Notes on the prime number theorem

Size: px

Start display at page:

Download "Notes on the prime number theorem"

Noah Cummings
5 years ago
Views:

1 Notes o the rime umber theorem Keji Kozai May 2, 24 Statemet We begi with a defiitio. Defiitio.. We say that f(x) ad g(x) are asymtotic as x, writte f g, if lim x f(x) g(x) =. The rime umber theorem tells us about the asymtotic behavior of the umber of rimes that are less tha a give umber. Let π(x) be the umber of rimes umbers such that x. For examle, π(2) =, π(3) = 2, π(4) = 2, etc. The statemet that we will try to rove is as follows. Theorem.2 (Prime Number Theorem,. 382). The asymtotic behavior of π(x) as x is give by π(x) x log x. This was origially observed as early as the 7s by Gauss ad others. The first roof was give i 896 by Hadamard ad de la Vallée Poussi. We will give a overview of simlified roof. Most of the material comes from various sectios of Gameli mostly XIV., XIV.3 ad XIV.5. 2 The Gamma Fuctio The gamma fuctio Γ(z) is a meromorhic fuctio that exteds the factorial! to arbitrary comlex values. For Rez >, we ca fid the gamma fuctio by the itegral: Γ(z) = e t t z dt, Re z >.

2 To see the relatio with the factorial, we itegrate by arts: Γ(z + ) = e t t z dt = t z e t + z = zγ(z). e t t z dt This holds wheever Re z >. We also otice that Γ() = e t dt =. By iductio, the Γ(2) =, Γ(3) = 2, Γ(4) = 6,..., Γ( + ) =!. This holds for all o-egative itegers. Remember that the itegral defiitio for Γ(z) oly works for Rez >. However, usig the relatio Γ(z + ) = zγ(z), we ca exted it for egative values of z by defiig it to be the fuctio that agrees with the itegral defiitio for Rez >, ad also satisfies Γ(z + ) = zγ(z) for all z. I articular, if we aly this relatio m times, we fid that We ca rewrite this as Γ(z + m) = (z + m ) (z + )zγ(z). Γ(z) = Γ(z + m) (z + m ) (z + )z. The fuctio Γ(z + m) is defied for Rez > m, so we defie Γ(z) for Re z > m by usig the above equatio. By doig this for larger ad larger value of m, we ca rogressively defie it for more of the comlex lae, ad i the limit, we get a fuctio defied o all of C. From this defiitio, we ca see that Γ(z) is meromorhic with simle oles at z =,, 2, 3, The Zeta Fuctio Before we give a roof of the theorem, we will eed to study the zeta fuctio, which is covered i XIV.3. The zeta fuctio is defied to be ζ(s) = s,re s >. = Here, s = σ+it is a comlex umber. Notice that if σ = Re s >, the series o the right i fact does coverge absolutely, as s = σ, 2

3 with the series covergig by the -test whe σ >. As σ, we σ obtai the harmoic series, which diverges. The zeta fuctio satisfies a roduct formula. Theorem 3. ((3.) o. 37). If Re s >, the ζ(s) = ( ) s, where the roduct is take over all rime umbers. Proof. The roof uses the geometric series formula s = + s + 2s + 3s Now, if, 2,..., m are m differet rimes, if we multily the geometric series for each of these rimes together, we obtai ( s ) ( s m ) = ( k km m ) s. k,...,k m= Every iteger ca be writte uiquely as a roduct of rimes, so each umber such that is a roduct of the rimes s,..., m aears exactly oce. I the limit, whe we take all rimes, we get all terms. s Hece, s = s = ζ(s). = Takig the recirocal the yields the formula i the theorem. It turs out that ζ(s) ca be exteded to be a meromorhic fuctio o C, with oly a simle ole at s = with residue. More details ca be foud i XIV.3. This extesio satisfies the equatio ζ(s) = 2 s π s si πs Γ( s)ζ( s). 2 I articular, the right had side is defied for Res <, so this tells us ζ(s) for all values of s excet for the critical stri Re s. A itegral formula described i XIV.3 is used to defie ζ(s) i this stri. The roduct formula for ζ(s) shows that ζ(s) has o zeros i the right half-lae Res >. The equatio above the imlies that the oly zeros of ζ(s) i the left half-lae Res < are the zeros of si πs 2, which are s = 2, 4, 6,.... The famous Riema hyothesis is the cojecture that all other zeros of ζ(s) lie o the lie Res = 2. 3

