Acta Mathematica Sinica, New Series
1996, Vol.12, No.4, pp. 433–445
The Greatest Prime Factor of the Integers
in a Short Interval (IV)
Jia Chaohua
Abstract.
Let P (x) denote the greatest prime factor of
prove that P (x) > x0.728 holds true for sufficiently large x.
1
1
x<n≤x+x 2
n. In this paper, we shall
Introduction
1 n. We are interested in the estimate
Let P (x) denote the greatest prime factor of
x<n≤x+x 2
for the lower bound of P (x).
The first results were given by K. Ramachandra[1],[2] who showed that
P (x) > xϕ
(1)
for sufficiently large x, where ϕ = 85 .
Using the estimate for trigonometrical sums and the sieve method, S.W. Graham[3] improved
on (1) to get ϕ = 0.662. Afterwards, Jia Chaohua[4] got ϕ = 0.692. Meanwhile, R.C. Baker[5]
got ϕ = 0.7. In papers [4] and [5], the main tools are Heath-Brown’s method on the estimate
for trigonometrical sums (see Lemma 4 in [6]) and Iwaniec’s sieve method (see [7]). Combining
these methods, Jia Chaohua[8] arrived at ϕ = 0.71. In Lemma 6 of [8] there is a discussion
which is not strict. If this lemma is abandoned and a small modification is made (see Section
2 below), then the exponent 0.71 could be obtained again.
In papers [9] and [10], Jia Chaohua has given a delicate treatment for trigonometrical sums,
with which the corresponding trigonometrical sums in our problem can also be treated. So, Jia
Chaohua[11] showed ϕ = 0.723. In [11], Lemma 6 of [8] was not used.
In this paper, we shall use the previous effective methods together with the new method of
Fouvry and Iwaniec[12] on the estimate for trigonometrical sums to prove the following result.
Theorem If x is sufficiently large, then we have P (x) > x0.728 .
Throughout this paper, we always assume that x is sufficiently large and that ε is a sufficiently small positive constant, δ = ε2 . Assume that c, c1 , c2 are positive constants which have
different values at different places. m ∼ M means that there arepositive constants c1 and c2
1
2
such that c1 M < m ≤ c2 M. We also assume that N (d) = x + x − x and ((θ)) = θ−[θ]− 12 .
d
d
Received December 28, 1994. Resived August 1, 1995.
Project supported by the Tian Yuan Item in the National Natural Science Foundation of China.
434
Acta Mathematica Sinica, New Series
Vol.12 No.4
Here the author would like to thank Profs. Wang Yuan and Pan Chengbiao for their
encouragement.
2
An Explanation
In [8], we abandon Lemma 6 and replace our result
Λ(d)N (d) =
43 1
1
x 2 log x + O(εx 2 log x)
72
Λ(d)N (d) =
1
13 1
x 2 log x + O(εx 2 log x).
22
43
d≤x 72 −ε
by the result in [5]
13
d≤x 22 −ε
Correspondingly, K1 =
3
5
