
1
1
η 2 (M 2 N + η −1 ) 2 (η −1 + N 2 η − 2 + N 4 (η −1 L) 2 ) 2 xε1 + η 2 x1−2ε1
η 2 x1−2ε1 .
In every region of (5), our conditions are satisfied. So Lemma 3 follows.
Lemma 4. Under the assumptions of Lemma 3, (5) being replaced by the
region
21
(6)
M X 40 ,
19
N X 160 ,
for T1 ≤ T ≤ X, we have
2T
1 \
min2 η,
|M (b + it)N (b + it)L(b + it)|2 dt η 2 x−2ε1 .
T
T
P r o o f. It is sufficient to show that
2T 2
1
1
1 \ 1
2
dt η 2 x1−2ε1 .
I = min η,
M
+
it
N
+
it
L
+
it
T
2
2
2
T
Short intervals containing prime numbers
27
Using the mean value estimate and Lemma 2, we have
2T
4 12
\ 1
1
2
M
I min η,
+ it dt
T
2
T
2T
4 4 21
\ 1
1
×
L 2 + it N 2 + it dt
T
1
1
1
2
2
2
min η,
(M + T ) T + N 4 T 2
T
21
12
5
T
2
2
−2
+N L T
T ε1
+ N 2 T min L,
L
1
1
1
5
1
η 2 (M 2 + η −1 ) 2 (η −1 + N 4 η − 2 ) 2 xε1 + η 2 M N 4 (η −1 L) 4 xε1
5
−1
7
+ N 4 η 2− 8 xε1 + η 2 T1 2 x1+ε1
η 2 x1−2ε1 ,
1
since min(L, T /L) ≤ T 2 . Thus Lemma 4 follows.
3. Mean value estimate (II)
Lemma 5. Suppose that M HK = X and M (s), H(s) and K(s) are
Dirichlet polynomials, and G(s) = M (s)H(s)K(s). Let b = 1 + 1/ log X,
B
B
T0 = log ε X. Assume further that for T0 ≤ |t| ≤ 2X, M (b + it) log− ε x
B
and H(b + it) log− ε x. Moreover , suppose that M and H satisfy one of
the following three conditions:
157
19
3
1) M H X 290 , X 110 H, M 29 /H X 10 , X 10 M , H 29 /M 57
X , X 10 M 12 H 11 ;
26
59
38
29
2) M H X 45 , M 29 H 19 X 4 , X 45 M 2 H, M 2 H 11 X 10 ,
57
X 2 M 58 H 49 ;
25
19
11
19
23
3) M H X 44 , X 100 H, M 6 H X 5 , X 70 M , M H 8 X 10 ,
57
X 20 M 6 H 5 .
31
5
Then for T0 ≤ T ≤ X, we have
2T
1 \
2
(7)
min η,
|G(b + it)|2 dt η 2 log−10B x.
T
T
P r o o f. Using the method of Lemma 1, we only show that for T = 1/η =
19
2X 20 −ε ,
\
2T
(8)
I=
T
|G(b + it)|2 dt log−10B x.
28
C. Jia
I. First, we assume condition 1). On applying the mean value estimate
and Halász method to M 3 (s), H 5 (s) and K 2 (s), we get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −6 (M 3 + T ), V −6 M 3 + V −18 M 3 T, W −10 (H 5 + T ),
W −10 H 5 + W −30 H 5 T, U −4 (K 2 + T ), U −4 K 2 + U −12 K 2 T }.
We discuss the following cases:
(a) F ≤ 2V −6 M 3 , 2W −10 H 5 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −10 H 5 , U −4 K 2 }
1
3
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 10 (W −10 H 5 ) 5 (U −4 K 2 ) 2
1
9
= V 5 M 10 HK x log−11B x.
(b) F ≤ 2V −6 M 3 , 2W −10 H 5 , F > 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −10 H 5 , U −4 T, U −12 K 2 T }
1
1
9
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −10 H 5 ) 5 (U −4 T ) 20 (U −12 K 2 T ) 60
7
1
= T 15 M HK 30 x1−ε1 .
(c) F ≤ 2V −6 M 3 , F > 2W −10 H 5 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −10 T, W −30 H 5 T, U −4 K 2 }
1
3
1
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −10 T ) 20 (W −30 H 5 T ) 60 (U −4 K 2 ) 2
1
1
= T 6 M KH 12 x1−ε1 .
(d) F ≤ 2V −6 M 3 , F > 2W −10 H 5 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 M 3 , W −10 T, W −30 H 5 T, U −4 T, U −12 K 2 T }
1
1
9
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −10 T ) 5 (U −4 T ) 20 (U −12 K 2 T ) 60
2
1
= T 3 M K 30 x1−ε1 .
(e) F > 2V −6 M 3 , F ≤ 2W −10 H 5 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −10 H 5 , U −4 K 2 }
17
1
1
1
≤ U 2 V 2 W 2 (V −6 T ) 60 (V −18 M 3 T ) 60 (W −10 H 5 ) 5 (U −4 K 2 ) 2
3
1
= T 10 M 20 HK x1−ε1 .
Short intervals containing prime numbers
29
(f) F > 2V −6 M 3 , F ≤ 2W −10 H 5 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −10 H 5 , U −4 T, U −12 K 2 T }
1
1
9
1
≤ U 2 V 2 W 2 (V −6 T ) 3 (W −10 H 5 ) 5 (U −4 T ) 20 (U −12 K 2 T ) 60
4
1
= T 5 HK 30 x1−ε1 .
(g) F > 2V −6 M 3 , 2W −10 H 5 , F ≤ 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −10 T, W −30 H 5 T, U −4 K 2 }
1
3
1
1
≤ U 2 V 2 W 2 (V −6 T ) 3 (W −10 T ) 20 (W −30 H 5 T ) 60 (U −4 K 2 ) 2
1
1
= T 2 H 12 K x1−ε1 .
(h) F > 2V −6 M 3 , 2W −10 H 5 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 , V −18 M 3 , W −10 , W −30 H 5 , U −4 , U −12 K 2 }T
1
17
1
1
≤ U 2 V 2 W 2 (V −6 ) 60 (V −18 M 3 ) 60 (W −10 ) 5 (U −4 ) 2 T
1
= T M 20 x1−ε1 ,
since M X.
II. Next, we assume condition 2). On applying the mean value estimate
and Halász method to M 2 (s)H(s), H 5 (s) and K 2 (s), we get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 W −2 (M 2 H + T ), V −4 W −2 M 2 H + V −12 W −6 M 2 HT,
W −10 (H 5 + T ), W −10 H 5 + W −30 H 5 T, U −4 (K 2 + T ),
U −4 K 2 + U −12 K 2 T }.
We consider several cases:
(a) F ≤ 2V −4 W −2 M 2 H, 2W −10 H 5 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −10 H 5 , U −4 K 2 }
3
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 8 (W −10 H 5 ) 8 (U −4 K 2 ) 2
1
3
= V 2 M 4 HK x log−11B x.
30
C. Jia
(b) F ≤ 2V −4 W −2 M 2 H, 2W −10 H 5 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −10 H 5 , U −4 T, U −12 K 2 T }
1
1
7
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −10 H 5 ) 10 (U −4 T ) 20 (U −12 K 2 T ) 20
2
1
= T 5 M HK 10 x1−ε1 .
(c) F ≤ 2V −4 W −2 M 2 H, F > 2W −10 H 5 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −10 T, W −30 H 5 T, U −4 K 2 }
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (U −4 K 2 ) 2
1
= W H 2 M K x log−11B x.
(d) F ≤ 2V −4 W −2 M 2 H, F > 2W −10 H 5 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −10 T, W −30 H 5 T, U −4 T, U −12 K 2 T }
1
1
9
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −30 H 5 T ) 30 (U −4 T ) 20 (U −12 K 2 T ) 60
1
2
1
= T 2 M H 3 K 30 x1−ε1 .
(e) F > 2V −4 W −2 M 2 H, F ≤ 2W −10 H 5 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −10 H 5 , U −4 K 2 }
7
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 20 (V −12 W −6 M 2 HT ) 20 (W −10 H 5 ) 10 (U −4 K 2 ) 2
2
1
11
= T 5 M 10 H 20 K x1−ε1 .
(f) F > 2V −4 W −2 M 2 H, F ≤ 2W −10 H 5 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −10 H 5 ,
U −4 T, U −12 K 2 T }
7
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 20 (V −12 W −6 M 2 HT ) 20 (W −10 H 5 ) 10 (U −4 T ) 2
9
1
11
= T 10 M 10 H 20 x1−ε1 .
Short intervals containing prime numbers
31
(g) F > 2V −4 W −2 M 2 H, 2W −10 H 5 , F ≤ 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −10 T,
W −30 H 5 T, U −4 K 2 }
9
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 20 (V −12 W −6 M 2 HT ) 60 (W −30 H 5 T ) 30 (U −4 K 2 ) 2
1
1
11
= T 2 M 30 H 60 K x1−ε1 .
(h) F > 2V −4 W −2 M 2 H, 2W −10 H 5 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 , V −12 W −6 M 2 H, W −10 , W −30 H 5 ,
U −4 , U −12 K 2 }T
9
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 ) 20 (V −12 W −6 M 2 H) 60 (W −30 H 5 ) 30 (U −4 ) 2 T
1
11
= T M 30 H 60 x1−ε1 .
III. Lastly, we assume condition 3). On applying the mean value estimate
and Halász method to M 3 (s), H 4 (s) and K 2 (s), we get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −6 (M 3 + T ), V −6 M 3 + V −18 M 3 T, W −8 (H 4 + T ), W −8 H 4
+ W −24 H 4 T, U −4 (K 2 + T ), U −4 K 2 + U −12 K 2 T }.
Consider the following cases:
(a) F ≤ 2V −6 M 3 , 2W −8 H 4 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −8 H 4 , U −4 K 2 }
1
1
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 4 (W −8 H 4 ) 4 (U −4 K 2 ) 2
3
1
= V 2 M 4 HK x log−11B x.
(b) F ≤ 2V −6 M 3 , 2W −8 H 4 , F > 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −8 H 4 , U −4 T, U −12 K 2 T }
1
1
3
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −8 H 4 ) 4 (U −4 T ) 8 (U −12 K 2 T ) 24
5
1
= T 12 M HK 12 x1−ε1 .
32
C. Jia
(c) F ≤ 2V −6 M 3 , F > 2W −8 H 4 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 M 3 , W −8 T, W −24 H 4 T, U −4 K 2 }
1
1
1
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −4 K 2 ) 2
1
1
= T 6 M KH 6 x1−ε1 .
(d) F ≤ 2V −6 M 3 , F > 2W −8 H 4 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 M 3 , W −8 T, W −24 H 4 T, U −4 T, U −12 K 2 T }
1
1
1
1
≤ U 2 V 2 W 2 (V −6 M 3 ) 3 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −4 T ) 2
2
1
= T 3 M H 6 x1−ε1 .
(e) F > 2V −6 M 3 , F ≤ 2W −8 H 4 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −8 H 4 , U −4 K 2 }
1
5
1
1
≤ U 2 V 2 W 2 (V −6 T ) 24 (V −18 M 3 T ) 24 (W −8 H 4 ) 4 (U −4 K 2 ) 2
1
1
= T 4 M 8 HK x1−ε1 .
(f) F > 2V −6 M 3 , F ≤ 2W −8 H 4 , F > 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −8 H 4 , U −4 T, U −12 K 2 T }
1
5
1
1
≤ U 2 V 2 W 2 (V −6 T ) 24 (V −18 M 3 T ) 24 (W −8 H 4 ) 4 (U −4 T ) 2
3
1
= T 4 M 8 H x1−ε1 .
(g) F > 2V −6 M 3 , 2W −8 H 4 , F ≤ 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 T, V −18 M 3 T, W −8 T, W −24 H 4 T, U −4 K 2 }
1
1
1
1
≤ U 2 V 2 W 2 (V −6 T ) 3 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −4 K 2 ) 2
1
1
= T 2 H 6 K x1−ε1 .
(h) F > 2V −6 M 3 , 2W −8 H 4 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −6 , V −18 M 3 , W −8 , W −24 H 4 , U −4 , U −12 K 2 }T
5
1
1
1
≤ U 2 V 2 W 2 (V −6 ) 24 (V −18 M 3 ) 24 (W −8 ) 4 (U −4 ) 2 T
1
= T M 8 x1−ε1 ,
2
25
19
since M X 5 (the latter follows from M H X 44 and X 100 H).
Combining the above, we obtain (8). Hence, Lemma 5 follows.
Short intervals containing prime numbers
33
Lemma 6. Under the assumption of Lemma 5, M and H lie in one of
the following regions:
19
3
6
3
19
M − 11 X 110 H M − 8 X 80 ;
247
X 110 H M − 8 X 80 ;
X 70 M X 10 ,
(ii)
X 10 M X 60 ,
(iii)
X 60 M X 770 ,
(iv)
X 770 M X 1100 ,
(v)
X 1100 M X 1450 ,
(vi)
(vii)
19
247
X
X
12
359
481
481
1450
443
1305
19
55
M X
M X
23
19
55
2143
5620
57
1
1
23
23
19
25
19
11
5
X 110 H M −1 X 44 ;
,
,
(viii)
X
(ix)
X 5620 M X 9860 ,
(x)
X 9860 M X 9860 ,
M X
1
19
359
443
1305
57
M − 5 X 100 H M − 8 X 80 ;
(i)
,
X 110 H M −6 X
X
X
M
19
110
19
110
H M −1 X
HM
− 58
49
58
X
57
98
2
− 11
;
157
290
X
;
29
110
2
− 11
HM
;
29
X 110 ;
29
2143
3963
M − 49 X 98 H M − 19 X 76 ;
57
3963
4331
X 85 H M − 19 X 76 .
9
29
59
59
Then (7) holds for T0 ≤ T ≤ X.
P r o o f. In the regions:
3
19
M − 11 X 110 H M 29 X 145 ;
19
19
X 110 H M −1 X 290 ,
X 10 M X 60 ,
X 60 M X 55 ,
12
57
1
19
31
157
we apply Lemma 5 with condition 1).
In the regions:
443
19
X 1305 M X 55 ,
19
2143
X 55 M X 5620 ,
157
2
29
M −1 X 290 H M − 11 X 110 ;
58
57
2
58
57
29
29
M − 49 X 98 H M − 11 X 110 ;
2143
3963
M − 49 X 98 H M − 19 X 76 ;
3963
4331
X 85 H M − 19 X 76 ,
X 5620 M X 9860 ,
X 9860 M X 9860 ,
9
29
59
59
we apply Lemma 5 with condition 2).
In the regions:
19
3
M − 5 X 100 H M − 8 X 80 ;
3
19
M 29 X 145 H M − 8 X 80 ;
247
M −1 X 290 H M − 8 X 80 ;
X 70 M X 10 ,
X 10 M X 60 ,
19
X 60 M X 770 ,
247
359
X 770 M X 1100 ,
359
481
X 1100 M X 1450 ,
6
1
57
1
23
31
1
23
157
1
23
157
25
157
11
5
M −1 X 290 H M −1 X 44 ;
M −1 X 290 H M −6 X
we apply Lemma 5 with condition 3).
