A Sieve Auxiliary FunctionDavid BradleyDedicated to Professor Heini Halberstam, on the occasion of his retirement.Abstract. In the sieve theories of Rosser-Iwaniec and Diamond-Halberstam-Richert, the upper and lower bound sieve functions (F and f ,respectively) satisfy a coupled system of di�erential-di�erence equations withretarded arguments. To aid in the study of these functions, Iwaniec introduceda conjugate di�erence-di�erential equation with an advanced argument, andgave a solution, q, which is analytic in the right half-plane. The analysis ofthe bounding sieve functions, F and f , is facilitated by an adjoint integralinner-product relation which links the local behaviour of F � f with that ofthe sieve auxiliary function, q. In addition, q plays a fundamental role indetermining the sieving limit of the combinatorial sieve, and hence in deter-mining the boundary conditions of the sieve functions, F and f . The sieveauxiliary function, q, has been tabulated previously, but these data were notsupported by numerical analysis, due to the prohibitive presence of high-orderpartial derivatives arising from the numerical quadrature methods used [15,17]. In this paper, we develop additional representations of q. Certain of theserepresentations are amenable to detailed error analysis. We provide this erroranalysis, and as a consequence, we indicate how q-values guaranteed to at leastseven decimal places can be tabulated.1. IntroductionIn his seminal paper, Rosser's Sieve [11], Iwaniec introduced a pair of di�erence-di�erential equations which have been studied more recently by Diamond, Hal-berstam, and Richert [3{10], and by Wheeler [17, 18]. The equations appearas auxiliary equations in connection with the problem of estimatingS(A ;P ; x) := #fa 2 A : gcd(a; Yp<xp2P p) = 1g;where P is a set of primes and A is a �nite set of integers. In the sievetheories of Rosser-Iwaniec and Diamond-Halberstam-Richert, the equationstake the form(1.1) (uq�(u))0 = �q�(u) + �q�(u+ 1)

2 DAVID BRADLEYand(1.2) (up�(u))0 = �p�(u)� �p�(u+ 1);where u; � are real and positive. The parameter � denotes the dimension ofthe sieve, or sifting density, and is a measure of the average number of residueclasses per prime in the sequence being sifted. Iwaniec gave solutions to (1.1)and (1.2) involving the so-called complementary exponential integral [14, p.40]de�ned by Ein(z) := Z z0 1� e�tt dt:The solutions are(1.3) q�(u) = �(2�)2�i Z z�2�euze�Ein(�z) dzand(1.4) p�(u) = Z 10 e�xu�� Ein(x) dx;where in (1.3), the contour starts at �1, hugging the negative real axis, thencircles the origin in the positive direction before returning to �1. In thispaper, we focus on the problems presented by the function q�, since p�, beingthe Laplace transform of a positive function, is relatively simple to deal with.In [7], it is shown that the solutions (1.3), (1.4) are unique, subject to mildpolynomial-like growth conditions at in�nity. In Section 2 below, we provethat (1.3) is the unique solution in a class of functions representable as aLaplace/Mellin transform. In Section 3, an asymptotic expansion is derived,and a few properties of the coe�cients are proved. In Section 4, we givea representation of q� in terms of an operator that arises in other contexts.Finally, it is of some interest to have values of q� tabulated. We take up thisproblem in Section 5. We remark that this paper is based in signi�cant parton the author's Ph.D. thesis [2].2. The Function q�(u)The di�erence-di�erential equation (1.1) can be rewritten in the form�u1��q�(u)�0 = �u��q�(u+ 1);so that the value of the function at u is given by an integral involving thefunction at larger values of the argument. Since integration is a smoothing

A SIEVE AUXILIARY FUNCTION 3operation, one expects repeated integrations to yield a C1 solution, givenonly mild assumptions on the behaviour of the function at in�nity. In fact, itis easy to see that Iwaniec's solution (1.3) is analytic in the right half-planeand that q�(u) as given by (1.3) is asymptotic to u2��1 as u tends to in�nity.In [7], the solution (1.3) is shown to be unique in the class of normalizedpolynomial-like functions. In other words, (1.3) is the unique solution to (1.1)which satis�es q�(u) � ub as u ! 1, for some constant b (and hence wemust have b = 2� � 1). In the sequel, we shall prove a uniqueness result of asomewhat di�erent kind, which shows that (1.3) is unique in a class of functionsrepresentable as a Laplace/Mellin transform. For this task, it is pro�table toview q�(s) as a function of the complex variable �, with s lying in the righthalf-plane, although for sieve applications, we are primarily concerned withpositive real values of the parameters. But �rst, we need to recast (1.3) as anintegral over the positive real axis.Proposition 2.1. Let n be a non-negative integer, and suppose <(n+1�2�) >0. Then(2.1) q�(s) = (�1)n�(n+ 1� 2�) Z 10 xn�2� � @@x�n e�sxe�Ein(x) dx; <(s) > 0:Remark. If � is real and small enough so that n = 0 or n = 1 is permissible(i.e. � < 1=2 in the former case, � < 1 in the latter) then one can use therepresentation (2.1) to compute q�(u) quite easily. However, as n increases,the higher order partial derivatives rapidly become cumbersome, and so forlarger values of �, the method of Section 5 is preferrable.Proof Sketch. The case n = 0 can be found in Iwaniec [11, p.184]. Te Riele[15, p.6] and Wheeler [17, p.73] derive (2.1) from the n = 0 case by performingrepeated integration by parts on the latter. One can also obtain (2.1) directlyfrom (1.3), integrating by parts n times. The integrated terms all vanish dueto the presence of esz as a factor in every derivative of esze�Ein(�z) . One canthen collapse the contour onto the negative real axis, and after some minorsimpli�cations, (2.1) results. See [2, p.12] for details.We are now ready to prove that (2.1) is the unique solution to (1.1) inthe class of functions representable as a Laplace/Mellin transform. For conve-nience, the following notation will be used. Let G denote the set of functions gsuch that g(n) 2 L1[0;1] for n = 0; 1; 2; 3 and for which g(n) vanishes at bothzero and in�nity for n = 0; 1; 2: Also, for each real B > 0, putFB := ff : y 7! e�Byf(y) 2 G g:

4 DAVID BRADLEYTheorem 1 (Uniqueness). Let n be the least non-negative integer such that<(2�) < n + 1: Then up to multiplication by an arbitrary function of �,(2.2) q�(s) := (�1)n�(n + 1� 2�) Z 10 xn�2� � @@x�n e�sxe�Ein(x) dx; <s > 0is the unique solution to the di�erence-di�erential equation(sq�(s))0 = �q�(s) + �q�(s+ 1)which satis�es q�(s) is an entire function of �;(2.3)9B > 0 such that q�(s) = Z 10 e�syf�(y) dy; <(2�) < �2; <(s) > 0;(2.4)for some f� 2 FB.Remark. Note that f� can be very general. Of course, if f�(x) � x�2��(1� 2�)as x ! 0+, then (2.4) would imply that q�(u) � u2��1 as u ! 1, but we donot assume this.Proof. Suppose that (2.4) holds for some B > 0. We shall show that neces-sarily, f�(x) = x�2�e�Ein(x) up to an arbitrary constant multiple (dependingon �), and that (2.2) follows. The Laplace inversion theorem guarantees thatfor any c > B, we have(2.5) f�(x) = 12�i Z c+i1c�i1 esxq�(s) ds; x > 0:Our approach is to di�erentiate (2.5) with respect to x and then use thedi�erence-di�erential equation satis�ed by q� to obtain a di�erential equationfor f� that can be solved explicitly. But �rst, we must justify di�erentiating(2.5) under the integral sign. We require the following lemmata.Lemma 2.1. Suppose (2.4) holds for some B > 0 and f� 2 FB. Let <(s) >B. Then q�(s) = 1sp Z 10 e�syf (p)� (y) dy; p = 0; 1; 2; 3:Proof. Integrate (2.4) by parts repeatedly. In each case, the integrated termvanishes by the hypotheses on f� and the de�nition of G .

