### 1-33

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Harrf.v.Ji.JDanuja:J Journal
Vol. i ')
(1~92)
1-:_\3
A Chebychev's Type of Prime Number Theorem in a Short
Inter"'IBl-ll
Shituo Lou and Qi Yao
§ 1. INTRODUCTION.
We shall investigate the number of primes in the interval (z - 1J, z] for
Jl = z8 with 1/2 < 0 ~ 7/12. In [1], we proved
Theorem A. Suppose z be a large numbe:-, then
' II ~ ll"(x)- ll"(z- Jl) ~ 0.991.011J
1ogz
1ogz
with y
===
(1.1)
z 6 , uniformly for
11
20
<0~
7
12'
(1.2)
Denote p{da) the smallest prime factor of tf.a. We write
Let
L an(k) =
nei•
Let the interval J1l
=
(1.4)
"t···41·=·
p(oi;)2:•,I\$i\$•
•er•
{z- y,z] with
zi/2
1
< Jl <
-z
-2
2
Shituo Loti and Qi Y.so
and the parameter
~ ~!l.:d:-yin~
where c is a positive integer that will be chosen later.
Let I;, 1 .'5 j .'5 r, be a set of integers, and I; c;;; [2,z] an.d H be the
"Direct Product" of sets I;, for 1 .'5 j .'5 r, it means dE H if and only if
d= dt···d.. with d; E 1;,1\$ j .'5 r, and dE 111. (1.5)
Suppose 8 be fixed in the inter~ (l/2,l),lf E [z 8 ,z ezp(-(log z)t/ 6 )].
Define the conditions (At) and (A 2 ) as following:
(At). Let It be an integer. H there exist some sets H,., 1 .'5 A: < A:', which
are collections of direct products H's and constants CH such that
(1.6)
then we call H,., 1 \$ A:< k', satisfy (At)·
(A 2 ).HH.,, 1 .'5 A:< A:', satisfy (At), there exists a subset H~, and a function
E~a(H, z) independent of If such that
L 1 = rE,.(H, z) + O(r ezp( -(log z )tf
7
)),
(1.7)
dEH•
uniformly for
z 8 \$y .'5 z exp(-(log z) 116 ),
then we call H~, 1 \$ k < k', satisfy (A 2 ). We call
H~ = H~c \ H~, a 'bad set', for 1 .'5 A:< k'.
In (3], we proved :
H~
a 'good set' and call
THEOREM B. Let z be a sufficient large number, 8 be fo;ed with 1/2 <
8 < l,z 6 .'511 < (1/2)z,JII = (z- y,z],ko be an integer which is dependent
on 8, and z be fixed in (ztl~, zt/ 5 ]. Let H~c, 1 \$ k < k', ruch that (At). If
there e:z:ists a rubset H~ of H, such that (A 2 ), and writing Hk = H,. \ H~,
then we have
~(z)- ~(z- y) = rE(z, z)
+ R(y) + O(r ezp( -(log z )t1 7 )
(1.8)
Primes in short intervals
3
z 9 ~ 11 ~ :e ezp(-(lo_q z) 116 ),
where E(z,z) independent ojJ, and
L {-1)'-tt-1 L
=
R(u)
19<Sio
L
11'(z)-1r{z-tf)=
ca
HeH~
L
I.
(-1)1 - 1 A:- 1 1':1+0(ya:-!).
tSAo:Sio
(1.9)
dEH
(1.10)
.es.
'tlleorem B is a generalization of Theorem 1 of Heath-Brown [2]. Take 1:' =
7, S1, . · · , Sr. as good sets and only S6 ·as a bad set i.e. Hf = St, · · · , H5 ==
Ss,H6 t/J; and Hq
= = t/J,H;; S6; then Theorem B become
Theorem 1 of [2]. We are not limited that the good set S~ or that the bad
set S% should to be whole of S1 . In fact, R(y) is the contribution of all bad
sets. In [1], we also proved:
=
n:
= ···
=
THEOREM C. Suppo6e that 8 ufized in (1/2, 1),1/o = z ezp( -(log z)~).'H1 , 1
k < 1:', satuf'J (At) and (A2 ). If there emu comtant6eq,e~,e'.f. and e~ 8UCh
that
(1.11)
and
(e;o+:)v <
g
L
(-1)A:-lk-1RA:(V)
< (e;-e)y;
1~~
!1.12)
~z
where e i3 a small positive constant. Then
(1-
ef- e~)v < 11'(z ) .- 11'(z - 'V ) < (1 + ei + e~)y
->------::-.._---=.!;.
-'----:-~---"!.;;.=_
log z
log z
(1.13)
uni/01-mly for z 9 \$ 11 \$ 1/o-
We will prove the main theorem of this paper :
Theorem 1. Let y = : 8 ,8 =6/11 + e, then
0.969y
!ryq z
( )
--<'li'X
(
-'li'X-V
)
<1.03ly
--.
log z
(1.14)
Shituo L;,u and QfYao
C si.n~
Theorem. B and Theorem C, we need t<> fl1'.t1 H.., aad H~. In [4] we
gllYe soii\1:! su.'lkient conditions that impiy z~oufe lUnd of "direct product" be
"good set". In § 2, we record those results from (4]. ln § 3, we use those
conditions to prove that Hi which will be defined in ~ 6 below ·be "good
r --~".
A criterion for good sets is extracted. However, the technical work needed to choose good sets and to make the size of the bad sets as small as possible, is precisely the main difference between our method an<! Heath-Brown's.
The new Theorem 1 will enable us to improve the results of Heath-Brown .
and lwaniec [5]. Moreover, we can improve (1.14) further but only at the
cost of much arduous computation.
§ 2 "GOOD SET"
Let eo be a constant that will be defined later on. Let / 0 be an interval
[ao,bo] which contains in [l,z] and 1;(1 ~ j ~ r) be a subset of interval
[a;, b;] contains in [zco, z] also. DenoteD= Io ···I,. be a direct product of 1;.
Let i; =log a;/log z and ij =log b;/log z and let d; = z 9J with i; ~ 8; ~ ij
and 0 ~ j ~ r. For convenience, we write d = {8o, 81 , • • ·, 9,.} E D, and a set
D
= {{8o,8t, · ··,8,.} : 1/2?:: 1-81- ·· ·- 8,. = 8o?:: 81 ~ ···?:: 9,.} .
(2.1)
For short, we denote {8;} = {80 ,81 , · · · ,9,.} .
Let D n 111 be a set of integers, d E D n 111 if and only if d E D and
d E 111 • d = c/ with d, t! E D n 111 means d = ~ ... dr and d' = dQ . .. c( with
· d; : rf; for 0 ~ j ~ r. We shall show the sufficient conditions for D n 111 be
a "good set", i.e. for a fixed z with z 1 / 5 > z = zc, there exists a function
En(z, z), independent of y, which satisfi.es that
L
1 = yEn(~. z)
+ 0(~ exp( -log117 x )),
(2.2)
dEDnl•
where En(z, z) and constant in "0" are uniformly for
z 9 ~ 11 ~ z ezp( - 4(log z)l(loglog z)-f).