Putting together the above regions, we get Lemma 6.
,
34
C. Jia
Lemma 7. Under the assumption of Lemma 5, suppose that M and H
also satisfy one of the following three conditions:
7
19
63
133
1) M H X 10 , X 110 H, M 12 H X 10 , X 290 M , H 19 /M 38
X , X 5 M 12 H 11 ;
52
59
19
29
57
2) M 2 H X 45 , M 58 H 9 X 2 , X 45 M , M H 5 X 20 , X 4 M 29 H 10 ;
51
19
63
19
13
3) M H X 70 , X 100 H, M 6 H X 20 , X 44 M , H 7 /M X 10 ,
19
X 5 M 6H 5.
19
5
Then (7) holds for T0 ≤ T ≤ X.
19
P r o o f. We only show that (8) holds for T = 1/η = 2X 20 −ε .
I. First, we assume condition 1). We apply the mean value estimate and
Halász method to M 2 (s), H 5 (s) and K 3 (s) to get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 (M 2 + T ), V −4 M 2 + V −12 M 2 T, W −10 (H 5 + T ), W −10 H 5
+ W −30 H 5 T, U −6 (K 3 + T ), U −6 K 3 + U −18 K 3 T }.
We consider several cases:
(a) F ≤ 2V −4 M 2 , 2W −10 H 5 , 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −10 H 5 , U −6 K 3 }
1
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −10 H 5 ) 6 (U −6 K 3 ) 3
5
1
= W 3 H 6 M K x log−11B x.
(b) F ≤ 2V −4 M 2 , 2W −10 H 5 , F > 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −10 H 5 , U −6 T, U −18 K 3 T }
1
1
17
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −10 H 5 ) 5 (U −6 T ) 60 (U −18 K 3 T ) 60
1
3
= T 10 M HK 20 x1−ε1 .
(c) F ≤ 2V −4 M 2 , F > 2W −10 H 5 , F ≤ 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −10 T, W −30 H 5 T, U −6 K 3 }
1
3
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −10 T ) 20 (W −30 H 5 T ) 60 (U −6 K 3 ) 3
1
1
= T 6 M KH 12 x1−ε1 .
Short intervals containing prime numbers
35
(d) F ≤ 2V −4 M 2 , F > 2W −10 H 5 , 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 M 2 , W −10 T, W −30 H 5 T, U −6 T, U −18 K 3 T }
1
3
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −10 T ) 20 (W −30 H 5 T ) 60 (U −6 T ) 3
1
1
= T 2 M H 12 x1−ε1 .
(e) F > 2V −4 M 2 , F ≤ 2W −10 H 5 , 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 H 5 , U −6 K 3 }
9
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 20 (V −12 M 2 T ) 60 (W −10 H 5 ) 5 (U −6 K 3 ) 3
7
1
= T 15 M 30 HK x1−ε1 .
(f) F > 2V −4 M 2 , F ≤ 2W −10 H 5 , F > 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 H 5 , U −6 T, U −18 K 3 T }
1
1
17
1
≤ U 2 V 2 W 2 (V −4 T ) 2 (W −10 H 5 ) 5 (U −6 T ) 60 (U −18 K 3 T ) 60
4
1
= T 5 HK 20 x1−ε1 .
(g) F > 2V −4 M 2 , 2W −10 H 5 , F ≤ 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 T, W −30 H 5 T, U −6 K 3 }
3
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 2 (W −10 T ) 20 (W −30 H 5 T ) 60 (U −6 K 3 ) 3
2
1
= T 3 H 12 K x1−ε1 .
(h) F > 2V −4 M 2 , 2W −10 H 5 , 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 , V −12 M 2 , W −10 , W −30 H 5 , U −6 , U −18 K 3 }T
1
1
17
1
≤ U 2 V 2 W 2 (V −4 ) 2 (W −10 ) 5 (U −6 ) 60 (U −18 K 3 ) 60 T
1
= T K 20 x1−ε1 ,
since K X.
II. Next, assume condition 2). We apply the mean value estimate and
Halász method to M 2 (s), H 5 (s) and K 2 (s)H(s) to get
I U 2 V 2 W 2 x−1 F logc x,
36
C. Jia
where
F = min{V −4 (M 2 + T ), V −4 M 2 + V −12 M 2 T, W −10 (H 5 + T ), W −10 H 5
+ W −30 H 5 T, U −4 W −2 (K 2 H + T ), U −4 W −2 K 2 H
+ U −12 W −6 K 2 HT }.
Consider the following cases:
(a) F ≤ 2V −4 M 2 , 2W −10 H 5 , 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −10 H 5 , U −4 W −2 K 2 H}
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (U −4 W −2 K 2 H) 2
1
= W H 2 M K x log−11B x.
(b) F ≤ 2V −4 M 2 , 2W −10 H 5 , F > 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 M 2 , W −10 H 5 , U −4 W −2 T, U −12 W −6 K 2 HT }
1
1
7
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −10 H 5 ) 10 (U −4 W −2 T ) 20 (U −12 W −6 K 2 HT ) 20
11
2
1
= T 5 M H 20 K 10 x1−ε1 .
(c) F ≤ 2V −4 M 2 , F > 2W −10 H 5 , F ≤ 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −10 T, W −30 H 5 T, U −4 W −2 K 2 H}
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (U −4 W −2 K 2 H) 2
1
= W H 2 M K x log−11B x.
(d) F ≤ 2V −4 M 2 , F > 2W −10 H 5 , 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 M 2 , W −10 T, W −30 H 5 T, U −4 W −2 T,
U −12 W −6 K 2 HT }
1
1
9
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −30 H 5 T ) 30 (U −4 W −2 T ) 20 (U −12 W −6 K 2 HT ) 60
1
11
1
= T 2 M H 60 K 30 x1−ε1 .
(e) F > 2V −4 M 2 , F ≤ 2W −10 H 5 , 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 H 5 , U −4 W −2 K 2 H}
7
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 20 (V −12 M 2 T ) 20 (W −10 H 5 ) 10 (U −4 W −2 K 2 H) 2
2
1
= T 5 M 10 HK x1−ε1 .
Short intervals containing prime numbers
37
(f) F > 2V −4 M 2 , F ≤ 2W −10 H 5 , F > 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 H 5 , U −4 W −2 T,
U −12 W −6 K 2 HT }
7
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 20 (V −12 M 2 T ) 20 (W −10 H 5 ) 10 (U −4 W −2 T ) 2
9
1
1
= T 10 M 10 H 2 x1−ε1 .
(g) F > 2V −4 M 2 , 2W −10 H 5 , F ≤ 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −10 T, W −30 H 5 T, U −4 W −2 K 2 H}
1
9
1
1
≤ U 2 V 2 W 2 (V −4 T ) 20 (V −12 M 2 T ) 60 (W −30 H 5 T ) 30 (U −4 W −2 K 2 H) 2
1
1
2
= T 2 M 30 H 3 K x1−ε1 .
(h) F > 2V −4 M 2 , 2W −10 H 5 , 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 , V −12 M 2 , W −10 , W −30 H 5 , U −4 W −2 ,
U −12 W −6 K 2 H}T
9
1
1
1
≤ U 2 V 2 W 2 (V −4 ) 20 (V −12 M 2 ) 60 (W −30 H 5 ) 30 (U −4 W −2 ) 2 T
1
1
= T M 30 H 6 x1−ε1 .
III. Lastly, we assume condition 3). We apply the mean value estimate
and Halász method to M 2 (s), H 4 (s) and K 3 (s) to get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 (M 2 + T ), V −4 M 2 + V −12 M 2 T, W −8 (H 4 + T ), W −8 H 4
+ W −24 H 4 T, U −6 (K 3 + T ), U −6 K 3 + U −18 K 3 T }.
Consider the following cases:
(a) F ≤ 2V −4 M 2 , 2W −8 H 4 , 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −8 H 4 , U −6 K 3 }
1
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −8 H 4 ) 6 (U −6 K 3 ) 3
2
2
= W 3 H 3 M K x log−11B x.
38
C. Jia
(b) F ≤ 2V −4 M 2 , 2W −8 H 4 , F > 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −8 H 4 , U −6 T, U −18 K 3 T }
1
1
5
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −8 H 4 ) 4 (U −6 T ) 24 (U −18 K 3 T ) 24
1
1
= T 4 M HK 8 x1−ε1 .
(c) F ≤ 2V −4 M 2 , F > 2W −8 H 4 , F ≤ 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −8 T, W −24 H 4 T, U −6 K 3 }
1
1
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −6 K 3 ) 3
1
1
= T 6 M KH 6 x1−ε1 .
(d) F ≤ 2V −4 M 2 , F > 2W −8 H 4 , 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 M 2 , W −8 T, W −24 H 4 T, U −6 T, U −18 K 3 T }
1
1
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −6 T ) 3
1
1
= T 2 M H 6 x1−ε1 .
(e) F > 2V −4 M 2 , F ≤ 2W −8 H 4 , 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −8 H 4 , U −6 K 3 }
3
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 8 (V −12 M 2 T ) 24 (W −8 H 4 ) 4 (U −6 K 3 ) 3
1
5
= T 12 M 12 HK x1−ε1 .
(f) F > 2V −4 M 2 , F ≤ 2W −8 H 4 , F > 2U −6 K 3 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −8 H 4 , U −6 T, U −18 K 3 T }
1
1
5
1
≤ U 2 V 2 W 2 (V −4 T ) 2 (W −8 H 4 ) 4 (U −6 T ) 24 (U −18 K 3 T ) 24
3
1
= T 4 HK 8 x1−ε1 .
(g) F > 2V −4 M 2 , 2W −8 H 4 , F ≤ 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −8 T, W −24 H 4 T, U −6 K 3 }
1
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 2 (W −8 T ) 8 (W −24 H 4 T ) 24 (U −6 K 3 ) 3
2
1
= T 3 H 6 K x1−ε1 .
Short intervals containing prime numbers
39
(h) F > 2V −4 M 2 , 2W −8 H 4 , 2U −6 K 3 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 , V −12 M 2 , W −8 , W −24 H 4 , U −6 , U −18 K 3 }T
1
1
5
1
≤ U 2 V 2 W 2 (V −4 ) 2 (W −8 ) 4 (U −6 ) 24 (U −18 K 3 ) 24 T
1
= T K 8 x1−ε1 ,
3
19
19
since X 5 M H (the latter follows from X 44 M and X 100 H).
Combining the above, we obtain (8). Hence, Lemma 7 follows.
Lemma 8. Under the assumption of Lemma 5, suppose that M and H
lie in one of the following regions:
19
19
29
57
1
29
29
57
1
29
M − 10 X 40 H M − 5 X 100 ;
(i)
X 45 M X 44 ,
(ii)
X 44 M X 1972 ,
(iii)
X 44 M X 1972 ,
(iv)
X 1972 M X 290 ,
(v)
X 1972 M X 290 ,
(vi)
X 290 M X 40 ,
(vii)
X 40 M X 110 ,
(viii)
X 110 M X 700 ,
(ix)
X 110 M X 700 ,
(x)
X 700 M X 100 ,
(xi)
X 700 M X 100 ,
(xii)
X 100 M X 2 ,
19
897
M − 10 X 40 H M − 5 X 100 ;
19
897
M − 5 X 25 H M 7 X 70 ;
6
19
1
897
133
X 85 H M − 5 X 100 ;
897
133
M − 5 X 25 H M 7 X 70 ;
133
19
53
19
9
13
29
1
6
19
9
1
1
13
13
X 85 H M 7 X 70 ;
9
51
X 85 H M −1 X 70 ;
53
339
X 85 H M − 9 X 18 ;
53
339
X 110 H M −1 X 70 ;
339
49
X 85 H M − 9 X 18 ;
339
49
X 110 H M −6 X 20 ;
1
49
9
58
19
59
51
9
58
59
19
63
19
7
X 110 H M −1 X 10 .
Then (7) holds for T0 ≤ T ≤ X.
P r o o f. In the regions:
133
19
X 290 M X 40 ,
19
1
X 40 M X 2 ,
12
38
1
1
M − 11 X 55 H M 19 X 5 ;
19
7
X 110 H M −1 X 10 ,
we apply Lemma 7 with condition 1).
In the regions:
19
897
X 45 M X 1972 ,
897
133
X 1972 M X 290 ,
133
19
X 290 M X 40 ,
29
57
1
29
M − 10 X 40 H M − 5 X 100 ;
9
1
9
12
29
X 85 H M − 5 X 100 ;
38
X 85 H M − 11 X 55 ;
40
C. Jia
19
53
9
X 40 M X 110 ,
53
19
X 85 H X 110 ;
49
9
58
59
X 110 M X 100 , X 85 H M − 9 X 18 ,
we apply Lemma 7 with condition 2).
In the regions:
19
133
X 44 M X 290 ,
133
19
X 290 M X 40 ,
19
339
X 40 M X 700 ,
339
49
X 700 M X 100 ,
6
19
1
13
M − 5 X 25 H M 7 X 70 ;
1
1
1
13
M 19 X 5 H M 7 X 70 ;
7
51
7
63
M −1 X 10 H M −1 X 70 ;
M −1 X 10 H M −6 X 20 ,
we apply Lemma 7 with condition 3).
Putting together the above regions, we get Lemma 8.
Lemma 9. Under the assumption of Lemma 5, suppose that M and H
lie in one of the following regions:
1843
1188
M − 59 X 295 H M − 49 X 98 ;
1188
3963
X 85 H M − 49 X 98 ;
1063
4331
M − 19 X 76 H M − 23 X 230 .
(i)
X 5280 M X 2975 ,
(ii)
X 2975 M X 9860 ,
(iii)
X 2640 M X 9860 ,
70
171
58
9
58
29
57
57
35
59
179
Then (7) holds for T0 ≤ T ≤ X.
19
P r o o f. First we show that (8) holds for T = 1/η = 2X 20 −ε , providing
M and H satisfy the following conditions:
25
179
19
M H X 44 , M 35 H 23 X 10 , X 22 M 2 H,
31
171
M 2 H 13 X 10 , X 5 M 70 H 59 .
We apply the mean value estimate and Halász method to M 2 (s)H(s),
H (s) and K 2 (s) to get
6
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 W −2 (M 2 H + T ), V −4 W −2 M 2 H + V −12 W −6 M 2 HT,
W −12 (H 6 + T ), W −12 H 6 + W −36 H 6 T, U −4 (K 2 + T ),
U −4 K 2 + U −12 K 2 T }.
Consider the following cases:
(a) F ≤ 2V −4 W −2 M 2 H, 2W −12 H 6 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −12 H 6 , U −4 K 2 }
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (U −4 K 2 ) 2
1
= W H 2 M K x log−11B x.
Short intervals containing prime numbers
41
(b) F ≤ 2V −4 W −2 M 2 H, 2W −12 H 6 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −12 H 6 , U −4 T, U −12 K 2 T }
1
1
3
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −12 H 6 ) 12 (U −4 T ) 8 (U −12 K 2 T ) 24
5
1
= T 12 M HK 12 x1−ε1 .
(c) F ≤ 2V −4 W −2 M 2 H, F > 2W −12 H 6 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −12 T, W −36 H 6 T, U −4 K 2 }
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (U −4 K 2 ) 2
1
= W H 2 M K x log−11B x.
