Math 229: Introduction to Analytic Number Theory The product formula for ξ(s) and ζ(s); vertical distribution of zeros

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1 Math 9: Introduction to Analytic Number Theory The product formula for ξs) and ζs); vertical distribution of zeros Behavior on vertical lines. We next show that s s)ξs) is an entire function of order ; more precisely: Lemma. There exists a constant C such that s s)ξs) expc s log s ), but no constant C such that s s)ξs) expc s ). Proof : The second part is easy, because ξs) already grows faster than expc s ) as s along the positive real axis. Indeed for s > we have ζs) >, while the factor s s)π s/ Γs/) of s s)ξs) grows faster than expc s ) by Stirling. For the first part, since s log s is bounded below we need only give an upper bound on s s)ξs) for large s, and the functional equation ξs) = ξ s) lets us assume that s = σ + it with σ /. Moreover, the sum and integral formulas for ζs) and Γs) yield ξσ + it) π σ/ Γσ/)ζσ) whenever σ >. By Stirling we readily conclude that s s)ξs) expc s log s ) holds if s is far enough from the critical strip, say in the half-plane σ. To extend this bound to σ /, it will be more than enough to prove that ζs) t for σ / and t ; and this follows from our formula ζs) = s + n= n+ n n s x s ) dx for ζs) as a meromorphic function on the half-plane σ > 0, in which we saw that the sum is O s ). Remarks: While the bound ζs) t on the critical strip away from s = ) is more than sufficient for our present purpose, we can do considerably better. To begin with, we can choose some N and leave alone the terms with n < N in ζs) = n= n s, using the integral approximation only for the n N terms: ζs) = N n= n s + N s s + n=n n+ n n s x s ) dx. For large t, N this is N σ + t N σ, uniformly at least for σ /. Taking N = t + O) we find ζσ + it) t σ for σ /, t >. At the end of this chapter we describe further improvements. Meanwhile, note our choice of N = t + O), which may seem suboptimal. We wanted to make the bound as good as possible, that is, to minimize N σ + t N σ. In calculus we learned to minimize such an expression by setting its derivative equal to zero. That would give N proportional to t, but we arbitrarily set the constant of proportionality to even though another choice would make N σ + t N σ somewhat smaller. In general when we bound some quantity by a sum OfN) + gn)) of an increasing and a decreasing function of some parameter N, we shall simply choose N so that fn) = gn) or, if N is constrained to be an integer, so that

2 fn) and gn) are nearly equal). This is much simpler and less error-prone than fumbling with derivatives, and is sure to give the minimum to within a factor of, which is good enough when we are dealing with O ) bounds. Product and logarithmic-derivative formulas. By our general product formula for an entire function of finite order we know that ξs) has a product expansion: ξs) = ea+bs s s/))e s/, ) s for some constants A, B, with the product ranging over zeros of ξ that is, the nontrivial zeros of ζ) listed with multiplicity. Moreover, ɛ < for all ɛ > 0, but = because ξs) is not Oexp C s ). The logarithmic derivative of ) is ξ ξ s) = B s s + s + ) ; ) since ξs) = π s/ Γs/)ζs) we also get a product formula for ζs), and a partial-fraction expansion of its logarithmic derivative: ζ ζ s) = B s + log π Γ Γ s + ) + s + ). 3) We have shifted from Γs/) to Γs/ + ) to absorb the term /s; note that ζs) does not have a pole or zero at s = 0.) Vertical distribution of zeros. Since the zeros of ξs) are limited to a strip we can find much more precise information about the distribution of their sizes than the convergence of ɛ and the divergence of. Let NT ) be the number of zeros in the rectangle σ [0, ], t [0, T ]; note that this is very nearly half of what we would call nt ) in the context of the general product formula for s s)ξs). Theorem von Mangoldt). As T, NT ) = T π log T π T + Olog T ). 4) π Proof : We follow chapter 5 of [Davenport 967], keeping in mind that Davenport s ξ and ours differ by a factor of s s)/. We may assume that T does not equal the imaginary part of any zero of ζs). Then NT ) = ξ πi CR s) ds = dlog ξs)) = dim log ξs)), ξ πi C R π C R where C R is the boundary of the rectangle σ [, ], t [ T, T ]. Since ξs) = ξ s) = ξ s), we may by symmetry evaluate the last integral by

