Mean value theorems for long Dirichlet polynomials and tails of Dirichlet series


 Joseph Ambrose Hawkins
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1 ACA ARIHMEICA LXXXIV Mean value theorems for long Dirichlet polynomials and tails of Dirichlet series by D. A. Goldston San Jose, Calif. and S. M. Gonek Rochester, N.Y. We obtain formulas for computing mean values of Dirichlet polynomials that have more terms than the length of the integration range. hese formulas allow one to compute the contribution of offdiagonal terms provided one knows the correlation functions for the coefficients of the Dirichlet polynomials. A smooth weight is used to control error terms, and this weight can in typical applications be removed from the final result. Similar results are obtained for the tails of Dirichlet series. Four examples of applications to the Riemann zetafunction are included.. Introduction and statement of results. Let {a n } n= be a sequence of real or complex numbers such that for any ε >, a n ε n ε as n. Let s = σ + it be a complex variable and let As = a n n s be a Dirichlet polynomial. By Montgomery and Vaughan s mean value theorem [4] we have As 2 dt = a n 2 n 2σ + On. It immediately follows that if N = o as, then As 2 dt a n 2 n 2σ. On the other hand, if N and σ <, the Oterms in can dominate so that we lose the asymptotic formula. he situation is similar for the 99 Mathematics Subject Classification: Primary M6; Secondary M26. he work of both authors was partially supported by grants from NSF. [55]
2 56 D. A. Goldston and S. M. Gonek meansquare of the tail Dirichlet series A s = a n n s n>n when σ >. Our purpose in this paper is to determine the meansquare behavior of As and A s even when N is significantly larger than. If we square out and integrate termwise in, we see that the Oterms on the righthand side come from offdiagonal terms. It is these we must carefully estimate therefore when N is large. We treat them by appealing to good uniform estimates for the coefficient correlation functions Ax, h = n x a n a n+h. Such estimates are available for a n, a n = dn the divisor function, a n = µ 2 n the square of the Möbius function, when a n is the nth Fourier coefficient of a modular form, and for a number of other arithmetical functions. Moreover, it is interesting to note that we can often formulate a conjectural estimate for Ax, h even when we cannot estimate Ax, h rigorously. In such cases we can then use our theorems to deduce conditional mean value formulae for the associated Dirichlet series. Since it is no more difficult to treat the more general means 2 and 3 AsBs dt A sb s dt, where Bs = b nn s and B s = n>n b nn s, we shall do so. he precise assumptions we shall make about the sequences {a n } n= and {b n } n= are: A For every ε > we have a n, b n ε n ε. A 2 If Ax = n x a n and Bx = n x b n, then for x we may write 4 Ax = M x + E x and 5 Bx = M 2 x + E 2 x, where 6 7 M x, M 2x ε x + ε, M x, M 2 x ε x + ε
3 and 8 for some θ [,. Mean value theorems 57 E x, E 2 x x + θ A 3 he coefficient correlation functions are of the form C x, h = n x a n b n+h and C 2 x, h = n x b n a n+h 9 C i x, h = M i x, h + E i x, h i =, 2 for x, where M i x, h i =, 2 is twice differentiable for each h =, 2,..., and E i x, h x + ϕ i =, 2 uniformly for h x η for some ϕ [, and some η,. Sometimes we shall also assume A 4 For every ε > we have uniformly for x and h =, 2,... M ix, h ε h ε x + ε i =, 2 Instead of estimating 2 and 3 directly, we find it more advantageous to estimate the integrals I = and 2 I = Ψ U t As Ψ U t A s N N M xx s dx Bs M xx s dx B s N N M 2 xx s dx dt M 2 xx s dx dt. Here M x and M 2 x are as in 4 and 5 and Ψ U t is a realvalued weight function satisfying the following conditions. Let B >, U log B, and C C 2, where C and C 2 are bounded but may be functions of U. hen Ψ U t is supported on [C U, C 2 + U ], 3 Ψ U t = if C + U t C 2 U, and 4 Ψ j U t U j for j =,,...
