Annals of Mathematics
An Elementary Proof of the Prime-Number Theorem
Author(s): Atle Selberg
Source: Annals of Mathematics, Second Series, Vol. 50, No. 2 (Apr., 1949), pp. 305-313
Published by: Annals of Mathematics
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ANNALS
OF MATHEMATICS
Vol. 50, No. 2, April,1949
AN ELEMENTARY PROOF OF THE PRIME-NUMBER
THEOREM
ATLE SELBERG
(Received October14, 1948)
1. Introduction
theorem,
whichis
In thispaperwillbe givena newproofoftheprime-number
in thesensethatit usespractically
no analysis,exceptthesimplest
elementary
properties
ofthelogarithm.
in theform
theorem
We shallprovetheprime-number
.lim =1(x) 1
(1.1)
Z__00
X
as usualby
whereforx > 0, tQ(x)is defined
#(X) =
(1.2)
logp,
E
theprimes.
p denoting
formula(2.8),
The basic new thingin the proofis a certainassymptotic
whichmaybe written
(1.3)
t(x) logx +
logpa
()
=2x logx + O(x).
thereare severalwaysto deducetheprime-number
theorem.
Fromthisformula
The wayI present??2-4ofthispaper,is chosenbecauseit seemsat thepresent
it has
way.1 But forcompleteness
to be themostdirectand mostelementary
thatthiswas not myfirstproof. The originalproofwas in
to be mentioned
and made use of the following
resultby P. Erdos,that
factratherdifferent,
foran arbitrary,
positivefixednumber6, thereexista K(a) > 0 and an xo = x(a)
suchthatforx > xo, therearemorethan
K(a) Xlog x
fromx to x + Ax.
primesin theinterval
thenotations
My first
proofthenranas follows:Introducing
lim.
t()
- x
= a,
fim,()
x;
-A
result
onecan easilydeducefrom(1.3),usingthewell-known
(1.4)
E logP =
,<?
X
logx + 0(1),
Because it avoids the conceptof lowerand upperlimit. It is in fact easy to modify
the proofin a fewplaces so as to avoid the conceptof limit at all, of course (1.1) would
thenhave to be stated differently.
305
1
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306
ATLE SELBERG
that
a+ A = 2.
(1.5)
Next,takinga largex, with
ax + o(x),
(x)
form
one can deducefrom(1.3) in themodified
()
(d(x) -ax) logx + E logp (
(1.6)
-A
x= 0(),
that,fora fixedpositivenumber6,onehas
> (A _ )x
s6(p
(1.7)
setofprimes? x with
exceptforan exceptional
log P
X).
O(lo_
Alsoone easilydeducesthatthereexistsan x' in therange-V1< x' < x, wit
6 (x') = Ax' + o(x').
andx' insteadofx,onededucesthe
Againfrom(1.6) witha andA interchanged,
(1.8)
(J;)< (a + 6)
*
-,
set ofprimes? x' with
exceptforan exceptional
a
lop
FromErd6s'resultitis thenpossibletoshowthatonecanchoseprimesp and p'
sets,with
to anyoftheexceptional
notbelonging
x<
P
x/ <
P
(l +
a) x.
P
Thenwe getfrom(1.7) and (1.8) that
(A-
O
<
<
(mu) < (a+6)
X
< (a+
so that
A
< (a + 6)(1 + 6).
or making6 tendto zero
A
a.
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X
307
THE PRIME-NUMBER THEOREM
Hencesincealso A ? a and a + A = 2 we havea = A = 1, whichprovesour
theorem.
Erd6s'resultwas obtainedwithoutknowledge
of mywork,exceptthatit is
based on myformula(2.8); and afterI had theotherpartsoftheaboveproof.
His proofcontainsideas relatedto thosein the above proof,at whichrelated
ideashe had arrivedindependently.
The methodcan be appliedalso to moregeneralproblems.For instance
one can provesome theorems
provedby analyticalmeansby Beurling,but
theresultsare notquiteas sharpas Beurlings.2Alsoone can provetheprimeforarithmetic
onehas thentouse inadditionideas
number
theorem
progressions,
and resultsfrommypreviouspaperon Dirichlets
theorem.
Ofknownresultswe use frequently
besides(1.4) also itsconsequence
t (x) = 0(x).
Throughoutthe paper p, q and r denoteprimenumbers.A(n) denotes
M6bius'number-theoretic
function,
r(n) denotesthe numberof divisorsof n.
and K to denoteabsolute
The letterc willbe usedto denoteabsoluteconstants,
are not carriedout
positiveconstants.Some of the moretrivialestimations
butleftto thereader.
(1.9)
2. Proofofthebasicformulas
whenx is a positivenumber
andd a positiveinteger,
We write,
(2.1)
Xd =
kd.. =
jut(d)
log d'
and ifn is a positiveinteger,
(2.2)
On=
=n,= =
Ed/n
Xd
Thenwe have
flog2
x, forn=1,
(2.3)
(2*3)
On
-
Jlog
forn== pa, aa
p logX2/p,forn
)2logplogq,
paql,
LO,forall othern.
