Functions of a Complex Variable

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1 MIT OenCourseWare htt://ocw.mit.edu 8. Functions of a Comle Variable Fall 8 For information about citing these materials or our Terms of Use, visit: htt://ocw.mit.edu/terms.

2 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.

3 The rime number theorem is aroached by use of the functions log Φ(s =, V( = log. rime 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. I = ( s n for Res >. ζ(s Proof: For each n ( s n = ms n. Putting this into the finite roduct m= N N ( s we obtain ( s n = n s k k=. Now let N. II ζ(s etends to a holomorhic function in Res >. s Proof: In fact for Res >, ζ(s = s n s n= n+ = n= n d s ( n s s d But dy s s = s ma, n s s y s+ n y y s+ R s n n e + 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 n n = ( + n = k= ( n k ( n n = e V(n V(n, n< n

4 Thus V(n V(n n log. ( 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 r+ r r- Use ( for the oints right of, ( V V ( r. ( V C, V ( C. r+ r Summing, we get ( V( V( V( V r+ C + + C, r so V( C( + O(. 3

5 IV ζ(s = and Φ(s is holomorhic for Res. s 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 = + function holomorhic near s. ζ(s s s We eloit the ositivity of each term in for ǫ >. It imlies so lim ǫφ( + ǫ + iα = µ. ǫ l Φ( + ǫ = og +ǫ log ( i + α +ǫ + iα, Φ( + ǫ + iα + Φ( + ǫ iα + Φ( + ǫ. (4 By II, s = is a simle ole of ζ(s with residue +, so lim ǫφ( + ǫ =. ǫց Thus (4 imlies so µ +, µ. 4

6 This is not good enough, so we try Putting log ( + +ǫ + iα iα 4. lim ǫφ( + ǫ ± iα = ν, ǫց 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( V d is convergent. Proof: The function V( is increasing with jums log at the oints =. Thus Φ(s = log s V( = s d s+ In fact, writing as i+ ( i i this integral becomes i= V( i s s i s 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 = V(e t e t, Φ(z + g(z =. z + z f(t is bounded by III and we have e T V( d = 5 T f(t dt. (5

7 Also, by IV, Φ(z + = where h is holomorhic in Rez, so + h(z, z is holomorhic in Rez. For Rez > we have g(z = Φ(z + h(z g(z = = z + z z + = Now we need the following theorem: e zt (f(t + e zt f(t dt. e zt 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 This will imly Part V by (5. Proof of Analytic Theorem will be given later. VI V(. Proof: Assume that for some λ > we have V( λ for arbitrary large. Since V( is increasing we have for such λ V(t t λ λ t λ λ s dt dt = ds = δ(λ >. t t s On the other hand, V imlies that to each ǫ >, K such that K V( d < ǫ for K, K > K. K 6

8 Thus the λ cannot eist. Similarly if for some λ <, V( λ for arbitrary large, then for t, V(t V( λ, so V(t t λ t λ s = ds = δ(λ <. λ t λ t λ s Again this is imossible for the same reason. Thus both V( β = lim su > and V( α = lim inf < are imossible. Thus they must agree, i.e. V(. Proof of Prime Number Theorem: so We have Secondly if < ǫ <, V( = log log = π( log, π( log V( lim inf lim inf =. V( ǫ ( ǫ log ǫ log = ( ǫ log ( π( + O( ǫ thus for each ǫ. Thus π( log V( lim su lim su ǫ π( log lim =. Q.E.D. 7

9 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 z = R Fi R and then take δ > small enough so that g(z is holomorhic on C and its interior. By Cauchy s formula δ C g( g T ( = ( z dz (g(z g T (z e zt +. (6 πi R z C On semicircle C + : C (Rez > integrand is bounded by B, where R B = su f(t. t In fact for Rez >, g(z g T (z = = T B f(te zt dt T Be RezT Rez e zt dt and ( e zt z + R z = e ezt Re (z = Re iθ. R z R 8

10 B So the contribution to the integral (6 over C + is bounded by, namely R Be RezT T z B e Rez Re πr =. Rez R π R Net consider the integral over C_ =. Look at g(z and g T (z searately. For g T (z which is entire, this contour can be relaced by C'_ =. B Again the integral is bounded by because R T g T (z = f(te zt dt = T B e zt dt T B e zt dt Be RezT Rez 9

11 and ( z + R z on C has the same estimate as before. There remains ( e zt z g(z + d z. C R z }{{} inde. of T On the contour, e zt and lim e zt for Rez <. T C_ = δ 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. 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 + ρ n {ρ} n n log π. He conjectured that Reρ = for all ρ. This is the famous Riemann Hyothesis.