We discuss those sequences d: {9o,91 , · · · ,fJ,.} = {fJ;} in D . For such
{8;}, we define a corresponding set 0 of all ofsequences {/Yo, 81. · · · , 8,., 8,.+1, · · ·, (},.+,.,}
(2.3)
By (2.3) and (2.1), we have that if r 1
= 0, then
(2.4)
=
For short, write {8;;}' {8~,81,· .. ,8,.,8,.+1• ... ,8,.+,.1 } = {8;}, and {8;}' E
0. Let 80 = log Xflog z,8; = log X,l(i)flog z(l ~ j ~ r) and 8,.H =
log Z;/1··:- 1:{15 j ~ r1). For each {OQ,8t, · · · ,8,.,8,.+1• · · · ,8,.+,.,}, we define
a product of Dirichlet series :
r
W(-')
,.,
= X(-')flx,Y>(,) Y(-')ll Z;(1)
j=l
(2.5)
i=l
where
L
X(-')=
n-•;
X<n~:rx
Z;(l)
L
=
Cti-',1 Ct I~ 1;
Zi<l\$2Zj
Y(,)=
L
p(t)v,t-•,!vtl~l.
Y<t\$2Y
withY= O(z 6 ),6 be a sufficient small number with 6 <e. Each {8;} ED
corresponds all of W(1,{8;}')'s for which {8;}' E 0. Define that W(D) is
a set of all of such W(-', {8;}'). For short, we write W(1, {8;}') W(1). In
[4], we proved that
=
Theorem A. If D sati..fiel one of foll0111ing conditions
(1) ao ~ zl/2;
(2) all ofW(-') E W(D) nu:h that
Shltuo .Lou and Oi Yao
6
,,r 1 .
1-r ! wr i + 1t): ott-< zlezp( -(log z)!(loglog z;
1
for
1
!-
';
(2.6)
zl-d
Tt-:;T-:; - - ,
!I
where .1. is any fixed positive constant, and
T1 == ezp((log z )l (log log z )-l ).
Then (1.2} holds i.e. D is a good set.
Let 80 , 8t, • •. , 81c be positive numbers. In (4], we discussed the sequence
{80 ,81 , • · · ,9~c} with positive number k such that
9o + 81
+ · · · + 81c = 1
(2.7}
defined a set E(IJ) of some {80 ,01 , · · · ,(h}'s and acutely proved that (4, §
5]).
Theorem B. Let {8;} ED. For each {8;}' E 0 define
W'(a) ==
r
rt
i=l
i=l
X(a)ITxY><,>IT Z;(a)
I/{9;}' E E(O), then
{2T
1
JT I W'(2 +it) I dt <
1 2
z 1 -E.
{2.8)
Moreover, (9.9} holds.
We now describe the set E(O) .
Suppose {a1, a 2, u} or {a 1, a 2, a3, u} be a complementary partial sum (it
means that each 8; belongs one and only one set and their sum in a set be
u or lli) of {8o, 81, · · ·, lh} with u = 9o or u-:; to/2, then
(2.9)
or
{2.10)
PrZJe6 iu short .iute.n-oW
1
Later on, we only detine two o! a1 , !12, c- if {2.9) holds; or define three of
a1.a,,a3 ,e7 if (2.10) hold&.
Let
(J
=6/11 +&,to== 1-9 + e/'2 and z = %c with c = to/10. Define
D == {{90 , 81, · · · ,9.. }: 8, ? · · ·-?: 8,.
Define Di(1
~
i
~
> e,On + 81 + · ·· +8,. == 1}.
{2.11)
7) be the subsets of D and
Di == {{9o, 81, · · · ,89} :Do ~ · · · ~ Og > to/5 and fJ1 + O, + Oa + 84 ~
8to/9};
Di == {{9o,l1t, · · · ,87}: 2to/7 ~ 9o ~ 8t ~···~Or~ to/5}.
Di ={{Do, 9t, · · · ,05}: 2to/5 ~ 9o ~ · · · ~ 85
D4
= {{9o,8t,···,Os}: Oo? 61
~
> to/5; 9a+9•+0s? to};
62 ~ Oa? 1- 20to/ll,to/3? 84 ?
to/4,
6r;? to/5,0o + 81
~
6to/7,82 + Oa? 4- 8to,8o
D5 = {{Oo,Ot,···,85}
~
8s/8 + 3to/8};
: to/2 ? Do ~ Dt ? 82 ~ Os ~ 8• ? 1~ to/10, Do + 81 ~ to/2};
2oto/11, Oa + o. ~ 4- 8to, to/5 ? 8&
D6 = {{Oo,81.···,84}: to/2 ~ 81
~
82
~
Oa? 84 ~ 1- 20to/11,0a+
84? 4- 8to}
D7 =
{{Oo,Ot,···,86}: 2to/5 ~ 8o ~ 81 ~ 82
20to/ll, to/5? 8s ? 86 ~ to/10, 84 + 06 > to/2}
~
83
~
84
~ 1-
and
(2.12)
In § 4, we shall prove :
Theorem 1. Suppose D' be a JJubset of D and
D'nD
= 0,
then D' sc.tisfies (1.2), i.e. D' is a good JJet.
{2.13)
Shituo Lou and Qi Yao
8
In [4], we gave that sot:l~ ~uffi<:i€nt ecnditions which imply that D i& a
good set. In this paper, in § 2, ...-e reeved those conditions from [4]. 'in§ 3,
we use them to show that D \ n- which is defined in (2.11) and (2.12) is
"good".
§ 3. THEOREM 2.
We disCWJB those sequences d
=
= {Oo,Ot,···,Or} {9;} in D. For such
{9i}, we define a corresponding set 0 of all sequences { 9fJ, 81, · · · , Or, Or+ 1, · · · , 9,. +r1 }
with 80::; Oo,
1/2
> 80= 8,.+1
~ · · · ~ 9,.+,. 1 ~ 81 ~ · · · ~ 8,. ~ log zflog
z > 8,.+1 ~ · · · ~ Or+r 1
(3.1)
and
~3 .2)
By (2.11) and (3.2), we have that if r 1 = 0, then
(3.3)
For short, write {8;}'
Write
r = r'
+ r",
with
(3 .4)
=
We now describe the set E(B). Suppose 8 6/ll+e and to= 5/ll+e/2.
