Most results in this section are stated without proof.

Transcription

1 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 even integers. Any other zeros ρ lie in the ritial strip 0 Re ρ. his means that ζ (s /ζ (s has simple poles at s =, residue, and at the negative even integers and the zeros in the ritial strip with residue. Lemma For > 0 we have { ( +i y s + O y 2πi i s ds = log y if y > 0 + O ( y log y if 0 < y <. I have written the result in this way to show that the integral is an approximation to the harateristi funtion, if y > and 0 if y <. Proof Apply Cauhy s heorem to y s /s integrated around the square i, + i, U + i and U i, with U > if y > and U < 0 if 0 < y <. Estimate the ontribution from the three edges other than [ i, + i ], and then let U + or U respetively. his is similar to a result in letures: if x > 0 and > 0 then { +i x s x 2πi i s (s + ds = if x > 0 if x <, and there is a variation where the integral is trunated at ±. he Lemma is applied in the proof of heorem Perron s formula.let f (s = n= a nn s, and suppose that n= a n n σ < for all σ >. Further assume that a n A (n, where A (x > 0 is a monotonially inreasing funtion of x, and n= ( a n n σ = O (σ α,

2 as σ +, for some α > 0. hen for >, and x = N + /2 for N N, n x a n = +i ( f (s xs 2πi i s ds + O n= x ( α + O ( xa (2x log x. ( Idea of Proof Without justifying the interhange of integration and summation we have +i f (s xs 2πi i s ds = ( +i ( x s ds a n = a n + R 2πi i n s n x by the Lemma, and where ( ( x R = O a n n log x. n= his sum is split into two parts. Problems might arise from n lose to x, for then x/n is lose to and log (x/n is small. Being on the denominator means its ontribution ould be large. Notie that in demanding x = N +/2 with N N means that x/n, whih otherwise would be a problem. Nevertheless the ontribution from n with /2 x/n 2 an be bounded by by the seond error in (. he remaining terms, for n satisfying either x/n < /2 or x/n > 2, an be bounded by the first error in (. We are going to apply this to f (s = ζ (s /ζ (s in whih ase a n = Λ (n. he pole of ζ (s at s = beomes a simple pole of the logarithmi derivative ζ (s /ζ (s, i.e. ζ (s ζ (s = g (s (s, for some funtion g, holomorphi at s =. hus the α will be. Reall that Λ (n = 0 unless n = p a, when Λ (n = log p log p a = log n. herefore A (x an be hosen as log x. As in most appliation hoose = + log x for then x = x +log x = xe log x/ log x = ex. n 2

3 hus ψ (x = n x Λ (n = +i 2πi i ζ (s x s ζ (s s ds + O ( x log 2 x. Now look upon the line [ i, + i ] as the right hand vertial line of a retangle, Γ say, with orners i, + i, /2 + i and /2 i. he idea is that +i = i Γ /2+i +i /2 i /2+i i /2 i We use Cauhy s residue theorem in this losed ontour Γ to get = x x ρ ρ ζ (0 ζ (0. Γ Im ρ. he first term omes from the pole of ζ (s /ζ (s at s =, the sum from the non-trivial zeros ρ of ζ (s, and the final term from the pole of /s whih appears in the integrand. We now have to estimate the integrals over the three sides [ + i, /2 + i ], [ /2, i, i ] and [ /2 + i, /2 i ]. here is a subtle point here onerning the horizontal lines of integration, [ + i, /2 + i ] and [ /2, i, i ]. hese ross the ritial strip, 0 < Re s <. his strip ontains the ritial zeros and it is important that the lines of integration do not go over any of these zeros, for otherwise the integrand will have a pole. Definition. Let N ( = {ρ : ζ (ρ = 0, 0 < Re ρ <, 0 < Im ρ < }. he first result gives an upper bound on the number of ritial zeros. Lemma For > 0 suffiiently large N ( + N ( log. (2 Proof not given. 3

4 Note this does atually say there are any non-trivial zeros satisfying 0 < Re ρ <, < Im ρ < +. It an be shown though that for suffiiently large we have N ( = ( 2π log 2π 2π + O ( log 2. his says quite aurately that there are many ritial zeros. But (2 is suffiient for us to say that given suffiiently large we an hoose [, + ] suh that the horizontal lines Im s = and Im s = are a distane log away from any non-trivial zero. his is suffiient to give bounds on ζ (s /ζ (s that lead to the error term in heorem Expliit formula Let 2 > < x. hen ψ (x = n x Λ (n = x Im ρ x ρ ρ + O ( x log 2 x Corollary Prime Number heorem with an error term. On the Riemann Hypothesis ψ (x = x + O ( x /2 log 2 x. Proof Reall that the Riemann Hypothesis means that Re ρ = /2 for all ritial zeros. hus x ρ ρ x Re ρ = x /2 ρ ρ Im ρ Im ρ Im ρ For this sum, split Im ρ into the union of n Im ρ < n+, for n <. he first ritial zero has imaginary part approximately so we only 4

5 need n 4 in this sum. hus Im ρ ρ = = n=4 n Im ρ <n+ n=4 ρ n=4 (N (n + N (n n n n Im ρ <n+ n=4 log n n using (2 log n=4 n log2. Combining ψ (x = x + O ( x /2 log 2 ( x log 2 x + O. Simply hoose as a large power of x, i.e. x 00, to get the stated result. Results on the position of the ritial zeros are most often given in terms of the regions not ontaining zeros, i.e. the zero-free regions. So the Riemann Hypothesis is that there are no zeros in /2 < Re s <. his has not been proved; it has not even been proved that there exists δ > 0 suh that there are no zeros in /2 + δ < Re s <. Most positive results are of the form that there exists a funtion η ( 0 as + for whih is zero-free for suffiiently large. {s : η ( < Re s <, Im s < } By the methods in letures it an be shown that we may hoose η ( = log 9, for some onstant > 0, due first to Landau. With a little more work we an take η ( = log, 5

6 due to Hadamard and de la Vallee Poussin, 896. he best possible result to date is η ( = (log 2/3 (log log, /3 due to Vinogradov and Korobov, 958. Whatever the zero-free region, the expliit formula gives ψ (x = x + O ( x η( log 2 ( x log 2 x + O. (3 Here is a parameter to be hosen and we do so by equalising the two error terms, i.e. setting log 2 x η( aking logarithms, = log2 x, i.e. log2 = x η( log 2 x. log + 2 log log = η ( log x + 2 log log x. he hoie of will be suh that log log and log log x are of approximately the same size, so we demand Example he zero free region log for all suffiiently large implies ψ (x = x + O log = η ( log x. (4 < Re s <, Im s < ( ( x exp log x, for some onstant > 0 and x suffiiently large. Solution Choose to satisfy log = ( log log x, i.e. = exp log x. Here I am using the standard notation that represents a onstant that may not be the same at eah ourrene. Note that log log = log log x + log. 2 6

7 Note also that for any ε > 0 we have (take logs. hus x log 2 x log 2 x < exp ( ε log x, ( ( = O x exp log x. here is no need to look at the other error term in (3 sine we have hosen to make these terms the same size. 7

8 Problems What error terms in the Prime number heorem arise from the following zero-free regions? a Re s > /2 + δ for any fixed δ > 0? b (Hard log 9 < Re s <, Im s <? < Re s <, Im s <? (log 2/3 /3 (log log 8