(d) F ≤ 2V −4 W −2 M 2 H, F > 2W −12 H 6 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −12 T, W −36 H 6 T, U −4 T, U −12 K 2 T }
1
1
11
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −36 H 6 T ) 36 (U −4 T ) 24 (U −12 K 2 T ) 72
1
2
1
= T 2 M H 3 K 36 x1−ε1 .
(e) F > 2V −4 W −2 M 2 H, F ≤ 2W −12 H 6 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −12 H 6 , U −4 K 2 }
3
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 8 (V −12 W −6 M 2 HT ) 24 (W −12 H 6 ) 12 (U −4 K 2 ) 2
5
1
13
= T 12 M 12 H 24 K x1−ε1 .
(f) F > 2V −4 W −2 M 2 H, F ≤ 2W −12 H 6 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −12 H 6 ,
U −4 T, U −12 K 2 T }
3
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 8 (V −12 W −6 M 2 HT ) 24 (W −12 H 6 ) 12 (U −4 T ) 2
11
1
13
= T 12 M 12 H 24 x1−ε1 .
42
C. Jia
(g) F > 2V −4 W −2 M 2 H, 2W −12 H 6 , F ≤ 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −12 T,
W −36 H 6 T, U −4 K 2 }
11
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 24 (V −12 W −6 M 2 HT ) 72 (W −36 H 6 T ) 36 (U −4 K 2 ) 2
1
1
13
= T 2 M 36 H 72 K x1−ε1 .
(h) F > 2V −4 W −2 M 2 H, 2W −12 H 6 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 , V −12 W −6 M 2 H, W −12 , W −36 H 6 ,
U −4 , U −12 K 2 }T
1
1
3
1
≤ U 2 V 2 W 2 (V −4 W −2 ) 2 (W −12 ) 12 (U −4 ) 8 (U −12 K 2 ) 24 T
1
= T K 12 x1−ε1 ,
2
171
since X 5 M H (the latter follows from X 5 M 70 H 59 ).
In every region, our conditions are satisfied. So the proof of Lemma 9 is
complete.
Lemma 10. Under the assumption of Lemma 5, suppose that M and H
lie in one of the following regions:
37
5541
(i)
X 100 M X 13940 ,
(ii)
X 13940 M X 2975 ,
(iii)
X 45 M X 9860 ,
5541
19
1188
4331
82
133
70
171
M − 69 X 230 H M − 59 X 295 ;
9
70
171
X 85 H M − 59 X 295 ;
35
179
41
421
M − 23 X 230 H M − 27 X 540 .
Then (7) holds for T0 ≤ T ≤ X.
19
P r o o f. First we show that (8) holds for T = 1/η = 2X 20 −ε , providing
M and H satisfy the following conditions:
73
421
57
M H X 130 , M 41 H 27 X 20 , X 65 M 2 H,
33
399
M 2 H 15 X 10 , X 10 M 82 H 69 .
We apply the mean value estimate and Halász method to M 2 (s)H(s),
H (s) and K 2 (s) to get
7
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 W −2 (M 2 H + T ), V −4 W −2 M 2 H + V −12 W −6 M 2 HT,
W −14 (H 7 + T ), W −14 H 7 + W −42 H 7 T, U −4 (K 2 + T ),
U −4 K 2 + U −12 K 2 T }.
Consider the following cases:
Short intervals containing prime numbers
43
(a) F ≤ 2V −4 W −2 M 2 H, 2W −14 H 7 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −14 H 7 , U −4 K 2 }
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (U −4 K 2 ) 2
1
= W H 2 M K x log−11B x.
(b) F ≤ 2V −4 W −2 M 2 H, 2W −14 H 7 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −14 H 7 , U −4 T, U −12 K 2 T }
1
1
11
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −14 H 7 ) 14 (U −4 T ) 28 (U −12 K 2 T ) 28
3
1
= T 7 M HK 14 x1−ε1 .
(c) F ≤ 2V −4 W −2 M 2 H, F > 2W −14 H 7 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −14 T, W −42 H 7 T, U −4 K 2 }
1
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (U −4 K 2 ) 2
1
= W H 2 M K x log−11B x.
(d) F ≤ 2V −4 W −2 M 2 H, F > 2W −14 H 7 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 M 2 H, W −14 T, W −42 H 7 T, U −4 T, U −12 K 2 T }
1
1
13
1
≤ U 2 V 2 W 2 (V −4 W −2 M 2 H) 2 (W −42 H 7 T ) 42 (U −4 T ) 28 (U −12 K 2 T ) 84
1
2
1
= T 2 M H 3 K 42 x1−ε1 .
(e) F > 2V −4 W −2 M 2 H, F ≤ 2W −14 H 7 , 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −14 H 7 , U −4 K 2 }
11
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 28 (V −12 W −6 M 2 HT ) 28 (W −14 H 7 ) 14 (U −4 K 2 ) 2
3
1
15
= T 7 M 14 H 28 K x1−ε1 .
(f) F > 2V −4 W −2 M 2 H, F ≤ 2W −14 H 7 , F > 2U −4 K 2 . Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −14 H 7 ,
U −4 T, U −12 K 2 T }
11
1
1
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 28 (V −12 W −6 M 2 HT ) 28 (W −14 H 7 ) 14 (U −4 T ) 2
13
1
15
= T 14 M 14 H 28 x1−ε1 .
44
C. Jia
(g) F > 2V −4 W −2 M 2 H, 2W −14 H 7 , F ≤ 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 T, V −12 W −6 M 2 HT, W −14 T,
W −42 H 7 T, U −4 K 2 }
13
1
≤ U 2 V 2 W 2 (V −4 W −2 T ) 28 (V −12 W −6 M 2 HT ) 84
1
1
× (W −42 H 7 T ) 42 (U −4 K 2 ) 2
1
1
5
= T 2 M 42 H 28 K x1−ε1 .
(h) F > 2V −4 W −2 M 2 H, 2W −14 H 7 , 2U −4 K 2 . Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 W −2 , V −12 W −6 M 2 H, W −14 , W −42 H 7 ,
U −4 , U −12 K 2 }T
1
1
11
1
≤ U 2 V 2 W 2 (V −4 W −2 ) 2 (W −14 ) 14 (U −4 ) 28 (U −12 K 2 ) 28 T
1
= T K 14 x1−ε1 ,
3
57
since X 10 M H (the latter follows from X 65 M 2 H).
In every region, our conditions are satisfied, so the proof of Lemma 10
is complete.
Lemma 11. Under the assumption of Lemma 5, suppose that M and H
lie in one of the following regions:
1843
37
M − 59 X 295 H M − 49 X 98 ;
5541
M − 69 X 230 H M − 49 X 98 ;
(i)
X 5280 M X 100 ,
(ii)
X 100 M X 13940 ,
(iii)
X 13940 M X 9860 ,
(iv)
X 2640 M X 45 ,
(v)
X 45 M X 9860 ,
37
5541
3963
1063
19
70
171
58
57
82
133
58
57
58
9
57
X 85 H M − 49 X 98 ;
19
M − 19 X 76 H M − 23 X 230 ;
29
59
35
179
4331
M − 19 X 76 H M − 27 X 540 .
29
59
41
421
Then (7) holds for T0 ≤ T ≤ X.
P r o o f. Putting together regions in Lemmas 9 and 10, we can get Lemma 11.
Lemma 12. Under the assumption of Lemma 5, suppose that M and H
lie in one of the following regions:
4331
2691
2691
897
(i)
X 9860 M X 5950 ,
(ii)
X 5950 M X 1972 ,
(iii)
X 480 M X 100 ,
(iv)
233
X
49
100
49
M X
2944
5950
35
57
29
9
29
57
X 85 H M − 10 X 40 ;
58
59
70
M − 9 X 18 H M − 11 X
,
Then (7) holds for T0 ≤ T ≤ X.
57
M − 12 X 40 H M − 10 X 40 ;
X
9
85
HM
− 70
11
X
179
55
.
179
55
;
Short intervals containing prime numbers
45
19
P r o o f. First we show that (8) holds for T = 1/η = 2X 20 −ε , providing
that M and H satisfy the following conditions:
25
179
19
M 2 H X 22 , M 70 H 11 X 5 , X 44 M,
31
171
M H 6 X 20 , X 10 M 35 H 12 .
We apply the mean value estimate and Halász method to M 2 (s), H 6 (s)
and K 2 (s)H(s) to get
I U 2 V 2 W 2 x−1 F logc x,
where
F = min{V −4 (M 2 + T ), V −4 M 2 + V −12 M 2 T, W −12 (H 6 + T ), W −12 H 6
+ W −36 H 6 T, U −4 W −2 (K 2 H + T ), U −4 W −2 K 2 H
+ U −12 W −6 K 2 HT }.
Consider the following cases:
(a) F ≤ 2V −4 M 2 , 2W −12 H 6 , 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −12 H 6 , U −4 W −2 K 2 H}
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (U −4 W −2 K 2 H) 2
1
= W H 2 M K x log−11B x.
(b) F ≤ 2V −4 M 2 , 2W −12 H 6 , F > 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 M 2 , W −12 H 6 , U −4 W −2 T, U −12 W −6 K 2 HT }
1
1
3
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −12 H 6 ) 12 (U −4 W −2 T ) 8 (U −12 W −6 K 2 HT ) 24
5
13
1
= T 12 M H 24 K 12 x1−ε1 .
(c) F ≤ 2V −4 M 2 , F > 2W −12 H 6 , F ≤ 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −12 T, W −36 H 6 T, U −4 W −2 K 2 H}
1
1
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (U −4 W −2 K 2 H) 2
1
= W H 2 M K x log−11B x.
(d) F ≤ 2V −4 M 2 , F > 2W −12 H 6 , 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 M 2 , W −12 T, W −36 H 6 T,
U −4 W −2 T, U −12 W −6 K 2 HT }
1
1
11
≤ U 2 V 2 W 2 (V −4 M 2 ) 2 (W −36 H 6 T ) 36 (U −4 W −2 T ) 24
1
× (U −12 W −6 K 2 HT ) 72
1
13
1
= T 2 M H 72 K 36 x1−ε1 .
46
C. Jia
(e) F > 2V −4 M 2 , F ≤ 2W −12 H 6 , 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −12 H 6 , U −4 W −2 K 2 H}
3
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 8 (V −12 M 2 T ) 24 (W −12 H 6 ) 12 (U −4 W −2 K 2 H) 2
5
1
= T 12 M 12 HK x1−ε1 .
(f) F > 2V −4 M 2 , F ≤ 2W −12 H 6 , F > 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −12 H 6 , U −4 W −2 T,
U −12 W −6 K 2 HT }
3
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 8 (V −12 M 2 T ) 24 (W −12 H 6 ) 12 (U −4 W −2 T ) 2
11
1
1
= T 12 M 12 H 2 x1−ε1 .
(g) F > 2V −4 M 2 , 2W −12 H 6 , F ≤ 2U −4 W −2 K 2 H. Then
U 2V 2W 2F
U 2 V 2 W 2 min{V −4 T, V −12 M 2 T, W −12 T, W −36 H 6 T, U −4 W −2 K 2 H}
11
1
1
1
≤ U 2 V 2 W 2 (V −4 T ) 24 (V −12 M 2 T ) 72 (W −36 H 6 T ) 36 (U −4 W −2 K 2 H) 2
1
1
2
= T 2 M 36 H 3 K x1−ε1 .
(h) F > 2V −4 M 2 , 2W −12 H 6 , 2U −4 W −2 K 2 H. Then
U 2 V 2 W 2 F U 2 V 2 W 2 min{V −4 , V −12 M 2 , W −12 , W −36 H 6 , U −4 W −2 ,
U −12 W −6 K 2 H}T
3
1
1
1
≤ U 2 V 2 W 2 (V −4 ) 8 (V −12 M 2 ) 24 (W −12 ) 12 (U −4 W −2 ) 2 T
1
= T M 12 x1−ε1 ,
179
3
since M X 5 (the latter follows from M 70 H 11 X 5 ).
In every region, our conditions are satisfied. So the proof of Lemma 12
is complete.
4. Mean value estimate (III)
Lemma 13. Suppose that P QRK = X and that P (s), Q(s), R(s) and
K(s) are Dirichlet polynomials. Define G(s) = P (s)Q(s)R(s)K(s). Let b =
B
1 + 1/ log X and T0 = log ε X. Assume further that for T0 ≤ |t| ≤ 2X,
B
B
P (b + it)Q(b + it) log− ε x and R(b + it) log− ε x. Moreover , assume
that
9
X 85 R Q
Short intervals containing prime numbers
47
and that P and Q lie in one of the following regions:
133
19
133
P −1 X 290 Q P ;
(i)
X 580 P X 80 ,
(ii)
X 80 P X 220 ,
(iii)
X 220 P X 580 ,
(iv)
X 580 P X 55 ,
(v)
X 55 P X 9860 ,
(vi)
X 9860 P X 100 ,
(vii)
X 100 P X 1700 ,
19
53
53
151
17
133
151
P −1 X 290 Q P −1 X 110 ;
133
17
P −1 X 1972 Q P −1 X 110 ;
37
53
897
3441
3441
19
P −1 X 290 Q P −1 X 40 ;
53
897
58
59
P −1 X 1972 Q P − 67 X 134 ;
37
9
58
59
X 85 Q P − 67 X 134 ;
653
9
49
X 85 Q P −1 X 100 .
Then (7) holds for T0 ≤ T ≤ X.
P r o o f. Let m = pq, h = r.
(a) On applying Lemma 8 with region (iv), we see that (7) holds under
the conditions
897
133
9
P −1 X 1972 Q P −1 X 290 ,
1
29
X 85 R Q (P Q)− 5 X 100 ,
which can be written as
897
133
9
P −1 X 1972 Q P −1 X 290 ,
1
29
9
X 85 Q P − 6 X 120 ,
X 85 R Q.
In the regions:
151
3441
897
3441
133
P −1 X 1972 Q P −1 X 290 ;
X 580 P X 9860 ,
1739
9
133
X 85 Q P −1 X 290 ,
X 9860 P X 4930 ,
the above conditions on P and Q are satisfied.
(b) On applying Lemma 8 with region (vi), we see that (7) holds under
the conditions
133
19
9
P −1 X 290 Q P −1 X 40 ,
1
13
X 85 R Q (P Q) 7 X 70 ,
which can be written as
133
19
9
P −1 X 290 Q P −1 X 40 ,
1
13
9
X 85 Q P 6 X 60 ,
X 85 R Q.
In the regions:
133
19
133
P −1 X 290 Q P ;
X 580 P X 80 ,
19
1739
X 80 P X 4930 ,
1739
251
X 4930 P X 680 ,
133
19
P −1 X 290 Q P −1 X 40 ;
9
19
X 85 Q P −1 X 40 ,
the above conditions on P and Q are satisfied.
(c) On applying Lemma 8 with region (vii), we see that (7) holds under
the conditions
19
53
P −1 X 40 Q P −1 X 110 ,
9
51
X 85 R Q (P Q)−1 X 70 ,
48
C. Jia
which can be written as
19
53
9
P −1 X 40 Q P −1 X 110 ,
1
51
9
X 85 Q P − 2 X 140 ,
X 85 R Q.