We define E(B) be a set which contain all of sequence { 00 , 81, · · · , 0•} with
(2. 7) which satisfies one of following four properties :
(I) There exists at least one complementary partial sum {at,a2,1T} of
{Bo, 81 , · · · , Bk} which satisfies one of following conditions :
( 3.5)
a1
:S t 0 , a2 < 4 - Bt0 (see Lemma 4.4 of (4 ));
(3.6) 1 - 20to/11 > IT > to/5 and to ~ a2 > Sto/9 or
a2 > Bto/9 (see ( 4.1.3) with i = 3 of [4]);
IT
> to/5, a1
~ to and
(3.7) a2 ~to, a, <; : 0 &nJ ~ < t ·- 2\lto/11 (sec (4.6.1) of [4j); ·
(3.8) u
> to/2. (~et> Leman 4.3 of 14]);
(3.9) a1?: to,a2 ~ t0 (11ee (4.1 .1) of[4]);
(3.10) 4t 2:: to,42
(3.11)
to
~
a2
above);
> 4.to/5, and u > to/3 (see (4.1.3) with
> 4t0 /5, and 1 - 20to/11 >
(3.12) 1- 20to/11
u
>
i
= 1 of [4]);
t 0 /3 (see (3.7) and (3.10)
> u and m,. < 41 < M, (see (4.4.4) and (4.6.1});
(3.13) 1/2 ~ 41 ~ to, and u
< 1/2- 8to/9 (see
(3.14) u ~ t 0 /4, a 1 ~to and 42
( 4.5.6) of [4]) ;
> 6t0 /7 (see (4.1.3) with i
= 2 of [4]);
{II) There exists at least one complementary partial sum {4t,a2,4s,u} of
{8 0 , 81, · · · ,81,} which satisfies
(3.15) a1 ~ to, a2 ~ to/2, as ~ to/4 and u
> 2to/7 (see (4.2.1) of [4]);
(3.16) a1 ~ to,42 ~ to/3,as ~ to/3 and u
> 2to/5 (see (4.2.2) of [4]}.
(ill) {80 , fh, · · ·, 8a} satisfies one of the following conditions:
(3.17) ko = 6,u
= 80 ~ to/2,to/5 <Drs~ · · · ~ 81
~
2to/7 (see (4.7.4) of (4]);
(3 .18) There exists at least one complementary partial sum {a 1 ,a2,a3 ,u} of
{80 , fit,·· · , 8.~:} which satisfies u = Oo, a1 < to, a2
(see Lemma 3.10 with to 5/11- £ of [4]).
< 1/3 and
aa
< to/5
=
(3.19) u
= 8o,a1
~ 8to/9,a3 ~ 4to/9,a3 ~ to/4,anda4
= 1-u-at - 42 - 43 ~
t0 /4 (see (4.7.3) of [4]).
(3.20) u = 8o, a1 ~ to/2, a2 ~ to/2, 43 ~ 4to/9, a4 ~ lo/4, and a5
a1 - 42- a3 ~ to/4 (see (4.7.7) of [4]).
(3.21) u
= Bo, 8, < to/5, and u > 3t0 /8 + 81/8 (see ( 4.7.9) of [4]) .
=1 -
u-
10 ··;
a .fixed tT < 1 -:: 20to/11, in !4\ WI! prqvt:d tbat there _exists a
numbers (m,.,M,) with the prope;:hi)J<
For
> fo/5 if tT ~ fo/5;
M, - m,
M,-
Pa!r
of
(3.22)
m.,. < tT if.tT < to/5;
(3.23)
M, ·>to> m,;
(3.24)
and
M,
+ m.,. + tT = 1
(IV) SuppOse {at,a2 1 t1'} or {at,112,113,tT} be a complementary
(3 .25)
BUm
of
{9o, Bt, · · · ,9.,} with
m.,. <IIi
<M.,., (i = 1 or 2),
(3.26)
(See Lemma 4.5 of [2]).
Moreover, for to/3 ~ 9 ~ to/4, we have that
3rr,./2 + 38
< 1.
(3.27)
Applying Theorem A and Theorem B, Theorem 1 follows from
Theorem 2. Suppo&e 8
= 6/11 + £, and D' &Uch that
DnD*
then fore~ {8;}
= 0,
e D', the All of coJTe.9J)Ofiding {8;}' e e contain in E(8).
§ 4. LEMMAS.
Let 8 = 6/11 +£and {8o,8t, · · · ,9,} with (2.7), i.e.
8o + 81
+ · · · + 9., = 1.
In this section, we shall show aome sufficient conditions for {IJo, 8t •.· · · ,8.,} E .
E(8). By the definition of E(8) we check that {0;} satisfies at:leaat one of
conditions (3.5) - (3.19), and (3.25). When 00 > t 0 /2, {8;}' e E by (3.8).
P rimt:.~ in ahcr! .in tenalll
11
'; ;::,n t L ·.:: 61 ~ l/2 and 00 :S t 0 /2, we hA.,~ thei r 1 f. 0 by {3.3) and
a1 :.c Bt,u = 9r+rt and a:~= 1- llt - u, then {B;Y E E by
(3 .13) .
.
fl, +•>. :: io/5, let
Lenuna 4.1. Suppo&e there mat two elements 8' anct 8'' of {80 , 81, · · · ,8~c} ·
with 8' ~ 1 - 20to/11 and 8" < t 0 f5. If there exists a partial sum 8 of
{8o,8t, · · · ,8.\}\{8',8''} such thata <to and8+8' ~ t 0 , then{8o,8t, .. · ,O~c} E
E(O).
Proof. We diaCUBB following three eases :
Case 1. to :S 8 + 8'
< M,,,
Let u = 8'' and a 1 =
and (3.26).
Case 2. s
+ 8' ~
8
+ 8', we have that
{80 ,81 ,· · · ,0~:} E E(8) by (3.24)
M,,.
By (3.25), we have
1-
8 -
8' - 9" :S 1 - M,. - 8'' = m 6 ,
and, by (3.23),
1- 1 -
8' :S 8" + m., < M,, .
Let a1 = 1- & - 8' and u = 8'', if a1 > m 8 ,, then {80 , 8t, · · · ,8,~:} E E(8)
by (3.26). H a1 :S 7716" :S t 0 , then {8o,8t,···,8~:} E E(8) by (3 .7) sinee
a2 = 1 -at -
u :S 7716" :S to and u = 8'' < 1.- 20to/ll.
Lemma 4.2. Suppo&e {a1 ,a2 ,u} be a complementary partial sum of{Oo, 81 , · · ·, 8,~:}
wit.~ a1 =
+ ... + tr,:,az = 81 + ... + fY.,al ~ a2,1- 20to/ll > u =
1 - a1 - a2 > 1/2 - 8to/9 and
8r
maz{8~' ....
then
{8o,8Io·r ,8~}E
,on- maz{8l .... ,oa < to/5;
(4.1)
E(9).
Proof. Ii a1 :S to, then {8o,Bt. · .. ,8~:} E E(8) by (3.7); if a2 ~to, then
{8o,8v · · ,8,~:} E E(8} by (3.6}; ifm.r <at< M.,.,{8o,8t, .. · ,8~:} E E(8) by
(3.26}.