4 58 D. A. Goldston and S. M. Gonek Note that 3 is vacuous if, for example, C = C 2. he removal of Ψ U from I and I is usually straightforward and will be demonstrated in the examples at the end of the paper. Before stating our results we introduce some more notation and useful estimates. We use ε to represent an arbitrarily small positive number which is fixed during the course of each proof. We then set τ = ε. We always assume that σ, the real part of s, is bounded above and below. he constants implied by the symbols O and may depend on ε, the upper and lower bounds for σ, and other parameters, but never on or parameters dependent on, like N and τ. hus, in particular, our Oterms hold uniformly for bounded σ. We define the Fourier transform of Ψ U t by Ψ U ξ = Ψ U teξt dt, where ex = e 2πix. It follows easily that t 5 Ψ U eξt dt = Ψ U ξ and, since Ψ U t is real, that 6 ΨU ξ = Ψ U ξ. Observe that Ψ U and Ψ U are trivially C 2 C + U. Also, integrating by parts j times and using 4, we see that Ψ U ξ and Ψ U ξ are C 2 C + U U/2πξ j if ξ. hus, for j arbitrarily large we have 7 ΨU ξ, Ψ U ξ C 2 C + U min, U/2πξ j. It follows that 8 ΨU ξ ξ D for ξ ε and that 9 ΨU ξ D 2 for ξ τ, where D and D 2 are arbitrarily large constants. We write K σ x, u = K σ x, u,, U = x 2σ + u σ ΨU x 2π log + u x and easily find by 7 that 2 K σ x, u x 2σ
5 Mean value theorems 59 and 2 x K σx, u x 2σ ε for u/x τ. By the mean value theorem of differential calculus and 7 we have 22 ΨU 2π log + u x = Ψ u U + O +2ε 2πx when u/x τ. Using this and 7 it is not difficult to deduce the approximation u 23 K σ x, u = x 2σ ΨU + Ox 2σ +2ε 2πx for u/x τ. We can now state our main results. heorem. Let σ < σ 2 <, let < ε < /2 be fixed, and let τ = ε. Suppose that the sequences {a n } n= and {b n } n= satisfy A, A 2, and A 3 and that 24 τ N τ / η, where η is as in A 3. Set 25 H = N/τ +. hen 26 I = Ψ U t As = Ψ U + H H h H τ τ N u N u N a n b n n 2σ + N h hτ M xx s dx Bs h H N h hτ M 2 x, hk σx, h dx M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du + ON 2σ+maxθ,ϕ+5ε + O 2ε uniformly for σ σ σ 2 and sufficiently large. N M 2 xx s dx dt M x, hk σ x, h dx
6 6 D. A. Goldston and S. M. Gonek he following weaker version of heorem is easier to apply and sufficient for many applications. Corollary. Let the hypotheses and notation be the same as in heorem except now assume that N and that A 4 also holds. Write 2 2σ C = 2π. 2π hen 27 I = Ψ U a n b n n 2σ + C h M h 2πv, h 2σ Ψ U vv 2σ 2 dv /2πN + C h 2πNv/ /2πN h 2πNv/ 2πNv/ M /2πτN 2C M 2 h h 2πv, h 2σ Ψ U vv 2σ 2 dv u u M 2 u 2σ du 2πv 2πv Re Ψ U vv 2σ 2 dv + O N 2 2σ+5ε + ON 2σ+maxθ,ϕ+5ε + ON 2ε uniformly for σ σ σ 2 and sufficiently large. heorem 2. Suppose that the sequences {a n } n= and {b n } n= satisfy A, A 2, and A 3. Let < σ < σ 2, let < ε < /2 be fixed, and set τ = ε. For σ σ σ 2 write λ = 2σ 2σ 2 ; let 28 τ N τ ε+η/λ η, where η is as in A 3, and set 29 = τ λ N λ/ ε. hen 3 I = Ψ U t A s = Ψ U N<n N M xx s dx B s a n b n n 2σ N M 2 xx s dx dt
7 + + h maxn,hτ h maxn,hτ maxn, maxn, Mean value theorems 6 M x, hk σ x, h dx M 2 x, hk σx, h dx M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du + ON 2σ+maxθ,ϕ+5ε + O ε/2 N 2σ uniformly for σ σ σ 2 and sufficiently large. A simpler form of heorem 2 is provided by Corollary 2. Let the hypotheses and notation be the same as in heorem 2 except now assume that N and that A 4 also holds. hen 3 I = Ψ U a n b n n 2σ N<n /2πN + C h h 2πv, h 2σ Ψ U vv 2σ 2 dv h M /2πN + C + C H + C H 2C H /2πN /2πN /2πN h M 2 h h 2πv, h 2σ Ψ U vv 2σ 2 dv 2πNv/ <h M M 2 /2πN 2πNv/ <h /2πN H M 2πNv/ h h 2πv, h 2σ Ψ U vv 2σ 2 dv h h 2πv, h 2σ Ψ U vv 2σ 2 dv u u M 2 u 2σ du 2πv 2πv Re Ψ U vv 2σ 2 dv + O N 2 2σ+5ε + ON 2σ+maxθ,ϕ+5ε + O ε/2 N 2σ uniformly for σ σ σ 2 and sufficiently large, where C is defined in Corollary.
