Lecture 21 and 22: The Prime Number Theorem

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Stanley Smith
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1 Lecture and : The Prime Number Theorem (New lecture, not in Tet) The location of rime numbers is a central question in number theory. Around 88, Legendre offered eerimental evidence that the number π() of rimes < behaves like / log for large. Tchebychev roved (848) the artial result that the ratio of π() to / log for large lies between 7/8 and 9/8. In 896 Hadamard and de la Valle Poussin indeendently roved the Prime Number Theorem that the limit of this ratio is eactly. Many distinguished mathematicians (articularly Norbert Wiener) have contributed to a simlification of the roof and now (by an imortant device by D.J. Newman and an eosition by D. Zagier) a very short and easy roof is available. These lectures follows Zagier s account of Newman s short roof on the rime number theorem. cf: () D.J.Newman, Simle Analytic Proof of the Prime Number Theorem, Amer. Math. Monthly 87 (98), () D.Zagier, Newman s short roof of the Prime Number Theorem. Amer. Math. Monthly 4 (997), The rime number theorem states that the number π() of rimes which are less than is asymtotically like : log π() / log as. Through Euler s roduct formula (I) below (tet.3) and esecially through Riemann s work, π() is intimately connected to the Riemann zeta function ζ(s) =, n s n= which by the convergence of the series in Res > is holomorhic there.

Tchebychev roved (848) the artial result that the ratio of π() to / log for large lies between 7/8 and 9/8.

2 I The rime number theorem is aroached by use of the functions Φ(s) = rime log, V() = log. = ( s ) for Res >. s rime Simle roerties of Φ will be used to show ζ(s) = and Φ( s) holomorhic s for Res. Deeer roerties result from writing Φ(s) as an integral on which Cauchy s theorem for contour integration can be used. This will result in the relation V() from which the rime number theorem follows easily. ζ(s) Proof: see tet.3. II ζ(s) etends to a holomorhic function in Res >. s Proof: In fact for Res >, But ζ(s) s = n= n d s s n+ = d n s s n= n n = s dy s s ma s s y s+ n y y s+ n n n n n n = ( + ) n = = e V(n) V(n), k n Res+, so the sum above converges uniformly in each half-lane Res δ (δ > ). III V() = O() (Sharer form roved later). Proof: Since the in the interval n < n divides (n)! but not n! we have k= n< n Thus V(n) V(n) n log. ()

This will result in the relation V() from which the rime number theorem follows easily. ζ(s) Proof: see tet.3. II ζ(s) etends to a holomorhic function in Res >.

3 If is arbitrary, select n with n < n +, then V() V(n + ) V(n + ) + (n + ) log (by ()) = V + + ( + ) log = V + log + + ( + ) log. Thus if C > log, V() V C for = (C). () Consider the oints Use () for the oints right of, V V C,. V V C. r r+ r Summing, we get V() V( ) V() V r+ C + + C, r so V() C() + O(). IV ζ(s) = and Φ( s) is holomorhic for Res. s 3

4 Proof: If Res >, art I shows that ζ(s) = and ζ (s) log log = = Φ(s) + ζ(s) s s ( s ). (3) The last sum converges for Res >, so by II, Φ(s) etends meromorhically to Res > with oles only at s = and at the zeros of ζ(s). Note that ζ(s) = = ζ( s) =. Let α R. If s = + iα is a zero of ζ(s) of order µ, then So ζ (s) µ ζ(s) = s s + function holomorhic near s. We eloit the ositivity of each term in for >. It imlies so lim Φ( + + iα) = µ. log Φ( + ) = + log iα iα + +, + Φ( + + iα) + Φ( + iα) + Φ( + ). (4) By II, s = is a simle ole of ζ(s) with residue +, so lim Φ( + ) =. Thus (4) imlies so µ +, µ. This is not good enough, so we try log + iα 4 + iα. + 4

Let α R. If s = + iα is a zero of ζ(s) of order µ, then So ζ (s) µ ζ(s) = s s + function holomorhic near s. We eloit the ositivity of each term in for >.