A SIEVE AUXILIARY FUNCTION 5Lemma 2.2. Suppose (2.4) holds for some B > 0. Let c > B. Then for allx > 0, the functionF (x) := Z c+i1c�i1 esxq�(s) ds satis�es F 0(x) = Z c+i1c�i1 esxs q�(s) ds:Proof. For positive integers n, de�neFn(x) := Z c+inc�in esxq�(s) ds; x > 0:It is clear that for each n, F 0n exists, is continous, and is given byF 0n(x) = Z c+inc�in esxs q�(s) ds; x > 0:Let G(x) := Z c+i1c�i1 esxs q�(s) ds; x > 0:We shall show that the sequence F 0n converges uniformly to G on compactintervals. Let 0 < a < M and for now, make the restriction x 2 [a;M ]: ByLemma 2.1,jG(x)� F 0n(x)j 6 ���� Z c+i1c+in esxs q�(s) ds����+ ���� Z c�inc�i1 esxs q�(s) ds����� ecM ���� Z c+i1c+in s q�(s) ds����= ecM ���� Z c+i1c+in Z 10 e�syf 000� (y) dy dss2 ����= ecM Z 1n ���� Z 10 e�cyf 000� (y)e�ity dy ���� dtc2 + t26 ecM Z 1n Z 10 je�cyf 000� (y)j dy dtc2 + t2 :(2.6)Now by the assumption f� 2 FB , we have that y 7! e�cyf 000� (y) 2 L1[0;1].Thus the inner integral in (2.6) is �nite, and as n!1, it follows that jG(x)�F 0n(x)j ! 0 independently of x 2 [a;M ]. In other words, F 0n ! G uniformly on[a;M ]; and hence G is continuous there. Furthermore, by uniform convergence,Z xa G(y) dy = limn!1 Z xa F 0n(y) dy = limn!1 (Fn(x)� Fn(a)) = F (x)� F (a):

6 DAVID BRADLEYBy continuity of G and the fundamental theorem of calculus, F 0(x) = G(x)holds for all x 2 [a;M ]. Finally, since 0 < a < M were arbitrary, the lemma isproved.Continuing with the proof of Theorem 1 we therefore havef 0�(x) = 12�i Z c+i1c�i1 esxs q�(s) ds= esxsq�(s)2�ix ����c+i1c�i1 � 12�ix Z c+i1c�i1 esx (s q�(s))0 ds:We claim the integrated term vanishes. Writing s = c+ iT , we havelimT!1 ��e(c+iT )x(c+ iT )q�(c+ iT )�� = ecx limT!1 ���� Z 10 e�(c+iT )yf 0�(y) dy���� = 0;by the Riemann-Lebesgue lemma. Therefore,xf 0�(x) = � 12�i Z c+i1c�i1 esx (sq�(s))0 ds= � �2�i Z c+i1c�i1 esxq�(s) ds� �2�i Z c+i1c�i1 esxq�(s+ 1) ds= ��f�(x)� �e�x Z c+1+i1c+1�i1 esxq�(s) ds= ��f�(x)� �e�xf�(x):Solving the separable �rst order di�erential equation for f� yields(2.7) f�(x) = A� x�2�e�Ein(x);where A� is a constant depending only on �. For convenience, put C� =A� �(1� 2�). Since the di�erence-di�erential equation (1.1) satis�ed by q� ishomogeneous, i.e. insensitive to multiplication by an arbitrary function of �,we may take C� = 1 without loss of generality. Also, note that f� as given by(2.7) does indeed lie in the set F for <(2�) < �2, as stipulated. Thus we haveshown that up to multiplication by an arbitrary function of �,(2.8) q�(s) = 1�(1� 2�) Z 10 x�2�e�sxe�Ein(x) dx; <(2�) < �2; <s > 0:However (2.8) is an analytic function of � for <(2�) < 1, so in fact, (2.8)provides an analytic continuation of q�(s) to the open left half-plane <(2�) < 1.To complete the proof of Theorem 1, we need to analytically continue q� to an

A SIEVE AUXILIARY FUNCTION 7entire function of �. We now setF�(x) = e�sxe�Ein(x); G�(x) = x�2kand integrate (2.8) by parts n times, using the formulaZ 10 F�(x)G�(x) dx =n�1Xj=0(�1)jF (j)� (x)G(�j�1)� (x) ����10+ (�1)n Z 10 F (n)� (x)G(�n)� (x) dx:We claim the integrated terms all vanish. To see this, note thatG(�j�1)� (x) = xj+1�2�(1� 2�)j+1 :Furthermore, since F�(x) is an entire function of x,limx!0+F (j)� (x)G(�j�1)� (x) = F (j)� (0) limx!0+ xj+1�2�(1� 2�)j+1 = 0;for <(2�) < 1 and j a non-negative integer. On the other hand, e�sx is a factorof every term in F (j)� (x), whereas for x > 0,����� @@x�r e�Ein(x)����� e"x;for every r > 0, " > 0. Thuslimx!1G(�j�1)� (x)F (j)� (x) = 0:It follows that for <s > 0,q�(s) = (�1)n(1� 2�)n �(1� 2�) Z 10 xn�2� � @@x�n e�sxe�Ein(x) dx= (�1)n�(n+ 1� 2�) Z 10 xn�2� � @@x�n e�sxe�Ein(x) dx; <(2�) < 1:(2.9)Now observe that (2.9) actually gives an analytic continuation of q�(s) to thehalf-plane <(2�) < n+ 1, which completes the proof.

8 DAVID BRADLEY3. The Asymptotic ExpansionRecalling (1.3) again, if 2� is a positive integer, then the branch point in theintegrand at z = 0 becomes a pole, and we can collapse the horizontal portionsof the contour onto the negative real axis, so thatq�(u) = �(2�)2�i Ijzj=1 z�2�euze�Ein(�z) dz = � @@z�2��1 euze�Ein(�z)����z=0:This suggests that when 2� is not a positive integer, we de�ne the fractionalderivative (@=@z)2��1 by means of� @@z�2��1 euze�Ein(�z)����z=0 := q�(u) = �(2�)2�i Z z�2�euze�Ein(�z) dz:In other words, for general �, it may be pro�table to view q�(u) as a fractionalderivative. Thus, in some sense, soon to be made precise,q�(u) � � @@z�2��1 euze�Ein(�z)����z=0:For example, applying Leibniz's rule to the above yields the formal expansionq�(u) � 1Xn=0�2�� 1n �� @@z�n e�Ein(�z)����z=0 � � @@z�2��1�n euz����z=0= 1Xn=0�2�� 1n �bn(�) u2��1�n;(3.1)which, in view of our previous remarks, gives a true equality when 2� is apositive integer. Here, bn(�) is the nth degree polynomial in � de�ned by(3.2) bn(�) := � @@z�n e�Ein(�z)����z=0and the fractional derivative of euz is given, of course, by� @@z�2��1�n euz����z=0 = �(2�� n)2�i Z zn�2�euz dz = u2��1�n :It turns out that (3.1) is a valid asymptotic expansion for u-values tending topositive in�nity.