Sbituo Lou and Qi Yao
\2
Now we auppose a1 ~ M,..
By (3.23), we have
8~
+ ... + B'Ll + 8/c =
8~
to
> MtT- S
+ ... + BZ + (8~- BZ)
~
m.,..
H
let a 1
= 8~ + · ·· + oz_ 1+ 8~, then {Bo,Blo · · · ,B~c} E E(O) by (3.25); if
and
8~
+ ... + oz_2 + BLl + 8~ < MtT,
repeating above process, let al = 8~ + ... + oz_2 + 8~-1 + 8~ we also have
{80 ,81 , · · · ,B~c} E E(O). And repeat it again, we have that, in all cases,
{8o,81 ,···,8~c} E E(O) since
~
+ · · · + ~ < to ~
M,..
LEMMA 4.3. ko = 3,u > 2to/5,a2 ~ to/3,a3 ~ to/3,a2
and a1 = 1 - t1 - a2 - a3; then { 8;} E E.
+ a3 < 4- Bto
Proof. H a 1 ~to, {8;} E E by (3.16). H a 1 <to, {B;} E E by (3.5).
§ 5. PROOF OF THEOREM 2.
We will prove that: if {8;} ED\ D*, then {B;} E E. By Theorem 1, it
is enough to show those { B; }' with { Bh + Br+l + · · · + Br+r•, 81, · · ·Or} f/ 'D*
belong to E.
Denote ko be the integer with
L
8; <to
(5.1)
1 \$i\$ko -1
and
L
l\$i:5ko
8; ~to.
(5.2)
By (2.14j, {Bj} e E if9~
.. ith
> to/2. By(~~} , we oniy ;~l;(:d t;:, iliacn111 the caaea
. to/2 ?: .DO·
If t't
(5.3)
=0, then
(5.4)
and
{8;}'
= {8;}.
to ?: 4 and r" + rt > 0, then {B;}E E.
By (5.1) and to?: •~ we have
LeiDDJa fi:l. Silppole
Proof.
B:s
~ ~ < 1-
'··
3
20
t0 •
lL
By Lemma 4.2 and 8r+rt < to/5 if r'' + rt > 0, we have {8;} E E . if
~ r rt. If to = r t rt, Jet Gt + · · · + , .. _.t < to,G2 = 8~ < to ~d
tr '=. 8.. < 1- 20to/ll,
have {8;} E E by (3.7).
<+
we
Lemma 5.2. S"f1110M r'?:
to+ 5, Uaen {8;} E E.
Proof. Let Gt = It + · · · + fJ~ce
then {8;} e E by {3.9). ' ·
LeiDDJa 5.3.
Suppo~e
thGt
> to, G2
= 8~ce+l + · · · 8r• > to and tr = llo,
to> r', then {8;} E E.
Proof. Let tr = Bo, Gt = 61 + · · · + S.O - t and t12
if to = r + rt, then {6;} E E by (3.5). If to
r
Lemma 4.1.
<
= 81ce < to/2 < 4 + r~,
8to
then {9;} E E by
Now may suppose that
ko
~ r' ~
ko + 4.
(5.5)
By (3.4), 9i ?: to/5 for; ~ ko (since ko ~ r'); then by (5.1), we have that
to' ~ · 5. By (5.2) and It ~ t0 , then we have that io ?: 2. Now we may suppose
that
(5.6)
2~ko~5.
We discuss the following cases :
Caae 1.
eo== .3.
By I.emma (.1, we.may auppose tha~ r''.+rt = 0. Then {8,;} ={fie,·· ·,Or}·
By (5.5) we may m_ppose that 5 ·~ r' ~ 9.
Ifr' ~ T,let ax= it+·· ·+B~ot-t < to,GJ = B~ot +8~ot+l +B._u ~ 3to/4 <,.
4 _.8to and IT= Bo, fuen {8;} e E by {3.5).
If r' = 8 let • = Bo,at = Bt + 62 < to; 112 = Bs +B,. < to/2,as = 8, +Be <
to/2,114 = 81 ~ to/4,~~& = 8r ~ to/4, then {B;} e E by (3.20).
If r' = 9, we have
Bt + 8, + Bs + 9,. > Bto/9
·'
since {8;} ;. Di. Let a1 = Bt + 82 +·~a+ '• and v = 85 > to/5, we have that
v < a1 /4 ~ to/4 < 1- 20to/11, fuen {8;} e E by (3.6).
Case 2. ko
= 4.
By Lemma 4.1, we may suppose that r" + r 1
suppose dlat 4 ~ r' ~ 8.
If r' ~ 5, let 111
{I;} e E by (3.5).
r'
= 0 again.
By {5.4), we may
=it + B2 + Ba < to, 111 = 8,. + Ba ~ 2fo/3 < 4 -
8fo, then
= 6.
When Bo
When Oo
> to/2, then {8;} e E by (3.8).
~
to/2, we discuss the following three cases :
{1) Bo + 81 +Be
< to.
=
Let a1
Bo + 85 +Be < to,a2 = Bt + 82 + B:s <to by (5.1) and
v 8,. ~ Ba < 1- 20to/11, then {B;} e E by (3.7).
=
(2) Bo
ko
= 4 and
+ B& + Be ~ to.
If Bi > 2to/7, by Bo ~ to/2, .we have 85 ~ to/4. Let a1 = Bo +Or.+ Be ·~
to,a2 = B:s + 8,. ~ to/2,a3 = 82 > to/4 and IT= 80 > 2to/7, then {8;} E E
by (3.15). IIBt ~ 2to/7,{9;} e E by (3.17).
r
= r' = 7.
15
(5.7)
Uo > 2to/1
siace {8o,···,8T} fl. n;.
Let a1 = 84 + 8& + 8e +81 ~ to,a2 = 82 + 83 ~ to/2,a3 = 81 > to/4 and
u = 9o > ?.to/7, then {8;} E E by (3·:15).
H9 4 +9&+86+87 <to, by Lemma 4.2 and 81 +9d8s+8" ~ to,{8;} E E
if 81- 81 < to/5. H 81-81 ~ to/5, then 81 > 2t0 /5, and, by (5.1) and ko = 4,
i.e. 82 - 81 < t 0 /5. By Lemma 4.2 again, we may suppose that 82 + 9s + 9" +
Dr. ~ to then 9.., + 8s ~ to/2, let a1 = 81 < to/2, a2 = 82 + 8s < to/2, as =
84 + 8s < to/2,a.. = 86 < to/4, and a5 = 81 < to/4, we have that {8;} E E
by (3.20).
r' = 8.
H8t~+86+9r+8s ~ to,leta1 = 8t+82+8s+84 ~to.~= 8r.+86+9r+9a ~
to, and u = 80 , then {8;} e E by (3.9).
