Zero-Free Region for ζ(s) and PNT

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1 Contents Zero-Free Region for ζs an PN att Rosenzweig Chebyshev heory ellin ransforms an Perron s Formula Zero-Free Region of Zeta Funtion 6. Jensen s Inequality Borel-Carathéoory Lemma ain Argument Prime Number heorem Notation: I will use the Vinograov notation. Chebyshev heory We let π enote the number of primes not eeeing, ϑ = p log p, ψ = n Λn. We reall some basi properties of Dirihlet series. Suppose that n = a k b m, km=n an let A, B, an C be the orresponing summatory funtions. hen C = heorem.. Chebyshev For, ψ. Proof. By öbius inversion, km a k b m = k Λn = n µ log n a k B k Set := n log n. hen we an write ψ = n Λn = n µ log n = µ n Disretizing, it is eivent that N loguu N N+ loguu for any N Z. Sine the antierivative of log is log this is a trivial appliation of integration by parts, we obtain N logn N N N + logn + N + Hene, if N =, then log = N N logn N + N log N N log We onlue that = log + Olog. We replae µ by a mok öbius funtion a that in some way is a trunate approimation to µ. Let D be a finite set of numbers, an let a be supporte in. hen by our preeing estimate, D a = log D a a log + Olog D

2 Sine we are trying to show that ψ, we only onsier a whih satisfiy the onition a D = with the hope that a log D is lose to. By efinition of, we see that a = a logn = a Λk = ΛkE k, D n n k n km a Λk = k where Ey := m y a = y a y. If y, then µ y = µ = µ = µ =, k y k y n y n where we use the fat that n µ = {n=}. Sine D =, we see that = y a = a y + { y } a = Ey + { y a D D D D whih implies that D a { y } = Ey. oreover, this shows that Ey is perioi with perio iviing lm D. Set a :=, a :=, an a := for >. Clearly, a D = an a log D = log. For y <, Ey =, an for y <, Ey = ; Ey is perioi with perio. Hene, ψ ψ = Λk ΛkE k Λk = ψ <k k k Sine k ΛkE k = D a = log + Olog, we obtain the lower boun We also obtain the upper boun a ψ log + Olog ψ ψ log + Olog Let C > be suh that ψ ψ log C log for. Let r be the maimal integer suh that. Note that r log = log r log r = Olog. r ψ ψ r r = ψ i ψ r i+ log + C log i log + Cr log log + C log i= We onlue the upper boun ψ log + Olog, whih ompletes the proof. i= Chebyshev obtaine better onstants than above, by setting a = a =, a = a = a 5 =, an a =, otherwise. His result was that ψ.9 + Olog, ψ.56 + Olog }, Corollary.. For, ϑ = ψ + O, ψ π = log + O log Proof. First, note that if Λn, then there eists a prime p an a positive integer k suh that n k = p. Hene, ϑ k = p log p an therefore p k ψ = Λn = log p = ϑ k n Clearly, ϑy ψy y, where the last inequality is an appliation of the preeing theorem. Hene, ψ ϑ = k= p k k= ϑ k + log

3 For the seon assertion, note that ϑ efines a Lebesgue-Stieltjes measure. Hene, by integration by parts, [ ] ϑ π = log u ϑu = log ϑ ϑu log + ulog u u For the integral, we have the estimate ϑu ulog u u log u u = u log u u= + u log u log + u ulog u log he following is immeiate from Chebyshev s theorem an the preeing orollary. Corollary.. For, ϑ an π log. ellin ransforms an Perron s Formula Lemma.. Let R >. Consier σ+i s πi σ i s s = > =, σ < Proof. Consier the integral over the retangular ontour γ with verties A±i, σ ±i. We use the resiue theorem to alulate the integral in the statement of the proposition. We have the following estimates: σ=σ σ= A σ σ ± i σ σ, > log t=+ t= A A t, A > A + it A he integral s πi γ s s = elog s s= = by Cauhy s integral formula. Suppose first that >. hen an then, we onlue that = lim [ πi t= t= σ+it t= it + lim σ + it A t= A+it ] σ A + it it + σ+± σ= A σ ± i t = σ+i s πi σ i s s Now suppose that <. he integrals over σ ± it, A σ σ still go to in the limit. We have the estimate t= σ+it σ + it it σ σ t= If we take σ larger, while keeping fie, then the RHS. Hene, for any A >, We onlue that If =, then lim πi lim t= A+it A + it it = σ+i s πi σ i s = it = lim σ + it πi logσ + it t=,