5 Putting V lim Φ( + ± iα) = ν, where ν is the order of ± iα as a zero of ζ(s), the same comutation gives 6 8µ ν, which imlies µ = since µ, ν. Now II and (3) imly Φ(s) holomorhic s for Res. V() d is convergent. Proof: The function V() is increasing with jums log at the oints =. Thus Φ(s) = log V() = s d s+ i+ In fact, writing as i this integral becomes s i i= V( i) s s i i+ which by V( i+ ) V( i ) = log i+ reduces to Φ(s). Using the substitution = e t we obtain Φ(s) = s e st V(e t ) dt Res >. Consider now the functions f(t) is bounded by III and we have e T T V() dt = f(t) dt. (5) s f(t) = V(e t )e t, Φ(z + ) g(z) =. z + z Also, by IV, Φ(z + ) = + h(z), z where h is holomorhic in Rez, so g(z) = Φ(z + ) = h(z) z + z z + 5

Thus Φ(s) = log V() = s d s+ i+ In fact, writing as i this integral becomes s i i= V( i) s s i i+ which by V( i+ ) V( i ) = log i+ reduces to Φ(s).

6 is holomorhic in Rez. For Rez > we have g(z) Now we need the following theorem: = e zt (f(t) + ) = e zt f(t) dt. Theorem (Analytic Theorem) Let f(t) (t ) be bounded and locally integrable and assume the function g(z) = e zt f(t) dt Re(z) > etends to a holomorhic function on Re(z), then T lim f(t) dt eists and equals g(). T Proof: Assume that for some λ > we have V() λ for arbitrary large. Since V() is increasing we have for such λ λ V(t) t λ t λ λ s dt dt = ds = δ(λ) >. t t s On the other hand, V imlies that to each >, K such that K V() d < for K, K > K. so V(t) t λ t = λ s ds = δ(λ) <. t t s e zt dt This will imly Part V by (5). Proof of Analytic Theorem will be given later. VI V(). Thus the λ cannot eist. K Similarly if for some λ <, V() λ for arbitrary large, then for t, V(t) V() λ, λ λ λ 6

equals g(). T Proof: Assume that for some λ > we have V() λ for arbitrary large. Since V() is increasing we have for such λ λ V(t) t λ t λ λ s dt dt = ds = δ(λ) >.

7 Again this is imossible for the same reason. Thus both β = lim su V() > and V() α = lim inf < are imossible. Thus they must agree, i.e. V(). Proof of Prime Number Theorem: so We have V() = log log = π() log, Secondly if < <, lim inf π() log lim inf V() =. V() ( ) log log = ( ) log π() + O( ) thus for each. Thus lim su π() log lim su V() π() log lim =. Q.E.D. Proof of Analytic Theorem: (Newman). Put T g T (z) = e zt f(t) dt, which is holomorhic in C. We only need to show lim g T () = g(). T 7

Proof of Prime Number Theorem: so We have V() = log log = π() log, Secondly if < <, lim inf π() log lim inf V() =.

8 Fi R and then take δ > small enough so that g(z) is holomorhic on C and its interior. By Cauchy s formula zt z dz g() g T () = (g(z) g T (z)) e +. (6) πi C R z On semicircle integrand is bounded by C + : C (Rez > ) B, where In fact for Rez >, R B = su f(t). t g(z) g T (z) = = f(t)e zt dt T B e zt T Be RezT dt Rez and zt z RezT Rez e + = e (z= Re iθ ). R z R B So the contribution to the integral (6) over C + is bounded by, namely R Be RezT Rez B e RezT πr =. Rez R π R Net consider the integral over 8

(6) πi C R z On semicircle integrand is bounded by C + : C (Rez > ) B, where In fact for Rez >, R B = su f(t).

9 Look at g(z) and g T (z) searately. For g T (z) which is entire, this contour can be relaced by B Again the integral is bounded by because R T g f(t)e zt T (z) = dt B B = T T e zt dt e zt dt Be RezT Rez and z + R z on C has the same estimate as before. There remains zt z e g(z) + dz. C R z inde. of T On the contour, zt e and lim e zt for Rez <. T By dominated convergence, the integral as T +, δ is fied. It follows that B lim su g() g T (). T R Since R is arbitrary, this roves the theorem. 9

= dt B B = T T e zt dt e zt dt Be RezT Rez and z + R z on C has the same estimate as before. There remains zt z e g(z) + dz.

10 Q.E.D. Remarks: Riemann roved an elicit formula relating the zeros ρ of ζ(s) in < Res < to the rime numbers. The imroved version by von Mangoldt reads V() = log ρ n = ρ + log π. n {ρ} n He conjectured that Reρ = for all ρ. This is the famous Riemann Hyothesis.

PRIME NUMBERS AND THE RIEMANN HYPOTHESIS

PRIME NUMBERS AND THE RIEMANN HYPOTHESIS CARL ERICKSON This minicourse has two main goals. The first is to carefully define the Riemann zeta function and exlain how it is connected with the rime numbers.