H 85 + 86 + 81 +8s <to, by Lemma 4.1, {8;} E E if 81- 8s < to/5. Now
may suppose 81 - 88 ~ to/5, then
83 + · · · + 8s < 1 -(Do+ 81 + 82) ~ 1 - 5(to/5) < 1 - Bto/9;
let a1 = 9" + IJr. + 86 + 81 + 81J ~ to, u
then {8;} E E by (3.6).
= 83 > to/5, a2 = 1 -
a1 -
u > Bto/9
Case 3. ko = 3.
By (5.1) and (5.2), we have that
(5.8)
and
(5.9)
We discu!fs the following cases :
Shituo Lou snd Qi Y110
16
~
Case 3.1. '.1:;
&, -'- fJ; >to.
By (3.2), (:U) Md (J.4), we have that, ifr
and, ifr
~
~
5
4,
+ 84 + 85 ~ 0.4 + to/5 < 1/2.
= 0•+•• < to/5, a1 = 83 + 84 + Bs
Ba
H r" + T1 > 0, let u
a2 = 1- a1 - u, then {8;}' E E by (3.13).
H r" +r 1 = 0, we have that r'
~
E [to, 1/2], and
5 since 83+ 84 + 05 > to and B4
~
03
~ t 0 /2.
H r' = 5, from {Oo, 81. .. ·, Os} ~ n;, then Bo > 2t 0 /5. In this case we have
83 ~ t 0 /3. Let a1 = 83 + (}4 + 85 ~ to, a2 = 82 =~ to/3, a3 = 81 ~ to/3, a11d
u = 1- a1 -- a2- a3 = Oo, then {8;} E E by (3.22).
Hr' > 5, u = 1-
a1-
a2-
a3
> 8o
+ 86 >
2t 0 /5, then {8j} E E by (3.16)
again.
Case 3.2. 83
+ 0• + 85
~
to.
By I.emma 4.2, we only need to discuss the cases with
fh - 8s > to/5.
(5.10)
Oo ~ 81 > 2to/5.
(5.11)
By (3.1), we have that
\Ve discuss the following cases :
Case 3.2.1. 82
< to/3.
By Lemma 4.1, we only need to discuss those cases with r" + r1 = 0. If
r' ~ 5, let a1 = 8o+81 ~ i 0 ,u = 82 < 1- 20to/ll, and u = 83+84 + 8s <to,
then {8;} E E by (3.7).
If r' ~ 6, by Lerruna 4.2, we only need to discuss those cases with (} 1
+
84 +8s ~to since fh +82+83 ~to and 82-85 < to/5. Let a1 = Bo+82 +83 ~
+ 05 ~ t0 ,u = 1-- a 1 - a 2 , then {85} E E ifr' ~ 6 by (3.9).
to,a2 = 81 +84
Primes in short intanals
11
Case 3.2.2. 62 -2: t.:J3 and Sa< J - 20to/ll.
Let
tT
= Be > 2to(ri, as = 82 ~ to/3, a2 = 81 2':: to/3 and
a1 = Oa + · · · + 0.,,
then {0;} E E by (3.16) if 83 + · · · + 8., 2':: t 0 . Now may suppose that
Da + · · · + 8..
< to.
Let tT = 8o,at = 8s + · · · + 8.. , and a2 = Bt
81 + 82 < 4 - Sto. Now may suppose that
+ 8,, then {0;} E
E by (3.5) if
(5.12)
also. Let a1 = 81 + 82 2':: 4- 8to > 4to/5,tT = 84, then {0;} E E if to/3 ~
83 < 1- 20fo/ll by (3.11).
By Lemma 4.1 and 83 < 1 - 20t0 /ll, we may suppose that r" + r 1 = 0.
We discuss the following cases :
(1) r' 2':: 7.
Ifr'? 7,letat == 8t+82+83 2':: to,a2 = 84+0s+IJ6+8r+ .. ·+B.. > 4(to/5)
and tT = Oo 2':: Ot? (Ot + 82 + Oa)/3 2':: t 0 /3, then {8;} E E by (3.10).
(2) r 1 = 6.
If 0s+IJ4 ~ to/2, in (3.20), take at = 81, a2 = 82, a3 = IJs+84, a4
and tT = 8o, then {8;} E E. If IJ3 + 84 > t 0 j2, let
at
a,
and
tT
=
= 8s, as = 86
+ Os + 84 ~ 2 - 4to + to/2 > Sfo/9,
= 8o + o, + Os 2':: 4- Sto - to/5 2':: to,
81
== 06 2':: t 0 /5, then {0;} E E by (3.6).
= 5.
Let tT = Oa, a2 =81 +82 2':: 4-8to > 4t0 /5 and a 1 = 1- tT- a 2 , then {0;} E E
(3) r'
Shituo Lou and Qi Yao
18
by (3.1u) if 83 ~ t 0 /3 and by (3.26~ i~" th ~ 81 > m e,. Now may suppose that
63 < 1.(./3 and
By (3.27), 83 ~ to/4, thus
80
Let u
80
+ 81
= 83 ,a2 =
= 1 - 82 - · · · - 85 ~ 1 - me3 /2 - 383 > me3 •
80
+ 81
and
a1
= 1- rr- a2, then {8;} E E
by (3.26) since
+ 81 < to < Me 3 •
(4) r' ~ 4, let a 1 = 8 1 + 82 <to by (5.3)
4- 8t0 , then {8;} E E by (2.5).
Case 3.2.3. 83 ~ 1-20to/11 and 84
if r"
a2
= 63
+ 84 < 2(1- 20t0 /11) <
< 1- 20to/ll. By Lemma 4.2, {8;}' E E
+ r1 > 0 and
Now we discuss the following cases
(1) r" + r 1
= 0.
We know that r'
~
3.
When r' = 3, let a 1 = 81
then {8;} E E by (3 .5).
When r'
84
= 4, let a1 = 8o+81 <to, a2 =82 + 83 < to, and u =
l - a 1 - a2 =
< 1 ·- 20to/ll, then {Oj} E E by (3.7).
When r'
~
5,
81
Let
fh
+ 82 < to, a2 = 83 < 1/3 < 4- Bt0 , and rr = 80 ,
a1
t 82
+ 82 + 84 > 2(1 -
20to/11)
+ t0 /5 > 8t0 /9.
= 81 + 82 + 84 >
+ 84 < Me.
8t 0 /9 and rr = 85 > to/5, then {0;} E E if
by (3.25). Now we may suppose that
When r' ~ 6, we have that
8o
+ 83 + 86 > 2(1 -
20to/11)
+ t 0 /5 > 8t0 /9
19
=
=
=
a j '" 81 t- 82 +e. > to, 42 == fJo + Ill • 9<) > 8;,jjS 3! 1.d (f Bs > to/5,
(9J t E E by (3.6). Now we dis::uaa tll<'SP <'. <1M'S wi th r'
5. When
0.; + 61 :> 6tc,/7 and 8,. > to/4, let 02 8o -t 9:J, !!" = 84 and a1 1- 02- u,
then {l'j} E E by (3.14). Now we may suppose that
a!- ;.
i.:.Ft
•.!;e;~
=
H fh + 8a < 4- Sto. Since 81 > 2to/5 (see (5.11) above), let u = 81,a2 ==
92 > to/ .J, aa == 8:s > to/3, and at =Bo + 8" + Os if Do+ 84 + Bs > to by (3 .6);
let u = 81. a, == 82 + Ba, and o1 == Bo + 8" + Os if Bo + 84 + Os ::; to by (3.5).