4 4 Chebyshev theta fuctio Istead of comarig the asymtotic behavior of π(x) with x will cosider the Chebyshev theta fuctio, log x directly, we Θ(x) = x. We will comare the theta fuctio with the statemet from the rime umber theorem. Lemma 4. (. 384). π(x) x log x if ad oly if Θ(x) x. Proof. Notice that log x, ad there are π(x) summads for Θ(x). Hece, Θ(x) π(x)log x, as log as x. Dividig by x gives us Θ(x) x π(x) log x x. Let ɛ >. The, We ca rearrage this to Θ(x) = x x ɛ < x x ɛ < x log x ɛ = ( ɛ)(log x)(π(x) π(x ɛ )) ( ɛ)(log x)(π(x) x ɛ ). Θ(x) ɛ (log x)(π(x) x ɛ ) Θ(x) ɛ + xlog x x ɛ (log x)π(x). The, dividig by x ad combiig with our first iequality, we ca see that Sice lim x log x x ɛ x. Θ(x) x π(x) log x x Θ(x) ɛ x + log x x ɛ. =, we see that π(x) Θ(x) x/ log x if ad oly if x. So i order to rove the rime umber theorem, it suffices to show Θ(x) 4

5 5 The Lalace trasform We will begi with a lemma. Lemma 5.. If f(x) is a icreasig fuctio of x such that exists, the f(x) x. lim R R ( f(x) ) dx x x Before rovig the lemma, we ote that Θ(x) is a icreasig fuctio i x, so if we ca rove that the limit above exists for Θ(x), the we would rove the rime umber theorem. Proof. We will rove this by the cotraositive. Sice f(x) is icreasig, the oly way that f(x) x fails is if there exists a ɛ > such that f(x) > ( + ɛ)x for sufficietly large values of x, or f(x) < ( ɛ)x for sufficietly large values of x. Let us assume the first case, that f(x) > ( + ɛ)x for large values of x. Suose that this is true for x x. The (+ɛ)x x ( ) f(x) dx (+ɛ)x x x = x (+ɛ)x x +ɛ ( ) f(x ) dx x x ( ) ( + ɛ)x dx x x ( ) ( + ɛ) dt t t. The last lie is by usig the chage of coordiates x = x t. I articular, the itegral i the last lie is ideedet of x ad is some o-zero c >. We ca reeat this for ifiitely may disjoit itervals, so lim R R ( f(x) ) dx x x diverges to, so it caot exist. Similarly, we ca show that if f(x) < ( ɛ)x for sufficietly large x, the the itegral diverges to. Therefore f(x) x. I order to rove that the limit exists for Θ(x), we will eed to itroduce oe more tool: the Lalace trasform. 5

6 Defiitio 5.2. Let h(s) be a cotiuous or iecewise fuctio o the ositive real axis s. The Lalace trasform of h(s) is the fuctio of z (Lh)(z) = rovided that the itegral coverges. e sz h(s)ds, We wo t eed to kow too much about the Lalace trasform, so we wo t go ito deth here. But if you wat to kow more, the book has a sectio o the Lalace trasform i XIV.2. The Lalace trasform is somewhat like the Fourier trasform. Notice that the Lalace trasform, evaluated at z =, is (Lh)() = e s() h(s)ds = Recall that what we are tryig to show is that lim R R ( Θ(x) ) dx x x h(s)ds. exists. If we do a chage of variages x = e s, the dx = e s ds = xds, so the itegral becomes lim T T (Θ(e s )e s )ds. This limit is exactly the exressio for (Lh)() above, for h(s) = Θ(e s )e s. Thus, out goal will be to show that (Lh)() exists. If we ca do that, we will have roved the rime umber theorem. We will do this by coectig the Lalace trasform with aother fuctio whose oles we uderstad, Φ(s). 6 Dirichlet Series Defiitio 6.. A Dirichlet series is a series of the form = a s = a + a 2 2 s + a 3 3 s The zeta fuctio ζ(s) is a Dirichlet series with a = a 2 = = a = =. 6

7 The roof will use a articular Dirichlet series defied by Φ(s) = s, which coverges ad is aalytic for Res >. Here, meas that we are summig oly over the rime umbers. We will take the roduct formula ζ(s) = ( ) s, ad take the logarithm of both sides ad differetiate. O the left had side, we obtai d ds log ζ(s) = d ds log ζ(s) = d ds log ζ(s) = ζ (s) ζ(s). The right had side becomes d ds log ( ) s = d ds = d ds log ( ) s s s = d log( s ) s ds = = s s s. Noticig that we coclude that s s = ζ (s) ζ(s) = Φ(s) + s ( s ), s ( s ). The left-had side is meromorhic o C, ad the sum o the right coverges to a aalytic fuctio for Res > 2 usig the -test. Thus, we ca exted 7