In the regions:
53
251
19
251
53
P −1 X 40 Q P −1 X 110 ;
X 220 P X 680 ,
703
9
53
X 85 Q P −1 X 110 ,
X 680 P X 1870 ,
the above conditions on P and Q are satisfied.
(d) On applying Lemma 8 with regions (viii) and (x), we see that (7)
holds under the conditions
53
9
49
P −1 X 110 Q P −1 X 100 ,
58
59
X 85 R Q (P Q)− 9 X 18 ,
which can be written as
53
9
49
P −1 X 110 Q P −1 X 100 ,
58
59
9
X 85 Q P − 67 X 134 ,
X 85 R Q.
In the regions:
17
37
53
37
703
59
53
49
P −1 X 110 Q P −1 X 100 ;
X 100 P X 1870 ,
703
58
P −1 X 110 Q P − 67 X 134 ;
X 55 P X 100 ,
653
9
49
X 85 Q P −1 X 100 ,
X 1870 P X 1700 ,
the above conditions on P and Q are satisfied.
Putting together the above regions, we get Lemma 13.
Lemma 14. Suppose that P QRL = X and that P (s), Q(s) and R(s) are
Dirichlet polynomials,
X 1
L(s) =
,
ls
l∼L
√
and F (s) = P (s)Q(s)R(s)L(s). Let b = 1 + 1/ log X and T1 = L. Assume
B
further that for T1 ≤ |t| ≤ 2X, P (b+it)Q(b+it) log− ε x and R(b+it) B
log− ε x. Moreover , assume that
9
X 85 R Q
and that P and Q lie in one of the following regions:
9
21
(i)
X 85 P X 100 ,
(ii)
X 100 P X 60 ,
(iii)
X 60 P X 49300 ,
(iv)
X 49300 P X 9860 ,
21
13
12613
13
12613
3441
9
X 85 Q P ;
9
2
7
9
1
29
X 85 Q P − 3 X 20 ;
X 85 Q P − 6 X 120 ;
9
897
X 85 Q P −1 X 1972 .
Short intervals containing prime numbers
49
Then for T1 ≤ T ≤ X, we have
(9)
min
2
1
η,
T
2T
\
|F (b + it)|2 dt η 2 log−10B x.
T
P r o o f. Let m = pq and n = r. An application of Lemma 3 yields that
(9) holds under one of the following conditions:
2
9
(a)
M X 32 ,
(b)
X 32 M X 160 ,
(c)
X 160 M X 32 ,
(d)
X 32 M X 45 ,
(e)
X 45 M X 1972 ,
1
N M 3 X 10 ;
9
53
53
23
N X 80 ;
2
13
13
19
19
61
N M − 3 X 120 ;
21
N M −2 X 20 ;
897
21
N M −2 X 20 .
Using the same discussion as in Lemma 8 with regions (i) and (ii), we
deduce that (9) holds under the condition
(f)
19
897
X 45 M X 1972 ,
21
1
29
M −2 X 20 N M − 5 X 100 .
In the regions:
9
9
9
477
X 85 P X 64 ,
X 64 P X 2720 ,
9
X 85 Q P ;
9
9
X 85 Q P −1 X 32 ,
condition (a) is satisfied.
In the regions:
9
53
X 64 P X 320 ,
53
477
X 320 P X 2720 ,
477
613
X 2720 P X 2720 ,
9
P −1 X 32 Q P ;
9
53
P −1 X 32 Q P −1 X 160 ;
9
53
X 85 Q P −1 X 160 ,
condition (b) is satisfied.
In the regions:
53
13
X 320 P X 64 ,
13
613
X 64 P X 2720 ,
613
817
X 2720 P X 2720 ,
condition (c) is satisfied.
53
P −1 X 160 Q P ;
53
13
P −1 X 160 Q P −1 X 32 ;
9
13
X 85 Q P −1 X 32 ,
50
C. Jia
In the regions:
13
21
13
P −1 X 32 Q P ;
X 64 P X 100 ,
21
13
P −1 X 32 Q P − 3 X 20 ;
13
817
P −1 X 32 Q P −1 X 45 ;
X 100 P X 60 ,
13
7
13
X 60 P X 2720 ,
817
2
242
19
9
19
X 85 Q P −1 X 45 ,
X 2720 P X 765 ,
condition (d) is satisfied.
In the regions:
13
12613
19
12613
242
19
897
P −1 X 45 Q P −1 X 1972 ;
X 49300 P X 765 ,
242
29
1
P −1 X 45 Q P − 6 X 120 ;
X 60 P X 49300 ,
3441
9
897
X 85 Q P −1 X 1972 ,
X 765 P X 9860 ,
conditions (e) and (f) are satisfied.
Putting together the above regions, we get Lemma 14.
Lemma 15. Under the assumption of Lemma 14, assume also that
9
X 85 R Q
and that P and Q lie in one of the following regions:
297
653
49
X 800 P X 1700 ,
(ii)
X 1700 P X 13120 ,
(iii)
X 13120 P X 13940 ,
653
5073
5073
19
P −1 X 100 Q X 160 ;
(i)
9
19
X 85 Q X 160 ;
5541
9
82
133
X 85 Q P − 69 X 230 .
Then (9) holds for T1 ≤ T ≤ X.
P r o o f. Let
m = pq
and
n = r.
An application of Lemma 4 yields that (9) holds under the conditions
21
Q P −1 X 40 ,
9
19
X 85 R Q X 160 .
In the regions:
297
653
X 800 P X 1700 ,
653
5073
X 1700 P X 13120 ,
5073
5541
X 13120 P X 13940 ,
49
19
P −1 X 100 Q X 160 ;
9
19
X 85 Q X 160 ;
9
82
133
X 85 Q P − 69 X 230 ,
the above conditions on P and Q are satisfied.
So the proof of Lemma 15 is complete.
Short intervals containing prime numbers
51
5. The remainder term in the sieve method
21
19
Lemma 16. Suppose that M X 40 , H X 160 and that a(m) = O(1),
b(h) = O(1). Then for real numbers x ∈ (X, 2X), except for a set the measure of which is O(X log−B X), we have
X
X
ηx
Σ=
a(m)b(h)
1−
= O(ηx log−B x).
mh
m∼M
h∼H
x<mhl≤x+ηx
1
P r o o f. If M H ≤ X 20 , the conclusion is obvious. In the following we
suppose
1
(10)
M H > X 20 .
Let b = 1 + 1/ log X and M HL
P = X. Suppose that M (s) and H(s) are
Dirichlet polynomials, L(s) = l∼L 1/ls and F (s) = M (s)H(s)L(s). Perron’s formula yields
X
x<mhl≤x+ηx
m∼M, h∼H
1
a(m)b(h) =
2πi
\
b+iX
F (s)
b−iX
(1 + η)s − 1 s
x ds + O(xε ).
s
If s = b + it and |t| ≤ c1 L, by Theorem 1 on page 442 of [18], we have
X
1
(c2 L)1−s − (c1 L)1−s
1
=
+O
.
s
l
1−s
L
c1 L<l≤c2 L
Moreover,
(1 + η)s − 1 s
x = ηxs + O(|s|η 2 x),
s
√
Let T1 = L. Then
1
2πi
\
b+iT1
b−iT1
η
=
2πi
F (s)
\
(1 + η)s − 1 s
x ηx.
s
(1 + η)s − 1 s
x ds
s
b+iT1
(c2 L)1−s − (c1 L)1−s s
x ds + O(S1 ) + O(S2 ),
1−s
M (s)H(s)
b−iT1
where
S1 =
ηx
L
\
T1
|M (b + it)H(b + it)| dt,
−T1
\
T1
S2 = η 2 x
−T1
|M (b + it)H(b + it)| dt.
52
C. Jia
A trivial estimate yields
ηx
S1 √ ηx1−ε
L
and
√
S2 η 2 x L ηx1−ε .
By Perron’s formula again,
η
2πi
\
b+iT1
(c2 L)1−s − (c1 L)1−s s
x ds
1−s
1+ε 1+ε X a(m)b(h)
ηx
ηx
= ηx
+O
+O
mh
T1
MH
M (s)H(s)
b−iT1
m∼M
h∼H
X a(m)b(h)
+ O(ηx1−ε ).
mh
= ηx
m∼M
h∼H
Now we have
(11)
X
Σ=
a(m)b(h) − ηx
m∼M
h∼H
x<mhl≤x+ηx
m∼M, h∼H
=
1
2πi
\
X a(m)b(h)
mh
b−iT1
F (s)%(s)xs ds +
b−iX
1
2πi
\
b+iX
F (s)%(s)xs ds + O(ηx1−ε ),
b+iT1
where
%(s) =
(1 + η)s − 1
.
s
If s = b + it and |Im(s)| ∼ T , then %(s) min(η, 1/T ). We proceed to
estimate
2X
2
\ b+iX
\
Ψ=
dx
F (s)%(s)xs ds .
X
b+iT1
We have
\
2X
2
Ψ log x max
T1 ≤T ≤X
Here
\
2X
X
b+2iT
2
\
dx
F (s)%(s)xs ds
b+iT
X
b+2iT
2
\
dx
F (s)%(s)xs ds .
b+iT
Short intervals containing prime numbers
\
2X
=
\
b+2iT
dx
X
b+2iT
ds1
b+iT
min2 η,
1
T
\
53
F (s1 )F (s2 )%(s1 )%(s2 )xs1 +s̄2 ds2
b+iT
b+2iT
\
x3 min2 η,
b+iT
1
T
\
2X
\
|F (s1 )F (s2 )| xs1 +s̄2 dx|ds2 |
b+2iT
|ds1 |
X
b+iT
b+2iT
\
\
b+2iT
|ds1 |
b+iT
b+iT
|F (s1 )|2 + |F (s2 )|2
|ds2 |
|1 + s1 + s̄2 |
2T
1 \
x log x min η,
|F (b + it)|2 dt.
T
2
3
T
By Lemma 4,
2T
1 \
|F (b + it)|2 dt η 2 x3−ε1 .
Ψ x log x max min η,
T1 ≤T ≤X
T
3
3
2
T
Hence, the measure of the set of x satisfying
b+iX
\
F (s)%(s)xs ds ≥ ηx log−B x
b+iT1
is O(X log−B X).
In the same way, we can deal with the integral
\
b−iT1
F (s)%(s)xs ds.
b−iX
So the proof of Lemma 16 is complete.
Lemma 17. Suppose that a(p), b(q), c(r) = O(1) and that
9
X 85 R Q.
Moreover , assume that P and Q lie in one of the following regions:
9
21
(i)
X 85 P X 100 ,
(ii)
X 100 P X 60 ,
(iii)
X 60 P X 49300 ,
(iv)
X 49300 P X 9860 ,
(v)
X 800 P X 1700 ,
(vi)
X 1700 P X 13120 ,
(vii)
X 13120 P X 13940 ,
21
13
12613
297
653
5073
13
12613
3441
653
5073
5541
9
X 85 Q P ;
9
2
7
9
1
29
X 85 Q P − 3 X 20 ;
X 85 Q P − 6 X 120 ;
9
897
X 85 Q P −1 X 1972 ;
49
19
P −1 X 100 Q X 160 ;
9
19
X 85 Q X 160 ;
9
82
133
X 85 Q P − 69 X 230 .
54
C. Jia
Let b = 1 + 1/ log X, P QRL = X and T1 =
T1 ≤ |t| ≤ 2X,
B
P (b + it)Q(b + it) log− ε x
and
√
L. Assume further that for
B
R(b + it) log− ε x.
Then for real numbers x ∈ (X, 2X), except for a set the measure of which
is O(X log−B X), we have
XXX
X
ηx
a(p)b(q)c(r)
1−
= O(ηx log−B x).
pqr
p∼P q∼Q r∼R
x<pqrl≤x+ηx
P r o o f. Using Lemmas 14, 15 and the discussion in Lemma 16, we get
the assertion.
6. Asymptotic formula
76
Lemma 18. Suppose that X 85 M X 1−δ and that 0 ≤ a(m) = O(1).
If m has a prime factor < X δ , then a(m) = 0. Then for real numbers
x ∈ (X, 2X), except for a set the measure of which is O(X log−B X), we
have
X
Σ=
a(m)
x<mp≤x+ηx
m∼M
1
=η 1+O
log x
X
a(m)
m∼M
X
1 + O(ηx log−B x).
x
2x
m <p≤ m
P r o o f. Let b = 1 + 1/ log X P
and M H = X. Suppose that M (s) is
s
a Dirichlet polynomial, H(s) =
h∼H Λ(h)/h and G(s) = M (s)H(s).
Perron’s formula yields
X
Σ1 =
x<mh≤x+ηx
m∼M
1
a(m)Λ(h) =
2πi
\
b+iX
b−iX
G(s)
(1 + η)s − 1 s
x ds + O(xε ).
s
B
Let T0 = log ε X. By (1) and the discussion in Lemma 16,
1
2πi
\
b+iT0
G(s)
b−iT0
η
=
2πi
(1 + η)s − 1 s
x ds
s
\
b+iT0
M (s)
b−iT0
where
S1 = ηx log−
2B
ε
\
(c2 H)1−s − (c1 H)1−s s
x ds + O(S1 ) + O(S2 ),
1−s
T0
x
−T0
\
T0
|M (b + it)| dt,
S2 = η 2 x
−T0
|M (b + it)| dt.
Short intervals containing prime numbers
55
A trivial estimate yields
S1 ηx log−2B x
and
S2 ηx log−2B x.
By Perron’s formula again,
η
2πi
\
b+iT0
M (s)
b−iT0
(c2 H)1−s − (c1 H)1−s s
x ds
1−s
X a(m)
ηx log2 x
ηx log2 x
= ηx
+O
+O
m
T0
M
m∼M
= ηx
X a(m)
+ O(ηx log−2B x).
m
m∼M
Hence,
Σ1 = ηx
X a(m)
+ O(ηx log−2B x)
m
m∼M
+
1
2πi
\
b−iT0
G(s)%(s)xs ds +
b−iX
1
2πi
\
b+iX
G(s)%(s)xs ds,
b+iT0
where
(1 + η)s − 1
.
s
By Lemma 1 and the discussion in Lemma 16, we have
%(s) =
1
2πi
and
1
2πi
\
b+iX
G(s)%(s)xs ds = O(ηx log−2B x)
b+iT0
\
b−iT0
G(s)%(s)xs ds = O(ηx log−2B x)
b−iX
for real numbers x ∈ (X, 2X), except for a set the measure of which is
O(X log−B X). So,
X a(m)
Σ1 = ηx
+ O(ηx log−2B x),
m
m∼M
X
1
a(m)
Σ = ηx 1 + O
+ O(ηx log−B x)
log x
m log(x/m)
m∼M
X
X
1
=η 1+O
a(m)
1 + O(ηx log−B x).
log x
x
2x
m∼M
The proof of Lemma 18 is complete.
m <p≤ m
56
C. Jia
9
76
Lemma 19. Suppose that X 85 HK X 85 , M HK = X and that
0 ≤ b(h) = O(1), 0 ≤ g(k) = O(1). If h has a prime factor < X δ , then
b(h) = 0, and similarly P
for g(k). Suppose that H(s) and K(s) are Dirichlet
s
polynomials, M (s) =
m∼M Λ(m)/m and G(s) = M (s)H(s)K(s). Let
B
b = 1 + 1/ log X and T0 = log ε X.