Now may suppose that
82
Let
01
t0 /3
+ Ba ~ 4 - Bto.
== 81 + 82
~ (}4
~ 4- 8to > 4to/5,u == 84, then {0;} E E by (3.11)
< 1 - 20to/11. Now may suppose that
if
e.< lo/3
also. H Oo > Bs/8 + 3to/8, then {81} E E by (3 .18). Thus {0;} '/. D\$ implies
04 < t0 /4, we have that {8;} E E by (3.20).
(2) r'
+ r1 > 0 and 81 + 82 + 84 + · .. + O,.+Tt-1
By (3.4), r' +r1
> 0 implies 8T+T 1
Let u = Be, a2 = 8,. ·i· O,.+Tt and
{9; }' E E by (3.5).
Case 3.2.4. ()._
H 81
and
~
<to.
~!f. By (3.4) B..+OT+T•
01
<
!f+~
< 4-Sto.
= f}r + 82 + 94 + ·· · + OT+"• -1 < to, then
1 - 20t 0 /11 .
> t 0 /2, by (3.1), we have that
r+"t
80 +_Eo;= 1.
i=1
Sb.ituo Lou and. Qi Yao
20
Thos
and
2
8:J+ 84 < (1- to)< 4- 8to.
3
By Lemma 4.3, we have that then {8;}' E E if there exists a partial sUm. of
{80,lit. 112 ,116 , • • • 11r+r1 } belong to (2to/5, to/2]. Now we only need to diScuss
those caaes with 9'o ~ 2to/5. By (3.1), we have that
8, + 8r+l
+ · ·· + 6r+r
1
~ 1 - 20fo/ll + 81 -
6Q > 2to/5.
Then there exists a partial sum of {fJ,, 8r+l, · · ·, 9r+r 1 } belong to (2to/5, to/2]
since 82 < to/5 and 8r+l < to/10 = to/2- 2to/5. Thus {8;}' E E in this
case.
Now we discuss the case : fJ1 ~ to/2.
When 83 + 84 ~ 4 - Sto,
Since we only need to cfu.cuss those {8;}' which corresponding {6;} ~ n;,
we may suppose r ~ 5. Thus
Since we only need to discuss those {fJ;}' which corresponding {8;} r;. D6,
we may suppose r ~ 6. Let at = 83 + (}4 + 8s + (}6 ~ 4- 8to + to/5 > to, a2 =
82 > to/3, a3 = 8, > to/3, tr = 1 - a 1 - a 2 - a 3 ~ 81 > 2t0/5, then {fJ;} E E_
by (3.24).
When 83
+ 84 < 4- 8t0 , by Lemma 4.3, we only need to discuss
cases with
8, ~ fJ1 ~ 2to/5,
9'o ~ 2t0 /5,
and
8[, + 8r+l
since
Br+l
~ to/10
= to/2 -
+ · · · + 8r+r
2to/5. Thus
1
~ 2to/5
those
Primel
jn
short iuterval6
and r ~ 6 diner. 63
+ f!4
21
+Or, <to. By (3.1),
8s = 1- 8~- · · ·- Ot --813-- · · ·- Br+rt
< 1- 5(1- 2<Jto/ll) ·· t:./10 < tc/5.
Since we only need to discuss those {0;}' which corresponding {19;} fl. D6,
we may ~uppose that 8-. + 86 < to/2. By Lemma 4.3 again, we have that
then {8;} E E.
Case 4.
ko
= 2.
We have 81 2:::: (ih + 82)/2 > to/2.
H 00> to/2, then {8;} E E by (3.4).
Now we may suppose that 80 ~ to/2, thus by 80 < 81 , we have that
> 0. Let at= 81 + 82,u = Or+r1 and a2 = 1- at- u, then {8;}' E E by
(3.13) if 81 + 82 < 1/2. Now may suppose that
TI
By (3.2), we have that 82 < 1/3. Let u = 8o, a2 = 82 and a1 = 81 + 83
· ·· + Br+rn then {8,;}' E E by {3.5) if a1 <to. Now may suppose that
+
(5.14)
H 83
< 1 - 20to/11, by Lemma 4.1, it is sufficient to discuss that case with
(5.15)
Let a1 = 81 + 83 + · · · + 8•+rt-l, a2
we have that {8;}' E E.
= 82, a3 = O.. +•t
and u
= Oo,
by (3.18)
:::0: l-20t0 /ll, by (3.2), 81 +83 ~ 1/2. By (3.19), {8;}' E E if Bt +0 3
Now may Sl!ppose that 81 + 83 < t 0. H r ~ 4,
If83
84
:::0: t0 .
< 1-(8b+8r+I+ .. ·+Br+rt)-Ot-82-8a
< 1- 3/4- (1- 20t 0 /11) < 1- 20to/ll.
By (5.14) and Lemma 4.2, we only need to discuss those case with (5.15)
again. Let a1 = 81 + 83 + · · · + O•+•t-1, a2 = 82, a3 = Or+r 1 and u
(3.18) we have that {0;}' E E again.
= 8o, by
Shituo Lou and Qi Y.ao
22
Theorem is complete u!!ei\li_;;.
§ 6. THE PROOF OF
TH.F~O.BJo:M 1.
Take c = to/10.
Define a direct product I = { Bo, · · · , 8.} be
I= {d: d
= do .. · d,., d, E I; ,I; =
[z 8', 2r 11') and p(di) 2': zcl 5 i 5 r }.
(6.1)
For I, we define, for 1 ~ k 5 r,
J(k) = {d : d
=do .. · d~cPio+l .. · Pr, doE I;(l 5 i 5 k),pi
E Ji, (k+l 5 j ~ r) and dE J}.
(6.2)
In this section we will choose H~ and to make it as large as possible.
First we write S~ to be a swn of some disjoint direct product of I which
is defined in (6.1). Set
H~, l = UmD=0{J} .
By Theorem 3,
Il~,l
satisfies (2.2) write
HZ,l
= s:. \ Hl.,l·
Thus
n~.l ~
n·.
We want to choose some "good set" from D*. We write
Di,t = {T(9) : J E Di}.
Then we have
D~ = Di',1
u (D~ \
nn,
where D~ is a collection of direct product J's with I= 10
...