8 Φ(s) to be meromorhic for Res > 2. The oles of Φ(s) are at the oles ad zeros of ζ(s). I articular, Φ(s) has a simle ole at s = with residue (from the loe simle ole of ζ(s)) ad a simle ole at s = s with residue m if ζ(s) has a zero or order m at s. Theorem 6.2 (. 383). The meromorhic fuctio Φ(s) has o oles o the vertical lie Re s = excet at s =. The zeta fuctio ζ(s) has o zeros o the lie Re s =. Proof. By the observatio before the theorem, the two statemets are equivalet. Let t >, ad suose that ζ(s) has a zero of order q at + it ad a zero of order q at + 2it. Here, if + it or + 2it are o-zero, the we use the covetio that q = ad q =, resectively. For a rime, ote that ( it/2 + it/2) 4 = 2it + 4 it it + 2it. Multily by ɛ() +ɛ to obtai 4 6 ɛ[ + + +ɛ 2it +ɛ it +ɛ Summig over all rimes the yields + 4 +ɛ+it + +ɛ+2it]. ɛ[φ(+ɛ+2it) +4Φ(+ɛ+it) +6Φ(+ɛ) +4Φ(+ɛ it) +Φ(+ɛ 2it)]. By Rule, if we take ɛ, we obtai Res[Φ(z), +2it]+4 Res[Φ(z), +it]+6 Res[Φ(z), ]+4 Res[Φ(z), it]+res[φ(z), 2it]. The residue at z = is, ad by assumtio, the residue at z = ± it is q ad at z = ± 2it the residue is q. Hece, the above becomes, 6 8q 2q, which ca oly be satisfied if q = q =. This roves the theorem. 7 Proof of the Prime Number Theorem Now, we come back to the Lalace trasform ad coect it to Φ(s). 8

9 Lemma 7.. We have that for Re s >. (LΘ(e t ))(s) = Φ(s) s Proof. The first art of the statemet is actually showig that the Lalace trasform exists for Re s >. After that, we will show that the formula holds. We first claim that Θ(x) (4 log 2)x. To see this, we will use the biomial coefficiet ( ) 2 < ( + ) 2 = 2 2. Every rime umber betwee ad 2 divides ( ) 2, hece their roduct also divides ( ) 2. This meas that ( ) 2 < 2 2. <<2 Takig logarithms yields <<2 2 log 2. This meas that m Θ(2 m ) = k= 2 k <<2 k m (2 k )log 2 k= < 2 m+ log 2. Now for ay x, choose m such that 2 m < x 2 m. The, Θ(x) Θ(2 m ) 2 m+ log 2 = 4(2 m )log 2 < (4 log 2)x. 9

10 Now, if we look at the Lalace trasform, (LΘ(e t ))(s) = = e ts Θ(e t )dt e ts Θ(e t ) dt e ts (4 log 2)e t dt (4 log 2)e t(re s ) dt. This itegral coverges whe Re s >, i.e. whe Re s >. To obtai the formula, we eumerate the rimes so that is the -th rime umber. The, Θ(e t ) is costat for < t < +. Therefore, + e st Θ(e t )dt = Θ( ) + = Θ( ) e st s + = s Θ( )( s e st dt s + ). Now, we break u the itegral e st Θ(e t )dt alog itervals [, + ] ad use the above idetity to show e st Θ(e t )dt = = s s Θ( )( s s + ) (Θ( ) Θ( )) s. Observe that Θ( ) Θ( ) =, ad the above becomes s s = s Φ(s). Notice that this idetity holds for all Res >. We have reviously show that the oly ole of Φ(s) o the lie Res = is the simle ole at s =. If we ca remove this ole, the we ca exted (LΘ(e t ))(s) ast the lie Re s =.

11 Now we cosider the fuctio h(t) = Θ(e t )e t. If we look at the Lalace trasform (Lh)(s), the effect of multilyig e t is to chage that e st i the itegrad to e (s+)t. Hece, the Lalace trasform will coverge for Re s + >, or i other words Res >. The Lalace trasform of is Hece, e st dt = e st s = s. (Lh)(s) = Φ(s + ) s + s, for Re s >. Now, Φ(s + ) has a simle ole with residue at s =, but the s art cacels that out. Hece, (Lh)(s) is aalytic for Res >, ad it has o oles o the lie Res =. This meas that we ca exted the fuctio to be aalytic o a domai that cotais the lie Re s =. I articular, (Lh)() exists, rovig the rime umber theorem. This last art of extedig the fuctio is ot trivial ad is show i the book o , but this gives a geeral idea of how the roof works. 8 Exercises. Exercise XIV Exercise XIV Exercise XIV Exercise XIV.5.5 (this is fairly hard, but see if you ca reduce it to the roblem of showig that lim log = )

Proposition 2.1. There are an infinite number of primes of the form p = 4n 1. Proof. Suppose there are only a finite number of such primes, say

Proposition 2.1. There are an infinite number of primes of the form p = 4n 1. Proof. Suppose there are only a finite number of such primes, say Chater 2 Euclid s Theorem Theorem 2.. There are a ifiity of rimes. This is sometimes called Euclid s Secod Theorem, what we have called Euclid s Lemma beig kow as Euclid s First Theorem. Proof. Suose to