Lecture 21 and 22: The Prime Number Theorem


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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.
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 halflane 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. ()
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
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
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
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
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
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
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.
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