I" r 2': 10,
l={Bo , · ·· ,Br},
where (Bo,·· · ,B8 ) satisfy the conditions same as in Di,IJ9
+ ··· + 17. E
(to / 5, lls), and dE I if and only if d = do··. d 8 pg . · · p., d; E 1i and Pi E Ii
and
D~ = D~ \ Di.
Primes in 1hort intervals
23
By Theorem 3, we have tha.t I in D~ aatisfies (2.2).
By the method tb.a~ we wlll use i.n. § 7 (see (7.4)), we have
I
nr i-= 0 Co:2z).
Thus we can replace Di by Di,1 in
Di
H;:.
Repeat it again, we might change
to
Dt = {d: d
= doPJ.···P9 E Di};
For Dj(l :5 i :5 2, orr :5 i :5 7), we can change Di to D;* as well. We only
need change 9 to 8 (i = 2),5(4 :5 i :5 5),4(i = 6} or 6(i = 7). For n;, we
only can change it to
D3* U D7*·
Now take
H~0
=Di,Hs = D2,H6 = D3 U D4 U Dij U D7,H~ = D~ and H~ = n:;.
Otherwise
In (1.9), take
CH
== 1, then
1r(z)-1r(z-y) = yE(z,z)-
L
(/(i)-1)!
1:9\$5
L
Ll+4!
DEDidED
L
2)+(6!-5!)
DED6dED
where /(1) = 10,/(2) = 8, and /(3) = /(4) = /(5) = 6.
Suppose e;(l S: i :5 7) and e0 be constants which satisfy :
(/(i)- 1)!
L
_L) :5 (/(i)- 1)! l;iYZ, 1 :5 i :5 5;
g
DEDidED
"'1 < er,y ·
4! )
L...J L.;
D<'D'dED
-
6
-
lo z'
g
L 2:)DED7dED
Shituo Lou and Qi Yao
24
wd
By (1.13), we have that
(1-eo)V <r(z)-r(z-v)< (l+eo)v .
log z
log z
We now estimate e,(l :5 i :5 7) . First, we estimate e3 . We have dE
implies d
= doPt· · · P5
D3
with do 2": Pt 2": • • • 2": 115, z - 11 :5 doPt · · · P5 :5
z,p(do) ;: :-: zto/10, and
Define
Then
I D*J<
s 271
:5 log z
j
"
L
n. ""P3P4 ""
(PI , ..·,P\$)EA 3 r •r~
f
ro
2tf
log
v
PIP2P3P4P5
· · · ]A, t1t2tat4ts(l- t1- t2- t3- t4- ls)'
(6.2)
where
a, =
1- ~
2to 1 - 2t 1
1 - 3t2
:5 t1 :5
5 , -4- :5 h :5 t1, - 3- :5 t3 :5 t2,
5
1- 4t3
- :5 t4 :5 t3, 1 - 5t, :5 t 5 :5 t 4}.
2
Estimate the integration of right hand-side of (6.2) (see Appendix), we have
{(t1, .. · ,ts): -
5!!D*I<
e3y
3 - log z'
'
whe.re
f3
Define :
< 0.00625 .
Prime6 in lhort iaterral.l
(1) H~- ~)~
25
th ~ .. . ~e.,~~ ;
(2) 1 - 81 - · · · - 8e ~
tb;
A 2 is a set of (t1. · · ·, t 1 } with the following conditions:
(1) ~to ~ 8t ~ ... ~ 8, ~
!f;
(2) 64 ~ · · · ~ Br ~ ~;
(3) 1 :_ 81 - ···-Or~
j;
A 4 is a set of (t 1 , • • • , t 5 ) with the following conditions :
(1) ~to ~ t1 ~ 2- 4to;
(2) t 1
~
t,
~
(3) t, ~ t:s ~
2 - 4to;
tnal:
{1-
w.•- 8to- t2};
(4) min {1f, -\'1- 2tt- t:a- t:s} ~ t4 ~ nuu: {!f,! (1- ~- t:a- ts)}
. (5) min {t 4 , 1- 2t1
-
tz- ts - t 4 } ~ t& ~ ma:.: { 1 - ~ - t2 - t 3
-
t 4 };
(6) ft \$ t5/8- 3to/8;
A 5 is a set of( t 1 , • • • , t 5 ) with the following conditions :
(1) 1- J(4- 8to)- ~ ~ t1 ~ 2- 4to;
(2) min{l- 4 + 8to- ~- 2ft,tt} ~ t 2 ~ 2- 4to;
fYi- 2ft- h,t2} ~ t3 ~ 2- 4to;
min {1-1t- 2tt- t, - t 3,t3} ~ t 4 ~ maz {4- 8to- t3, 1- ~};
(3) min {1- .(1- ~)(4)
(5) min {1- 2tt- t,- t:s- t., '} ~ t 5 ~ ~;
A 11 is a set of (t1 , • • ·, t 4 ) with the following conditions :
Sbit~to
(1) ~ ?' 6l
Lou and Qi Yao
> 2 - 4l{.;
(2) min{Bt, 1- {4- 8to)- 281} ~ 82 ~ 2- 4to;
(3) min{IJ2,1-
(1- ~)- 281- 82} ~ 83 ~ 2- 4tc,;
(4) min{83, 1-281-82- 83} ~ 84 ~ maz {1- ~,4 '- 8to- 83};
t..1 is a set of (t1 , ···,tel) with the following conditions :
(1) ,. ~ tl ~ t2 ~ t3 ~ t4 ~ 1 - ~;
(2) min {1 - .,. - 2tt - t2 - t3, '} ~
ts
~
., -
(3) min{1 - 2tt - t2 - t3 - f4 - t5, t5} ~ tel ~
t4;
-\l- t.,;
With same reason we have
e,<(/(i)-l)!( 2)
I
where,.= r(i), r(1) =9, r(2)
ha-,e that (see Appendix)
···
t
h;
dtt···dt,.
tt···t,.(1-tt-···-t,.)'
= 1, r(3) = r(5) = r(6) = 5 and r(7) = 6. We
e1
:S 2 (In
H~-'i~)) g (10) < 1.1(10)-:- 5 ;
e2
< 2 ((In
l) 7 (8) +(Inn· (In ~)3 3![!,,) < 0.002.
e4
< 0.00276;
e5
< 0.007817;
·ecs < 0.01351
and
e7
< 0.0002071.
In Theorem C, take
5
er =e; = Ee• :S o.o1n.
i=l
and
e~ ::.::
e2
=e6 + er < 0.01372 .
then we have that.
0.969lf
- < 11' ( 2: ) -11' ( z
log z
Theorem 3 follows .
-11
)
1.031y
<-.
log z
S!>i~l1!1
26
Lou llll.d Qi Yao
REFERENCES
(l] Lou, S. and Qi Yao, A Cheb71chev's Type of Prirr:e Number Theorem
m a Short Interval, -! (to appear).
(2] Heath-Brown, D.R. The Number of Primes in a Short Interval, J.
Reine Angew, Math 389 (1988) 22-63.
