# Guide The Distribution of Prime Numbers: Large Sieves and Zero-density Theorems

The corresponding functional equation for L s, x contains the sum over d explicitly. We shall not need this case again. A number of proofs of the functional equation can be found in Chapter 2 of Titchmarsh In proving the prime-number theorem, Hadamard studied integral functions of finite order, that is, functions f s regular over the whole. Hadamard showed that an integral function o f finite order can be written as an infinite product containing a factor s p corresponding to each zero p o f the function.

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This generalizes the theorem that a polynomial can be written as a product of linear factors. Weierstrass s definition Note that By Polyas theorem 4. Stirling s formula To bound s we recall eqn 5.

As in the derivation of I f s is real and positive then First we prove a lemma. Then 2. We can write the left-hand side of We shall prove later that 0, x is non-zero, so that When we examine the formulae The converse is not true, because L s, x has extra zeros at negative integer values to cancel the poles o f the gamma function in Riemann stated the hypothesis for s , but it is difficult to conceive a proof o f the hypothesis for s that would not generalize to s, x.

For later use we now prove a result more precise than This proves 12,19 , and We shall not need the following more accurate formula.

## Download The Distribution Of Prime Numbers: Large Sieves And Zero Density Theorems 1972

We can now prove the product formula. Let f s be s, x or s , and let p run through all zeros of f s. By P s is a regular function with the same zeros as f s. We should like f s jP s to be a constant or some other simple function. Since We can substitute the equation.

It can be shown that. You remember how he discovered the North P ole; well, he was so proud of this that he asked Christopher Robin if there were any other Poles such as a Bear of Little Brain might discover. To achieve these, the numbers to-1? To make this argument rigorous, we use For all real 9,. Since s has a pole at 1, there is a circle centre 1 and some radius r, within which s is non-zero. We now write down three inequalities.

When we substitute I f we constrain c to be less than f r then, by For twenty years, This was proved in by Korobov and by I. Vinogradov independently. Intermediate improvements on Vinogradov W hat do you think youll answer? I shall have to wait till I catch up with it, said Winniethe-Pooh.

In this chapter and the next we prove results like We use the product formula in the differentiated form The first complication is the elimination o f B x from eqn We subtract from The inequality I f xa is non-trivial, substitution of The absolute constant c3 in When The third and greatest difficulty is to deal with zeros close to unity when x2 is trivial.

First we show that there is at most one. Since p2 is also a zero when L s, x has real coefficients, if p2 fails to satisfy We devote the next chapter to the case of an exceptional zero ft on the real axis. Piglet said that the best place would be somewhere where a Heffalump was, just before he fell into it, only about a foot farther on.

Just as before we saw that L s ,x cannot have two zeros both close to 1, we shall now see that two functions L s, x corresponding to different proper characters cannot both have zeros close to I. Suppose Xi is proper m od?! In place of To prove the prime-number theorem for an arithmetic progression with common difference q, we need to know that neither s nor any. The proof is simpler if we have p explicitly bounded away from unity.

We have to deal only with the case x real, p real. One method is to interpret L 1, x as the density of ideals in a quadratic number field. This gives a very weak bound. We shall prove Siegels theorem. The constant c e in Siegel s theorem is ineffective; that is, the proof does not enable us to calculate it. All previous constants in upper bounds, such as c2 in Since F s is zero at 0, 2, 4, To prove the more precise result We have. This is a lower bound for the product A of the three L-functions at 1. Since - l,X'a is a known constant, we seek an upper bound far L l , x x Xt From eqn 5.

When we integrate round a circle radius i lo g g -1 to find L ' s ,x , we have for. Hence, if L s, xi has a zero j8x in the range The clock slithered gently over the mantelpiece, collecting vases on the way. WE can now prove the prime-number theorem in the form 5. Com bining 5. We now suppose that x is an integer plus one-half. Then a-f-i T. We must now find an approximate value for the integral on the left of eqn We can now estimate the integral round C.

The contour has been moved past several poles of the integrand. These are all zeros o f s , and by inequality We can quickly deduce from Clearly, only the O-constants in We have approached the prime-number theorem by the classical route of Riemanns functional equation and Hadamards product formula. Neither o f these is necessary for the proof o f eqn There are elementary proofs of eqn 5.

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The elementary proofs are disappointing in that much extra effort is needed to prove the prime-number theorem with an explicit error term, rather than as an asymptotic equality. An analytic proof that does not use the functional equation is given by Titchmarsh , Chapter 3. The contour C consists of Gx: the line segment [aiT, f i? As in We now treat the sum in The value o f c3 depends on that of c2 in Our characters can at last abandon propriety.

The estimate This error term absorbs easily into that o f I f a, q is not unity, only the powers o f primes that divide q can occur in the sum We have proved The result For a given q we may have to choose x larger still to ensure that the second term in Thus From the proof o f Siegel s theorem it follows that, if we knew one L-function with an exceptional zero, we should be able to find the c7 corresponding to one particular value of N in It wasnt what Christopher Robin expected, and the more he looked at it, the more he thought what a Brave and Clever Bear Pooh was.

This is known as W eyls criterion for uniform distribution. Franels theorem o f Chapter 9 fits this scheme. We were concerned there with the distribution of the F Farey fractions in the interval [0,1]. The appropriate functions a m are e bfm , where b is an integer, and a calculation with Ramanujan sums shows that F. Franels theorem relates a measure of the uniform distribution of Farey fractions in the interval to the mean square of the sums in eqn Many results in number theory rest on the uniform distribution of some sequence among certain classes.

This is the final step in the proof of the prime-number theorem for arithmetic progressions. Often the uniformity result is needed at an intermediate stage, and in these cases it is usually sufficient to know in a quantitative sense that An assertion that, for a given sequence m t , and set o f classes, eqn A common form of large-sieve result is that the mean square o f the left-hand side o f The sieve 8.

These correspond to distribution among reduced residue classes, and distribution of the sequence within intervals. The proof o f a large-sieve result depends on an upper bound for an average of some convenient auxiliary function. Let u m be any complex coefficients and The stun on the left of When we restrict our attention to proper characters we can sum over q on the left of A third sieve result for characters is possible if the coefficients u m are 0 whenever to has any factor smaller than Q; for then to,?

This give sanother proof that at most one x has an exceptional zero , blit the range for q is not as large as in Let f x. To obtain large-sieve results from Gallaghers lemma, we arrange the sum on the left o f eqn Like the relation 7.

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Whereas 7. The first integral on the right of Kanga saicl to Roo, Drink up your milk first, dear, and talk afterwards. So Roo, who was drinking his milk, tried to say that he could do both at once. In the last chapter we obtained sieve results in the form of inequalities with two terms on the right-hand side, one of which corresponded to the maximum size of the summands, the other to the mean square of the function times the number of summands. Montgomery a sieved over both x and s and obtained a hybrid result, which is better than the one we should obtain by sieving over the one and summing over the other.

Gallagher has found a simple method of obtaining hybrid sieve results, which we describe here. Plancherels identity. We shall use eqns We can verify from first principles that the interchange o f integrations in the proof of We now deduce Gallaghers second lemma. A more realistic application is the following. Clearly we need Gallaghers first lemma We turn to Cauchys identity.