2du = 0.0055555 is replaced by
43
72
3
5
13
22
2du = 0.0181819 and
35
2 − 3u
2 − 3u
du = 0.0000955 is replaced by
du = 0.0010197.
log
log
L1 =
43
13
2u − 1
2u − 1
72
22
At the same time, we have
3
5
kT3 (1, ek )
( 13
22 −ε) log x<k≤0.676 log x
1
2
≥ x log x
+
+
θdθ
5
5
3−2θ
2−3θ
θ− 12
1
2 (θ−α1 )
dα1
α1
1
2 (θ−α1 )
dα1
13
10
6 − 3 θ
31
31
75 θ− 300
θdθ
2−3θ
0.676
8
13
θdθ
3
5
8
13
11
18
11
18
α1
dα2
α1 α2 (θ − α1 − α2 )
1
2 (θ−α1 )
dα1
α1
dα2
α1 α2 (θ − α1 − α2 )
dα2
α1 α2 (θ − α1 − α2 )
1
2
≥ 0.01073x log x.
Then, we can arrive at ϕ = 0.71 again.
3
An Inverse Formula
In this section we state an inverse formula for the trigonometrical sum, which can be found
in [13]. We only replace the condition that f (x) is an algebraic function by a rigorous condition
on the third derivative of f (x).
A (U ≥
Lemma 1 Let ψ(x) and η(x) be real functions. Suppose that i) ψ(x) ∼ A, |ψ (x)| ∼ U
1 , |η (x)| ∼ 1 ; iii) η(0) = η (0) = 0.
1); ii) |η (x)| ∼ R
RU
Then
ξ
ψ(0)
1
ψ(x)e(η(x))dx = e
8
2 η (0)
0
√
AR
1
1
+O U log(2 + U ) + log 2 + R
R, + O A min
.
|η (ξ)|
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
We might as well suppose η (x) > 0.
ξ
ξ
ψ(x)e(η(x))dx =
(ψ(x) − ψ(0))e(η(x))dx
0
0
ξ
1
1
−
η (x)e(η(x))dx
+ψ(0)
η
(x)
2η (0)η(x)
0
ξ
1
ψ(0)
+
· η (x)e(η(x))dx.
2η (0) 0
η(x)
Proof
435
(2)
By the Taylor’s expansion, we have
η(x) =
η (ξ1 ) 2
x ,
2
(3)
η (0) 2 η (ξ2 ) 3
x +
x ,
2
6
η (ξ3 ) 2
x .
η (x) = η (0)x +
2
On applying (3) and (4) to η(ξ), we find that ξ U.
η(x) =
(4)
(5)
1
1
2η (0)η(x) − (η (x))2
−
=
.
η (x)
2η (0)η(x)
η (x) 2η (0)η(x)( 2η (0)η(x) + η (x))
3
2η (0)η(x)( 2η (0)η(x) + η (x)) is an increasing function which ∼ x 3 .
R
Since (2η (0)η(x) − (η (x))2 ) = 2η (0)η (x) − 2η (x)η (x) = −2η (x)η (ξ4 )x, 2η (0)η(x) −
3
(η (x))2 is a monotonic function which x2 by (4) and (5).
R U
√
When b = min( R, ξ) ≤ V ≤ ξ,
V
1
R
1
−
η (x)e(η(x))dx V
η
(x)
U
2η (0)η(x)
2
Obviously, η (x)
by using the second mean value theorem twice.
By the trivial estimate,
b
1
R
1
−
η (x)e(η(x))dx .
η (x)
U
2η (0)η(x)
0
Hence,
0
ξ
1
1
−
η (x)
2η (0)η(x)
Similarly,
ξ
0
ψ(x) − ψ(0)
η (x)
η (x)e(η(x))dx η (x)e(η(x))dx AR
U
R
U
1
log(2 + U ) + log 2 +
.
R
1
log(2 + U ) + log 2 +
.
R
As for the main term, we have
ξ
η(ξ)
∞
∞
1
e(t)
e(t)
e(t)
√ dt =
√ dt −
√ dt.
· η (x)e(η(x))dx =
t
t
t
η(x)
0
0
0
η(ξ)
(6)
(7)
(8)
436
Acta Mathematica Sinica, New Series
Vol.12 No.4
By the second mean value theorem,
∞
η(ξ)
√
1
1
1
e(t)
√ dt min 1, .
R, √ min
|η (ξ)|
t
R
η(ξ)
(9)
The complex integration yields
0
∞
e(t)
1
√ dt = √ e
t
2
1
.
8
(10)
So, we get the conclusion from (2), (6), (7), (8), (9), (10), and the proof of Lemma 1 is