If (7) holds for T0 ≤ T ≤ X, then
X
b(h)g(k)
x<hkp≤x+ηx
h∼H, k∼K
1
=η 1+O
log x
X
X
b(h)g(k)
h∼H, k∼K
1 + O(ηx log−B x).
2x
x
hk <p≤ hk
7. Buchstab’s function. We define w(u) as the continuous solution of
the equations
w(u) = 1/u,
1 ≤ u ≤ 2,
(12)
0
(uw(u)) = w(u − 1), u > 2.
w(u) is called Buchstab’s function and plays an important role in finding
asymptotic formulas in the sieve method. In particular,
w(u) =
w(u) =
1 + log(u − 1)
,
u
1 + log(u − 1) 1
+
u
u
\
u−1
2
1 + log(u − 1) 1
+
w(u) =
u
u
+
1
u
\
u−1
3
dt
t
\
t−1
2
2 ≤ u ≤ 3,
log(t − 1)
dt,
t
\
u−1
2
3 ≤ u ≤ 4,
log(t − 1)
dt
t
log(s − 1)
ds,
s
4 ≤ u ≤ 5.
Lemma 20. We have the following bounds:
(i) w(u) ≥ 0.5607 for u ≥ 2.47,
(ii) w(u) ≤ 0.5644 for u ≥ 3,
(iii) 0.5612 ≤ w(u) ≤ 0.5617 for u ≥ 4.
P r o o f. It is easy to see that 0.5 ≤ w(u) ≤ 1 for 1 ≤ u ≤ 2. Then we
employ induction.
Suppose that 0.5 ≤ w(u) ≤ 1 for 1 ≤ k ≤ u ≤ k + 1. If k + 1 ≤ u ≤ k + 2,
Short intervals containing prime numbers
57
then (12) yields
\
u−1
(13)
uw(u) = (k + 1)w(k + 1) +
w(t) dt.
k
Hence, 0.5 ≤ w(u) ≤ 1 for k + 1 ≤ u ≤ k + 2. By induction, we obtain
0.5 ≤ w(u) ≤ 1 for u ≥ 1.
If u > 2, (12) yields
w(u − 1) − w(u)
.
u
If 2 ≤ u ≤ 3, by calculation, we have
w0 (u) =
(14)
max w(2 + 10−4 k) ≤ 0.56716.
0≤k≤104
From (14) and 0.5 ≤ w(u) ≤ 1 for u ≥ 1, it follows that |w0 (u)| ≤ 14 if
u > 2. Using Lagrange’s mean value theorem, we have w(u) ≤ 0.5672 for
2 ≤ u ≤ 3. By induction, we obtain 0.5 ≤ w(u) ≤ 0.5672 for u ≥ 2.
If 3 ≤ u ≤ 4, by calculation, we have
max w(3 + 10−4 k) ≤ 0.56439,
0≤k≤104
min w(3 + 10−4 k) ≥ 0.56081.
0≤k≤104
From (14) and 0.5 ≤ w(u) ≤ 0.5672 for u ≥ 2, it follows that |w0 (u)| ≤
0.0224 if u > 3. The above discussion implies that 0.5607 ≤ w(u) ≤ 0.5644
for u ≥ 3. By the same discussion we can also get 0.5607 ≤ w(u) for u ≥ 2.47.
If 4 ≤ u ≤ 5, by calculation, we have
max w(4 + 10−4 k) ≤ 0.5616,
0≤k≤104
min w(4 + 10−4 k) ≥ 0.5613.
0≤k≤104
The above discussion and the fact that |w0 (u)| ≤ 0.0224 for u > 3 imply
that 0.5612 ≤ w(u) ≤ 0.5617 for u ≥ 4.
Gathering together the above discussion, we get Lemma 20.
Lemma 21. Suppose that E = {n : t < n ≤ 2t} and z ≤ t. Let
Y
P (z) =
p.
p<z
Then for sufficiently large t and z, we have
X
log t
t
S(E, z) =
1= w
+ O(ε)
.
log z
log z
t<n≤2t
(n,P (z))=1
1
P r o o f. See Lemma 5 of [10]. If (2t) 2 < z ≤ t, it is the prime number
theorem.
58
C. Jia
8. Sieve method. We proceed to show that
π(x + ηx) − π(x) ≥ 0.011ηx log−1 x
(15)
for real numbers x ∈ (X, 2X), except for a set the measure of which is
O(X log−B X).
Let
(16)
A = {n : x < n ≤ x + ηx},
Y
X
P (z) =
p, S(A, z) =
p<z
1.
a∈A
(a,P (z))=1
Then
1
(17)
π(x + ηx) − π(x) = S(A, (2X) 2 ).
Buchstab’s identity yields
(18)
1
X
9
S(A, (2X) 2 ) = S(A, X 85 ) −
X
= S(A, X
9
85
9
85
S(Ap , p)
1
<p≤(2X) 2
X
)−
9
S(Ap , X 85 )
1
9
X 85 <p≤(2X) 2
X
+
X
9
85
X
1
<p≤(2X) 2
X
9
85
1
<q<min(p,(2X/p) 2
S(Apq , q).
)
The following lemmas always concern real numbers x ∈ (X, 2X), except
for a set the measure of which is O(X log−B X).
Let
(19)
B = {n : x < n ≤ 2x}.
Lemma 22.
9
S(A, X 85 ) ≥ 5.300221ηx log−1 x.
P r o o f. We have
(20)
9
S(A, X 85 ) = S(A, X δ ) −
X
X δ <p≤X
Let
re(A, d) = |Ad | −
ηx
,
d
where γ is Euler’s constant.
W (z) =
S(Ap , p).
9
85
Y
e−γ
1
= (1 + O(ε))
,
1−
p
log
z
p<z
Short intervals containing prime numbers
59
21
Let z = X δ and D = X 40 . Applying Iwaniec’s sieve method (see Theorem 1 of [8]), we have
log D
ηx
δ
f
− O(εηx log−1 x) − R− ,
S(A, X ) ≥
log z
log z
where
X
R− =
m≤X
Lemma 16 yields R
we have
−
a(m)e
r(A, m).
21
40
= O(ηx log−5 x). By Theorem 8 on page 181 of [17],
log D
f
= e−γ + O(ε2 ),
log z
where γ is Euler’s constant. Thus,
S(A, X δ ) ≥
e−γ
ηx log−1 x + O(εηx log−1 x).
δ
In the same way,
ηx
log D
S(A, X ) ≤
F
+ O(εηx log−1 x) +
log z
log z
X
δ
m≤X
e
b(m)e
r(A, m)
21
40
−γ
ηx log−1 x + O(εηx log−1 x).
δ
So, we have the asymptotic formula
≤
S(A, X δ ) =
(21)
Now,
e−γ
ηx log−1 x + O(εηx log−1 x).
δ
X
S(Ap , p) =
9
X δ <p≤X 85
X
1,
x<pq≤x+ηx
9
where X δ < p ≤ X 85 and the least prime factor of q is greater than p.
Using Lemmas 18 and 21 and the prime number theorem, we have
X
(22)
S(Ap , p)
9
X δ <p≤X 85
X
=η
S(Bp , p) + O(εηx log−1 x)
9
X δ <p≤X 85
= ηx
X
9
X δ <p≤X 85
1
log(x/p)
w
+ O(δηx log−1 x)
p log p
log p
60
C. Jia
\
9
85
= ηx log
−1
x
δ
\
1
1−t
w
dt + O(δηx log−1 x)
t2
t
1
δ
= ηx log
−1
x
w(u − 1) du + O(δηx log−1 x)
85
9
1
85
85
1
− w
+ O(δηx log−1 x)
= ηx log x w
δ
δ
9
9
e−γ
85
85
=
ηx log−1 x − w
ηx log−1 x + O(δηx log−1 x),
δ
9
9
−1
since w(1/δ) = e−γ + O(ε2 ) (see Lemma 12 on page 179 of [17]). Hence,
9
85
85
(23)
S(A, X 85 ) =
w
ηx log−1 x + O(δηx log−1 x).
9
9
By Lemma 20, we get
9
85
· 0.5612ηx log−1 x + O(δηx log−1 x) ≥ 5.300221ηx log−1 x.
S(A, X 85 ) ≥
9
So the proof of Lemma 22 is complete.
Lemma 23.
X
9
S(Ap , X 85 ) ≤ 8.234757ηx log−1 x.
9
1
X 85 <p≤(2X) 2
P r o o f. Buchstab’s identity yields
X
9
S(Ap , X 85 )
9
1
X 85 <p≤(2X) 2
=
X
9
85
X
X
S(Ap , X δ ) −
1
<p≤(2X) 2
X
9
85
1
<p≤(2X) 2
X
X δ <q<X
S(Apq , q).
9
85
Using Lemma 16, in the same way as in Lemma 22, we have
X
S(Ap , X δ )
9
1
X 85 <p≤(2X) 2
X
=
9
1
e−γ 1
· · ηx log−1 x + O(εηx log−1 x)
δ
p
X 85 <p≤(2X) 2
e−γ
=
· ηx log−1 x
δ
\
1
2
9
85
dt
+ O(εηx log−1 x).
t
Short intervals containing prime numbers
61
Using Lemmas 18 and 21, in the same way as in Lemma 22, we have
X
X
9
85
X
1
<p≤(2X) 2
S(Apq , q)
9
X δ <q<X 85
X
=η
X
9
1
S(Bpq , q) + O(εηx log−1 x)
9
X 85 <p≤(2X) 2 X δ <q<X 85
X
= ηx
X
9
1
9
log(x/(pq))
1
w
pq log q
log q
X 85 <p≤(2X) 2 X δ <q<X 85
+ O(δηx log−1 x)
\
1
2
= ηx log
−1
x
dt
t
9
85
\
1
2
= ηx log
−1
x
9
85
\
1
1−t−u
du + O(δηx log−1 x)
w
u2
u
9
85
δ
dt
t(1 − t)
1
= · ηx log−1 x
δ
\
1
2
9
85
w(r − 1) dr + O(δηx log−1 x)
85
9 (1−t)
1
1
w
(1 − t) dt
t
δ
85
−
· ηx log−1 x
9
e−γ
· ηx log−1 x
=
δ
\
1
δ (1−t)
\
1
2
9
85
\
1
2
9
85
1
85
w
(1 − t) dt + O(δηx log−1 x)
t
9
dt 85
−
· ηx log−1 x
t
9
\
1
2
9
85
1
85
w
(1 − t) dt
t
9
+ O(δηx log−1 x).
Gathering together the above discussion and applying Lemma 20, we
have
X
9
9
S(Ap , X 85 )
1
X 85 <p≤(2X) 2
85
=
· ηx log−1 x
9
\
1
2
9
85
1
85
w
(1 − t) dt + O(δηx log−1 x)
t
9
62
C. Jia
≤ 0.5617 ·
85
· log
9
85
ηx log−1 x + O(δηx log−1 x)
18
≤ 8.234757ηx log−1 x.
So the proof of Lemma 23 is complete.