A SIEVE AUXILIARY FUNCTION 9Theorem 2. Let bn(�) be de�ned as in (3.2), � real. Then the asymptoticexpansion q�(u) � 1Xn=0�2�� 1n �bn(�) u2��1�n; u!1is valid.Proof Sketch. The asymptotic expansion with remainder after n terms ap-pears in essentially the above guise in both [11, p.183] and [17, p.37]. Themain idea is to expand e�Ein(�z) into its Taylor series, and then use Hankel'scontour formula for the reciprocal of the gamma function. For an analysis ofthe size of the remainder term in (3.1) see [2, Chapter 8, Theorem 6].It is interesting to deduce some properties of the coe�cients bn(�) of theasymptotic expansion provided by Theorem 2. Rewriting (3.2) in the form(3.3) e�Ein(�z) = 1Xn=0 znn! bn(�);it is immediate that b0(�) = 1. For positive integers n, we have the followingrecurrence formula which gives bn(�) in terms of b0(�); b1(�); : : : ; bn�1(�).Proposition 3.1. If n is a positive integer and bn(�) is given by (3.3), thenbn(�) = ��n n�1Xj=0 �nj�bj(�):Proof. Applying the operator z � d=dz to (3.3), one obtains1Xn=0nbn(�)znn! = �� (ez � 1) e�Ein(�z) :The result now follows on comparing coe�cients of zn=n!.For concreteness, the �rst few b polynomials are listed below.b0 = 1; b1 = ��; b2 = �2 � �=2; b3 = ��3 + 3�2=2� �=3:The following result shows a connection with the Bernoulli numbers, de�nedby zez � 1 = 1Xn=0 znn!Bn; jzj < 2�:

10 DAVID BRADLEYTheorem 3. Let � be an arbitrary constant. Then for all non-negative integersn, n�2Xj=0 �nj�bj+1(�)Bn�j = (n=2� �) bn(�)� bn+1(�):Remark. The sum on the left vanishes when n = 0; 1. For general n, bn(�) isa polynomial in � of degree n. Thus the coe�cients of �n and of �n+1 on theright hand side both must vanish.Proof. Let [zn=n!]F (z) denote the coe�cient of zn=n! in the analytic functionF (z). We proceed by developingQn(u; �) := � @@z�n euze�Ein(�z)����z=0 = [zn=n!] euze�Ein(�z)as a Fourier series in u on ]0; 1[: Note that bn(�) = Qn(0; �). We require thefollowingLemma 3.1. Let � be an arbitrary constant. Then for all non-negative inte-gers n, Qn(1; �) = (1� n=�)Qn(0; �) = (1� n=�) bn(�):Proof. By de�nition,Qn(1; �) = [zn=n!]eze�Ein(�z)= [zn=n!] (ez � 1) e�Ein(�z) + [zn=n!] e�Ein(�z)= n � zn�1(n� 1)!��ez � 1z � e�Ein(�z) + bn(�)= �n� � zn�1(n� 1)!� ddz e�Ein(�z) + bn(�)= (1� n=�) bn(�):Returning to the proof of Theorem 3, we will compute the Fourier coe�-cients of Qn(u; �). We note that Qn(u; �) is su�ciently smooth to be repre-sentable as the sum of its Fourier series for 0 < u < 1. Since� Z 10 euze�Ein(�z) du = ��ez � 1z � e�Ein(�z) = � ddz e�Ein(�z)

A SIEVE AUXILIARY FUNCTION 11and � Z 10 euze�Ein(�z)e�2�imu du = � e�Ein(�z)z � 2�im (ez � 1)= � zz � 2�im � ddz e�Ein(�z) ;it follows that for 0 < u < 1,�Qn(u; �) = � �znn! � ddz e�Ein(�z) � Xm6=0 e2�imu �znn! � zz � 2�im � ddz e�Ein(�z)= �bn+1(�) + Xm6=0 e2�imu n�1Xj=0 �nj�bj+1(�) (n� j)!(2�im)n�j :When u = 0, the Fourier series converges to the arithmetic mean of the functionevaluated at the two end-points u = 0; 1. Thus applying Lemma 3.1, we have�Xm6=0 n�1Xj=0 �nj�bj+1(�) (n � j)!(2�im)n�j = ��2Qn(0; �)� �2Qn(1; �)� bn+1(�)= ��2 bn(�)� �2 �1� n�� bn(�)� bn+1(�)= (n=2� �) bn(�)� bn+1(�):The result now follows on applying Euler's famous evaluation of the Riemannzeta function in the form�Xm6=0 (n� j)!(2�im)n�j = (Bn�j ; n � j > 2;0; n � j = 1:Next, we analyze the behaviour of bn(�) for large n. Since the gener-ating function e�Ein(�z) is entire, it follows by Hadamard's root test thatlim supn!1 npj bn(�)=n! j < 1=R for every R > 0, i.e. lim supn!1 npj bn(�)=n! j = 0:However, since this information does not reveal how quickly or how slowlyj bn(�)=n! j tends to zero, we seek a concrete upper bound.Theorem 4. Let � > 0. Then, for all non-negative integers n, we have(3.4) ����bn(�)n! ���� 6 � elog (1 + n=�)�n :A somewhat more precise inequality is given by(3.5) ����bn(�)n! ���� 6 exp�� Z L0 t�1 �et � 1� dt��� log (1 + n=�)�n;

12 DAVID BRADLEYwhere L := log (1 + n=�) :Proof. For any r > 0, Cauchy's inequality gives����bn(�)n! ���� 6 r�nmaxjzj=r ��e�Ein(�z)��:Now on the circle jzj = r,��e�Ein(�z)�� = ���� exp�� � 1Xn=1 znn!n����� 6exp�� 1Xn=1 rnn!n� = exp�� Z r0 et � 1t dt�:Thus(3.6) ����bn(�)n! ���� 6 infr>0 r�n exp�� Z r0 t�1 �et � 1� dt� = infr>0 r�ne��Ein(�r):Minimizing with respect to r > 0, we �nd on taking the logarithmic derivativethat �r�1 (er � 1) � r�1n = 0 so that r = log (1 + n=�) = L: The inequality(3.5) now follows on substituting the optimal value of r into (3.6). If we usethe fact that Z r0 et � 1t dt = 1Xn=1 rnn!n 6 er � 1;then (3.5) with r = L yields����bn(�)n! ���� 6 expf� (1 + n=�� 1)glogn (1 + n=�) = � elog (1 + n=�)�n ;which is (3.4).Improvements. Put M(r) := maxjzj=r ��e�Ein(�z)��: Our proof of (3.4) used theinequality(3.7) M(r) = maxjzj=r ���� exp�� � 1Xn=1 znn!n����� 6 exp�� 1Xn=1 rnn!n�;which at �rst glance may appear wasteful, since at no point on the circle jzj = rdo we have arg(zn) = � for every positive integer n. However, an attractiveargument of Andrew Odlyzko [13] shows that the estimate (3.7) cannot besubstantially improved. Consider z = x + �i, where x > 0 is large. The main