8 62 D. A. Goldston and S. M. Gonek Although we could make the next theorem more precise by arguing along the lines of the proofs of heorems and 2, the version below is usually all that we require. heorem 3. Assume that the sequences {a n } n= and {b n } n= satisfy A and 6 and that N. Let σ < σ 2 <, < σ < σ 2, s = σ + it, s = σ + it, and let < ε < /2 be arbitrary. hen J = Ψ U t As N 2 σ σ +5ε N M xx s dx B s N M 2 xx s dx dt uniformly for σ σ σ 2 and σ σ σ 2 and sufficiently large. One measure of the strength of our results is how much larger than we may take N and still retain an asymptotic formula. his is determined by the parameters θ, ϕ, and η as can be seen, for example, from 24 and the error term 32 ON 2σ+maxθ,ϕ+5ε in 26 of heorem. It turns out that this term comes from using the pointwise upper bounds for E i x and E i x, h i =, 2 given in A 2 and A 3 to estimate various expressions involving these functions. It is worth noting that if E i x and E i x, h i =, 2 act like random variables in x and behave independently as functions of h, then one might expect to be able to replace 32 by 33 O /2 N /2 2σ+maxθ,ϕ+5ε. his observation makes it easy to conjecture the mean values of very long Dirichlet polynomials as we shall illustrate in Example 3 of Section 5. We would similarly expect 33 to replace the nexttolast error term in heorem 2 and Corollaries and 2. It is also worth noting that one can sometimes exploit averages of E i x, h over h to improve Proof of heorem and its corollary. Multiplying out in, we obtain t I = Ψ U t AsBs dt Ψ U N As M 2 xx s dx dt Ψ U t Bs N M xx s dx dt
9 or Mean value theorems 63 t + Ψ U N N M xx s dx M 2 yy s dy dt, 34 I = I I 2 I 3 + I 4. First consider I. By 5 and 6 we have or I = Ψ U a n b n n 2σ + a n b m mn σ ΨU 2π log m n n<n n<m N + b n a m mn σ Ψ U 2π log m, n n<n n<m N 35 I = Ψ U a n b n n 2σ + I 2 + I 3 for short. In I 2 we set m = n + h and note that by A and 9 the total contribution of those terms with h > n/τ is no more than O, say. It follows that I 2 = n<n h minn/τ,n n + O = n<n h minn/τ,n n a n b n+h n 2σ + h σ ΨU n a n b n+h K σ n, h + O. 2π log Changing the order of summation, we obtain I 2 = a n b n+h K σ n, h + O. h H hτ h By 9 and Stieltjes integration this becomes I 2 = h H + h H N h hτ N h M x, hk σ x, h dx hτ K σ x, h de x, h + O. + h n
10 64 D. A. Goldston and S. M. Gonek he second term equals N h 36 E x, hk σ x, h N h hτ E x, h x K σx, h dx. h H o bound this we use, but first we must check that h x η whenever h H and x [hτ, N h]. his will be the case if h hτ η for every h H, or if hτ H τ η/ η. But this follows from 24 and 25. By, 2, and 2 we now find that 36 is ε hτ ϕ 2σ+ε/2 + N ϕ 2σ+ε/2. h H Here we have appealed to the estimate 37a B A x λ dx A +λ+δ + B +λ+δ, which holds uniformly for A B and bounded λ, where δ > is arbitrarily small, and where the implied constant depends at most on δ. We also note for later use that the δ is unnecessary if λ is bounded away from. Next, using the discrete analogue of this, namely 37b h λ A +λ+δ + B +λ+δ, A h B we see that the sum above is since ε τ ϕ 2σ+ε H +ϕ 2σ+ε + + ε HN ϕ 2σ+ε ε τ N +ϕ 2σ+ε + τ +ϕ 2σ+ε N +ϕ 2σ+5ε + τ +ϕ 2σ+5ε, τ ε < ε < τ 2ε < N 2ε when < ε < /2. If + ϕ 2σ + 5ε, this is and in the opposite case it is N +ϕ 2σ+5ε, because τ N. hus, 36 is N +ϕ 2σ+5ε + and it follows that I 2 = h H N h hτ M x, hk σ x, h dx + O N +ϕ 2σ+5ε + O. reating I 3 in the same way, we obtain I 3 = h H N h hτ M 2x, hk σ x, h dx + O N +ϕ 2σ+5ε + O.
11 Mean value theorems 65 Combining these results with 35, we now find that 38 or I = Ψ U + h H + h H a n b n n 2σ N h hτ N h hτ Next we treat I 2. By 5 we have I 2 = = a n N n σ a n N n σ M x, hk σ x, h dx M 2 x, hk σx, h dx + ON +ϕ 2σ+5ε + O. M 2 xx σ ΨU 2π log x n 39 I 2 = I 2 + I 22. n + n dx M 2 xx σ ΨU 2π log x dx n In I 2 we set x = n + u and note as before that by A, 6, and 9, that portion of the integral with u > n/τ contributes a negligible amount. hus we find that I 2 = minn/τ,n n a n n σ M 2 n + un+u σ ΨU 2π log + u du n + O = a n minn/τ,n n M 2 n + uk σn, u du + O, say. Changing the order of summation and integration, we find that I 2 = τ + H u τ a n M 2 n + uk σn, u du u a n M 2 n + uk σn, u du + O. he first term is τ n ε 2σ du τ N 2σ+2ε + N 2σ+4ε + 2ε
12 66 D. A. Goldston and S. M. Gonek by A, 6, 2 and 37a. By 4 and Stieltjes integration the second term equals 4 H τ N u M xm 2 x + uk σx, u dx du + H τ N u M 2 x + uk σx, u de x du. Integrating by parts and using 6 8, 2, 2, and 37b, we see that the second term is = H M 2 x + uk σx, ue x N u du τ H N u M 2 x + uk σx, u + M 2 x + u x K σx, u E x dx du τ ε H ε τ N τ θ 2σ+ε/2 + N θ 2σ+ε/2 du v θ 2σ+ε/2 dv + ε HN θ 2σ+ε/2 ε τ N +θ 2σ+ε + N +θ 2σ+5ε + 2ε. 4 hus we have I 2 = H τ N u M xm 2 x + uk σx, u dx du + O N +θ 2σ+5ε + O +2ε. We treat I 22 similarly. Setting x = n u, we see that I 22 = n a n n σ M 2 n un u σ ΨU 2π log u du. n Using A, 6, and 9 for that part of the integral for which u > n/τ +, we find that minn/τ+,n a n n σ M 2 n un u σ ΨU 2π log u du n + O = 2 a n n/τ+ M 2 n uk σn u, u du + O.