>1
>
>
1,
followreadilyfrom(2.2) and (2.1), the
The firstthreeof thesestatements
n squareis easilyprovedby induction.Clearlyit is enoughto consider
fourth
free,thenifn = PlP2 * Pk ,
On,,
=
On/rPi,z -
On1Pkl/Pk
Fromthistheremaining
partof(2.3) follows.
2 A. BEURLING: Analysede la loi asymptotique
de la distribution
des nombrespremiers
gneralisks,Acta Math., vol. 68, pp. 255-291(1937).
3 These Annalsthisissue,pp. 237-304.
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308
ATLE
SELBERG
Now considerthe expression
(2.4)
zn!
On
=
n<z din
F1=x
d
XdXd
d
d<z
=X E A(d) og2 Z + o
d
d
d<z
E
I'd
dx
+
d
OZ
[
u2 )=xE(d)
d
\dx
]
d_
d
j
oz
Xd)
d
X
d
lo2
+
O(X).
This on the otherhand is equal to, by (2.3),
2
ORn-log2x
n<x
+
log p log -
E
Pa<X
P
+ 2
log p logq=
VcaqB?x
v<q
+ E logplogq+O
pq?z
(2.5)
+0(
Z
p<:
E
paq<
a>1
x>l
+ log2x =
log2p++
<X
p log)
(1log
\p~z
og2x) +O(
jpa <
log2p
P1
logplogq)
z
Vpq~z
logp logq + O(x).
The remainderterm being obtained by use of (1.4) and (1.9).
(2.4) and (2.5),
(2.6)
(2.6)
E
log2p + E
pq<_
Hence from
logp logq = x 2L?(dQ)log2x + O(x).
d
dz
d
It remainsnow to estimatethe sum on the right-hand-side.To this purpose
we need the formulas
(2.7)
=
log z + c1 + 0(Z-*),
and
(2.7)
og2 Z +
=
c2log z + C3+ O(Z )
where the c's are absolute constants,(2.7) is well known,and (2.7') may be
easily derivedby partial summationfromthe well-knownresult
>T(v)
=
zlogz+c4z+0(A/).
From (2.7) and (2.7') we get
log2z = 2
2
p:z X
(G) +
V
c5
I+
w'VXV
C6+ O(z1).
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309
THEOREM
PRIME-NUMBER
THE
By takingherez = x/d,we get
d?xd
1,,ud
2X
d
2 E jAxd),
=
d
d<z
+ c6
r(v) +C5
Z ,8(d)
+
d
O(x
d~x
+ c6 E
1d)
dv
dax
n
=2
+
Z dt4)
d
dz_
2Z
=
/(d)T(v)
dv
dv-z<
) + 0(1)
+ 0(1) = 2 j
(d)
+ c5+ 0(1) = 2 logx + 0(1).
-
,(d) r(n/d) = 1,and thewell-known d
Weused herethatEd/n
Now (2.6) yields
E log2p + E
(2.8)
v~xld V
+ c5 Z
r
d
d
d~z
C6 E
E
iAg(d)
d~z
V
a xld
(,u(d))/d= 0(1).
logp logq = 2x logx + O(x).
pqz
pz
<z
in theformgivenin theintroduction
mayalso be written
Thisformula
Z logp 6
t6(x)logx +
(2.9)
that
by noticing
Z
vZ
= 2x logx + O(x),
-
log2p = t(x) log x + O(x).
we getfrom(2.8)
By partialsummation
E logp logq
VQ_: logpq
E log p+
V:5.x
(2.10)
2x + 0 (x)
logx
Thisgives
E
Pq_$
logp logq =
-E
pE~z:
I
Pz
logp E
q~zIp
logp E
P
0 E
q gllogrg
log qr
qr~z/p
=
z
logq = 2x P-z logP
2x logx - E log qlogr
logqr
qr~z
log
dp (1+log-))
V
(f)
\qr/
+ O(x loglogx).
thisforthesecondtermin (2.8) we get
Inserting
(2.11)
t,(x)logx = E log p log q 6
V.<
log pq
(
-
)
\pq/
+ O(x log logx).
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\
310
ATLE
SELBERG
now
Writing
6(x) = x + R(x)
,
-Z
R(x) logx =
(2.12)
(2.9) easily gives
logp R (
posx
p
+ O(x),
and (2.11) yieldsin thesamemanner
lo p
log qR
+ O(x log log x),
,~log pq
\pq/
R(x) logx
(2.13)
since
lo1glo1gq =~logx + O(logO
logx),
x pq logpq
from
by partialsummation
whichfollows
logP logq =
E
x + O(logx),
log2
pq
vq <
whichagainfollowseasilyfrom(1.4).