complete.
Lemma 2 is called an inverse formula for the trigonometrical sum, which is essentially
Theorem 2.2 on page 36 of [13].
Lemma 2 Let ϕ(x) and f (x) be real functions. Suppose that f (x) has the third continuous
1
1 (U ≥ 1),
derivative and ϕ (x) is a monotonic function. Suppose that |f (x)| ∼ R
, |f (x)| ∼ RU
Φ . Let [α, β] be the image of [a, b] under the transformation y = f (x).
ϕ(x) ∼ Φ, |ϕ (x)| ∼ U
Then
ϕ(n )
1
ν e (f (nν ) − νnν )
ϕ(n)e(f (n)) = e
8
f (nν )
a<n≤b α<ν≤β
R + 1 log(2 + U ) + log 2 + 1
+O Φ U
R
√
√
+ O Φ min
,
R, 1
R, 1
+O Φ min
α
β
where nν is the solution of the equation f (x) = ν and x denotes the nearest distance from x
to integers.
By Lemma 1 and the discussion in Theorem 2.2 on page 36 of [13], we can get Lemma 2.
4
Some Preliminary Lemmas
Lemma 3
Suppose a(n) = O(1). Let 0 < L ≤ M < N ≤ cL, L 1, T ≥ 1. Then
T 1
a(l) N it − M it
dt
a(n) =
·
2πi −T
t
lit
M <n≤N
L<l≤cL
L log(2 + L)
L
L
+O min 1,
+ O min 1,
+O
.
T
T M T N This is the Perron’s formula.
Lemma 4 Suppose that f (x) P and f (x) ∆. Then
1
1
1
(P + 1) D +
log 2 +
.
min D,
f (n)
∆
∆
n∼N
This is Lemma 3 in [10].
Define w(u) as the continuous solution of the equations:
1,
w(u) = u
1≤u≤2
(uw(u)) = w(u − 1), u > 2.
(11)
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
437
w(u) is called Buchstab’s function which plays an important role in finding the asymptotic
formulas in sieve method. In particular,
w(u) =
1 + log(u − 1)
,
u
w(u) =
1 + log(u − 1)
1
+u
u
2 ≤ u ≤ 3;
u−1
2
log(t − 1)
dt,
t
3 ≤ u ≤ 4.
Lemma 5 For the function w(u), we have the following bounds: i) when u ≥ 2.47, w(u) ≥
0.5607; ii) when u ≥ 3, w(u) ≤ 0.5644; iii) when u ≥ 1, w(u) ≥ 0.5.
This is Lemma 13 in [14].
5
A Leading Step
From the deduction in §2 of [4] and §1 of [5], it is known that
1
1
x 2 log x =
Λ(d)N (d) +
N (p) log p + O(x 2 ).
d≤x0.6−ε
We have proved in [11]
(12)
x0.6−ε <p≤P (x)
1
1
Λ(d)N (d) = 0.6x 2 log x + O(εx 2 log x).
d≤x0.6−ε
Once
1
N (p) log p ≤ 0.39999x 2 log x
(13)
x0.6−ε <p≤x0.728
is proved, we can get the theorem by comparing (12) with (13).
In the following, we always assume that
x0.6−ε ≤ v ≤ x0.73 , H =
v
1
x 2 −6δ
, v = 2M N.
We shall mainly deal with sums such as
Ω=
g(d)f (d),
(14)
(15)
v<d≤ev
1
x .
x + x2
−
d
d
From the discussion in §4 of [4] and §3 of [5], we know that in order to get
where g(d) = O(1), f (d) =
1
Ω x 2 −5δ ,
(16)
it is sufficient to prove: When 21 < J ≤ H, it is true that
1
hy
x 12 −12δ .
max √ max g(d)e
H x≤y≤x+ x v<u≤ev d J<|h|≤2J v<d≤u
For simplicity, we only prove: When 21 < J ≤ H, it is true that
1
hx
1 x 2 −12δ .
g(d)e
H
d
J<h≤2J v<d≤ev
(17)
438
