The Prime Number Theorem without Euler products

Transcription

1 The Prime Number Theorem without Euler products Michael Müger Istitute for Mathematics, Astrophysics ad Particle Physics Radboud Uiversity Nijmege, The Netherlads October, 27 Abstract We give a simple proof of the Prime Number Theorem (PNT, the mai aims beig to miimize ad isolate the iput from umber theory ad to avoid the Euler factorizatio of the zeta fuctio. We will use some very basic complex ad harmoic aalysis, provig all we eed of the latter. The Prime Number Theorem Part (ii of the followig lemma is the oly place where the fudametal theorem of arithmetic ad related umber theoretic reasoig play a rôle. P deotes the set of primes.. Lemma (i There is a uique arithmetic fuctio Λ : N R satisfyig Λ(d log N. (. (ii This solutio is give by the vo Magoldt fuctio defied by d { log p if p Λ( k, p P, k N else (.2 (iii We have Λ( log for all N. Proof. (i For we have Λ( d Λ(d log. For 2 we are forced to iductively defie Λ( log Λ(d. (.3 d d< This shows that (. has a uique solutio. (ii By the fudametal theorem of arithmetic, every N has a uique prime factorizatio p k pkm m, ad the divisors of are of the form p l plm m where l i k i. The oly divisors d of for which the Λ of (.2 is o-zero are {p k i i,..., m, k,..., k m}. Thus m k i m m Λ( log p i k i log p i log p ki i log, d i k i i provig that (.2 is a, thus the, solutio of (.. Now (iii is a obvious cosequece of (ii. For use i the followig proofs, we recall oe fact from basic complex aalysis: If f : Ω C is meromorphic, the for every z Ω there is a uique (z Z such that f(z (z z (z g z (z, where g z is holomorphic i some eighborhood of z ad g z (z. Furthermore, the fuctio z f (z f(z is meromorphic o Ω, has a pole of order oe ad residue (z wheever f has a zero or pole at z, ad is holomorphic elsewhere. Thus (z ε ε(f /f(z + ε.

2 .2 Propositio The series ζ(z z (.4 defies a holomorphic fuctio o the ope halfplae Re z >. It there satisfies ζ(z ad ζ (z ζ (z ζ(z log z, (.5 Λ( z. (.6 Proof. I view of z Re z ad the covergece of /α for α >, it is clear that (.4 coverges absolutely for Re z > ad uiformly for Re z α >. It therefore defies a holomorphic fuctio o the ope half plae Re z >. By basic complex aalysis, we may differetiate uder the sum ad obtai (.5. Usig the cosequece Λ( log of Lemma.(iii, we similarly see that Λ( defies a holomorphic fuctio o Re z >. Sice the series (.4 ad (.6 z coverge absolutely for Re z >, we may compute ζ(z Λ( z z d m Λ(m m z z Λ(d,m log z Λ(m (m z ζ (z, where we used (. ad (.5. Sice ζ clearly does ot vaish idetically, it has oly isolated zeros, ad for ζ(z we have prove (.6. But sice ζ /ζ has poles at all zeros of ζ, whereas the r.h.s. of (.6 is holomorphic o Re z >, we have prove that ζ(z for Re z >..3 Lemma There is a holomorphic fuctio g : {z Re z > } C such that ζ(z g(z + z whe Re z >. Thus ζ has a meromorphic extesio, which we also deote ζ, to {z Re z > }\{}. Proof. For Re z > we have t z dt z. Thus g(z ζ(z z ( + z t z dt + ( z t z dt. (.7 Now, + ( z t z dt z + t du u z+ dt sup u [,+] z z u z+, Re(z+ so that (.7 coverges, ad thus defies g(z as a holomorphic fuctio o Re z >..4 Propositio (i We have ζ( + it for all t R. (ii The fuctio z ζ ζ (z z exteds cotiuously to Re z. Proof. (i We have ζ(z R for real z >, implyig ζ(z ζ(z, thus ( + it ( it. By Lemma.3 we have ε ε ζ ζ ( + ε ( ad, assumig t, ε εζ ( + ε ± it ( ± it : µ, ζ ε εζ ( + ε ± 2it ( ± 2it : ν, ζ 2