A SIEVE AUXILIARY FUNCTION 13contribution to the sum comes from terms with n close to x, and for suchterms, zn = xn �1 + �ix �n � xne�i = �xn; x!1:Thus it would seem that any signi�cant improvement on Theorem 4 cannot bebased on Cauchy's inequality. However, if one considers the Cauchy integralformula bn(�)n! = 12�i Ijzj=r z�n�1e�Ein(�z) dz;one observes that ��e�Ein(�z)�� is close to its maximum M(r) for only a smallportion of z-values on the circle. This suggests that Theorem 4 can be improvedusing saddle-point asymptotics. Using this approach, the author was able toimprove on Theorem 4 by the factor (2�(�+ n) log (1 + n=�))�1=2. See [2,Chapter 9, Theorem 8] for details.4. An Operator RepresentationWe begin this section with an informal argument which should help motivatewhat follows. Let f(z) be a formal power series in z and let D = d=du: SinceDneuz = zneuz for all non-negative integers n, it follows by linearity thatthe equation f(z) euz = f(D) euz holds, at least in the formal sense. If wenow apply this observation to the contour representation (1.3) with f(z) :=e�Ein(�z) , we obtainq�(u) = �(2�)2�i Z z�2�e�Ein(�D)euz dz= e�Ein(�D) �(2�)2�i Z z�2�euz dz= e�Ein(�D) u2��1:(4.1)Pulling the di�erential operator outside the integral requires justi�cation, andwe shall do this shortly. But for the moment, a few remarks about (4.1) are inorder. Recalling the power series representation for Ein, we may writeEin(�D) = � 1Xn=1 Dnn!n

14 DAVID BRADLEYand hence expanding the operator into powers of D, we formally obtain theexpression(4.2) q�(u) � 1Xn=0 bn(�)Dnn! u2��1 = 1Xn=0�2�� 1n �bn(�) u2��1�nin agreement with Theorem 2. In particular, the n = 0 term gives the knownasymptotic formula q�(u) � u2��1 as u!1.Next, we point out that the di�erential operator Ein(�D) can be recast inthe form of an integral operator. For the sake of brevity, we put T := Ein(�D).Recalling the integral representation for Ein, we haveT := Ein(�D) = Z 10 1� etDt dt:Now if f is analytic in a disk centred at u with radius r > 0, and jtj < r, thenTaylor's theorem givesetDf(u) = 1Xn=0 tnn!Dnf(u) = 1Xn=0 tnn!f (n)(u) = f(u+ t):Thus for those functions f which are analytic in a disk centred at u with radiusr > 1,(4.3) Tf(u) = Z 10 �1� etD� f(u)dtt = Z 10 f(u)� f(u+ t)t dt:Now the integral on the far right of (4.3) makes sense if f is integrable on[u; u+ 1] and for some " > 0, we have jf(u)� f(u + t)j � t" as t ! 0+. Forsuch f , we can de�ne Tf(u) by (4.3), and if T is de�ned this way, then Tf(u)makes sense for a larger class of functions than merely those functions whichare analytic in a suitably large disk centred at u. Thus there is no need to viewT as a power series in D in order to determine Tf(u). In the case of interest,f(u) = u2��1 and (4.1) becomes(4.4) q�(u) = e�T u2��1 = 1Xn=0 �nn! Tnu2��1and so it makes sense to study the iterated integral operator Tn. We shall takethis up after �rst proving the representation (4.4) rigorously.Theorem 5. Let D := d=du and T := Ein(�D). Then for any complexnumber �, and real u > 0,q�(u) = e�T u2��1 = 1Xn=0 �nn! Tnu2��1:

A SIEVE AUXILIARY FUNCTION 15Aside. In sieve applications, we are concerned primarily with positive realvalues of �, and � > 1 in particular.Proof. Fix � and consider the function of the complex variable w de�ned bygw(u) := Z z�2�euzewEin(�z) dz; u > 0:By Taylor's theorem,gw(u) = 1Xn=0 wnn! � @@w�n gw(u) ����w=0:On the other hand, from (1.3) and the de�nition of gw(u),q�(u) = �(2�)2�i gw(u) ����w=�:Thus, q�(u) = �(2�)2�i 1Xn=0 �nn! � @@w�n gw(u) ����w=0and it remains only to show that for all non-negative integers n,(4.5) Tnu2��1 = �(2�)2�i � @@w�n gw(u) ����w=0:For n = 0, we have�(2�)2�i g0(u) = �(2�)2�i Z z�2�euz dz = u2��1 = T 0u2��1:

16 DAVID BRADLEYSuppose now that (4.5) holds up to n� 1, where n is a positive integer. Then�(2�)2�i � @@w�n gw(u) ����w=0= �(2�)2�i Z z�2�euz (Ein(�z))n dz= �(2�)2�i Z z�2�euz (Ein(�z))n�1 Z 10 1� etzt dt dz= Z 10 1t �(2�)2�i Z z�2�euz �1� etz� (Ein(�z))n�1 dz dt= Z 10 t�1fTn�1u2��1 � Tn�1(u+ t)2��1g dt= Tnu2��1;by induction.Having proved the representation (4.4), it is natural to ask how rapidlythe series of T -iterates converges to the function q�. To this end, we prove thefollowingTheorem 6. Let � > 0, u > 0, and let n be a non-negative integer satisfyingn > 2�� 1. De�ne Sn := q�(u)� n�1Xj=0 �jj! T j u2��1:Then as n!1, we have, with c := Ein(1) = 0:796599 : : : ,Sn �� �nn! + �n �ecu �n= logn � log nn �n �� nlogn + 1� 2�� u2��1:Remark. If 2� > 1, then Stirling's formula provides the simpli�cationSn �� �nn! + �n � ecue�n= logn � lognn �n(1�1= logn+1=2n) u2��1:

A SIEVE AUXILIARY FUNCTION 17Proof. We haveSn = 1Xj=n �jj! T j u2��1 = 1Xj=n �(2�)2�i � @@w�j gw(u)����w=0= 1Xj=n �jj! �(2�)2�i Z z�2�euz Einj(�z) dz= �(2�)2�i Z z�2�euz 1Xj=n �jj! Einj(�z) dz:Since n > 2� � 1 and 1Xj=n �jj! Einj(�z) � jzjn as z ! 0; we may collapse thecontour onto the real half-line, obtainingSn = ��1�(2�) sin(2��) Z 10 x�2�e�ux 1Xj=n �jj! Einj(x) dx:Let In := Z 10 x�2�e�ux 1Xj=n �jj! Einj(x) dx;Jn := Z 11 x�2�e�ux 1Xj=n �jj! Einj(x) dx:Note that for any y > 0,1Xj=n yjj! = ynn! �1 + yn + 1 + y2(n+ 1)(n+ 2) + � � �� 6 ynn! ey :Thus, as 0 6 Ein(x) 6 x, we have(4.6) In 6 �nn! Z 10 xn�2�e�uxe�x dx 6 �nn! � e�n+ 1� 2�:With In now satisfactorily estimated, we turn to Jn. Recall the formula [14,p.40] Ein(x) = log x+ + E1(x); x > 0;where E1 denotes the exponential integral de�ned byE1(x) = Z 1x e�t dtt ; x > 0;

18 DAVID BRADLEYand denotes Euler's constant. To estimate Jn, we use Ein(x) 6 log x+ +E1(1) 6 log x+ c, where c := Ein(1) = 0:7965995 : : :6 1: Thus,(4.7) Jn 6 Z 11 x�2�e�ux 1Xj=n (��)jj! dx;where � := �(x) = c+ log x. We now estimate the sum 1Xj=n (��)jj! . Let r > �.Then(4.8) 1Xj=n (��)jj! = 1Xj=n��r�j (�r)jj! 6 ��r �n 1Xj=0 (�r)jj! = ��r�n e�r:Substituting (4.8) into the estimate (4.7) for Jn yieldsJn 6 ��r �n Z 11 x�2�e�uxe�r dx= ��r �n ecr Z 11 xr�2�e�ux dx6 ��r �n ecr �(r + 1� 2�)ur+1�2� :(4.9)For �xed r > max(�; 2� � 1), it is clear from (4.9) that limn!1 Jn = 0: Forexample, taking r = 2� and using c = Ein(1) < 1, we see that Jn 6 e2�u�12�n.But one can do better by choosing r more carefully, say by minimizing the righthand side of (4.9) with respect to r. By taking the logarithmic derivative, theoptimal r is seen to satisfyc� n=r+ (r+ 1� 2�)� log u = 0;or(4.10) n = rf (r+ 1� 2�) + c� log ug;where as customary, = �0=�: Inverting (4.10) and using the fact that (r+1� 2�) � log r as r!1, we �nd that r is approximately n= logn. With thischoice, (4.9) yields(4.11) Jn 6 �n �ecu �n= log n � lognn �n �� nlogn + 1� 2�� u2��1:The theorem now follows on combining (4.6) with (4.11) and absorbing thefactors which are independent of n into the �� constant.