Acta Mathematica Sinica, New Series
Vol.12 No.4
Others can be proved in the same way.
6
The Estimate for Trigonometrical Sums
Lemma 6
3
Assume that 12 < J ≤ H, M x 8 −ε and that ρ(m) = O(1). Then
1
1 hx
x 2 −20δ .
ρ(m)
e
H
mn
v
ev
J<h≤2J M <m≤2M
m <n≤ m
This is Lemma 9 of [11].
5
1
Lemma 7 Assume that 12 < J ≤ H, v x 8 −ε , v 4 x−2+ε N x 2 −ε and that a(n), b(m) =
O(1). Then
1
1 hx
x 2 −20δ .
a(n)b(m)e
H
mn
J<h≤2J N <n≤2N M <m≤2M
Lemma 7 is Lemma 9 of [5] which is obtained by Heath-Brown’s method on the estimate
for trigonometrical sums. One may also refer to Lemma 6 of [4].
3
2
1
1
1
Lemma 8 Assume that 21 < J ≤ H, x 5 v x 3 −ε , vx− 2 +ε M v 7 x 14 −ε and that
a(n), b(m) = O(1). Then
1
H
a(n)b(m)e
J<h≤2J N <n≤2N M <m≤2M
hx
mn
1
x 2 −20δ .
Proof The trigonometrical sum here is the one appearing in the proof of Theorem 7 of [12]
1
with y = x 2 and m, n being interchanged. It could be estimated by the Theorem 6 of [12]. In
the proof of Theorem 6 of [12], we may replace Lemmas 5 and 6 of [12] by our Lemmas 2 and
3 respectively. Then we use Lemma 4 to deal with the corresponding error terms so that the
3
condition M 4 N 2 x 2 −ε is canceled and the limit of v is extended. So Lemma 8 holds.
Lemma 9 Assume that 12 < J ≤ H and that a(n), b(m) = O(1). Furthermore, one of fol437
1469
1
67
22
1469
lowing conditions is satisfied: i) x 680 v x 2228 , vx− 2 +ε M v 129 x− 129 −ε ; ii) x 2228 49
1
77
31
49
31
1
73
32
v x 72 , vx− 2 +ε M v 115 x− 115 −ε ; iii) x 72 v x 44 −10ε , vx− 2 +ε M v 95 x− 95 −ε .
Then
1
1 hx
x 2 −20δ .
a(n)b(m)e
H
mn
J<h≤2J N <n≤2N M <m≤2M
Proof
Let (k, l) be any exponent pair. By Lemma 14 of [5], we have that for
1
1+k−l
1+4k−2l
vx− 2 +ε M v 1−k+l x− 4(1−k+l) −ε
with (M 2 N )1+ε ≤ x, the estimate for trigonometrical sum holds true.
(18)
57 , BA2
Taking (k, l) as one of following exponent pairs: BABABA2 BAB(0, 1) = 13
,
49 98
13 , BA3 BA2 B(0, 1) = 11 , 11 , we can assert that the conclusion
BABA2 B(0, 1) = 11
,
32 24
28 21
is true.
7
Sieve Method
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
Assume that
P(z) =
p, A = {n : v < n ≤ ev, n repeats N (n) times}, S(A, z) =
p<z
439
1.
(19)
n∈A
(n,P(z))=1
1
1
2
2
Let X = x , ω(d) = 1, r(d) = |Ad | − x , W (z) =
d
e−ν
1
=
(1 + O(ε)), where ν
1−
p
log z
p<z
is the Euler constant.
1
Lemma 10 Suppose that for vx− 2 +ε M W and 21 < J ≤ H,
1
1 hx
x 2 −20δ .
a(n)b(m)e
H
mn
J<h≤2J N <n≤2N M <m≤2M
3
1
8 −2ε
Let D = W xH
. If z ≤ D 2 , then
1
x2
F
S(A, z) ≤
log z
log D
log z
+O
1
εx 2
log z
,
(20)
where F (s) is a standard function the definition and properties of which could be found in
Chapter 8 of [15].
In particular,