3 where µ, ν sice ζ(z is holomorphic at + it for t. With (.6 we have 2 r 2 ( r ζ ζ ( + ε + irt Λ( +ε (it/2 + it/2 4, sice Λ ad it/2 + it/2 R. Multiplyig by ε ad takig ε gives 6 8µ 2ν. (The 5th row ( 4 of Pascal s triagle is, 4, 6, 4,. Thus ( + it µ, which meas ζ( + it. (ii By Lemma.3, ζ is holomorphic o Re z > with the exceptio of the simple pole at z. By (i, ζ( + it for t. Thus ζ /ζ is holomorphic at + it for t ad has a simple pole of residue at z. The latter is cacelled by the subtractio of the term (z..5 Lemma Defiig ψ(x x Λ(, we have z e zx ψ(e x dx Λ( z ζ (z ζ(z for Re(z >. (.8 Proof. I view of Lemma., we have ψ(x O(x log x O(x +ε, thus the itegral coverges absolutely. Now z e zx ψ(e x dx z e zx Λ( dx z Λ( e zx dx e x log [ ] Λ( z Λ( z e zx z log proves the first idetity of (.8. The secod idetity was prove i Propositio.2. The followig Tauberia theorem, due to Ikehara ad ot ivolvig umber theory, will be prove i the Appedix:.6 Theorem Let f : R + R + be o-decreasig. If there are a, c such that the itegral F (z f(xe zx dx, coverges absolutely for Re(z > a [thus automatically defies a cotiuous fuctio there] ad F (z has a cotiuous extesio to the half-plae Re(z a the c z a.7 Propositio ψ(x x as x. f(x ce ax as x +. Proof. By Lemma.5, the itegral e zx ψ(e x dx coverges for Re(z > ad equals ζ (z zζ(z. By Propositio.4(ii, the fuctio z ζ (z ζ(z z has a cotiuous extesio to Re(z, thus also ζ (z zζ(z z z ( ζ (z ζ(z z. I view of Λ, the fuctio ψ(x x Λ( is o-decreasig. Now Theorem.6 gives ψ(e x e x as x, ad therefore ψ(x x..8 Theorem (Prime Number Theorem Let π(x #(P [, x], ad let p deote the -th prime. The π(x x ad p log as x,. log x 3

4 Proof. Usig Lemma.(ii, we compute ψ(x Λ( log p log x log p log x π(x log x. log p x p x p x If < y < x the p k x π(x π(y y<p x y<p x Thus π(x y + ψ(x/ log y. Takig y x/(log x 2 this gives ψ(x x π(x log x ψ(x x x log p log y ψ(x log y. log x log(x/(log x 2 + log x. I view of log(x/(log x 2 log x 2 log log x log x ad ψ(x x from Propositio.7 we have π(x x/log x. Takig logarithms of π(x x/log x, we have log π(x log x log log x log x ad thus π(x log π(x x. Takig x p ad usig π(p gives log p..9 Remark We avoided the Euler factorizatio of the zeta-fuctio (which at ay rate just is a aalytic restatemet of the uique prime factorizatio ad the power series expasio of the logarithm. Istead we directly arrived at the Dirichlet series (.6 for the logarithmic derivative ζ /ζ, usig the very basic theory of Λ ad the relatio betwee Dirichlet covolutio of arithmetic fuctios ad poitwise multiplicatio of the associated Dirichlet series. Arithmetic (or umber-theoretic reasoig is cofied to the proof of part (ii of Lemma.. That result is used oly via its cosequece (iii. More precisely, to prove ζ(z for Re z > we oly eeded the slow growth of Λ. The positivity of Λ gave the mootoicity of ψ, eeded to apply Theorem.6, ad wet ito the proof of ζ( + it. Actually, Λ( log is clearly implied by the elemetary idetity (.3 together with Λ. Thus the latter statemet is the oly arithmetic iput eeded for the proof. But eiatig the use of Λ or provig it without use of the fudametal theorem of arithmetic seems impossible: Eve the simplest elemetary (i.e. ζ-free proof of the PNT uses Λ. Circumvetig this by takig (.2 as the defiitio of Λ is o solutio, sice the oe eeds the fudametal theorem of arithmetic for provig that Λ satisfies Λ log. A Proof of Ikehara s Tauberia theorem The proof of Theorem.6 will be give usig the Fourier trasform o R, cocerig which we oly eed the defiitio ad a easy special case of the Riema-Lebesgue Lemma: A. Lemma Let f : R C be cotiuous with compact support. The the Fourier trasform f(ξ : f(xe iξx dx (A. (itegrals without bouds always exted over R satisfies f(ξ as ξ. Proof. For ξ, we have f(ξ f(xe iξ(x π ξ dx f(x + π ξ e iξx dx, 4