A SIEVE AUXILIARY FUNCTION 19We now return to study the integral operator T and its iterates in moredetail. Recall that for f 2 L1[u; u+ 1] satisfying some Lipschitz condition atu, we may de�ne(4.12) Tf(u) = Z 10 f(u)� f(u+ t)t dt;which is consistent with the de�nition of T as a di�erential operator for fanalytic at u. We also note that the representation (4.12) has occurred inimportant contexts outside sieve theory. For example,T (log u) = � Z 10 log(u+ t)� log ut dt = � Z 10 log(1 + t=u) dtt ; w = �t=u= � Z �1=u0 log(1� w) dww= Li2 (�1=u) ;where Li2(z) := � Z z0 log(1� w) dww = 1Xn=1 znn2 ; jzj < 1;is the ubiquitous dilogarithm. For f 2 C1[u; u+ 1], we can integrate (4.12) byparts. Thus(4.13) Tf(u) = ff(u)� f(u+ t)g log t ����10 + Z 10 f 0(u+ t) log t dt:Butlimt!0+ff(u)� f(u+ t)g log t = limt!0+ f(u)� f(u+ t)t � t log t = f 0(u) limt!0+ t log t= 0:Thus (4.13) becomes(4.14) Tf(u) = Z 10 f 0(u+ t) log t dt:Since R 10 log t dt = �1, it follows that Tf can be viewed as a weighted averageof �f 0 . More generally, for n a positive integer, one can show thatTnf(u) = Z[0;1]n f (n)(u+ t1 + � � �+ tn) log t1 � � � log tn dt1 � � �dtn

20 DAVID BRADLEYif f 2 Cn[u; u+ 1].The integrands in the representations (4.12) and (4.14) both contain ap-parent singularities at the origin. We develop an additional representation inwhich the corresponding singularity is hidden. Again, assume f possesses therequisite derivatives.Proposition 4.1. Let f 2 C1[u; u+1]: Then Tf(u) = � R 10 R 10 f 0(u+st) ds dt:More generally, if n is any positive integer and f 2 Cn[u; u+ 1], thenTnf(u) = (�1)n Z[0;1]n Z[0;1]n f (n)(u+ s1t1 + � � �+ sntn) ds1 � � �dsn dt1 � � �dtn:Proof. By de�nitionTf(u) = Z 10 f(u)� f(u+ t)t dt = � Z 10 Z u+tu f 0(x) dx dtt= � Z 10 Z t0 f 0(u+ r) dr dtt= � Z 10 Z 10 f 0(u+ st) ds dt:(4.15)The general case is handled inductively.Note that the apparent singularity at the origin has been hidden. Also,(4.15) shows that Tf is a weighted average of �f 0 over the unit square.5. Numerical AnalysisHere, we give a detailed numerical analysis of a Simpson's rule-based schemefor computing q�(u) for the range 1 6 � < 3, 1 6 u 6 6. Following Iwaniec[11], we set e�Ein(�z) = nXj=0 bj(�) zjj! + Sn(�z)and rewrite (1.3) (via Hankel's formula for 1=�) in the formq�(u) = nXj=0�2�� 1j � bj(�)u2��1�j + �(2�)2�i Z z�2�euz Sn(�z) dz:Next, since jSn(�z)j � jzjn+1 as z ! 0, the integral around the circularportion tends to 0 with the radius, as long as n+ 1 > 2�� 1. Here 1 6 � < 3,

A SIEVE AUXILIARY FUNCTION 21so we require n > 4. Then the contour may be collapsed onto the negative realaxis, and after routine calculations, one obtains(5.1) q�(u) = 4Xj=0�2�� 1j � bj(�) u2��j�1+ sin(2��)� �(2�) Z 10 x�2�e�uxfe�Ein(x) � P (x)g dx;where P (x) := 4Xj=0(�1)j bj(�) xjj! is the fourth-degree Maclaurin polynomial fore�Ein(x). We use the representation (5.1) to compute q�(u) for various values of�, u. We have written C code to perform these computations, but to guaranteetheir accuracy, an accompanying error analysis is necessary. Since �2��1j � bj(�)is a polynomial in � for non-negative integers j, we regard computations ofthe summation term in (5.1) as exact and henceforth focus exclusively on thecomputational problems that the integral in (5.1) presents. To avoid additionalcomplications in the corresponding error analysis, we do not make use of anynumerical integration packages. For the same reason, we avoid using built-in special functions such as the incomplete gamma function. The only non-elementary function we do use is the complete gamma function, since we aresatis�ed that its implementation gives the correct answers to within the numberof signi�cant digits speci�ed.Let I = I1 + I2 + I3 � I4 denote the integral in (5.1), whereI1 := Z 10 x�2�e�uxfe�Ein(x) � P (x)gdx; I2 := Z 201 x�2�e�uxe�Ein(x) dxI3 := Z 120 x�2�e�uxe�Ein(x) dx; I4 := Z 11 x�2�e�uxP (x) dx:In I3, we have chosen 20 as the lower limit of integration so as to make theintegral negligible in the rectangle of interest, namely 1 6 � < 3, 1 6 u 6 6.See Theorem 7 below for details. In I1, we have chosen 1 as the upper limit ofintegration in order to make computation via Taylor series expansion feasible.It turns out (Theorem 9) that at least 35 terms are needed to make the errornegligible. Accordingly, we approximate the braced expression in I1 by itsTaylor polynomial of degree 35. By de�nition of P (x), the lowest degree term inthis polynomial has x to the �fth power. We then approximate I1 by incompletegamma functions. Complementary incomplete gamma functions are used toapproximate I4. This leaves I2, which we treat by splitting up the interval[1, 20] into appropriate subintervals and then applying Simpson's rule.

22 DAVID BRADLEYTheorem 7. Again, letE1(x) := Z 1x e�tt dt; x > 0denote the exponential integral [1, p.228], [14, p.40]. Then,(5.3) I3 6 u�1 20��e� e�E1(20)e�20u :Furthermore, on any rectangle 1 6 u1 6 u 6 u2 6 6, 1 6 �1 6 � 6 �2 6 3, wehave(5.4)I3 6 u�11 20��1e�1 e�1 E1(20)e�20u1 6 0:184� 10�9; with u1 = �1 = 1:Proof. We apply once more the relationship [14, p.40]Ein(x) = log x+ + E1(x); x > 0;where denotes Euler's constant. Thus,I3 = Z 120 x�2�e�uxe�Ein(x) dx = Z 120 x��e�uxe�( +E1(x)) dx:Note that the factors x�� and e�( +E1(x)) in the integrand are decreasing func-tions of x. Therefore,I3 6 20��e�( +E1(20)) Z 120 e�ux dx;which gives (5.3). Since +E1(20)� log 20 < 0, the expression on the right in(5.3) is decreasing in � and u, from which (5.4) follows.We can compute values of Ein(x) for x > 1 using Ein(x) = log x+ +E1(x),and a continued fraction which can approximate E1 to within any speci�edamount. Our approximation is based on a continued fraction expansion forthe complementary incomplete gamma function [1, p.263], [16, p.356](5.5) �(a; u) := Z 1u ta�1e�t dt = e�uuau+ 1� a1 + 1u+ 2� a1 + 2u+ 3� a1 + 3u+...