We now set
X
X
(24)
Ω=
X
≥
9
85
1
<p≤(2X) 2
94
X
X
X
9
85
S(Apq , q)
<q<min(p,( 2X
p
S(Apq , q) =
i=1 (p,q)∈Di
94
X
1
)2
)
Ωi ,
i=1
where
9
21
9
D1 = {(p, q) : X 85 < p ≤ X 100 , X 85 < q < p},
21
13
9
2
7
133
9
1
29
133
19
9
1
29
133
19
D2 = {(p, q) : X 100 < p ≤ X 60 , X 85 < q < p− 3 X 20 },
13
D3 = {(p, q) : X 60 < p ≤ X 580 , X 85 < q < p− 6 X 120 },
D4 = {(p, q) : X 580 < p ≤ X 80 , X 85 < q < p− 6 X 120 },
133
D5 = {(p, q) : X 580 < p ≤ X 80 , p−1 X 290 < q < p},
19
53
19
53
1
9
29
D6 = {(p, q) : X 80 < p ≤ X 220 , X 85 < q < p− 6 X 120 },
133
19
D7 = {(p, q) : X 80 < p ≤ X 220 , p−1 X 290 < q < p−1 X 40 },
53
12613
53
12613
9
1
29
D8 = {(p, q) : X 220 < p ≤ X 49300 , X 85 < q < p− 6 X 120 },
133
53
D9 = {(p, q) : X 220 < p ≤ X 49300 , p−1 X 290 < q < p−1 X 110 },
12613
151
12613
151
9
897
D10 = {(p, q) : X 49300 < p ≤ X 580 , X 85 < q < p−1 X 1972 },
133
53
D11 = {(p, q) : X 49300 < p ≤ X 580 , p−1 X 290 < q < p−1 X 110 },
151
19
151
19
9
897
D12 = {(p, q) : X 580 < p ≤ X 70 , X 85 < q < p−1 X 1972 },
897
53
D13 = {(p, q) : X 580 < p ≤ X 70 , p−1 X 1972 < q < p−1 X 110 },
19
3
9
19
3
19
3
6
3
17
9
897
D14 = {(p, q) : X 70 < p ≤ X 10 , X 85 < q < p−1 X 1972 },
897
53
D15 = {(p, q) : X 70 < p ≤ X 10 , p−1 X 1972 < q < p−1 X 110 },
57
1
23
D16 = {(p, q) : X 70 < p ≤ X 10 , p− 5 X 100 < q < p− 8 X 80 },
897
D17 = {(p, q) : X 10 < p ≤ X 55 , X 85 < q < p−1 X 1972 },
Short intervals containing prime numbers
3
17
3
17
12
17
19
9
17
19
17
19
19
247
19
247
19
247
63
897
53
D18 = {(p, q) : X 10 < p ≤ X 55 , p−1 X 1972 < q < p−1 X 110 },
57
23
1
D19 = {(p, q) : X 10 < p ≤ X 55 , p− 11 X 110 < q < p− 8 X 80 },
897
D20 = {(p, q) : X 55 < p ≤ X 60 , X 85 < q < p−1 X 1972 },
897
58
59
D21 = {(p, q) : X 55 < p ≤ X 60 , p−1 X 1972 < q < p− 67 X 134 },
12
57
1
23
D22 = {(p, q) : X 55 < p ≤ X 60 , p− 11 X 110 < q < p− 8 X 80 },
9
897
D23 = {(p, q) : X 60 < p ≤ X 770 , X 85 < q < p−1 X 1972 },
897
58
59
D24 = {(p, q) : X 60 < p ≤ X 770 , p−1 X 1972 < q < p− 67 X 134 },
19
1
23
D25 = {(p, q) : X 60 < p ≤ X 770 , X 110 < q < p− 8 X 80 },
247
359
247
359
247
359
9
897
D26 = {(p, q) : X 770 < p ≤ X 1100 , X 85 < q < p−1 X 1972 },
58
897
59
D27 = {(p, q) : X 770 < p ≤ X 1100 , p−1 X 1972 < q < p− 67 X 134 },
19
25
D28 = {(p, q) : X 770 < p ≤ X 1100 , X 110 < q < p−1 X 44 },
359
481
9
359
481
359
481
19
481
443
9
481
443
481
443
897
D29 = {(p, q) : X 1100 < p ≤ X 1450 , X 85 < q < p−1 X 1972 },
897
58
59
D30 = {(p, q) : X 1100 < p ≤ X 1450 , p−1 X 1972 < q < p− 67 X 134 },
D31 = {(p, q) : X 1100 < p ≤ X 1450 , X 110 < q < p−6 X
11
5
},
897
D32 = {(p, q) : X 1450 < p ≤ X 1305 , X 85 < q < p−1 X 1972 },
58
897
59
D33 = {(p, q) : X 1450 < p ≤ X 1305 , p−1 X 1972 < q < p− 67 X 134 },
19
157
D34 = {(p, q) : X 1450 < p ≤ X 1305 , X 110 < q < p−1 X 290 },
443
19
443
19
443
19
9
897
D35 = {(p, q) : X 1305 < p ≤ X 55 , X 85 < q < p−1 X 1972 },
897
58
59
D36 = {(p, q) : X 1305 < p ≤ X 55 , p−1 X 1972 < q < p− 67 X 134 },
19
2
29
D37 = {(p, q) : X 1305 < p ≤ X 55 , X 110 < q < p− 11 X 110 },
19
3441
19
3441
19
3441
9
897
D38 = {(p, q) : X 55 < p ≤ X 9860 , X 85 < q < p−1 X 1972 },
897
58
59
D39 = {(p, q) : X 55 < p ≤ X 9860 , p−1 X 1972 < q < p− 67 X 134 },
58
57
2
29
D40 = {(p, q) : X 55 < p ≤ X 9860 , p− 49 X 98 < q < p− 11 X 110 },
3441
1843
9
3441
1843
58
58
59
D41 = {(p, q) : X 9860 < p ≤ X 5280 , X 85 < q < p− 67 X 134 },
57
2
29
D42 = {(p, q) : X 9860 < p ≤ X 5280 , p− 49 X 98 < q < p− 11 X 110 },
64
C. Jia
1843
37
9
58
1843
37
70
171
1843
37
58
57
59
D43 = {(p, q) : X 5280 < p ≤ X 100 , X 85 < q < p− 67 X 134 },
57
58
D44 = {(p, q) : X 5280 < p ≤ X 100 , p− 59 X 295 < q < p− 49 X 98 },
2
29
D45 = {(p, q) : X 5280 < p ≤ X 100 , p− 49 X 98 < q < p− 11 X 110 },
37
297
9
49
37
297
82
133
37
297
58
57
D46 = {(p, q) : X 100 < p ≤ X 800 , X 85 < q < p−1 X 100 },
58
57
D47 = {(p, q) : X 100 < p ≤ X 800 , p− 69 X 230 < q < p− 49 X 98 },
2
29
D48 = {(p, q) : X 100 < p ≤ X 800 , p− 49 X 98 < q < p− 11 X 110 },
297
2143
297
2143
297
2143
82
133
297
2143
58
57
9
49
D49 = {(p, q) : X 800 < p ≤ X 5620 , X 85 < q < p−1 X 100 },
49
19
D50 = {(p, q) : X 800 < p ≤ X 5620 , p−1 X 100 < q < X 160 },
58
57
D51 = {(p, q) : X 800 < p ≤ X 5620 , p− 69 X 230 < q < p− 49 X 98 },
2
29
D52 = {(p, q) : X 800 < p ≤ X 5620 , p− 49 X 98 < q < p− 11 X 110 },
2143
653
9
49
2143
653
2143
653
82
133
2143
653
58
57
D53 = {(p, q) : X 5620 < p ≤ X 1700 , X 85 < q < p−1 X 100 },
49
19
D54 = {(p, q) : X 5620 < p ≤ X 1700 , p−1 X 100 < q < X 160 },
58
57
D55 = {(p, q) : X 5620 < p ≤ X 1700 , p− 69 X 230 < q < p− 49 X 98 },
29
59
D56 = {(p, q) : X 5620 < p ≤ X 1700 , p− 49 X 98 < q < p− 19 X 76 },
653
5073
9
19
653
5073
82
133
653
5073
58
57
D57 = {(p, q) : X 1700 < p ≤ X 13120 , X 85 < q < X 160 },
58
57
D58 = {(p, q) : X 1700 < p ≤ X 13120 , p− 69 X 230 < q < p− 49 X 98 },
29
59
D59 = {(p, q) : X 1700 < p ≤ X 13120 , p− 49 X 98 < q < p− 19 X 76 },
5073
5541
9
5073
5541
82
133
5073
5541
58
57
82
133
D60 = {(p, q) : X 13120 < p ≤ X 13940 , X 85 < q < p− 69 X 230 },
58
57
D61 = {(p, q) : X 13120 < p ≤ X 13940 , p− 69 X 230 < q < p− 49 X 98 },
29
59
D62 = {(p, q) : X 13120 < p ≤ X 13940 , p− 49 X 98 < q < p− 19 X 76 },
5541
3963
9
5541
3963
58
58
57
D63 = {(p, q) : X 13940 < p ≤ X 9860 , X 85 < q < p− 49 X 98 },
57
29
59
D64 = {(p, q) : X 13940 < p ≤ X 9860 , p− 49 X 98 < q < p− 19 X 76 },
3963
1063
9
1063
19
9
1063
19
29
29
59
D65 = {(p, q) : X 9860 < p ≤ X 2640 , X 85 < q < p− 19 X 76 },
29
59
D66 = {(p, q) : X 2640 < p ≤ X 45 , X 85 < q < p− 19 X 76 },
59
35
179
D67 = {(p, q) : X 2640 < p ≤ X 45 , p− 19 X 76 < q < p− 23 X 230 },
Short intervals containing prime numbers
65
19
19
9
59
19
19
29
59
41
19
19
29
57
1
19
4331
9
19
4331
29
59
41
19
4331
29
57
1
19
4331
6
29
D68 = {(p, q) : X 45 < p ≤ X 44 , X 85 < q < p− 19 X 76 },
421
D69 = {(p, q) : X 45 < p ≤ X 44 , p− 19 X 76 < q < p− 27 X 540 },
29
D70 = {(p, q) : X 45 < p ≤ X 44 , p− 10 X 40 < q < p− 5 X 100 },
29
59
D71 = {(p, q) : X 44 < p ≤ X 9860 , X 85 < q < p− 19 X 76 },
421
D72 = {(p, q) : X 44 < p ≤ X 9860 , p− 19 X 76 < q < p− 27 X 540 },
29
D73 = {(p, q) : X 44 < p ≤ X 9860 , p− 10 X 40 < q < p− 5 X 100 },
19
1
13
D74 = {(p, q) : X 44 < p ≤ X 9860 , p− 5 X 25 < q < p 7 X 70 },
4331
2691
35
57
29
4331
2691
29
57
1
4331
2691
6
2691
897
9
2691
897
29
2691
897
6
57
D75 = {(p, q) : X 9860 < p ≤ X 5950 , p− 12 X 40 < q < p− 10 X 40 },
29
D76 = {(p, q) : X 9860 < p ≤ X 5950 , p− 10 X 40 < q < p− 5 X 100 },
19
1
13
D77 = {(p, q) : X 9860 < p ≤ X 5950 , p− 5 X 25 < q < p 7 X 70 },
29
57
D78 = {(p, q) : X 5950 < p ≤ X 1972 , X 85 < q < p− 10 X 40 },
1
57
29
D79 = {(p, q) : X 5950 < p ≤ X 1972 , p− 10 X 40 < q < p− 5 X 100 },
19
1
13
D80 = {(p, q) : X 5950 < p ≤ X 1972 , p− 5 X 25 < q < p 7 X 70 },
897
133
9
897
133
6
1
29
D81 = {(p, q) : X 1972 < p ≤ X 290 , X 85 < q < p− 5 X 100 },
19
1
13
D82 = {(p, q) : X 1972 < p ≤ X 290 , p− 5 X 25 < q < p 7 X 70 },
133
19
9
1
13
D83 = {(p, q) : X 290 < p ≤ X 40 , X 85 < q < p 7 X 70 },
19
53
9
51
D84 = {(p, q) : X 40 < p ≤ X 110 , X 85 < q < p−1 X 70 },
53
339
9
53
339
19
339
233
9
339
233
19
233
49
9
233
49
58
233
480
49
100
19
110
58
59
D85 = {(p, q) : X 110 < p ≤ X 700 , X 85 < q < p− 9 X 18 },
51
D86 = {(p, q) : X 110 < p ≤ X 700 , X 110 < q < p−1 X 70 },
58
59
D87 = {(p, q) : X 700 < p ≤ X 480 , X 85 < q < p− 9 X 18 },
63
D88 = {(p, q) : X 700 < p ≤ X 480 , X 110 < q < p−6 X 20 },
58
59
D89 = {(p, q) : X 480 < p ≤ X 100 , X 85 < q < p− 9 X 18 },
59
70
D90 = {(p, q) : X 480 < p ≤ X 100 , p− 9 X 18 < q < p− 11 X
D91 = {(p, q) : X
49
<p≤X
, X
2944
9
< q < p−6 X
70
63
20
D92 = {(p, q) : X 100 < p ≤ X 5950 , X 85 < q < p− 11 X
179
55
},
179
55
},
},
66
C. Jia
49
2944
19
7
D93 = {(p, q) : X 100 < p ≤ X 5950 , X 110 < q < p−1 X 10 },
2944
1
19
7
D94 = {(p, q) : X 5950 < p ≤ X 2 , X 110 < q < p−1 X 10 }.
9. Bounds of the sieve functions
Lemma 24.
Ω16 + Ω19 + Ω22 + Ω25 + Ω28 + Ω31 + Ω34 + Ω37 + Ω40 + Ω42 + Ω45
+ Ω48 + Ω52 + Ω56 + Ω59 + Ω62 + Ω64 + Ω65 + Ω66 + Ω68 + Ω71
≥ 0.394443ηx log−1 x.
P r o o f. We have
X
Ω16 =
X
19
70
X
<p≤X
19
3
10
6
p− 5
X
57
100
6
3
S(Apq , q) =
1
<q<p− 8
X
1,
x<pqr≤x+ηx
23
80
1
57
X
23
where X 70 < p ≤ X 10 , p− 5 X 100 < q < p− 8 X 80 and the least prime factor
of r is greater than q.
Let h = q and k = r. By Lemma 6 with region (i), (7) holds. Then
Lemmas 19 and 21 yield
X
X
Ω16 = η
S(Bpq , q) + O(εηx log−1 x)
19
3
57
6
23
1
X 70 <p≤X 10 p− 5 X 100 <q<p− 8 X 80
X
= ηx
X
19
3
6
57
1
23
X 70 <p≤X 10 p− 5 X 100 <q<p− 8 X 80
1
log(x/(pq))
w
pq log q
log q
+ O(εηx log−1 x)
\
3
10
= ηx log
−1
x
19
70
= ηx log−1 x
\
3
10
+
71
260
3
10
+
\
71
260
dt
t
71
260
\
19
70
1
3 (1−t)
\
\
23
t
80 − 8
57
6
100 − 5 t
dt
t
1
1−t−u
w
du + O(εηx log−1 x)
u2
u
\
23
t
80 − 8
57
6
100 − 5 t
du
u(1 − t − u)
1
1−t
1 + log
−2
du
u(1 − t − u)
u
57
6
100 − 5 t
!
23
−t
dt 80 \ 8
du
+ O(ε)
t 1
u(1 − t − u)
dt
t
3 (1−t)
≥ (0.000516 + 0.010796 + 0.011351)ηx log−1 x
= 0.022663ηx log−1 x.
Short intervals containing prime numbers
67
Using the above discussion and Lemma 20, we have
Ω19 + Ω22 + Ω25 + Ω28 + Ω31 + Ω34 + Ω37 + Ω40 + Ω42 + Ω45
+ Ω48 + Ω52 + Ω56 + Ω59 + Ω62 + Ω64 + Ω65 + Ω66 + Ω68 + Ω71
= ηx log
−1
x
19
60
\
dt
t
3
10
\
247
770
+
19
60
\
481
1450
+
359
1100
\
443
1305
+
481
1450
\
19
55
+
443
1305
\
2143
5620
+
19
55
\
3963
9860
+
2143
5620
\
4331
9860
+
\
23
t
80 − 8
dt
t
3963
9860
≥ ηx log
19
110
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
−1
x
\
19
110
157
290 −t
19
110
19
110
\
29
2
110 − 11 t
58
57
98 − 49 t
\
59
29
76 − 19 t
57
58
98 − 49 t
59
29
76 − 19 t
19
60
\
3
10
\
19
60
+
3
10
dt
t
\
23
t
80 − 8
1
3 (1−t)
247
770
dt
t
\
25
44 −t
19
110
1−t−u
1
w
du
u2
u
1
1−t−u
w
du
u2
u
\
9
85
\
359
1100
1
1−t−u
w
du
u2
u
29
2
110 − 11 t
\
57
12
110 − 11 t
1
1−t−u
w
du
u2
u
1−t−u
1
w
du +
u2
u
11
5 −6t
\
\
23
t
80 − 8
1−t−u
1
w
du
u2
u
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
+
O(ε)
u2
u
dt
t
\
1
3 (1−t)
57
12
110 − 11 t
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
u(1 − t − u)
68
C. Jia
\
247
770
+
19
60
\
247
770
+
19
60
\
359
1100
+
247
770
\
359
1100
+
247
770
\
28
85
+
359
1100
\
28
85
+
359
1100
\
481
1450
+
28
85
\
443
1305
+
481
1450
\
19
55
+
443
1305
\
65
183
+
19
55
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
\
1
3 (1−t)
19
110
\
23
t
80 − 8
1
3 (1−t)
du
u(1 − t − u)
\
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
3 (1−t)
19
110
\
25
44 −t
du
u(1 − t − u)
1
3 (1−t)
\
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
3 (1−t)
19
110
\
11
5 −6t
du
u(1 − t − u)
1
3 (1−t)
\
11
5 −6t
19
110
\
157
290 −t
19
110
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
29
2
110 − 11 t
19
110
\
29
2
110 − 11 t
57
58
98 − 49 t
\
2143
5620
+ 0.5607
1
1−t
1 + log
−2
du
u(1 − t − u)
u
65
183
dt
t
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
1
4 (1−t)
57
58
98 − 49 t
du
u2
Short intervals containing prime numbers
\
2143
5620
+
65
183
dt
t
\
2
29
110 − 11 t
1
4 (1−t)
\
3963
9860
+ 0.5607
2143
5620
\
3963
9860
+
2143
5620
dt
t
\
59
29
76 − 19 t
1
4 (1−t)
\
40
97
+ 0.5607
3963
9860
\
40
97
+
3963
9860
dt
t
dt
t
\
29
59
76 − 19 t
1
4 (1−t)
\
4331
9860
+ 0.5607
dt
t
40
97
dt
t
69
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
1
4 (1−t)
58
57
98 − 49 t
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
1
4 (1−t)
9
85
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
59
29
76 − 19 t
9
85
du
+ O(ε)
u2
≥ (0.034609 + 0.009183 + 0.009870 + 0.002518
+ 0.012892 + 0.003183 + 0.006792 + 0.000789
+ 0.004547 + 0.012016 + 0.008083 + 0.015437
+ 0.021797 + 0.049656 + 0.056577 + 0.028631
+ 0.038798 + 0.004151 + 0.052251)ηx log−1 x
= 0.371780ηx log−1 x.