4 where we take the branh of the logarithm with argument in [ π, π. Hene, [ ] = lim πi i artan artan σ As, artan σ π an artan σ π. We onlue that lim πi σ σ + it it = πi π i = heorem.. Perron s Formula Let αs be a Dirihlet series with absissa of onvergene σ. If σ > ma, σ an >, then n σ+i a n = lim αs s πi σ i s s, where iniates that if is an intege, then the last term is to be ounte with weight. Proof. Fi R an hoose N so large that N > +, an write αs = n N a n n s + n>n a n n s = α s + α s Applying Lemma. an observing that n for all n with equality if an only if Z, we have σ+i lim α s s πi σ i s s = σ+i s a lim n πi n σ i s s = a n n As for α s, observe that α s = n u s Au AN = s Sine Au AN u θ for θ > ma, σ, we have α s s N u θ s u s N N Au ANu s u u θ σ u = s N θ σ, σ θ for σ > θ > ma, σ, where the impliit onstants may epen on the a n. hus, ±i σ ±i an α s s s s σ +i i s θ σ σ N s N σ θ s θ σ θ σ σ N θ σ = N σ θ log N α s s N θ s s θ t = N θ N θ N θ σ N σ X = N θ σ N N σ θ log N, for large. ake θ so that σ > θ > ma, σ. Sine α s s s is holomorphi in the half plane σ > maσ,, by Cauhy s theorem, σ+i i +i σ+i α s s σ i s s = α s s σ i s s + α s s i s s + α s s +i s s σ N θ σ Combining our estimates, we onlue that lim sup a n σ+i αs s πi σ i s s σ N θ σ n Sine this hols for arbitrarily large N, letting N, we onlue that lim sup is, whih ompletes the proof. 4

5 Define the sine integral si by Lemma.. si min, for >. Proof. Integrating by parts, we have si = osu + u si = sinu u u osu u u = os + osu u u Also note that Lemma.4. for σ >. si + si = sinu u = π u + O σ+i y s s yσ y πi σ i s = + π si log y + O σ π si log y + O y σ y O yσ y Proof. Suppose first that y, an let C be the pieewise linear path from infty i to σ i to σ + i to + i. It follows from our work in Lemma. that y s s πi s = Furthermore, σ±i ±i C y s s σ s = y σ±i σ ± i σ σ σ = yσ log y yσ If y, then we take the analogous pieewise path to the right of σ an note that the resiue of the integran is zero. Suppose that y. Let C be the lose retangular path from σ i to σ + i to i to i with a semiirular inentation of raius ɛ > to the right. By Cauhy s theorem, Observe that σ±i ±i πi C y s s s = y s s s σ y σ σ = logσ yσ σ he semiirle tens to as ɛ sine the integran has a pole with resiue at s =. For the remaining integral, observe that i πi lim y s iɛ ɛ iɛ s y s s = i s πi lim y it y it t ɛ ɛ t = π lim sinlog yt t ɛ ɛ t ake the hange of variable u = log yt to obtain = [ log y sinu u π u = sinu u π u sinu u ] log y u sinu u = u π he ase y is ompletely analogous. [ π + si log y π ] = si log y π 5

6 Proposition.5. Perron Formula Error If σ > ma, σ a an >, then n a n = σ+i πi σ i αs s s + R, s where R = π a n si log n π <n< <n< a n si log n 4 σ + σ a n + O n σ n Proof. Sine the series is absolutely onvergent on the interval [σ i, σ + i ], we an interhange the orer of summation an integration to obtain σ+i αs s πi σ i s s = n= he rest of the proof follows from the preeing lemma. Corollary.6. If σ > ma, σ a an >, then Proof. R <n< n Definition.7. If F s is given by σ+i s s a n πi σ i n s + a n min, + 4σ σ n F s = f s, then we say that F s is the ellin transform of f. When f = πi σ+i σ i hols, we say that f is the inverse ellin transform of F. Zero-Free Region of Zeta Funtion. Jensen s Inequality F s s s Lemma.. Jensen s Inequality If fz is analyti in a omain ontaining D; R, if fz for all z R, an if f, then for r < R, the number of zeroes of f in D; R oes not eee log f Proof. Sine f f, f has finitely many zeroes z,, z K in the isk D; R. Set gz := fz log R r K k= R zz k Rz z k Eah fator z z k anels the zero of f at z = z k, so gz is analyti in D; R. I laim that gz = fz for z = R. Inee, for eah k, R zz k Rz z k = R zz k Rz z k R zz k Rz z k = R4 R zz k + zz k + R z k R z z k = z zz k + zz k + z k z z k = n= a n n σ 6