2,

2 ≤ u ≤ 3,
u
u−1
(21)
F (u) = 2
log(t − 1)

dt , 3 ≤ u ≤ 5.
u 1+
t
2
Proof
By Iwaniec’s bilinear sieve method (see Theorem 1 of [7]), we have


 
1
log D
S(A, z) ≤ x 2 eν W (z) F
+ O(ε) + O 
c(k)b(m)r(km)

,
log z
3
−2ε
k≤ x 8H
where r(d) =
v<ld≤ev
1
x + x2
ld
−
x ld
m≤W
.
By the explanation in Section 5, in order to prove
3 −2ε
k≤ x 8H
1
c(k)b(m)r(km) x 2 −5δ ,
m≤W
it is sufficient to prove
1
=
H
3
J<h≤2J K<k≤2K M <m≤2M
v
ev
km <l≤ km
c(k)b(m)e
hx
lkm
1
x 2 −14δ ,
8 −2ε
where K ≤ x H , M ≤ W.
1
1
x 2 −14δ by the supposition in the lemma.
If vx− 2 +ε M W, letting kl = n, then
1
3
1
6,
x 2 −14δ .
If M vx− 2 +ε , letting km = s, then s x 8 −ε. By Lemma
1
1
2
2
log D
. Lemma 10 then follows.
Hence, we have S(A, z) ≤ x F
+ O εx
log z
log z
log z
440
8
Acta Mathematica Sinica, New Series
Asymptotic Formula
Lemma 11

1
For v ε ≤ p ≤ v 2 −ε , we have
v <n≤ ev
p
p
(n,P(p))=1
Vol.12 No.4
v
1
ε
p 1

= w
+
O
, where

n
log p log p
log p
log
w(u) is the function as defined in (11).
1−ε
Proof If t, z are
large
and z ≤ t , by Lemma 4 on page 240 of [16], then
sufficiently
t
log t
+ O(ε)
. This yields, for v ε ≤ z ≤ v 1−ε ,
1= w
log
z
log
z
n≤t
(n,P(z))=1


1
1  
=
d
1
n
t
v
v<n≤ev
n≤t
(n,P(z))=1
(n,P(z))=1 ev 1
ε
log t
1
1
log t
dt
+
O
=
w
+ w
t
log z log z
log z log2 z
log z
v
v
1
log
+
log z
log z
ε
w(r)dr + O
=
log v
log
z
log z
log v
1 +O
ε
=w
.
log z log z
log z
ev
So, Lemma 11 follows.
Lemma 12 Suppose that when U M W,
a(n)b(m)G(mn) K,
(22)
n∼N m∼M
and that when N T,
c(n)G(mn) K.
(23)
n∼N m∼M
Let
Then when
v
1
(24)
W ≤ v 2 , W ≤ , U 2 T v, U v W 2 T.
T
T
A U
,
Φ(a)Λ(b)G(ab) Kv 2δ . Here Λ(b) is the Mangoldt function
v
W2
a∼A ab∼v
and Φ(a), a(n), b(m), c(n) are arbitrary complex sequences which are not greater than one in
modulus.
Proof
Φ(a)Λ(b)G(ab) =
Φ(a)Λ(b)µ(c)G(abcd)
a∼A ab∼v
=
abcd∼v
a∼A
cd≤W
Φ(a)Λ(b)µ(c)G(abcd) −
abcd∼v
a∼A
c≤W
Φ(a)Λ(b)µ(c)G(abcd)
abcd∼v
a∼A
c≤W
cd>W
= Σ1 − Σ2 .
Σ1
=
abcd∼v
a∼A
c≤U
Φ(a)Λ(b)µ(c)G(abcd) +
abcd∼v
a∼A
U <c≤W
Φ(a)Λ(b)µ(c)G(abcd) = Σ3 + Σ4 .
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
441
By (22), we have Σ4 Kv 2δ . If the indexes of the sums
do not satisfy
our requirements,
we can apply Lemmas 3 and 4 to deal with them. Σ3 =
α(m)
G(ml) log l, which
mAU
ml∼v
could be estimated by (23).
Σ2 =
Φ(a)Λ(b)µ(c)G(abcd) +
Φ(a)Λ(b)µ(c)G(abcd) = Σ5 + Σ6 .
abcd∼v
a∼A
c≤W
d> v
T
Σ5
=
β(m)
mT
abcd∼v
a∼A
c≤W
cd>W
d≤ v
T
G(md),
md∼v
which could be dealt with by (23).
Φ(a)Λ(b)µ(c)G(abcd) +
Φ(a)Λ(b)µ(c)G(abcd) = Σ7 + Σ8 .
Σ6 =
abcd∼v
a∼A
c≤W
cd>W
W <d≤ v
T
abcd∼v
a∼A
c≤W
cd>W
d≤W
v
v
We note that in Σ7 , W
AW ad Av
T U , which implies U bc W. Hence, Σ7
could be estimated by (22).
In Σ8 , if one of c, d > U, we can apply (22) to get the estimate. So, we need only to deal
Φ(a)Λ(b)µ(c)G(abcd).
with Σ9 =
abcd∼v
a∼A
c≤U
cd>W
d≤U
v
AW acd AU 2 U T Uv , which implies U b W. So, we can
Note that W
get Σ9 Kv 2δ .
Combining all of the above discussions, we get the proof of Lemma 12.
9
The Proof of the Theorem
We discuss in six cases.
27
I.
x0.6−ε ≤ v = ek ≤ x 44 .
2−ε
1
3
Let U = vx− 2 +ε , W = xv3 , T = x 8 −ε .
By the Buchstab’s identity, we have
1
N (p) = S(A, (ev) 2 ) = S(A, U ) −
S(Ap , p).
(25)
1
U <p≤(ev) 2
v<p≤ev
By Lemmas 7 and 10, we have
1 1
2
2
log D
x
S(A, U ) ≤
F
+ O εx
log U
log U
log x