A SIEVE AUXILIARY FUNCTION 23de�ned for all a and u > 0. Note that E1(x) = �(0; x) for x > 0.We now wish to determine the error committed in using an approximationE�1 in place of E1 in the evaluation of I2.Theorem 8. Let 0 < " < 0:048 be given and suppose that jE�1(x)�E1(x)j < "for 1 6 x 6 20. Let�I2 := ����I2 � Z 201 x��e�uxe�( +E�1(x)) dx����:Then for all u > 1, � > 1,(5.6) �I2 6 (e�" � 1)e�"fe�( +E�1(1))e�u + e�( +E�1(2)�log 2)e�2u=(1� e�u)g:Furthermore, on any rectangle 1 6 u1 6 u 6 u2 6 6, 1 6 �1 6 � 6 �2 6 3, wehave�I2 6 (e�2" � 1)e�2"fe�2( +E�1(1))e�u1 + e�1( +E�1(2)�log 2)e�2u1=(1� e�u1 )g6 0:127� 10�12; with " = 10�14; u1 = �1 = 1; �3 = 3:(5.7)Proof. For (5.6) we have�I2 = ���� Z 201 x��e�uxe� (e�E�1(x) � e�E1(x))dx����6 Z 201 x��e�uxe� e�E1(x) ��e�(E�1(x)�E1(x)) � 1�� dx6 (e�" � 1) 19Xj=1 j��e�jue� e�E1(j);since the rest of the integrand is decreasing in x. Thus�I2 6 (e�" � 1) 19Xj=1 j��e�jue� e�E�1(j)e�(E1(j)�E�1(j))6 (e�" � 1)e�" 19Xj=1 j��e�jue� e�E�1(j):It is easy to show that + E1(x) � log x < �0:048 if x > 2. Thus if x > 2,then + E�1(x)� log x < 0. It follows that�I2 6 (e�" � 1)e�"fe�( +E�1(1))e�u + e�( +E�1(2)�log 2)(e�2u + e�3u + :::)g= (e�" � 1)e�"fe�( +E�1(1))e�u + e�( +E�1(2)�log 2)e�2u=(1� e�u)g;

24 DAVID BRADLEYwhich completes the proof of (5.6). For (5.7), we note that the expressionabove is majorized by the expression in the statement of (5.7) over the givenrectangle. The reason for this is that since " < , + E�1(x) � log x > 0 ifx = 1, and + E�1(x)� log x < 0 if x > 2 as we have seen.Theorem 9. Let n > 6, and let Rn(x) be the Taylor polynomial of degree nfor e�Ein(x) � P (x). Put I�1 = Z 10 x�2�e�uxRn(x) dx. Then(5.8) jI1 � I�1 j 6 e��Ein(�r)rn(r � 1) Z 10 xn�5e�ux dxholds for every r > 1. Furthermore, on any rectangle 1 6 u1 6 u 6 u2 6 6,1 6 �1 6 � 6 �2 6 3, we have(5.9) jI1 � I�1 j 6 e��2 Ein(�r)rn(r � 1) Z 10 xn�5e�u1x dxfor every r > 1.Note that the integrals can be evaluated exactly in terms of elementaryfunctions. For the full rectangle (i.e. u1 = 1; u2 = 6; �1 = 1; �2 = 3 ), it ispossible to take r = 2:6; n = 35 and thereby obtain jI1� I�1 j 6 0:173:::� 10�8.Proof. Let r > 1. By de�nition of I�1 and Rn(x), we havejI1 � I�1 j = ���� Z 10 x�2�e�ux 1Xj=n+1(�1)j bj(�) xjj! dx����:We now invoke a bound on j bj(�)=j! j which was developed in the course ofproving Theorem 4. Thus, by inequality (3.6) with r > 1, we havejI1 � I�1 j 6 Z 10 x�2�e�ux 1Xj=n+1 e��Ein(�r) �xr�j dx6 e��Ein(�r) Z 10 x�6e�uxxn+1 1Xj=n+1 r�j dx= e�� Ein(�r)rn(r � 1) Z 10 xn�5e�ux dxwhich completes the proof of (5.8). Since ��Ein(�r) = � 1Xn=1 rnn!n > 0, (5.9)follows from (5.8).

A SIEVE AUXILIARY FUNCTION 25In practice, we take n = 35 in Theorem 9 and we computeI�1 = 35Xj=5(�1)j bj(�)j! Z 10 xn�2�e�ux dx= 35Xj=5(�1)j bj(�)j! u2��n�1 Z u0 tn�2�e�t dtusing the incomplete gamma function (a; u) := Z u0 ta�1e�t dt = ua 1Xn=0 (�u)nn! (n+ a) ;which is de�ned for all u > 0 and <(a) > 0. We calculate values of (a; u)using a truncation of the series.Theorem 10. Let 6 6 L be an integer and let r > 1. Then the error committedin computing I�1 using L terms of the series representation of (a; u) is lessthan 6L+1e��Ein(�r)(L+ 1)! (L+ 1) r4 (r � 1):When L = 41; r = 32 , the expression above is 0:3381:::� 10�17.Proof. Recall I�1 = 35Xj=5 (�1)j bj(�)j! u2��j�1 (j � 2�+ 1; u). Put�I 1 = 35Xj=5 (�1)j bj(�)j! LXn=0 (�u)nn! (n+ j � 2�+ 1) :Then jI�1 � �I 1j 6 35Xj=5 ����bj(�)j! 1Xn=L+1 (�u)nn! (n+ j � 2�+ 1)����:Since 1 6 u 6 6 and L > 6, the terms unn! (n+ j � 2�+ 1) are decreasing tozero, and so by the alternating series testjI�1 � �I 1j 6 35Xj=5 ����bj(�)j! ���� uL+1(L+ 1)! (L+ 1 + j � 2�+ 1) :

26 DAVID BRADLEYOnce again invoking inequality (3.6) with r > 1, we havejI�1 � �I 1j 6 6L+1(L+ 1) (L+ 1)! 1Xj=5 e��Ein(�r)r�j= 6L+1e��Ein(�r)(L+ 1)! (L+ 1) r4 (r� 1) :Recall that P (x) = 4Xj=0 (�1)j bj(�) xjj! andI4 = Z 11 x�2�e�uxP (x) dx = 4Xj=0 (�1)j bj(�)j! Z 11 xj�2�e�ux dx= 4Xj=0 (�1)j bj(�)j! u2��j�1 Z 1u tj�2�e�t dt:Therefore, we can compute I4 using the complementary incomplete gammafunction (5.5). We have implemented a continued fraction algorithm whichcomputes ��(a; u), an approximation to �(a; u) := euu�a�(a; u), to withinany speci�ed tolerance.Theorem 11. Let �(a; u) := euu�a�(a; u) and let �� be a function whichsatis�es j�(a; u) � ��(a; u)j < 10�11 for all values of a = j � 2� + 1, 1 6� 6 3, and 1 6 u 6 6. Put I�4 := 4Xj=0 (�1)j bj(�)j! e�u��(j � 2�+ 1; u). ThenjI4 � I�4 j < 4:016:::� 10�11 on the full rectangle 1 6 u 6 6; 1 6 � 6 3.Proof. We haveI4 = 4Xj=0 (�1)j bj(�)j! u2��j�1�(j � 2�+ 1; u)= 4Xj=0 (�1)j bj(�)j! e�u�(j � 2�+ 1; u):Thus, jI4 � I�4 j 6 e�u 4Xj=0 ����bj(�)j! ���� 10�11. The remainder of the proof consistsin improving our bounds on j bj(�)=j! j for j = 0; 1; 2; 3; 4.