13 Mean value theorems 67 If we change the order of summation and integration we obtain I 22 = τ + H 2 τ a n M 2 n uk σn u, u du + a n M 2 n uk σn u, u du + O. As in the case of I 2, the first term is easily seen to be N 2σ+4ε + 2ε. Hence we have I 22 = H τ + a n M 2 n uk σn u, u du + O N 2σ+4ε + O +2ε. By 4 and Stieltjes integration we may write this as I 22 = H τ + H N + τ N + M ym 2 y uk σy u, u dy du M 2 y uk σy u, u de y du + O N 2σ+4ε + O +2ε. If we estimate the second term as was done for the corresponding term in 4, we see that it also is N +θ 2σ+5ε + 2ε. In the first term we replace y by x + u. We then obtain 42 I 22 = H τ N u M x + um 2 xk σx, u dx du + O N +θ 2σ+5ε + O +2ε. Combining this with 4 in 39 we now find that I 2 = H τ + H N u τ M x + um 2 xk σx, u dx du N u M 2 x + um xk σ x, u dx du + ON +θ 2σ+5ε + O 2ε. Clearly I 3 is the complex conjugate of I 2, but with Bs instead of As
14 68 D. A. Goldston and S. M. Gonek and M x instead of M 2x. It therefore follows from 42 that 43 I 3 = H τ + H N u τ M 2 x + um xk σ x, u dx du N u M x + um 2 xk σx, u dx du + ON +θ 2σ+5ε + O 2ε, which is identical to the expression for I 2. Finally, we come to I 4. By 5 and 6 we see that 44 I 4 = N N x N N + M xm 2 yxy σ ΨU 2π log y x x = I 4 + I 42, M 2 xm yxy σ Ψ U 2π log y x dy dx dy dx say. In I 4 we set y = x + u and use 6 and 9 for u > x/τ to obtain I 4 = = N N minx/τ,n x Ψ U 2π log + u x minx/τ,n x M xm 2 x + ux 2σ + u σ x du dx + O M xm 2 x + uk σx, u du dx + O. Next we change the order of integration and find that I 4 = τ + H N u τ N u M xm 2 x + uk σx, u dx du By 6, 2, and 37a the first term is τ N M xm 2 x + uk σx, u dx du + O. x ε 2σ dx +ε N 2σ+2ε + N 2σ+4ε + ε.
15 hus, I 4 = H τ N u Mean value theorems 69 M xm 2 x + uk σx, u dx du + O N 2σ+4ε + O +ε. Since I 42 is I 4 with M and M 2 interchanged, we also have I 42 = hus we find that I 4 = H τ N u M 2xM x + uk σx, u dx du + O N 2σ+4ε + O +ε. H τ + H N u τ M x + um 2 xk σx, u dx du N u M 2 x + um xk σ x, u dx du + ON 2σ+4ε + O ε. On combining 34, 38, and 42, 43, and 45, we obtain I = Ψ U + h H + H H h H τ τ N u N u a n b n n 2σ N h hτ N h hτ M x, hk σ x, h dx M 2 x, hk σx, h dx M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du + ON 2σ+maxθ,ϕ+5ε + O 2ε. his agrees with 26 so the proof of heorem is complete. We now deduce Corollary from heorem. In the second term on the right in 46 we replace N h by N and H by N/τ. hen by A 4 and 2
16 7 D. A. Goldston and S. M. Gonek this results in a change of at most hhn ε/2 hτ 2σ + N 2σ h H + N/τ+<h N/τ h ε/2 N Nτ/τ+ x 2σ+ε/2 dx N ε/2 τ 2σ H 2 2σ+ε + + N 2σ H 2+ε + + τ 3 N 2 2σ+ε N ε/2 τ 2σ N/τ 2 2σ+ε + N 2σ N/τ 2+ε + τ 3 N 2 2σ+ε τ 2 N 2 2σ+2ε N 2 2σ+4ε since σ < and N. Hence the second term on the righthand side of 46 equals N h N/τ hτ M x, hk σ x, h dx + O N 2 2σ+4ε. By 23 we may replace K σ x, h by x 2σ ΨU h 2πx with a total error of at most 2ε N ε/4 2ε N ε/4 N h N/τ hτ h N/τ x 2σ+ε/4 dx hτ 2σ+ε/2 + N 2σ+ε/2 2ε N ε/4 τ 2σ+ε/2 N/τ 2 2σ+ε + + τ N 2 2σ+ε/2 +3ε N ε/4 N 2 2σ+ε + τ 2 2σ+ε N 2 2σ+5ε. hus, the expression above equals N h N/τ hτ M x, hx 2σ ΨU h 2πx dx + O N 2 2σ+5ε. If we write v for h /2πx and then change the order of summation and integration, we get C /2πτ /2πN h 2πNv/ M h h 2πv, h 2σ Ψ U vv 2σ 2 dv + O N 2 2σ+5ε. Finally, by A 4, 8, and 9 if we extend the interval of integration to infinity we change our term by a negligible amount. hus, the second term