7 he onlusion of the laim follows immeiately. Hene, gz = fz for all z = R. By the maimum moulus priniple, g. But g = f = f Observe that eah fator in the prout is, an if z k r, then the fator is R r. Fi r < R, an let L equal the number of zeroes of f in D; r. hen f K k= K k= R z k L R R z k f log r f L log R r L log f log R r. Borel-Carathéoory Lemma Lemma.. Borel-Carathéoory Lemma Suppose that hz is analyti in a omain ontaining the lose isk D; R, that h =, an that Re hz for z R. If z r < R, then hz r R r, h z R R r Proof. If we show that h k k!, k, Rk then sine hz = h k k= k! z k for z R, for all z r < R, hz h k r k r k! z k = R R r k= k= an h z h k+ k! z k R k= r k R k + = R R r = R R r k= Note that by Cauhy s formula, hre πiθ θ = hz z πi z =R z = h = oreover, if k >, then sine the funtion hzz k is analyti in a omain ontaining D; R, we have by Cauhy s formula that hre πiθ e πikθ θ = R k πi By Cauhy s integral formula for erivatives, Let φ R. hen hre πiθ + osπkθ + φθ = z =R hzz k z = hre πiθ e πikθ θ = Rk hzz k z = Rk h k πi z =R k! hre πiθ θ + e πiφ hre πiθ e πikθ θ + e πiφ hre πiθ e πikθ θ = e iφ R k h k k! 7

8 By taking real parts, it follows that Re Rk e πiφ h k k! [ + osπkθ + φ]θ = for k >. Sine the above inequality hols for any φ R, we an hoose φ suh that e πiφ h k = h k, whih yiels the esire inequality. If P z = K k= z z k, then P K P z = We now generalize this to arbitrary analyti funtions. f K f z = k= k= z z k + O z z k Lemma.. Suppose that fz is analyti in a omain ontaining the isk D;, that fz in this isk, an that f. Let r, R be fie, < r < R <. hen for z r, we have log, where the sum runs over all zeroes z k of f in D; R. f Proof. If fz vanishes on the irle z = R, then sine the zeroes of f are isolate, we an replae R by a slightly larger R > R, suh that fz for z = R. So without loss of generality, we assume that fz for z = R. Set gz = fz By Lemma. Jensen s Inequality, we know that K log log R K k= f R zz k Rz z k log f If z = R, then eah fator in the prout giving gz has moulus. hus, gz = fz for z = R. By the maimum moulus priniple, gz for all z R. Also observe that g = f Sine gz has no zeroes in D; R, we may efine hen h =, an K k= hz := log R z k f gz g Re hz = log gz log g log log f = log From the Borel-Carathéoory lemma, we obtain h z R R r log f f, But h z = g g z = f K f z K + z z k z R k= k= z k z Note that R z k R, so that if z r, then R z k R r. Hene, for z r, f K f z z z k R R r log K f + R r = R R r log f + k= k= z R K R r R + R r log f 8