5 which leads to f(ξ 2 (f(x f(x + πξ e iξx dx. (A.2 Beig cotiuous with compact support, f is uiformly cotiuous, thus sup x f(x f(x + ε as ε. Isertig this i (A.2 ad keepig i mid that supp(f is bouded, the claim follows. Proof of Theorem.6. As a preparatory step, pick ay cotiuous, compactly supported eve fuctio g : R R such that g( ad ĝ(ξ ξ, where ĝ is as i (A.. (A possible choice is the tet fuctio g(x max( x,. By a easy computatio, ĝ(ξ 2( cos ξ/ξ 2, so that all assumptios are satisfied. The the Fourier iversio theorem applies, thus g(x (2π ĝ(ξe iξx dξ, ad i particular we have ĝ(ξdξ 2πg( 2π. (This is our oly use of Fourier iversio, ad it ca be avoided at the expese of showig ĝ(ξdξ 2π by had, which ca be doe for the above choice of g, if somewhat arduously. Defiig g (x g(x/, it is immediate that ĝ (ξ ĝ(ξ ad ĝ 2π for all. For every δ > we have ξ δ ĝ(ξ u δ ĝ(udu, which teds to zero as sice ĝ is itegrable ad o-egative. Thus the ormalized fuctios h ĝ /2π costitute a approximatio of uity. (I.e. fh f( for sufficietly ice f. Let G(z be the cotiuous extesio (clearly uique of F (z c z a to Re(z a. Defiig φ(x e ax f(x, we have for all t R, ε > : G(a + ε + it F (a + ε + it c ε + it (φ(x ce (ε+itx dx ψ ε (xe itx dx, (A.3 where we abbreviate ψ ε (x e εx (φ(x c to make the followig computatio readable. Multiplyig (A.3 by e iyt g (t ad itegratig over t, we have ( e iyt g (tg(a + ε + itdt e iyt g (t ψ ε (xe itx dx dt ( ψ ε (x g (te it(y x dt dx ψ ε (xĝ (y xdx e εx φ(xĝ (y xdx c e εx ĝ (y xdx. (A.4 (The secod equality holds by Fubii s theorem, sice ψ ε ad g are itegrable. The third idetity uses that g is eve. We ow cosider the it ε of (A.4. Sice G by assumptio is cotiuous o Re(z a, we have G(a+ε+it G(a+it, uiformly for t i the compact support of g. Thus ε e iyt g (tg(a + ε + itdt e iyt g (tg(a + itdt. Sice φ ad ĝ are o-egative, the mootoe covergece theorem gives ε ε e εx φ(xĝ (y xdx e εx ĝ (y xdx Thus the ε it of (A.4 is e iyt g (tg(a + itdt φ(xĝ (y xdx, ĝ (y xdx y ĝ (xdx. y φ(xĝ (y xdx c ĝ (xdx. Let a, b >. Twofold partial itegratio gives e ax a cos bx dx. Applyig c db gives a 2 +b 2 ax si cx e dx arcta c. Applyig f dc gives ax cos fx e dx f arcta f a log ( + ( f 2. (The exchages of itegratio order are justified by absolute covergece ad Fubii. Takig f ad the it a we x a x 2 a 2 a obtai cos x dx π. x (A.5

6 Sice t g (tg(a + it is cotiuous ad compactly supported, Lemma A. gives e iyt g (tg(a + itdt, y + with which the complex umbers leave the stage. Obviously, y ĝ (xdx ĝ (xdx 2π. y + Thus for y +, (A.5 becomes (i terms of h ĝ /2π y + φ(xh (y xdx c. (A.6 It remais to show that this implies x + φ(x c, usig that {h } is a approximate uit ad that f(x e ax φ(x is o-decreasig. For δ > we have φ(x + δ yh (ydy f(x + δ ye a(x+δ y h (ydy δ δ f(x + δ ye a(x+δ y h (ydy f(xe a(x+2δ h (ydy δ φ(xe 2aδ h (ydy φ(xe 2aδ I (δ, where we write I (δ δ h (ydy. By (A.6, the l.h.s. teds to c as y, so that sup φ(x c e2aδ x + I (δ N, δ >. Recallig that I (δ for each δ >, takig gives sup x + φ(x ce 2aδ for every δ >. Takig δ we obtai sup x + φ(x c. This also implies that φ is bouded above: φ(x M for some M >. O the other had, φ(x δ yh (ydy δ δ φ(x δ yh (ydy + M y δ h (ydy f(x δ ye a(x y h (ydy + M φ(xe 2aδ I (δ + M( I (δ. ( δ h (ydy This gives With (A.6 we obtai φ(x δ yh (ydy M( I (δ φ(x e 2aδ. I (δ if x φ(x c M( I (δ e 2aδ. I (δ Takig we fid if x φ(x c, ad δ gives if e 2aδ x φ(x c. We have thus prove φ(x c as x +, which is equivalet to f(x ce ax. 6