(3) Lou, S. and Qi Yao, The Number of Primes in a Short Interval, HardyRamanujan J., vol. 16 (1993), (to appear).
(4) Lou, S. and Qi Yao, Estimate of sums of Dirichlet's series, HardyRamanujan J., vol. 17 (1994), (to appear) .
[5] Heath-Brown, D.R., and Iwaniec, H., On the difference between con-
secutive primes, Invent. Math. 55 (1979), 49-69.
Lou, S. and Qi Yao
Department of Math.
Dalhousie University
Halifax, N.S. B3H 3J5
APPENDIX
Estimate of e1 :
Define
Then
Estimate of e2 :
Define
Then
e~
~ 2(7!)
J ··· ft.. ,, ...,,~!.:;:~~ ..
tr)
< 7!(2) ((In .) fy + if.h (In ~) (In
7
Estimate of e3
Define
:
4
!) 3 )
< 2.6(10)- 4 .
Shituo Lou and Qi 1'.w
30
Then
e3
~
~
~
~
~
~
~
~
Estimation of e4.
We have, for {9;} E
D4
Thus
and
9,.
+ 9s
4
- 5
>.
-
6t 0
-
5
- .
Moreover
80 < 1 - 2 ( -2to - -3to) - 3(2 - 4to) :::: -66to - -29 < -1
5
5
5
5
5'
and
1 ( 1-2 (2- -3t-0 ) - 2(2- 4to) ) <
23to
19 < 21.
81 <- - 10
- 2
5
5
5
llO
Primee in short intervals
31
'then
Estimat'~n
')f "~
to/5? Bs ? to/10,8o
+ 85
\$ to/2}.
We have that
Bo
< 5/22- Bs/2 < 9/44,
and
1/22 < 85 < 5/11-280
s 5/11-281.
Thus
es \$ (5!)(2)(5.5)
j ... ~. dtt ... tsdts
} 6..
t1 · · ·
where
as
=
{(tJ, ... ,ts) : fr 5 h 5
;fi, fi s t2 s min { Pz -
2tt, t!}'
fr S ts S min {t2, ~~- 2tl- t2} ,max {A- t3, 1 2~} 5 t4 5 t3,
f2 S ts S min { fr -- 2tt, fr }}.
2
Shituo Lou and Qi Yao
32
We have
es
:s:
:s:
:s:
:s:
. { 13 2 11 11 }
44
Jmm
t!)dt
fr f i - ' dt 2 rfr dt 3 rfir ln(l0t, totJf<
4
9
(5!)(2)(5.5)J:rr dt
fr
1
I.
Jm.in{
1320 JIT
2 dt1
2
IT
I.
3
1320(5.5) f~r dh
1320(5.5) 3 [ ] ;
a
-2!1 ,!1}
dt2
1!2 tTT
1
(
(5 .5 ) ( t3 2
TT
123
55
( ;;
(tt --
trl + ~ (tl- f-r)
I{t (565 (!2 - 2tt)
:s:
4 + ll (..L) 5) + ~
219625[ 56·5 (..!...
12 (..L)
66
5 66
2
+
5.5
6
'S
0.007817.
3
+ ~ (~- 2t1)
(..!...
(..!...)4 + ii
(..L)5) + ~
(..!...
24 66
10 66
2
18
We estimate e6 now.
We have
where
-
· · ·-
22)
+ Inli
~ 9- 44tt
I
rYnin{ a-2tl ,t}} ( 5 5
2 2
22
2 )
Jfr
;-- (t2- rd
+ (In 21
) (t2- rr)
+
Then 1- 01
2 )
-11
84 = Bo 2: 1/5, and
2
)
2
)
(9- 44ti)dtl
(9- 44tt)dttJ
3 + 11 (..!...1 4)
(.!.9 (..!...)
66
3 66/
(..!...)3
66
+ ii8 r\66
LJ4)l
)
dt3
(9- 44tt)dt2
Primes in short intervals
33
Thus
f6
~
240ff,; dt Imin{f,-2t,,ti} dt
H
1
H
1
li-.!J-
2
*-i-
2
~
240ff,; dt Jmin{f,- 2t,,tt} dt
~
240(5 5) ff, dt
~
It'fr - 1-t ((Inill;:_)
+Inll) dt ·
2
21
3
!} dt It, +Inll) dt
2 f. t
t3
2
21
3
t 1t 2 3
fmin{f,- 211 ' 1
1
1113
2
- (
H 1 if-!j.
1t2
1 (5.5(t 2 - IT2)2 + (In 22)(t2- n2))dt2
240 (5.5 )J11ob dt1 fmin{f,-2t,,tJ}
, 7 .,
tT
2
1
21
12
·
-
~
I fr1' dt 3 f:fir3 ~
t,t,tat•
.k..
88.
1320Jf dt1
33
u-T
7
IJJ~~~
66
3
-n
-fs dt1 ffr--2tt
+1320 f ll
ll_!l.
88
66
•
('/-!t) ( 525 (t2- fir+ (Inm (t211
3
tl
1
(5.5
2 )2
(!L!i.)
2 (t2- IT
66
-3"
~ 132 o(ll)(l2!1)f-Ars.s
_ 2t 1 )3
5
12
l! T (.1.
11
+! (-fr- 2t1)
2
_ s.5 ( s _ !.1.)3
66
6
3
(InH)- ~ (~ -1-)Jdtl
* (5-5 (t -
2.)3- 5.5
1320(178)(!!!!)J*
~
31944
+
~ (lk) 3 - ~ (is) 3 ] < 0.01351.
7
2 )) dt2
-IT
In 22
+
37
+ (In22)
21 (t2
fr.)) dt2
6
1
11
6
(.1.!l.)3 + (Inll)(tJ--f;-1'- ~ (.1.66
3
21
2
2
66
(~-~ (-Ja)4- ¥ (~) (rk)4 + f2 (InH) (-1J)3- !::f- (rk)3)
(..l.)4- 24
5.5 (..1.)4
+ 16.5 ( 1 )4- 16.5 ( 1 )4 + ~ (..l.)3- ~ (..l.)3
+ 29937(5.5
24 33
88
24
198
24 ~
6
33
6
88
Finally, we estimate e 7 • We have
f7 ~
121
(5!)(5)(2)(-)
21
I 1
. ..
47
dt1 ... dt6
t1 · · · l6
---
where
f17 = {{t1, · · · ,ta): 2/11 2: t1 2: · .. 2: t4 2: 21/121,1/11 2: ts 2: t 6 2: 1/22}.
Thus
e7
1
22
~ 41(In )4 (1- In2)
21
< 0.0002071.
EDITORIAL NOTE. Editors came to know (by private communication)
that Theorem A of this paper wa.s proved independently by Professor D.R.
HEATH-BROWN long ago. He had a lot of unpublished material dating
back to 1983 regarding Theorem A.
!.1.)2)
dt
3
I
```