The proof of Lemma 24 is complete.
Lemma 25.
Φ = Ω70 + Ω73 + Ω74 + Ω76 + Ω77 + Ω79 + Ω80 + Ω81 + Ω82 + Ω83
+ Ω84 + Ω85 + Ω86 + Ω87 + Ω88 + Ω89 + Ω91 + Ω93 + Ω94
≥ 0.321616ηx log−1 x.
P r o o f. On applying Lemmas 8, 19, 20 and 21, in the same way as in
Lemma 24, we have
70
C. Jia
Φ = ηx log
−1
x
19
44
\
dt
t
19
45
\
897
1972
+
19
44
\
133
290
+
897
1972
\
19
40
+
133
290
\
339
700
+
53
110
\
49
100
+
339
700
\
1
2
+
49
100
≥ ηx log
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
−1
\
29
t
100 − 5
29
57
40 − 10 t
\
9
85
9
85
\
59
58
18 − 9 t
9
85
\
59
58
18 − 9 t
9
85
\
19
110
x
\
+
131
308
\
19
44
+
131
308
\
47
106
+
19
44
\
47
106
+
19
44
dt
t
dt
t
dt
t
dt
t
25 − 5 t
44
\
133
290
13
t
70 + 7
1972
19
6
25 − 5 t
\
53
110
19
40
dt
t
\
1
1−t−u
w
du +
u2
u
339
700
1
1−t−u
w
du +
u2
u
49
100
53
110
\
339
700
\
51
70 −t
dt
t
dt
t
9
85
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
\
51
70 −t
19
110
1
1−t−u
w
du
u2
u
\
63
20 −6t
19
110
1
1−t−u
w
du
u2
u
1
1−t−u
w
du + O(ε)
u2
u
131
308
\
19
45
19
44
897
+ 7t
1972
70 \
\ dt 13
1
1
1−t−u
1−t−u
w
du+
w
du
u2
u
t 19 6 u2
u
19
1
1−t−u
w
du +
u2
u
13
t
70 + 7
7
10 −t
57
29
40 − 10 t
1
1−t−u
w
du
u2
u
\ dt
1
1−t−u
w
du+
u2
u
t
897
29
t
100 − 5
\
\
t
29
100 − 5
\
1
3 (1−t)
57
29
40 − 10 t
\
29
t
100 − 5
1
3 (1−t)
\
1
3 (1−t)
57
29
40 − 10 t
\
29
t
100 − 5
1
3 (1−t)
dt
t
\
29
t
100 − 5
57
29
40 − 10 t
du
u(1 − t − u)
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
u(1 − t − u)
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1972
\ dt
du
+ 0.5607
u(1 − t − u)
t
47
897
106
\
1
4 (1−t)
57
29
40 − 10 t
du
u2
Short intervals containing prime numbers
\
897
1972
+
47
106
\
897
1972
+
47
106
dt
t
dt
t
\
1
3 (1−t)
1
4 (1−t)
\
29
t
100 − 5
1
3 (1−t)
\
133
290
+ 0.5607
dt
t
897
1972
\
133
290
+
897
1972
\
133
290
+
897
1972
dt
t
dt
t
\
1
3 (1−t)
1
4 (1−t)
\
29
t
100 − 5
1
3 (1−t)
\
19
40
+ 0.5607
133
290
\
19
40
+
133
290
\
19
40
+
133
290
\
53
110
+
19
40
\
53
110
+
19
40
\
339
700
+
53
110
\
339
700
+
53
110
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
\
1
3 (1−t)
1
4 (1−t)
\
13
t
70 + 7
1
3 (1−t)
\
1
3 (1−t)
1
4 (1−t)
\
51
70 −t
1
3 (1−t)
\
59
58
18 − 9 t
1
4 (1−t)
\
51
70 −t
19
110
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
+
u(1 − t − u)
\
1
4 (1−t)
9
85
\
897
1972
19
44
\
13
t
70 + 7
dt
t
19
6
25 − 5 t
du
u(1 − t − u)
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
+
u(1 − t − u)
\
1
4 (1−t)
9
85
\
133
290
897
1972
\
13
t
70 + 7
dt
t
19
6
25 − 5 t
du
u(1 − t − u)
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
110
\ dt
du
+ 0.5607
u(1 − t − u)
t
19
53
\
1
4 (1−t)
9
85
40
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
700
\ dt
du
+ 0.5607
u(1 − t − u)
t
53
339
\
1
4 (1−t)
9
85
110
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
223
\ dt
du
+ 0.5607
u(1 − t − u)
t
339
109
700
\
1
4 (1−t)
9
85
du
u2
71
72
C. Jia
\
109
223
+
339
700
dt
t
\
58
59
18 − 9 t
1
4 (1−t)
\
49
100
+ 0.5607
109
223
\
1
2
+
49
100
dt
t
\
dt
t
7
10 −t
19
110
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
59
58
18 − 9 t
9
85
\
49
100
du
+
u2
dt
t
339
700
du
+ O(ε)
u(1 − t − u)
\
63
20 −6t
19
110
du
u(1 − t − u)
≥ (0.000892 + 0.001922 + 0.002834 + 0.015864
+ 0.005547 + 0.013560 + 0.025626 + 0.006139
+ 0.014504 + 0.009587 + 0.008357 + 0.002151
+ 0.004454 + 0.038219 + 0.036373 + 0.038613
+ 0.014188 + 0.015100 + 0.016708 + 0.004887
+ 0.004744 + 0.005821 + 0.008584 + 0.003871
+ 0.001900 + 0.010238 + 0.010933)ηx log−1 x
= 0.321616ηx log−1 x.
The proof of Lemma 25 is complete.
Lemma 26.
Ω5 + Ω7 + Ω9 + Ω11 + Ω13 + Ω15 + Ω18 + Ω21 + Ω24 + Ω27 + Ω30
+ Ω33 + Ω36 + Ω39 + Ω41 + Ω43 + Ω46 + Ω49 + Ω53
≥ 0.277632ηx log−1 x.
P r o o f. We have
X
(25) Ω5 =
X
133
580
=
<p≤X
X
133
X
19
80
p−1 X
133
290
S(Apq , q)
<q<p
X
19
S(Apq , X δ )
133
X 580 <p≤X 80 p−1 X 290 <q<p
X
−
X
133
580
X
133
580
−
X
<p≤X
19
80
<p≤X
19
80
X
p−1 X
133
290
p−1 X
133
290
X
<q<p X δ <r<X
X
<q<p X
9
85
S(Apqr , r)
9
85
X
<r<min(q,( 2X
pq
S(Apqr , r)
1
)2
)
= Φ1 − Φ2 − Φ3 .
21
Let z = X δ and D = D(p, q) = X 40 /(pq). Applying Iwaniec’s sieve
Short intervals containing prime numbers
method, we have
1
log D
F
pq
log z
X
X
ηx
Φ1 ≤
log z
133
19
133
X 580 <p≤X 80 p−1 X 290 <q<p
X
X
+
X
133
580
<p≤X
+ O(εηx log
19
80
−1
133
p−1 X 290
19
X
<q<p r<X
21
40
a(r)e
r(A, pqr)
/(pq)
x).
Let m = pqr. Then Lemma 16 yields
X
X
X
133
73
133
a(r)e
r(A, pqr) = O(ηx log−5 x).
21
X 580 <p≤X 80 p−1 X 290 <q<p r<X 40 /(pq)
Hence,
Φ1 ≤
X
e−γ
ηx log−1 x
δ
133
19
133
1
+ O(δηx log−1 x)
pq
X 580 <p≤X 80 p−1 X 290 <q<p
X
=η
X
X
133
S(Bpq , X δ ) + O(δηx log−1 x).
133
19
X 580 <p≤X 80 p−1 X 290 <q<p
In the same way, we can get the lower bound
X
X
S(Bpq , X δ ) + O(δηx log−1 x).
Φ1 ≥ η
133
19
133
X 580 <p≤X 80 p−1 X 290 <q<p
Now we have the asymptotic formula
X
X
S(Bpq , X δ ) + O(δηx log−1 x).
(26)
Φ1 = η
19
133
133
X 580 <p≤X 80 p−1 X 290 <q<p
Using Lemma 18, we have
X
(27)
Φ2 = η
X
133
580
<p≤X
19
80
X
p−1 X
133
290
X
<q<p X δ <r<X
S(Bpqr , r)
9
85
+ O(δηx log−1 x).
By Lemma 13 with the region (i), (7) holds. Then Lemma 19 with a
small modification yields
X
X
X
(28) Φ3 = η
S(Bpqr , r)
133
19
133
9
1
2
X 580 <p≤X 80 p−1 X 290 <q<p X 85 <r<min(q,( 2X
pq ) )
+ O(δηx log−1 x).
74
C. Jia
Gathering together (25)–(28), we have
X
X
Ω5 = η
S(Bpq , q) + O(δηx log−1 x)
133
19
133
X 580 <p≤X 80 p−1 X 290 <q<p
\
19
80
= ηx log
−1
x
133
580
\
19
80
= ηx log
−1
x
133
580
dt
t
dt
t
\t
133
290 −t
\t
133
290 −t
1
1−t−u
w
du + O(δηx log−1 x)
u2
u
1
1−t
1 + log
−2
du
u(1 − t − u)
u
+ O(δηx log−1 x)
≥ 0.003001ηx log−1 x.
In the same way, it can be shown that
Ω7 + Ω9 + Ω11 + Ω13 + Ω15 + Ω18 + Ω21 + Ω24 + Ω27 + Ω30
+ Ω33 + Ω36 + Ω39 + Ω41 + Ω43 + Ω46 + Ω49 + Ω53
53
220
\ dt
= ηx log−1 x
t
19
80
\
151
580
+
53
220
\
3441
9860
+
17
55
\
653
1700
+
37
100
≥ ηx log
\
53
110 −t
dt
t
133
290 −t
dt
t
dt
t
−1
\
897
1972 −t
49
100 −t
9
85
\
151
580
+
53
220
dt
t
\
53
110 −t
133
290 −t
\
17
55
151
580
dt
t
\
53
110 −t
897
1972 −t
37
100
\ dt
1
1−t−u
w
du
+
u2
u
t
3441
9860
\
1
1−t−u
w
du
u2
u
59
58
134 − 67 t
9
85
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
+
O(ε)
u2
u
53
220
\ dt
x
t
19
80
133
290 −t
1
1−t−u
w
du
u2
u
1
1−t−u
w
du +
u2
u
59
58
134 − 67 t
\
\
19
40 −t
\
19
40 −t
133
290 −t
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
1−t
1 + log
−2
du
u(1 − t − u)
u
Short intervals containing prime numbers
\
404
1479
+
151
580
\
53
110 −t
dt
t
897
1972 −t
\
17
55
+ 0.5607
404
1479
\
17
55
+
404
1479
dt
t
\
53
110 −t
dt
t
1
4 (1−t)
\
3441
9860
+ 0.5607
17
55
\
653
1700
+ 0.5607
37
100
dt
t
dt
t
75
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
1
4 (1−t)
897
1972 −t
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
59
58
134 − 67 t
897
1972 −t
\
49
100 −t
100
\ dt
du
+
0.5607
u2
t
3441
37
9860
du
+ O(ε)
u2
9
85
\
59
58
134 − 67 t
9
85
du
u2
≥ (0.002490 + 0.020848 + 0.018080 + 0.033046
+ 0.027398 + 0.105118 + 0.055429 + 0.012222)ηx log−1 x
= 0.274631ηx log−1 x.
The proof of Lemma 26 is complete.
Lemma 27.
Φ = Ω44 + Ω47 + Ω51 + Ω55 + Ω58 + Ω61 + Ω63 + Ω67 + Ω69 + Ω72
≥ 0.033767ηx log−1 x.
P r o o f. On applying Lemmas 11, 19, 20 and 21, in the same way as in
Lemma 24, we have
−1
Φ = ηx log x
\
37
100
1843
5280
\
5541
13940
+
37
100
\
3963
9860
+
5541
13940
dt
t
dt
t
dt
t
\
57
58
98 − 49 t
133
82
230 − 69 t
\
57
58
98 − 49 t
9
85
\
57
58
98 − 49 t
171
70
295 − 59 t
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
76
C. Jia
\
19
45
+
1063
2640
\
4331
9860
+
19
45
dt
t
dt
t
\
35
179
230 − 23 t
59
29
76 − 19 t
\
421
41
540 − 27 t
59
29
76 − 19 t
1
1−t−u
w
du
u2
u
1
1−t−u
w
du + O(ε)
u2
u
389
1105
\ dt
≥ ηx log−1 x
t
1843
5280
\
65
183
+ 0.5607
\
65
183
+
389
1105
dt
t
+ 0.5607
\
\
+ 0.5607
+
1063
2640
\
29
59
76 − 19 t
\
+ 0.5607
dt
t
40
97
\
27
65
+
40
97
dt
t
\
27
65
\
dt
t
4331
9860
19
45
100
\
57
58
98 − 49 t
133
82
230 − 69 t
du
u2
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
29
59
76 − 19 t
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
4 (1−t)
19
45
+ 0.5607
9
85
\
179
35
230 − 23 t
+ 0.5607
\
1
4 (1−t)
\
13940
\ dt
du
+
0.5607
u2
t
37
57
58
98 − 49 t
179
35
230 − 23 t
27
65
du
u2
5541
171
70
295 − 59 t
dt
t
5541
13940
dt
t
\
57
58
98 − 49 t
dt
t
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
1−t
1 + log
−2
du
u(1 − t − u)
u
1
4 (1−t)
3963
9860
\
171
70
295 − 59 t
\
65
183
40
97
\
57
58
98 − 49 t
37
100
171
70
295 − 59 t
1
4 (1−t)
dt
t
389
1105
\
58
57
98 − 49 t
\
179
35
230 − 23 t
29
59
76 − 19 t
dt
t
du
u2
\
421
41
540 − 27 t
59
29
76 − 19 t
du
+ O(ε)
u2
Short intervals containing prime numbers
77
≥ (0.000517 + 0.000285 + 0.000279 + 0.002997
+ 0.013589 + 0.001413 + 0.002113 + 0.000371
+ 0.000359 + 0.001830 + 0.010014)ηx log−1 x
= 0.033767ηx log−1 x.
The proof of Lemma 27 is complete.
Lemma 28.
Φ = Ω75 + Ω78 + Ω90 + Ω92 ≥ 0.019054ηx log−1 x.