17 C /2πN h 2πNv/ Mean value theorems 7 on the righthand side of 46 is h M h 2πv, h 2σ Ψ U vv 2σ 2 dv + O N 2 2σ+5ε. Similarly, we see that the third term on the righthand side of 46 is h M 2 h 2πv, h 2σ Ψ U vv 2σ 2 dv C /2πN h 2πNv/ + O N 2 2σ+5ε. In much the same way we find that the fourth term on the right in 46 equals 47 N/τ τ N u M x + um 2 xx 2σ ΨU dx du 2πx + O N 2 2σ+5ε. Now by 7 and the mean value theorem of differential calculus we have 48 M x + u = M x + Ox + ε/2 τ for u/x τ. Hence, replacing M x+u by M x and using the estimates σ < and N, and 7 we change the above by at most hus, 47 equals N/τ τ N τ N/τ τ τ N/τ N τ τ N x ε 2σ dx du τ 2σ+2ε + N 2σ+2ε du y 2σ+2ε dy + τ N 2 2σ+2ε +2ε N 2 2σ+3ε + N 2 2σ+5ε. u M xm 2 xx 2σ ΨU dx du + O N 2 2σ+5ε. 2πx Substituting v for u /2πx and then changing the order of integration, we
18 72 D. A. Goldston and S. M. Gonek find that this equals C /2πτ /2πτN 2πNv/ τ M u u M 2 u 2σ du Ψ U vv 2σ 2 dv 2πv 2πv + O N 2 2σ+5ε. Now by 6, M i u 2πv N ε i =, 2 in the rectangle [, τ ] [/2πNτ, /2πτ]. Using this and 7, we find that if we begin the u integral at zero, the first term changes by /2πτN N ε /τ 2 2σ /2πτ /2πNτ Ψ U v v 2σ 2 dv N ε /τ 2 2σ /τ 2σ + /Nτ 2σ N ε τ + N 2σ N 2ε + N 2σ+2ε. Moreover, if we then extend the v integral to infinity, this changes our expression by a negligible amount because of 6 and 8. hus, 47 equals C 2πNv/ u u M M 2 u 2σ du Ψ U vv 2σ 2 dv 2πv 2πv /2πτN + O N 2 2σ+5ε + ON 2ε. reating the fifth term in 46 in exactly the same way, we find that it equals C 2πNv/ u u M M 2 u 2σ du Ψ U vv 2σ 2 dv 2πv 2πv Combining all our results, we now obtain I = Ψ U + C + C /2πN a n b n n 2σ h 2πNv/ /2πN h 2πNv/ 2πNv/ M /2πτN 2C M M 2 + O N 2 2σ+5ε + ON 2ε. h h 2πv, h 2σ Ψ U vv 2σ 2 dv h h 2πv, h 2σ Ψ U vv 2σ 2 dv u u M 2 u 2σ du 2πv 2πv Re Ψ U vv 2σ 2 dv
19 Mean value theorems 73 + O N 2 2σ+5ε + ON 2σ+maxθ,ϕ+5ε + ON 2ε, which is the same as 27. hus, the proof of Corollary is complete. 3. Proof of heorem 2 and its corollary. Multiplying out in 2, we have t I = Ψ U t A sb s dt Ψ U A s M 2 xx s dx dt or + t Ψ U B s M xx s dx dt N t Ψ U M xx s dx M 2 yy s dy dt, N 49 I = I I 2 I 3 + I 4. In I we multiply the two series and note by A and our assumption that σ > that the resulting double series is absolutely convergent. We may therefore integrate termwise. Using 5 and 6, we then find that or I = Ψ U a n b n n 2σ N<n + a n b m mn σ ΨU 2π log m n N<n n<m + b n a m mn σ Ψ U 2π log m, n N<n n<m 5 I = Ψ U N<n a n b n n 2σ + I 2 + I 3. Setting m = n + h in I2 and using A and 9 for h n/τ, we see that I2 = a n b n+h n 2σ + h σ ΨU n 2π log + h n N<n h<n/τ + O N 2σ = a n b n+h K σ n, h + O N 2σ, N<n h<n/τ say. Changing the order of summation, which is permissible by absolute N N
20 74 D. A. Goldston and S. M. Gonek convergence, and then splitting the sum over h at, we obtain I2 = a n b n+h K σ n, h + a n b n+h K σ n, h h maxn,hτ<n + O N 2σ. By A, 2, 37b, and 29 the second term is n 2σ+ε/42σ 2 <h hτ<n τ 2σ+ε2σ 2 2 2σ ε = τ ε N 2σ ε/2 N 2σ. <h hus, by 9 and Stieltjes integration we have I 2 = h maxn,hτ + h maxn,hτ M x, hk σ x, h dx he second term equals 5 E x, hk σ x, h maxn,hτ h h hτ<n hτ 2σ+ε/22σ 2 K σ x, h de x, h + O ε/2 N 2σ. maxn,hτ E x, h x K σx, h dx. We may replace E x, h here by Ox ϕ if we can show that h x η for all h and x > maxn, hτ. his condition will be met if τ η/ η. But this follows immediately from 28 and 29, so we find that 5 is ε maxn, hτ ϕ 2σ h H = ε N ϕ 2σ + ε h N/τ N/τ<h hτ ϕ 2σ ε τ N 2σ+ϕ N 2σ+ϕ+4ε by 2, 2, and 37b. Combining our estimates, we see that I 2 = h maxn,hτ + O ε/2 N 2σ. M x, hk σ x, h dx + O N 2σ+ϕ+4ε