3 integrating over a quarter of C R and multiplying by 4. We use the top right quarter, going from to + it to /) + it. Because log ξs) is real at s =, we have πnt ) ) = Im log ξ +it ) = Imlog Γ 4 + it )) T log π+imlog ζ +it )). By Stirling, the first term is within OT ) of it Im ) it log 4 + 4)) T = T log it Im log it + ) T 4 = T T log ) + O). Thus 4) is equivalent to Im log ζ + it ) log T. 5) We shall show that for s = σ + it with σ [, ], t > we have ζ ζ s) = Ims ) < + Olog t ), 6) s and that the sum comprises at most Olog t ) terms, from which our desired estimate will follow by integrating from s = + it to s = / + it. We start by taking s = + it in 3). At that point the LHS is uniformly bounded use the Euler product), and the RHS is + it + ) + Olog t ) by Stirling. Thus the sum, and in particular its real part, is Olog t ). But each summand has positive real part, which is at least /4 + t Im ) ). Our second claim, that t Im < holds for at most Olog t ) zeros, follows immediately. It also follows that Now by 3) we have Ims ) ζ ζ s) ζ ζ + it) = log t. Ims ) s ) + O). + it LHS and RHS are the left-hand side and right-hand side of an equation. 3

4 The LHS differs from that of 6) by O), as noted already; the RHS summed over zeros with Ims ) < is within Olog t ) of the RHS of 6); and the remaining terms are σ) Ims ) s ) + it ) Ims ) log t. Ims ) This proves 6) and thus also 5); von Mangoldt s theorem 4) follows. For much more about the vertical distribution of the nontrivial zeros of ζs) see [Titchmarsh 95], Chapter 9. Further remarks Recall that as part of our proof that s s)ξs) has order we showed that for each σ there exists νσ) such that ζσ + it) t νσ) as t. Any bounded νσ) suffices for this purpose, but one naturally asks how small νσ) can become. Let µσ) be the infimum of all such νσ); that is, µσ) := lim sup t log ζσ + it). log t We have seen that µσ) = 0 for σ >, that µ σ) = µσ) + σ by the functional equation so in particular µσ) = σ for σ < 0), and that µσ) for σ <, later improving the upper bound from to σ. For σ 0, ) an even better bound is obtained using the approximate functional equation for ζs) usually attributed to Siegel, but now known to have been used by Riemann himself) to show that µσ) σ)/; this result and the fact that µσ) 0 for all σ, also follow from general results in complex analysis, which indicate that since µσ) < for all σ the function µ ) must be convex. For example, µ/) /4, so ζ + it) ɛ t 4 +ɛ. The value of µσ) is not known for any σ 0, ). The Lindelöf conjecture asserts that µ/) = 0, from which it would follow that µσ) = 0 for all σ / while µσ) = σ for all σ /. Equivalently, the Lindelöf conjecture asserts that ζσ+it) ɛ t ɛ for all σ / excluding a neighborhood of the pole s = ), and thus by the functional equation that also ζσ +it) ɛ t / σ+ɛ for all σ /. We shall see that this conjecture is implied by the Riemann hypothesis, and also that it holds on average in the sense that T 0 ζ + it) dt T +ɛ. However, the best upper bound currently proved on µ/) is only a bit smaller than /6; when we get to exponential sums later this term we shall derive the upper bound of /6. Exercises. Show that in the product formula ) we may take A = 0. Prove the formula γ = lim ζs) ) s s 4

5 for Euler s constant, and use it to compute ξ B = lim s 0 ξ s) + s ) s = lim ξ s ξ s) + s ) s = log 4π γ = Show also starting by pairing the and terms in the infinite product) that B = Re)/, and thus that Im) > 6 for every nontrivial zero of ζs). [From [Davenport 967], Chapter. It is known that in fact the smallest zeros have real part / and) imaginary part ± ]. By the functional equation, ξs) can be regarded as an entire function of s s; what is the order of this function? Use this to obtain the alternative infinite product ξs) = ξ/) 4s s ) + [ ) ] s /, 7) / the product extending over zeros of ξ whose imaginary part is positive. [This symmetrical form eliminates the exponential factors e A+Bs and e s/ of ). We nevertheless use ) rather than 7) to develop the properties of ξ and ζ, because ) generalizes to arbitrary Dirichlet L-series, whereas 7) generalizes only to Dirichlet series associated to real characters χ since in general the functional equation relates Ls, χ) with L s, χ), not with L s, χ).] 3. Let f be any analytic function on the vertical strip a < σ < b such that M f σ) := lim sup t log fσ + it) log t is finite for all σ a, b). Prove that M f is a convex function on that interval. [Hint: Apply the maximum principle to αf for suitable analytic functions αs).] It follows in particular that M f is continuous on a, b). While ζs) is not analytic on vertical strips that contain s =, we can still deduce the convexity of µ : R R from µσ) = M f σ) for fs) = ζs) /s )). Much the same argument proves the three lines theorem : if f is actually bounded on the strip then log sup t fσ + it) is a convex function of σ. The name of this theorem alludes to the equivalent formulation: if a < σ < σ < σ 3 < b then the supremum of fs) on the line s = σ +it is bounded by a weighted geometric mean of its suprema on the lines s = σ + it and s = σ 3 + it. Reference [Titchmarsh 95] Titchmarsh, E.C.: The Theory of the Riemann Zeta-Function. Oxford: Clarendon, 95. [HA / QA35.T49; nd ed. revised by D.R. Heath-Brown 986, QA46.T44] 5