23
1 1
 8 log x − 4k 
εx 2 .
x2
F
+
O
=


1
1
log x
(k − log x)
k − log x
2
2
39
If x0.6−ε ≤ v ≤ x 64 −2ε , then
S(Ap , p) ≥
1
U <p≤(ev) 2
U <p≤W
S(Ap , p) +
vx8ε
W2
T
<p≤ U
S(Ap , p).
(26)
442
Acta Mathematica Sinica, New Series
39
Vol.12 No.4
27
If x 64 −2ε ≤ v ≤ x 44 , then
S(Ap , p) ≥
1
U <p≤(ev) 2
S(Ap , p),
U <p≤W
U <p≤W
v<np≤ev
(n,P(p))=1
S(Ap , p) =
U <p≤W
N (np).
By Lemmas 7, 11 and 5, the explanation in Section 5, we have
U <p≤W
v<np≤ev
(n,P(p))=1
S(Ap , p) =
U <p≤W
U <p≤W
1
1
x2
+ O(x 2 −δ )
np
v
1 1
log
εx 2
p  x2

+O
w

log p p log p
log x

=
k
− 2)
log p
≥ 0.5606
v
p log
1
1
U <p≤v 4
v 4 <p≤W
p
k
1
v4
W 1 + log(
− 2)
1
dt
1
log t
2
+
x
≥ 0.5605x 2
dt
v
1
t log2 t
v4
U
t log t log
t
k4
2 log x−3k 1 + log( k − 2)
1
du
1
u
2
du.
≥ 0.5604x 2
2 +x
1
k
u(k
−
u)
u
k− 2 log x
4
1
1
x2
+ x2
p log p
1 + log(
(27)
39
By Lemmas 6, 7 and 12, we have that for x0.6−ε ≤ v ≤ x 64 −2ε ,
1 1
εx 2
x2
S(Ap , p) =
v + O log x .
vx8ε
T
vx8ε
T p log
<p≤ U
<p≤ U
p
W2
W2
(28)
By the above discussion and the prime number theorem,
27
x0.6−ε <p≤x 44
1
2
≤ x log x
27
44
0.6
1
2tdt
23
( − 4t)
8 27
1
2
−x log x
−x log x
tdt
0.6
0.6
39
64
2−3t
t
4
7
8 −t
tdt
0.6
1
2
≤ 0.024328x log x.
II.
x 44 ≤ v ≤ x 680 .
1
1
1
Let U = vx− 2 +ε , W = v 7 x 14 −ε .
23−32t
8t−4 −1
2
44
27
44
1+
tdt
1
2
437
(k + 1)
(0.6−ε) log x<k≤ 27
44 log x
−0.5603x 2 log x
27
N (p) log p ≤
7t−4
t
4
t− 12
t
− 2)
u
du
u(t − u)
du
u(t − u)
ek <p≤ek+1
log(u − 1)
du
u
du
u2
1 + log(
N (p)
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
By Lemmas 8, 10 and 5, we have
1
S(A, (ev) 2 ) ≤ S(A, U ) −
S(Ap , p)
v
1 1
εx 2
p  x2