21 reating I 3 in the same way, we obtain I3 = h maxn,hτ + O ε/2 N 2σ. Hence, by 5 we have Mean value theorems 75 M 2x, hk σ x, h dx + O N 2σ+ϕ+4ε 52 I = Ψ U N<n + a n b n n 2σ h maxn,hτ + h maxn,hτ M x, hk σ x, h dx M 2 x, hk σx, h dx or + ON 2σ+ϕ+4ε + O ε/2 N 2σ. Next consider I2. By 5 and absolute convergence we have I2 = M 2 xx σ ΨU 2π log x dx n N<n N = a n n n σ + M 2 xx σ ΨU 2π log x dx, n N<n n a n n σ N 53 I 2 = I 2 + I 22. In I 2 we write x = n + u and use A, 6, and 9 for u n/τ to obtain I 2 = N<n a n n σ n/τ + O N 2σ = N<n a n n/τ M 2 n + un + u σ ΨU 2π log + u du n M 2 n + uk σn, u du + O N 2σ, say. Changing the order of summation and integration by absolute convergence and then splitting the u integral at, we find that
22 76 D. A. Goldston and S. M. Gonek I 2 = + maxn,<n <n a n M 2 n + uk σn, u du a n M 2 n + uk σn, u du + O N 2σ. By A, 6, 2, 37b, and 29, the second term is <n n 2σ+ε/42σ 2 du τ 2σ+ε2σ 2 2 2σ ε = τ ε N 2σ ε/2 N 2σ. hus, by 4 and Stieltjes integration, we have 54 I 2 = he second term is + maxn, 2σ+ε/22σ 2 du M xm 2 x + uk σx, u dx du maxn, M 2 x + uk σx, u de x du + O ε/2 N 2σ. M 2 x + uk σx, ue x maxn, du maxn, M 2 x + uk σx, u + M 2 x, u x K σx, u E x dx du H ε maxn, θ 2σ+ε/2 du ε N/τ H N θ 2σ+ε/2 du + ε N/τ ε τ N 2σ+θ+ε N 2σ+θ+5ε θ 2σ+ε/2 du
23 Mean value theorems 77 by A 2, 2, 2, and 37a and 37b. hus we find that 55 I 2 = maxn, M xm 2 x + uk σx, u dx du + O N 2σ+θ+tε + O ε/2 N 2σ. In I 22 we set x = n u and obtain I 22 = N<n a n n σ n N By 9 for u > n/τ +, this equals N<n M 2 n un u σ ΨU 2π log u du. n minn N,n/τ+ a n n σ M 2 n un u σ ΨU 2π log u du n = N<n a n minn N,n/τ+ + O N 2σ M 2 n uk σn u, u du + O N 2σ. As in I 2 we interchange the order of summation and integration and split the resulting integral at to obtain I 22 = + maxn+u,+<n +<n a n M 2 n uk σn u, u du a n M 2 n uk σn u, u du + O N 2σ. Estimating the second term as we did in the case of I 2, we see that it is ε/2 N 2σ. hus, by 4 and Stieltjes integration I 22 = + maxn+u,+ maxn+u,+ + O ε/2 N 2σ. M ym 2 y uk σy u, u dy du M 2 y uk σy u, u de y du he second term is estimated just like the corresponding term in 54 with the result that it also is N 2σ+θ+5ε. In the first term we make the
24 78 D. A. Goldston and S. M. Gonek substitution x = y u and find that I 22 = maxn, M 2 xm x + uk σ x, u dx du + O N 2σ+θ+5ε + O ε/2 N 2σ. Combining 53, 55, and 56, we now have I 2 = + maxn, maxn, M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du + ON 2σ+θ+5ε + O ε/2 N 2σ. Since I 3 is the complex conjugate of I 2 with B s replacing A s and M x replacing M 2x, it follows from 57 that 58 I 3 = + maxn, maxn, M 2 x + um xk σ x, u dx du M x + um 2 xk σx, u dx du + ON 2σ+θ+5ε + O ε/2 N 2σ. Note that this is identical to the expression for I2. Next we treat I4. By absolute convergence, 5, and 6 we have I4 = M xm 2 yxy σ ΨU 2π log y dy dx x N + x N x = I 4 + I 42, M 2 xm yxy σ Ψ U 2π log y x dy dx say. In I 4 we set y = x + u and use 6 and 9 for u > x/τ to obtain I 4 = N x/τ M xm 2 x + ux 2σ + u σ ΨU x 2π log + u du dx x + O N 2σ