9 . ain Argument Lemma.4. If t 7 8 an 5 6 σ, then ζ ζ s = ρ + Olog τ, s ρ where τ = t + 4 an the sum is etene over all zeroes ρ of ζs for whih ρ + it 5 6. Proof. We apply the preeing lemma to the funtion fz = ζz + + it, with R = 5 6 an r =, Lemma.5. If σ >, then Re ζ ζ σ 4ζ ζ ζ σ + it σ + it ζ Proof. For σ >, the LHS above is Λnn δ + 4 ost log n + ost log n n= Realling the trigonometri ientity + 4 osθ + osθ = + osθ ompletes the proof. heorem.6. Zero-Free Region here is an absolute onstant > suh that ζs for σ log τ. Proof. We know that for σ >, ζs, so it suffies to onsier σ. Reall from the etension of ζs to the half plane σ >, we have the estimate ζs s s = s {u} u s u s σu σ u = s σ σ If ζs = for σ > s, then ζs s s = s s > s σ, whih is a ontraition. Note that σ > s = σ + t = σ σ + + t σ > + t In partiular, ζs in the retangle 8 9 σ, t 7 8, sine = = 8 8 < 8 9 Now suppose that ρ = β + iγ is a zero of ζs with 5 6 β an γ 7 8. Sine Re ρ for all zeroes ρ of ζs, we have that Re s ρ >, whenever σ >. Applying Lemma.4 with s = + δ + iγ, we have Re ζ ζ + δ + iγ + log γ δ β Applying Lemma.4 with s = + δ + iγ, we see that Re ζ ζ + δ + iγ log γ + 4, where > is some onstant, inepenent of δ an ρ. Sine ζs has is meromorphi in the halfplane σ >, with only a simple pole at s =, we have that Re ζ + δ = ζ ζ ζ + δ = δ + O 9

10 Putting these estimates together an applying Lemma.5, we obtain that whih implies that δ log γ log γ O, + δ β δ 4 + log γ + 4, + δ β for some onstant > inepenent of δ, ρ. Choose δ = log γ + 4. hen the above yiels or 4 δ β 6 log γ log γ + 4 = 7 log γ + 4, β 4 7 log γ + 4 log γ + 4 = 4 log γ + 4 heorem.7. Let be a onstant suh that ζs is zero-free in the region an t 7 8, then ζ ζ s log τ, log ζs log log τ + O, log τ ζs If log τ < σ an t 7 8, then ζ ζ s = s + O, logζss, s ζs 4 Prime Number heorem heorem 4.. here is a onstant > suh that ψ = + O ep, ϑ = + O log where li = log uu is the logarithmi integral. ep log { } σ log τ. If σ > log τ, π = li + O ep log Proof. We first show that ψ = + O ep for some onstant >. Sine ψ = ϑ + O by log Chebyshev s estimates, it follows that ϑ = + O ep. log By Perron s formula with error term, for σ > an, ψ = σ+i ζ πi σ i ζ ss s + R, R Λn min, + 4σ Λn s n n σ <n< n= Observe that the series in the seon term is just ζ ζ σ, whih is σ for < σ. We now estimate the first sum. First, note that Λn log n log. For the n, < n <, losest to we replae the the min fator by the first argument an the for the remaining n, we replae the min fator by the seon argument. Hene, Λn min <n<, n log + k log + log k Suppose now that an set σ = σ := + log. hen putting together the preeing estimates, we have R Λn min <n<, + 4σ n Λn log + n σ log + 4+ n= log log + log,

11 Let > be a small positive onstant suh that +ɛ log, for some ɛ >, is a zero-free region for ζs. Set σ = σ := log. Let C enote the positively oriente retangular ontour with verties σ ±i, σ ±i. By heorem.6, ζ ζ s is meromorphi on an insie C with only a simple pole at s = with resiue. By Cauhy s resiue theorem, ζ πi C ζ ss s s = Using the estimates of heorem.7, we have that σ±i ζ σ ζ ss s s log σ σ log σ ±i σ σ σ σ = log + log log + log For the piee of the ontour integral from σ i to σ + i, we nee to onsier 7 8 an > 7 8, separately. Applying heorem.7 again, we have the estimate σ+i ζ σ i ζ ss s s ζ σ + it σ log ζσ + it t σ + it + t σ t + + O σ + it σ t σ σ σ log + σ σ log Combining these estimates, we have ψ = + O log + log We seek suh that = log. akig the logarithm of both sies, we have log = log log log = log = ep log is the esire hoie of. We onlue that the error term is log ep log ep log, sine replaing by a smaller if neessary, we may assume that < <. We have alreay ealt with the seon assertion. For the thir assertion, we note that ϑ efines a Lebesgue-Stieltjes measure, an we use integration by parts to obtain π = ϑu = li + log u ϑu u = li + log u log u ϑu u + ϑu uu ulog u Sine ϑ = + O ep log, the onlusion follows immeiately.

Most results in this section are stated without proof.

Most results in this section are stated without proof. Leture 8 Level 4 v2 he Expliit formula. Most results in this setion are stated without proof. Reall that we have shown that ζ (s has only one pole, a simple one at s =. It has trivial zeros at the negative