log
D
x
+O
w
≤
F
−

log U
log U
log p p log p
log x
U <p≤W


6
53
1
1
log x − k 
x2
2

x
56
7  − 0.5606
F 
≤
1
1
p log p
1
k − log x
k − log x
U <p≤v 4
2
2
k
−
2
1
+
log
1
log p
−x 2
.
v
p log
1
v 4 <p≤W
p
1
2
U <p≤W

log
Thus,
1
N (p) log p ≤ x 2 log x
27
−0.5603x log x
1
−x 2 log x
tdt
437
680
27
44
437
680
27
44
437
680
27
44
437
x 44 <p≤x 680
1
2

t
1
7 + 14
t
4
t− 12

53 6
−
t
t


F 56 7  dt
1
1 
t−
t−
2
2
du
u2
1 + log
t
−2
u
u(t − u)
tdt
t
4
du
53−48t
56t−28 −1 log(u − 1)
2tdt
du
≤ x log x
53 6 1 +
27
u
2
44
− t
56 7
437
680
1
2tdt
1
+x 2 log x
53 6 − 0.016616x 2 log x
137
216
− t
56 7
1
≤ 0.074467x 2 log x.
1
2
437
137
216
1469
III.
x 680 ≤ v ≤ x 2228 .
1
67
22
Let U = vx− 2 +ε , W = v 129 x− 129 −ε .
By Lemmas 9, 10 and 5, we have
437
1469
2228
2tdt
727
62 −
t
1032 129
67
22
1469
129
t− 129
2228
du
1
−0.5603x 2 log x
tdt
437
1
u2
t−
680
2
1
N (p) log p ≤ x 2 log x
1469
x 680 <p≤x 2228
437
680
1
≤ 0.051479x 2 log x.
1469
49
IV.
x 2228 ≤ v ≤ x 72 .
1
77
31
Let U = vx− 2 +ε , W = v 115 x− 115 −ε .
443
444
Acta Mathematica Sinica, New Series
Vol.12 No.4
In the same way as in the case III, we have
1469
49
72
2tdt
38 −
t
920 115
49
72
1
tdt
−0.5603x 2 log x
1
N (p) log p ≤ x 2 log x
49
x 2228 <p≤x 72
1469
2228
557
1469
2228
77
31
115 t− 115
t− 12
du
u2
1
2
≤ 0.071633x log x.
49
31
x 72 ≤ v ≤ x 44 −10ε .
V.
3
1
73
32
8 −2ε
Let U = vx− 2 +ε , W = v 95 x− 95 −ε , D = W xH
.
By Lemmas 9, 10 and 5, we have
1
S(A, (ev) 2 ) ≤ S(A, U ) −
1
≤ S(A, D 3 ) −
S(Ap , p)
U <p≤W
≤
1
2
3F (3)x
−
log D
S(Ap , p)
U <p≤W
U <p≤W
v
1 1
εx 2
p  x2

+O
w

log p p log p
log x

1
2
≤ 409 2x
22 − x
log x − k)
(
760
95
log
1
2
U <p≤W
k
1 − 2)
εx 2
log p
+O
.
v
log x
p log
p
1 + log(
Thus,
49
31
x 72 <p≤x 44 −10ε
31
44
2tdt
409 22 − t
760 95
t
73
32
31
t−
−
2
1
+
log
44
95
95
1
1
u
−x 2 log x
du + O(εx 2 log x)
tdt
49
1
u(t − u)
t− 2
72
1
N (p) log p ≤ x 2 log x
49
72
1
≤ 0.087253x 2 log x.
31
VI.
x 44 −10ε ≤ v ≤ x0.728 .
1
Let U = W = vx− 2 +ε .
In the same way as in the case V, we have
31
x 44 −10ε <p≤x0.728
1
2
N (p) log p ≤ x log x
0.728
31
44
1
1
16t
dt + O(εx 2 log x) ≤ 0.089600x 2 log x.
3
Combining the conclusions in all of the above cases, we can get
1
0.39877x 2 log x. So, (13) is proved and our Theorem follows.
x0.6−ε <p≤x0.728
N (p) log p ≤
Jia Chaohua
The Greatest Prime Factor of the Integers in a Short Interval (IV)
445
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Jia Chaohua
Institute of Mathematics
Academia Sinica
Beijing, 100080
China