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 Stable URL: http://www.jstor.org/stable/1969455 . Accessed: 04/02/2014 11:21

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This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions ANNALS OF MATHEMATICS Vol. 50, No. 2, April,1949

AN ELEMENTARY PROOF OF THE PRIME-NUMBER THEOREM

ATLE SELBERG (Received October14, 1948) 1. Introduction In thispaper will be givena newproof of the prime-number theorem, which is elementaryin thesense that it uses practicallyno analysis,except the simplest propertiesof the logarithm. We shallprove the prime-number theorem in theform

(1.1) .lim =1(x) 1 Z__00 X wherefor x > 0, tQ(x)is definedas usualby

(1.2) #(X) = E logp, p denotingthe primes. The basic new thingin the proofis a certainassymptotic formula (2.8), whichmay be written

(1.3) t(x) logx + logpa () =2x logx + O(x).

Fromthis formula there are severalways to deducethe prime-number theorem. The wayI present??2-4 of this paper, is chosenbecause it seemsat thepresent to be themost direct and mostelementary way.1 But forcompleteness it has to be mentionedthat thiswas not my firstproof. The originalproof was in fact ratherdifferent, and made use of the followingresult by P. Erdos,that foran arbitrary,positive fixed number 6, thereexist a K(a) > 0 and an xo = x(a) suchthat for x > xo, thereare morethan K(a) Xlog x primesin theinterval from x to x + Ax. My firstproof then ran as follows:Introducing the notations

lim. t() = a, fim,() -A - x x; one can easilydeduce from (1.3), usingthe well-known result

(1.4) E logP = logx + 0(1), ,

1 Because it avoids the conceptof lower and 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

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions 306 ATLE SELBERG that (1.5) a+ A = 2. Next,taking a largex, with (x) ax + o(x), one can deducefrom (1.3) in themodified form

(1.6) (d(x) -ax) logx + E logp ( () -A x= 0(), that,for a fixedpositive number 6, onehas

(1.7) s6(p > (A _ )x exceptfor an exceptionalset ofprimes ? x with

log P O(lo_ X).

Alsoone easilydeduces that there exists an x' in therange -V1 < x' < x, wit 6 (x') = Ax' + o(x'). Againfrom (1.6) witha andA interchanged,and x' insteadof x, onededuces the

(1.8) * (J;)< (a + 6) -, exceptfor an exceptionalset ofprimes ? x' with

lop a FromErd6s' result it is thenpossible to showthat one can choseprimes p and p' notbelonging to anyof the exceptional sets, with

x< x/ < (l + a) x. P P P Thenwe getfrom (1.7) and (1.8) that X (A- O < < (mu) < (a+6) < (a+ X so that A < (a + 6)(1 + 6). or making6 tendto zero A a.

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions THE PRIME-NUMBER THEOREM 307

Hencesince also A ? a and a + A = 2 we have a = A = 1, whichproves our theorem. Erd6s' resultwas obtainedwithout knowledge of mywork, except that it is based on myformula (2.8); and afterI had theother parts of the above proof. His proofcontains ideas relatedto thosein the above proof,at whichrelated ideashe had arrivedindependently. The methodcan be appliedalso to moregeneral problems. For instance one can provesome theoremsproved by analyticalmeans by Beurling,but theresults are notquite as sharpas Beurlings.2Also one can provethe prime- numbertheorem for arithmetic progressions, one has thento use in additionideas and resultsfrom my previous paper on Dirichletstheorem. Ofknown results we use frequentlybesides (1.4) also its consequence (1.9) t (x) = 0(x). Throughoutthe paper p, q and r denoteprime numbers. A(n) denotes M6bius'number-theoretic function, r(n) denotesthe numberof divisorsof n. The letterc willbe usedto denoteabsolute constants, and K to denoteabsolute positiveconstants. Some of the moretrivial estimations are not carriedout but leftto thereader. 2. Proofof the basic formulas We write,when x is a positivenumber and d a positiveinteger,

(2.1) Xd = kd.. = jut(d)log d'

and ifn is a positiveinteger,

(2.2) On= =n,= = Ed/n Xd Thenwe have flog2x, forn=1,

(2.3) - Jlog log X2/p, forn = pa, a >1 (2*3) On )2logplogq,p forn = paql, a > > 1, LO,for all othern. The firstthree of thesestatements follow readily from (2.2) and (2.1), the fourthis easilyproved by induction.Clearly it is enoughto considern square- free,then ifn = PlP2 * Pk ,

On,, = On/rPi,z - On1Pkl/Pk Fromthis the remaining part of (2.3) follows.

2 A. BEURLING: Analysede la loi asymptotiquede la distributiondes nombrespremiers gneralisks,Acta Math., vol. 68, pp. 255-291(1937). 3 These Annalsthis issue,pp. 237-304.

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions 308 ATLE SELBERG

Now considerthe expression

= Xd I'd OZd Xd) On d Xd F1=x E + [ ] (2.4) zn! n

j X =X E A(d) og2 Z + o u2 )=xE(d) lo2 + O(X). d

2 ORn-log2x + E log p log - n

(2.5) + E logplogq+O (1log p log) pq?z \p~z P1

+0( Z og2x) +O( E logplogq) jpa < paq< z x>l a>1

+ log2x = log2p++ logp logq + O(x).

(2.6) E log2p + E logp logq = x 2L?(dQ)log2 x + O(x). (2.6) pq<_ dz d 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 summationfrom the well-knownresult

>T(v) = zlogz+c4z+0(A/).