25 = N x/τ Mean value theorems 79 M xm 2 x + uk σx, u du dx + O N 2σ. he double integral converges absolutely so we may change the order of integration. After doing so and splitting the resulting u integral at, we obtain I 4 = + maxn, M xm 2 x + uk σx, u dx du M xm 2 x + uk σx, u dx du + O N 2σ. By 6, 2, 37a, and 29, the second term is x 2σ+ε/42σ 2 dx du τ 2σ+ε2σ 2 2 2σ ε τ ε N 2σ ε/2 N 2σ. hus we find that I 4 = maxn, 2σ+ε/22σ 2 du M xm 2 x + uk σx, u dx du + O ε/2 N 2σ. Since I 42 is I 4 with M and M 2 interchanged, we see that I 42 = maxn, hus, we find that 59 I 4 = + M 2xM x + uk σx, u dx du + O ε/2 N 2σ. maxn, maxn, + O ε/2 N 2σ. M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du By 49, 52, 57, 58, and 59, we now see that
26 8 D. A. Goldston and S. M. Gonek 6 I = Ψ U N<n + + a n b n n 2σ h maxn,hτ h maxn,hτ maxn, maxn, M x, hk σ x, h dx M 2 x, hk σx, h dx M x + um 2 xk σx, u dx du M 2 x + um xk σ x, u dx du + ON 2σ+maxθ,ϕ+5ε + O ε/2 N 2σ. his is 3 so the proof of heorem 2 is complete. he proof of Corollary 2 is along the same lines as that of Corollary so we leave out most of the details. Replacing K σ x, u by x 2σ u ΨU 2πx and M i x + u by M i x i =, 2, we see from 23, 48, and the fact that σ > and N, that the righthand side of 6 changes by no more than O N 2 2σ+5ε. herefore we have 6 I = Ψ U N<n a n b n n 2σ h maxn,hτ h maxn,hτ maxn, h M x, hx 2σ ΨU dx 2πx h M 2 x, hx 2σ ΨU dx 2πx M xm 2 xx 2σ Re Ψ U u 2πx + ON 2σ+maxθ,ϕ+5ε + O ε/2 N 2σ + O N 2 2σ+5ε. N/τ<h N M x, hx 2σ ΨU h 2πx dx du Next consider the second term on the righthand side of 6. If we replace the lower limit of integration by N, then this changes the term by the amount hτ dx.
27 h N Mean value theorems 8 For this range of h and x we have h/2πx τ, so by 9 this term is negligible. We may therefore take the second term to be dx. M x, hx 2σ ΨU h 2πx We now set v = h/2πx and find that this equals h /2πN C h 2σ M h h 2πv, h v 2σ 2 ΨU v dv. Changing the order of summation and integration, we find that this is /2πN h C h 2πv, h 2σ Ψ U vv 2σ 2 dv + C H h M /2πN /2πN 2πNv/ <h M Similarly, the third term on the righthand side of 6 is C /2πN + C H h M 2 /2πN /2πN h h 2πv, h 2σ Ψ U vv 2σ 2 dv 2πNv/ <h M 2 h h 2πv, h 2σ Ψ U vv 2σ 2 dv. h h 2πv, h 2σ Ψ U vv 2σ 2 dv. Finally, the same basic analysis applied to the fourth term on the righthand side of 6 leads to 2C H /2πN H 2πNv/ M u u M 2 u 2σ du Re 2πv 2πv Ψ U vv 2σ 2 dv. Combining our expressions, we obtain 3, so the proof of Corollary 2 is complete. 4. Proof of heorem 3. We have t J = Ψ U As N M xx s dx B s N M 2 dx dt xx s
28 82 D. A. Goldston and S. M. Gonek = + Ψ U t Ψ U t AsB s dt As N N M 2 xx s dx dt t Ψ U B s t Ψ U N M xx s dx = J J 2 J 3 + J 4, say. By 5 we see that J = N<m M xx s dx dt N M 2 xx s dx dt a n b m n σ m σ ΨU 2π log m n Now if either m N/N or N n/n is greater than τ, then logm/n τ so that by 9, Ψ U 2π log m D 2 for any D 2. n Hence, we have J = N τ N<m N+τ + O N 2 σ σ, say. By A, 7, and 37b, this is N /τ N<m N+/τ τ 2 N 2 σ σ +ε + N 2 σ σ +2ε N 2 σ σ +ε N 2 σ σ +5ε.. a n b m n σ m σ ΨU 2π log m n n ε/2 σ m ε/2 σ + N 2 σ σ he integrals J 2, J 3, and J 4 are treated similarly with the same result. hus, J N 2 σ σ +5ε and the proof of the theorem is complete.