A SIEVE AUXILIARY FUNCTION 27From Proposition 3.1, we infer thatb0 = 1; b1 = ��; b22! = �22 � �4 ; b33! = ��36 + �24 � �18 ;b44! = �424 � �38 + 25�2288 � �96 :Thus jb0j = 1; jb1j = j�j 6 3. Since b02(�) = 2� � 1=2 > 0 on [1, 3], it followsthat b2(1) 6 b2 6 b2(3) whence jb2=2!j 6 15=4: A similar argument givesjb3=3!j 6 29=12: Unfortunately, b4 is not monotone on [1, 3]. However, it is notdi�cult to show that b4 has precisely one extremum in ]1, 3[. The extremumis a local minimum and lies in [1.5, 2.0]. The graph of b4(�)=4! is easily shownto be convex on this subinterval and so we construct tangents at � = 1:5 and� = 2:0 and take the height of their intersection as a possible lower bound onb4(�)=4!. In this way it is easily shown that jb4=4!j 6 b4(3)=4! = 3=4 on [1, 3].From this the theorem follows immediately.It now remains to show howI2 = Z 201 x�2�e�uxe�Ein(x) dxcan be e�ectively computed. For this, we use Simpson's rule. Recall that theerror committed in using Simpson's rule to estimate R ba r(x)dx is no more thanM(b� a)5=2880n4, where M := maxfjr(4)(x)j : x 2 [a; b]g, and a = x0 < x1 <::: < x2n = b are evenly spaced. For our application, we partition the interval[1, 20] into 15 subintervals chosen so that the number of function evaluationsneeded to commit an error 6 10�10 for each subinterval is approximately thesame. Thus approximately 60 (n = 30 above) function evaluations are neededfor each subinterval. If for given �, u, we put r(x) = x�2�e�uxe�Ein(x), theneach pair of parameter values for �, u requires a separate computation of I2.For each �, u, and subinterval [a, b], we estimate M and hence determinen so that M(b� a)5=2880n4 < 10�10. The estimation of the various M valuesis not as di�cult as it may appear, since it so happens that r(4) is decreasing inx for x > 1; � > 0; u > 0. In fact, r(5)(x) = �x�2�e�uxe�Ein(x)S(x)=x5, whereS(x) is a polynomial in x; �; u; e�x consisting entirely of non-negative terms,a result that can be most easily veri�ed using MAPLE's symbolic capabilities.For a more elegant approach, see [2, p.64]. Thus, it is easy to estimate Mand hence to determine the number of function evaluations needed to keep thequadrature error 6 10�10 for each subinterval. With 15 subintervals, the totalquadrature error is then 6 15� 10�10. If we combine this error with the errorbounds of Theorems 7, 8, 9, and 11, we see that q�(u) can be computed withan absolute error 6 0:5� 10�8 if 1 6 u 6 6, and 1 6 � < 3.

28 DAVID BRADLEY6. Table of q� ValuesBased on the methods of Section 5, we have written C code to evaluatethe function q�(u) in the rectangle 1 6 u 6 6, 1 6 � < 3. The following tableof q� values, rounded to seven decimal places, was created using this code. Inview of the analysis we have given, the accuracy of these data is guaranteed towithin 10�7.We note that various sieve functions, including q�, have been tabulatedpreviously [12, 15, 17]. Although these previous calculations were not sup-ported by numerical analysis, they appear to be in good agreement with ourtables.

A SIEVE AUXILIARY FUNCTION 29un� 1:0 1:1 1:2 1:3 1:41:0 0:0000000 �0:2463298 �0:4517757 �0:5827670 �0:60703621:1 0:1000000 �0:1554682 �0:3861985 �0:5584687 �0:63687351:2 0:2000000 �0:0599213 �0:3103561 �0:5184864 �0:64721841:3 0:3000000 0:0397676 �0:2252552 �0:4640109 �0:63897131:4 0:4000000 0:1431639 �0:1317106 �0:3960188 �0:61288121:5 0:5000000 0:2499139 �0:0303940 �0:3153254 �0:56958191:6 0:6000000 0:3597240 0:0781317 �0:2226219 �0:50961791:7 0:7000000 0:4723479 0:1933882 �0:1185025 �0:43346251:8 0:8000000 0:5875756 0:3149645 �0:0034833 �0:34153211:9 0:9000000 0:7052267 0:4425035 0:1219827 �0:23419622:0 1:0000000 0:8251444 0:5756920 0:2574941 �0:11178562:1 1:1000000 0:9471912 0:7142531 0:4026928 0:02540162:2 1:2000000 1:0712460 0:8579405 0:5572571 0:17709472:3 1:3000000 1:1972009 1:0065331 0:7208959 0:34304652:4 1:4000000 1:3249594 1:1598317 0:8933446 0:52303042:5 1:5000000 1:4544349 1:3176559 1:0743612 0:71683722:6 1:6000000 1:5855488 1:4798409 1:2637237 0:92427332:7 1:7000000 1:7182301 1:6462364 1:4612274 1:14515872:8 1:8000000 1:8524137 1:8167041 1:6666826 1:37932552:9 1:9000000 1:9880400 1:9911164 1:8799134 1:62661653:0 2:0000000 2:1250544 2:1693554 2:1007557 1:88688433:1 2:1000000 2:2634064 2:3513115 2:3290561 2:15998973:2 2:2000000 2:4030493 2:5368827 2:5646710 2:44580193:3 2:3000000 2:5439398 2:7259739 2:8074655 2:74419673:4 2:4000000 2:6860376 2:9184959 3:0573122 3:0550567Table 1. Values of q�(u)

30 DAVID BRADLEYun� 1:0 1:1 1:2 1:3 1:43:5 2:5000000 2:8293050 3:1143652 3:3140913 3:37827013:6 2:6000000 2:9737070 3:3135030 3:5776892 3:71373073:7 2:7000000 3:1192105 3:5158354 3:8479984 4:06133723:8 2:8000000 3:2657846 3:7212924 4:1249166 4:42099263:9 2:9000000 3:4134002 3:9298079 4:4083469 4:79260464:0 3:0000000 3:5620300 4:1413192 4:6981965 5:17608444:1 3:1000000 3:7116478 4:3557668 4:9943774 5:57134704:2 3:2000000 3:8622293 4:5730942 5:2968052 5:97831074:3 3:3000000 4:0137512 4:7932477 5:6053993 6:39689694:4 3:4000000 4:1661916 5:0161759 5:9200823 6:82703014:5 3:5000000 4:3195293 5:2418298 6:2407803 7:26863744:6 3:6000000 4:4737447 5:4701628 6:5674220 7:72164854:7 3:7000000 4:6288186 5:7011300 6:8999392 8:18599554:8 3:8000000 4:7847330 5:9346886 7:2382662 8:66161294:9 3:9000000 4:9414706 6:1707975 7:5823395 9:14843705:0 4:0000000 5:0990149 6:4094173 7:9320981 9:64640665:1 4:1000000 5:2573501 6:6505100 8:2874832 10:15546205:2 4:2000000 5:4164612 6:8940393 8:6484378 10:67554545:3 4:3000000 5:5763335 7:1399700 9:0149071 11:20660105:4 4:4000000 5:7369531 7:3882684 9:3868379 11:74857425:5 4:5000000 5:8983068 7:6389019 9:7641788 12:30141215:6 4:6000000 6:0603815 7:8918391 10:1468799 12:86506345:7 4:7000000 6:2231651 8:1470496 10:5348929 13:43947805:8 4:8000000 6:3866455 8:4045041 10:9281711 14:02460725:9 4:9000000 6:5508113 8:6641742 11:3266691 14:6204036Table 1. Values of q�(u) (continued)