The (2.12) and (2.13) yield
2j
R(x)IlogzxS E logp RI-)
E logp logq |R (X)
logpq
\pq
Fromthis,by partialsummation,
+
I
2j R(x)logx
<
pE
n
vq<
logp+
{
E
R
log Pq
-
}
R)
+ }+O(x
orby (2.10)
2
logz
R(x)I
2
2n{R(n)
+0
1
logn
=2
IR )
-
)
(n
+ O(x log logx).
l)|
+- O(x loglogx)
+ (Xloglog x),
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log logx),
311
THEOREM
PRIME-NUMBER
THE
or
(2.14)
IR(x)I
< log1X n?Ei| R(?)
\l
+0(
loglogXx)
\logx
/
whichis the resultwe will use in the following.4
3. Some propertiesof R(x)
From (1.4) we get by partial summationthat
E
n~x
or
6(n) = log x + 0(1),
fl2
z R(n)
0(1).
This means thereexistsan absolute positiveconstantK1, so that forall x > 4
and x' > x,
(3.1)
()
E
< K1.
Accordinglywe have, if R(n) does not change its sign betweenx and x', that
thereis a y in the intervalx < y ? x', so that
<
(3.2)
K221.
19xi
This is easily seen to hold trueifR(n) changesthe signalso.'
Thus for an arbitraryfixedpositive a < 1 and x > 4, there will exist a y
y 5_ e K2/6 x, with
in the intervalx
(R(y) <6y.
(3.3)
From (2.10) we see that fory < y',
0 <
log p ? 2(y' - y) + ?
E
Y
fromwhichfollowsthat
R(y')
-
R(y)
y'
-
y +
o (I'Y,)
4 Apparentlywe have herelost somethingin the orderof the remainder-term
compared
to (2.8). Actuallywe could insteadof (2.14) have used the inequality
I5
IR(x)
1
2
1:
Lon
IR (n
+
? (lga
whichcan be provedin a similarway.
5 Because therewill then be a I R(y) I < log y.
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312
ATLE
SELBERG
Hence, if y/2 ? y' < 2y,y > 4,
y' - y +
R(y') -R(y)
o
y
or
IR(y') I _ R(y)1+ I y'Now consider an interval (x,
in this intervalwith
y) + O (l'y,).
x), according to (3.3) there exists a y
eK2I
I R(y) I < by.
Thus forany y' in the intervaly/2 ? y' < 2 y, we have
I R(y')
I _ by+ IY
R~~~y')
g
-
y I + K3l '
x'
or
R(y')
< 26 + 1
Hence ifx > e "'/ and e~812) ? y'/yS
I + l3
we get
R(y') 2+(e812-1+_45
+ 3 < 43.
I~ < 25 + (e21)
Thus forx > e K/8 the interval (x, e K2/ZX) will always contain a sub-interval
such that JR(z) J< 48zif z belongs to this sub-interval.
(yi, e8/2y1),
4. Proofof the prime-number
theorem
We are now goingto prove the
THEOREM.
limx
Zx-*oo
X
= 1.
Obviouslythis is equivalentto
lim =()
(4.1)
_-*oO
We knowthat forx > 1,
(4.2)
x
jR(x) I <
0.
K4x.
Now assume that forsome positivenumbera < 8,
(4.3)
j R(x) I < ax,
holds for all x > x0. Taking 5
=
a/8, we have accordingto the preceding
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THE
313
THEOREM
PRIME-NUMBER
section (since we may assume that xo > eK3I8),that all intervals of the type
(x, eK2/8X)withx > xo, containan interval(y, e812y) such that
| R(z) I
(4.4)
< az/2,
fory < z ? e6/2y.
The inequality(2.14) thengives,using (4.2),
+ ? (+Ixc
R (n
iR(x) I <! lo x E
<K4-X '<
z
N 0 oxx
log1x (Ixo)<n~zxn
0o)n
log1x n?(z
writingnow p =
using (4.3)
eK2/8,we get further,
R~)I<ax
i R(x) < log X
n?(zlxo)
y,<n!zy~e(8-2)
p-<yr~v
a2
>
n
2 log x
n
\/logx)X
and (4.4),
(\/log x)
__
(
(zxz0)/logp)
?<?<(log
ax-
=
2 log x ifv (log(xzlo)logp)2
4 log p
x
VAnalog
forx > xi.
x
ax
l
n
a
Rx\
~~x+ 0(
?
256K2)
(
+
V
Jx<
log x)
<
a(13of~)
300K )2
t(
Since the iteration-process
an+1 = an
(
11 -
2
_
obviouslyconvergesto zero if we startforinstancewitha, = 4 (one sees easily
that thenan < K5/V/n),thisproves (4.1) and thus our theorem.
FINAL REMARK. As one sees we have actually never used the full force of
(2.8) in the proof,we could just as well have used it with the remainderterm
o(x log x) instead of O(x). It is not necessaryto use the full force of (1.4)
o(log x) but in additionknowingthat
either,ifwe have herethe remainder-term
t~(x) > Kx for x > 1 and some positive constantK, we can still prove the
theorem. However,we have thento make some changein the argumentsof ?3.
THE INSTITUTE FOR ADVANCED STUDY
AND
SYRACUSE UNIVERSITY
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