29 Mean value theorems Four examples. he following examples illustrate the application of some of our results. Example. Let a n = b n = for n =, 2,... hen so that we may take and Ax = Bx = C x, h = C 2 x, h = [x] M x = M 2 x = M x, h = M 2 x, h = x E x = E 2 x = E x, h = E 2 x, h. In Corollary we may therefore take θ = ϕ = and η = ε, where < ε < /2 is arbitrarily small. Also, taking U = log, N, and σ = /2, we find that t I = Ψ U n /2 it x /2 it dx 2 dt = Ψ U [ ] 2πN n + 2 v Re Ψ U vv dv 2 /2πNτ + O N +5ε. /2πN N 2πN v Re Ψ U vv dv Since Ψ U v, the lower limit of integration in the third term may be replaced by /2πN with an error of O. hus we may rewrite the above as I = Ψ U [ ] 2πN 2πN log N 2 v v Re 2 Ψ U vv dv + 2 /2πN [ 2πN /2πN Re Ψ U vv dv ] v + O + O N +5ε. For v we have 2πN v Ψ U vv dv
30 84 D. A. Goldston and S. M. Gonek 62 ΨU v = Ψ U + OC 2 C + U v and Ψ U is real. hus, the second term on the righthand side above is 2 Ψ U /2πN he third term equals Ψ U /2πN 2πN = 2 Ψ U = O. Finally, by 7 the fourth term is hus we find that U [ ] 2πN v v v dv + O 2 2πN/ y [y] /2y dy + O v dv + O = Ψ U log/n + O. v dv + U U v 2 dv log log. I = Ψ U + o log + O N +5ε. If in the definition of Ψ U t we take C = U and C 2 = U and Ψ U t, then Ψ U t is a minorant for the characteristic function of the interval [,]. On the other hand, taking C = U and C 2 = + U, we obtain a majorant. Since in either case it follows that 63 Ψ U = + OU, n /2 it N x /2 it dx 2 dt log for N 2/+5ε. Notice that if N, then the meansquare of N x /2 it dx is N, so by the lefthand side of 63 is log N. We conclude this example by remarking that since θ = ϕ = and η = ε, a straightforward but tedious application of heorem would allow us to prove that 63 is in fact valid for N A for any fixed A. It is interesting to note that this and the simple approximation see itchmarsh [5; p. 49]
31 ζ 2 + it = n /2 it N Mean value theorems 85 x /2 it dx + O 2 + it N /2 + O 2 + it with N = 2, gives the classical mean value formula 2dt ζ 2 + it log. Example 2. Let a n = enα, b n = enβ for n =, 2,..., with < α, β <. Consider first the case where α = β. hen Ax = Bx = n x enα,, so we may take M i x = and E i x i =, 2. Also, C i x, h = e hα[x], so we may take M i x, h = e hαx and E i x, h i =, 2. hus θ = ϕ = and we may take η = ε. aking U = log 2, N, and σ = /2 in Corollary, we find that t I = Ψ U enαn /2 it 2 dt = Ψ U n + 2 Re + O N +5ε /2πN = Ψ U log N + γ + O/N [ e α 2πN 2 Re /2πN + O N +5ε. By 62 the middle term equals 2 Ψ U Re eα v] eα /2πN h 2πNv/ Ψ U vv dv e hα ΨU vv dv [ ] 2πN e α v v dv + OC 2 C + U + O Ψ U v v dv.
32 86 D. A. Goldston and S. M. Gonek he second error term here is C 2 C + U U C 2 C + U log log v dv + U U v 2 dv by 7. hus, changing variables in the remaining integral, we find that the middle term above equals 2πN/ 2 Ψ U Re e α[y] y dy eα Now 2πN/ = e α[y] y dy [2πN/ ] k= k+ e αk = log/n + O k y dy + e α + OC 2 C + U log log. [ 2πN ] log 2πN/ 2πN log [2πN/ ] by partial summation. Hence, our middle term is equal to 2 Ψ U Re logn/ + OC 2 C + U log log. eα Observing that Reeα = /2 and combining our results, we find that 64 I = Ψ U log + OC 2 C + U log log + O N +5ε. If we remove the weight function Ψ U t/ as in the last example, we deduce that 65 enαn /2 it 2 dt log for N 2/+5ε. Note that by the lefthand side is log N when N <. Had we used heorem rather than Corollary, we could have shown with more work that 65 in fact holds for N A for any A. Now consider the case when α β. Here we may take M x = M 2 x = M x, h = M 2 x, h =
33 and Mean value theorems 87 E x, E 2 x, E x, h, E 2 x, h. It is particularly easy to use heorem in such a case. We take θ = ϕ =, η = ε, σ = /2, U = log 2, and ε N ε/5ε, and find that t I = Ψ U enαn /2 it e nβn /2+it dt = Ψ U enα βn + ON 5ε = Ψ U log eα β + O ε. If we are interested in the unweighted integral I = enαn /2 it e nβn /2+it dt instead, we can proceed as follows. We take C = U and C 2 = + U in the definition of Ψ U in I and use the Cauchy Schwarz inequality to see that { +2U I I + enαn /2 it 2 } /2 dt 2U { +2U enβn /2 it 2 } /2. dt 2U + hese integrals can be estimated by using 64 with C = 3U, C 2 = 3U in the definition of Ψ U and then with C = 3U, C 2 = + 3U. his gives I I U log = o provided that N 2 ε/+5ε. hus, for α β and for the same range of N, we have 66 enαn /2 it e nβn /2+it dt = + o log eα β. Had we proved a longerrange version of 64 by appealing to heorem instead of Corollary, 66 would also hold in such a range.
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