From (2.7) and (2.7') we get

log2z = 2 2 (G) + c5 I + C6 + O(z1). p:z X V w'VXV

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions THE PRIME-NUMBER THEOREM 309

By takinghere z = x/d,we get

1,,ud 2 X = 2 E jAxd), r(v) +C5 iAg(d) E d?xd d d

+ c6 Z ,8(d) + O(x Z dt4) = 2 Z /(d)T(v) d~x d d~z dv-z< dv

+ c6 E 1d) + C6 E ) + 0(1) dax dv dz_ d

=2 n d r + c5 Z (d)

+ 0(1) = 2 j - + c5+ 0(1) = 2 logx + 0(1).

Weused herethat Ed/n ,(d) r(n/d) = 1, and thewell-known d

(2.9) t6(x)log x + Z logp 6 - = 2x logx + O(x), by noticingthat Z log2p = t(x) log x + O(x). vZ By partialsummation we getfrom (2.8)

(2.10) E log p+ E logp logq 2x + 0 (x) V:5.x VQ_: log pq logx Thisgives

E logp logq = I logp E log q = 2x z logP Pq_$ Pz q~zIp P-z P

-E logp E q gllogrg 0 E log \ pE~z: qr~z/p log qr V dp (1+log-))

= 2x logx - E log qlogr (f) + O(x log logx). qr~z log qr \qr/ Insertingthis for the second term in (2.8) we get

(2.11) t,(x)log x = E log p log q 6 ( - ) + O(x log logx). V.< log pq \pq/

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions 310 ATLE SELBERG

Writingnow 6(x) = x + R(x) , (2.9) easily gives (2.12) R(x) logx = -Z logp R ( + O(x), posx p and (2.11) yieldsin thesame manner lo (2.13) R(x) logx p log qR + O(x log log x), ,~log pq \pq/ since lo1glo1g q =~log x + O(logOlog x pq logpq x), whichfollows by partialsummation from E logP logq = log2x + O(logx), vq < pq whichagain follows easily from (1.4). The (2.12) and (2.13) yield 2 j R(x)I logzx S E logp RI-)

+ E logp log q |R (X) + O(x log logx). vq< log pq \pq Fromthis, by partialsummation,

2 j R(x)Ilog x < pE n logp + E log Pq }

{ R - R) + }+O(x log logx), or by (2.10)

2 R(x)Ilogz 2 2n{R(n) - (n l)|

+0 1 logn ) +- O(x loglog x)

=2 IR ) + (Xloglog x),

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions THE PRIME-NUMBER THEOREM 311 or

(2.14) IR(x)I < 1 Ei| R(?) +0( loglogXx) log X n? \l \logx / whichis the resultwe will use in the following.4 3. Some propertiesof R(x) From (1.4) we get by partial summationthat

E 6(n) = log x + 0(1), n~x fl2 or z R(n) 0(1).

This means thereexists an absolute positiveconstant K1, 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 true if R(n) changesthe sign also.' Thus for an arbitraryfixed positive a < 1 and x > 4, there will exist a y in the intervalx y 5_ e K2/6 x, with (3.3) (R(y) <6y. From (2.10) we see that fory < y',

0 < E log p ? 2(y' - y) + ? Y fromwhich follows that

R(y') - R(y) y' - y + o (I'Y,)

4 Apparentlywe have herelost somethingin the orderof the remainder-termcompared to (2.8). Actuallywe could instead of (2.14) have used the inequality 2 Lon IR(x)I 5 1 1: IR (n + ? (lga whichcan be proved in a similarway. 5 Because therewill then be a I R(y) I < log y.

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions 312 ATLE SELBERG

Hence, if y/2 ? y' < 2y, y > 4,

R(y') -R(y) y' - y + o y

or

I R(y') I _ R(y)1 + I y'- y) + O (l'y,).

Now consider an interval (x, eK2I x), according to (3.3) there exists a y in this intervalwith

I R(y) I < by. Thus forany y' in the intervaly/2 ? y' < 2 y, we have

I R(y') I _ by+ - y I + K3l ' ~~~log R~~~y') IY x' or

R(y') < 26 + 1 I + l3

Hence if x > e "'/ and e~812) ? y'/yS we get R(y') 2+(e812-1+_45 I~ < 25 + (e21) + 3 < 43.

Thus for x > e K/8 the interval (x, e K2/ZX) will always contain a sub-interval (yi, e8/2y1),such that JR(z) J< 48z if z belongs to this sub-interval. 4. Proofof the prime-numbertheorem We are now goingto prove the THEOREM.

limx = 1. Zx-*oo X Obviouslythis is equivalent to

(4.1) lim =() 0. _-*oO x We know that forx > 1, (4.2) jR(x) I < K4x. Now assume that forsome positivenumber a < 8, (4.3) j R(x) I < ax, holds for all x > x0. Taking 5 = a/8, we have accordingto the preceding

This content downloaded from 130.39.169.231 on Tue, 4 Feb 2014 11:21:29 AM All use subject to JSTOR Terms and Conditions THE PRIME-NUMBER THEOREM 313 section (since we may assume that xo > eK3I8),that all intervals of the type (x, eK2/8X)with x > xo, containan interval(y, e812y) such that (4.4) | R(z) I < az/2, fory < z ? e6/2y. The inequality(2.14) then gives,using (4.2),

iR(x) I l ax R~)I

__ = ax- y,

VAnalogx 4 log p + (

a ~~x+ 0( Jx< a(13of~) ( 256K2) ? V log x) < t( 300K )2 forx > xi. Since the iteration-process ( 2 an+1 = an 11 - _

obviouslyconverges to zero if we startfor instance with a, = 4 (one sees easily that then an < K5/V/n),this proves (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) either,if we have here the remainder-termo(log x) but in additionknowing that 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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