P r o o f. On applying Lemmas 12 and 19–21, in the same way as in
Lemma 24, we have
Φ = ηx log
−1
x
2691
5950
\
4331
9860
\
897
1972
+
2691
5950
\
49
100
+
\
2944
5950
+
49
100
9
85
\
70
179
55 − 11 t
dt
t
233
480
\
57
29
40 − 10 t
dt
t
59
58
18 − 9 t
\
9
85
\
≥ ηx log
−1
141
320
x
4331
9860
\
47
106
+ 0.5607
141
320
\
47
106
+
141
320
dt
t
\
57
29
40 − 10 t
1
4 (1−t)
\
2691
5950
+ 0.5607
dt
t
47
106
dt
t
57
35
40 − 12 t
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
1
1−t−u
w
du
u2
u
179
70
55 − 11 t
dt
t
\
57
29
40 − 10 t
dt
t
1
1−t−u
w
du + O(ε)
u2
u
\
57
29
40 − 10 t
dt
t
35
57
40 − 12 t
\
1
4 (1−t)
57
35
40 − 12 t
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
\
57
29
40 − 10 t
57
35
40 − 12 t
1972
\ dt
du
+
0.5607
u2
t
2691
897
5950
\
57
29
40 − 10 t
9
85
du
u2
78
C. Jia
\
109
223
+
dt
t
233
480
\
70
179
55 − 11 t
59
58
18 − 9 t
\
49
100
+ 0.5607
\
49
100
+
dt
t
109
223
\
1
4 (1−t)
\
dt
t
49
100
\
661
1345
+
49
100
dt
t
\
1
4 (1−t)
9
85
\
1
4 (1−t)
\
661
1345
dt
t
du
u2
1
1−t
1 + log
−2
du
u(1 − t − u)
u
179
70
55 − 11 t
2944
5950
+ 0.5607
58
59
18 − 9 t
179
70
55 − 11 t
661
1345
+ 0.5607
\
1
4 (1−t)
dt
t
109
223
1
1−t
1 + log
−2
du
u(1 − t − u)
u
du
u2
1
1−t
−2
du
1 + log
u(1 − t − u)
u
\
179
70
55 − 11 t
9
85
du
+
O(ε)
u2
≥ (0.000612 + 0.000691 + 0.000645 + 0.005586
+ 0.001031 + 0.002927 + 0.000331 + 0.000980
+ 0.002633 + 0.000435 + 0.003183)ηx log−1 x
= 0.019054ηx log−1 x.
The proof of Lemma 28 is complete.
Lemma 29.
Φ = Ω1 + Ω2 + Ω3 + Ω4 + Ω6 + Ω8 + Ω10 + Ω12 + Ω14 + Ω17 + Ω20
+ Ω23 + Ω26 + Ω29 + Ω32 + Ω35 + Ω38 + Ω50 + Ω54 + Ω57 + Ω60
≥ 1.902585ηx log−1 x.
P r o o f. The main idea in this lemma is adopted from [20]. We have
X
X
(29) Ω1 ≥
S(Apq , q)
21
9
X 85 <p≤X 100 −10
−8
X
=
9
21
X 85 <p≤X 100 −10
X
−
X
9
85
<p≤X
9
X 85 <q<p
X
−8
S(Apq , X δ )
9
X 85 <q<p
21 −10−8
100
X
X
9
85
X
<q<p X δ <r<X
S(Apqr , r)
9
85
Short intervals containing prime numbers
X
−
X
9
85
X
21 −10−8
100
<p≤X
9
85
X
79
X
<q<p X
9
85
S(Apqr , r)
1
2
<r<min(q,( 2X
pq )
)
= Φ1 − Φ2 − Φ3 .
21
Let z = X δ and D(p, q) = X 40 /(pq). Using Iwaniec’s sieve method, in
the same way as in Lemma 26, we have
X
X
(30)
Φ1 = η
S(Bpq , X δ ) + O(δηx log−1 x).
9
21
X 85 <p≤X 100 −10
Lemma 18 yields
(31) Φ2 = η
X
9
9
−8
X 85 <q<p
X
21
X 85 <p≤X 100 −10
Hence,
X
9
−8
9
X 85 <q<p X δ <r<X 85
X
X
(32) Φ1 − Φ2 = η
21
9
= ηx log
85
x·
9
X 85 <q<p
\
−8
21
100 −10
9
85
+ O(δηx log−1 x)
21
100
\ dt
85
≥ ηx log−1 x
9 9 t
85
≥ ηx log
−1
9
S(Bpq , X 85 ) + O(δηx log−1 x)
9
−8
X 85 <p≤X 100 −10
−1
S(Bpqr , r) + O(δηx log−1 x).
\t
9
85
85
x 0.5612 ·
9
dt
t
\t
1
85
w
(1 − t − u) du
u
9
9
85
1
85
−6
w
(1 − t − u) du − 10
u
9
\
21
100
9
85
dt
t
\t
9
85
du
− 10−6
u
≥ 1.242693ηx log−1 x.
1
1
Note that if q < (2X/p) 3 , then q < (2X/(pq)) 2 . We have
X
X
X
9
(33) Φ3 ≤
S(Apqr , X 85 )
9
21
X 85 <p≤X 100 −10
−8
X
=
9
21
X 85 <p≤X 100 −10
X
−
X
9
85
<p≤X
= Φ4 − Φ5 .
−8
9
9
X
X
9
9
X 85 <q<p X 85 <r<q
S(Apqr , X δ )
X 85 <q<p X 85 <r<q
21 −10−8
100
X
X
9
85
<q<p X
X
9
85
X
<r<q X δ <s<X
S(Apqrs , s)
9
85
80
C. Jia
−8
Let z = X δ and D(p, q, r) = X 10 . Using Lemma 17 with region (i), in
the same way as in Lemma 26, we have
X
X
X
(34)
Φ4 = η
S(Bpqr , X δ )
9
21
9
9
X
X
−8
X 85 <p≤X 100 −10
X 85 <q<p X 85 <r<q
+ O(δηx log−1 x).
Lemma 18 yields
(35)
X
Φ5 = η
X
9
85
<p≤X
21 −10−8
100
+ O(δηx log
−1
X
9
85
<q<p X
9
85
X
<r<q X δ <s<X
S(Bpqrs , s)
9
85
x).
We therefore have
X
Φ3 ≤ η
9
21
X 85 <p≤X 100 −10
−8
X
X
9
9
9
S(Bpqr , X 85 )
X 85 <q<p X 85 <r<q
+ O(δηx log−1 x)
≤ ηx log
−1
85
x·
9
\
21
100
9
85
dt
t
\t
9
85
85
−1
≤ ηx log x 0.5617 ·
9
85
+ 0.5617 ·
9
85
+ 0.5644 ·
9
\
21
100
49
255
\
21
100
49
255
dt
t
dt
t
\
\
49
255
9
85
1
85
w
(1 − t − u − v) dv
v
9
u
du
u
9
85
dt
t
\
1 49
2 ( 85 −t)
9
85
\t
1 49
2 ( 85 −t)
\t
9
85
du
u
du
u
du
u
\
u
9
85
\
u
9
85
\
u
9
85
dv
v
dv
v
dv
v
≤ (0.187169 + 0.077309 + 0.019528)ηx log−1 x
= 0.284006ηx log−1 x.
Hence,
(36)
Ω1 ≥ 0.958687ηx log−1 x.
In the same way, it can be shown that
Ω2 + Ω3 + Ω4 + Ω6 + Ω8 + Ω10 + Ω12 + Ω14 + Ω17 + Ω20 + Ω23
Short intervals containing prime numbers
81
+ Ω26 + Ω29 + Ω32 + Ω35 + Ω38 + Ω50 + Ω54 + Ω57 + Ω60
≥ ηx log
85
−
9
85
+
9
85
−
9
85
+
9
85
−
9
85
+
9
85
−
9
85
+
9
85
−
9
85
+
9
−1
85
x
9
\
13
60
21
100
dt
t
\
12613
49300
13
60
\
12613
49300
13
60
\
3441
9860
12613
49300
\
3441
9860
12613
49300
\
653
1700
297
800
\
653
1700
297
800
\
5073
13120
653
1700
\
5073
13120
653
1700
\
5541
13940
5073
13120
\
13
60
21
100
\
7
2
20 − 3 t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
9
85
\
9
85
\
t
29
120 − 6
9
85
\
897
1972 −t
9
85
\
897
1972 −t
9
85
\
49
100 −t
\
19
160
49
100 −t
\
19
160
9
85
\
19
160
9
85
9
85
\
1
85
w
(1 − t − u) du
u
9
85
1
w
(1 − t − u − v) dv
v
9
u
du
u
29
t
120 − 6
19
160
\
2
7
20 − 3 t
dt
t
9
85
85
1
w
(1 − t − u) du
u
9
\
u
du
u
9
85
1
85
w
(1 − t − u − v) dv
v
9
1
85
w
(1 − t − u) du
u
9
\
u
du
u
9
85
1
85
w
(1 − t − u − v) dv
v
9
1
85
w
(1 − t − u) du
u
9
\
u
du
u
9
85
1
85
w
(1 − t − u − v) dv
v
9
1
85
w
(1 − t − u) du
u
9
du
u
\
\
u
9
85
133
82
230 − 69 t
9
85
85
1
w
(1 − t − u − v) dv
v
9
1
85
w
(1 − t − u) du
u
9
82
C. Jia
85
−
9
\
5541
13940
5073
13120
dt
t
\
82
133
230 − 69 t
9
85
− 0.000100
≥ ηx log
−1
85
x 0.5612 ·
9
85
− 0.5644 ·
9
85
+ 0.5612 ·
9
85
− 0.5617 ·
9
85
− 0.5644 ·
9
85
+ 0.5612 ·
9
85
− 0.5617 ·
9
85
− 0.5644 ·
9
85
− 0.5644 ·
9
85
+ 0.5612 ·
9
\
13
60
21
100
dt
t
\
12613
49300
13
60
\
12613
49300
13
60
\
12613
49300
13
60
\
3441
9860
12613
49300
\
1643
4930
12613
49300
\
1643
4930
12613
49300
\
3441
9860
1643
4930
\
653
1700
297
800
dt
t
dt
t
dt
t
dt
t
dt
t
dt
t
1
85
w
(1 − t − u − v) dv
v
9
9
85
\
13
60
21
100
\
2
7
20 − 3 t
dt
t
dt
t
\
u
du
u
9
85
dt
t
du
u
\
t
29
120 − 6
\
1 49
2 ( 85 −t)
9
85
\
29
t
120 − 6
1 49
2 ( 85 −t)
\
897
1972 −t
9
85
\
9
85
\
897
1972 −t
1 49
2 ( 85 −t)
9
85
\
19
160
49
100 −t
\
u
9
85
dv
v
du
u
du
u
\
u
9
85
\
u
9
85
dv
v
dv
v
du
u
1 49
2 ( 85 −t)
\
9
85
du
u
du
u
9
85
897
1972 −t
\
7
2
20 − 3 t
du
u
du
u
du
u
du
u
\
u
9
85
\
u
9
85
\
u
9
85
dv
v
dv
v
dv
v
Short intervals containing prime numbers
85
− 0.5644 ·
9
85
+ 0.5612 ·
9
85
− 0.5644 ·
9
85
+ 0.5612 ·
9
85
− 0.5644 ·
9
\
653
1700
297
800
\
5073
13120
653
1700
\
5073
13120
653
1700
\
5541
13940
5073
13120
\
5541
13940
5073
13120
dt
t
dt
t
dt
t
dt
t
dt
t
\
19
160
49
100 −t
\
19
160
9
85
\
19
160
9
85
\
u
du
u
9
85
83
dv
v
du
u
du
u
\
u
9
85
\
133
82
230 − 69 t
9
85
\
133
82
230 − 69 t
9
85
dv
v
du
u
du
u
\
u
9
85
dv
v
− 0.000100
≥ (0.111673 − 0.037863 + 0.570625 − 0.100014
− 0.085420 + 0.599832 − 0.063002 − 0.072362
− 0.000822 + 0.010098 − 0.000782 + 0.004011
− 0.000232 + 0.008592 − 0.000335 − 0.000100)ηx log−1 x
= 0.943899ηx log−1 x.
Hence,
Φ ≥ 1.902585ηx log−1 x.
The proof of Lemma 29 is complete.
10. The proof of the Theorem. From Lemmas 24–29, it follows that
Ω ≥ 2.949096ηx log−1 x.
Then using Lemmas 22 and 23, we obtain
1
π(x + ηx) − π(x) = S(A, (2X) 2 ) ≥ 0.011ηx log−1 x.
Now (15) holds and the proof of the Theorem is complete.
References
[1]
H. C r a mé r, On the order of magnitude of the difference between consecutive prime
numbers, Acta Arith. 2 (1937), 23–46.
84
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
[19]
[20]
[21]
C. Jia
J.-M. D e s h o u i l l e r s and H. I w a n i e c, Power mean values of the Riemann zetafunction, Mathematika 29 (1982), 202–212.
G. H a r m a n, Primes in short intervals, Math. Z. 180 (1982), 335–348.
—, On the distribution of αp modulo one, J. London Math. Soc. (2) 27 (1983), 9–18.
D. R. H e a t h-B r o w n, Finding primes by sieve methods, Proc. 1982 ICM, Warsaw,
1983, Vol. 1, PWN–Polish Sci. Publ., Warszawa, 1984, 487–492.
D. R. H e a t h-B r o w n and H. I w a n i e c, On the difference between consecutive
primes, Invent. Math. 55 (1979), 49–69.
M. N. H u x l e y, On the difference between consecutive primes, ibid. 15 (1972),
164–170.
H. I w a n i e c, A new form of the error term in the linear sieve, Acta Arith. 37
(1980), 307–320.
H. I w a n i e c and J. P i n t z, Primes in short intervals, Monatsh. Math. 98 (1984),
115–143.
C. J i a, On Pjateckiı̆-Šapiro prime number theorem (II ), Sci. China Ser. A 36 (1993),
913–926.
—, Goldbach numbers in short interval, Sci. China Ser. A 24 (1994), 1233–1259 (in
Chinese); I. Sci. China Ser. A 38 (1995), 385–406; II. Sci. China Ser. A 38 (1995),
513–523.
—, On the difference between consecutive primes, Sci. China Ser. A 25 (1995),
785–804 (in Chinese); Sci. China Ser. A 38 (1995), 1163–1186.
—, On the exceptional set of Goldbach numbers in the short interval, Acta Arith.,
to appear.
H. L i, On the Goldbach numbers in short interval, Sci. China Ser. A 38 (1995),
641–652.
—, Primes in short intervals, preprint.
H. L. M o n t g o m e r y, Topics in Multiplicative Number Theory, Lecture Notes in
Math. 227, Springer, Berlin, 1971.
Chengdong P a n and Chengbiao P a n, Goldbach Conjecture, Science Press, Beijing,
1981 (in Chinese).
—, —, The Basis of Analytic Number Theory, Science Press, Beijing, 1991 (in
Chinese).
A. S e l b e r g, On the normal density of primes in short intervals, and the difference
between consecutive primes, Arch. Math. Naturvid. 47 (1943), 87–105.
N. W a t t, Short intervals almost all containing primes, Acta Arith. 72 (1995), 131–
167.
—, Kloosterman sums and a mean value for Dirichlet polynomials, J. Number Theory, to appear.
Institute of Mathematics
Academia Sinica
Beijing, 100080, China
Received on 29.3.1995
(2766)

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