A SIEVE AUXILIARY FUNCTION 31un� 1:5 1:6 1:7 1:8 1:91:0 �0:5000000 �0:2527189 0:1194642 0:5692400 1:01304701:1 �0:5900000 �0:3981341 �0:0624276 0:3857664 0:87756671:2 �0:6600000 �0:5279768 �0:2392354 0:1907910 0:70958281:3 �0:7100000 �0:6407309 �0:4075077 �0:0103330 0:51559971:4 �0:7400000 �0:7350924 �0:5642137 �0:2127803 0:30170261:5 �0:7500000 �0:8099226 �0:7066537 �0:4121459 0:07363831:6 �0:7400000 �0:8642136 �0:8323939 �0:6043663 �0:16312561:7 �0:7100000 �0:8970635 �0:9392178 �0:7856610 �0:40335461:8 �0:6600000 �0:9076573 �1:0250893 �0:9524867 �0:64201321:9 �0:5900000 �0:8952530 �1:0881243 �1:1015022 �0:87423662:0 �0:5000000 �0:8591695 �1:1265684 �1:2295403 �1:09530912:1 �0:3900000 �0:7987782 �1:1387787 �1:3335854 �1:30064582:2 �0:2600000 �0:7134953 �1:1232092 �1:4107544 �1:48577822:3 �0:1100000 �0:6027764 �1:0783990 �1:4582820 �1:64634142:4 0:0600000 �0:4661112 �1:0029623 �1:4735077 �1:77806422:5 0:2500000 �0:3030198 �0:8955804 �1:4538653 �1:87676002:6 0:4600000 �0:1130491 �0:7549941 �1:3968734 �1:93831942:7 0:6900000 0:1042299 �0:5799982 �1:3001275 �1:95870372:8 0:9400000 0:3492247 �0:3694364 �1:1612938 �1:93393932:9 1:2100000 0:6223233 �0:1221966 �0:9781029 �1:86011253:0 1:5000000 0:9238962 0:1627929 �0:7483444 �1:73336553:1 1:8100000 1:2542974 0:4865664 �0:4698631 �1:54989233:2 2:1400000 1:6138667 0:8501242 �0:1405542 �1:30593533:3 2:4900000 2:0029301 1:2544352 0:2416395 �0:99778233:4 2:8600000 2:4218014 1:7004388 0:6787316 �0:6217639Table 1. Values of q�(u) (continued)

32 DAVID BRADLEYun� 1:5 1:6 1:7 1:8 1:93:5 3:2500000 2:8707826 2:1890474 1:1726943 �0:17425063:6 3:6600000 3:3501654 2:7211480 1:7254611 0:34834913:7 4:0900000 3:8602314 3:2976040 2:3389294 0:94959093:8 4:5400000 4:4012534 3:9192567 3:0149622 1:63299633:9 5:0100000 4:9734951 4:5869266 3:7553904 2:40205474:0 5:5000000 5:5772126 5:3014147 4:5620143 3:26022484:1 6:0100000 6:2126543 6:0635033 5:4366053 4:21093584:2 6:5400000 6:8800616 6:8739578 6:3809072 5:25758904:3 7:0900000 7:5796692 7:7335268 7:3966378 6:40355914:4 7:6600000 8:3117055 8:6429433 8:4854900 7:65219504:5 8:2500000 9:0763932 9:6029258 9:6491329 9:00682104:6 8:8600000 9:8739492 10:6141785 10:8892130 10:47073794:7 9:4900000 10:7045851 11:6773922 12:2073550 12:04722384:8 10:1400000 11:5685076 12:7932452 13:6051631 13:73953494:9 10:8100000 12:4659185 13:9624034 15:0842214 15:55090665:0 11:5000000 13:3970153 15:1855212 16:6460949 17:48455365:1 12:2100000 14:3619907 16:4632418 18:2923302 19:54367155:2 12:9400000 15:3610338 17:7961978 20:0244564 21:73143665:3 13:6900000 16:3943295 19:1850117 21:8439855 24:05100705:4 14:4600000 17:4620588 20:6302960 23:7524132 26:50552345:5 15:2500000 18:5643994 22:1326540 25:7512194 29:09810885:6 16:0600000 19:7015255 23:6926798 27:8418686 31:83187005:7 16:8900000 20:8736077 25:3109589 30:0258108 34:70989755:8 17:7400000 22:0808138 26:9880684 32:3044815 37:73526625:9 18:6100000 23:3233083 28:7245773 34:6793027 40:9110356Table 1. Values of q�(u) (continued)

A SIEVE AUXILIARY FUNCTION 33un� 2:0 2:1 2:2 2:3 2:41:0 1:3333333 1:3908464 1:0499836 0:2185501 �1:10010991:1 1:3043333 1:5243262 1:3821575 0:7439733 �0:45131991:2 1:2213333 1:5881626 1:6432278 1:2199458 0:19975991:3 1:0903333 1:5852899 1:8299174 1:6340082 0:82994061:4 0:9173333 1:5193991 1:9406981 1:9763980 1:41915511:5 0:7083333 1:3948118 1:9755306 2:2397103 1:95017991:6 0:4693333 1:2163857 1:9356654 2:4186357 2:40841721:7 0:2063333 0:9894412 1:8234878 2:5097531 2:78172051:8 �0:0746667 0:7197033 1:6423938 2:5113625 3:06025231:9 �0:3676667 0:4132552 1:3966896 2:4233480 3:23636552:0 �0:6666667 0:0765007 1:0915091 2:2470638 3:30450452:1 �0:9656667 �0:2838678 0:7327443 1:9852380 3:26112002:2 �1:2586667 �0:6608950 0:3269873 1:6418900 3:10459582:3 �1:5396667 �1:0473851 �0:1185195 1:2222605 2:83518532:4 �1:8026667 �1:4359209 �0:5959269 0:7327492 2:45495522:5 �2:0416667 �1:8188800 �1:0968160 0:1808610 1:96773582:6 �2:2506667 �2:1884493 �1:6122308 �0:4248418 1:37907692:7 �2:4236667 �2:5366372 �2:1327077 �1:0747824 0:69620762:8 �2:5546667 �2:8552849 �2:6483007 �1:7584074 �0:07199972:9 �2:6376667 �3:1360758 �3:1486047 �2:4642212 �0:91506213:0 �2:6666667 �3:3705447 �3:6227762 �3:1798173 �1:82091583:1 �2:6356667 �3:5500851 �4:0595517 �3:8919068 �2:77594423:2 �2:5386667 �3:6659566 �4:4472651 �4:5863441 �3:76500313:3 �2:3696667 �3:7092911 �4:7738629 �5:2481510 �4:77144373:4 �2:1226667 �3:6710987 �5:0269181 �5:8615382 �5:7771352Table 1. Values of q�(u) (continued)

34 DAVID BRADLEYun� 2:0 2:1 2:2 2:3 2:43:5 �1:7916667 �3:5422730 �5:1936439 �6:4099264 �6:76248423:6 �1:3706667 �3:3135958 �5:2609050 �6:8759644 �7:70645463:7 �0:8536667 �2:9757418 �5:2152288 �7:2415469 �8:58658483:8 �0:2346667 �2:5192823 �5:0428162 �7:4878312 �9:37900453:9 0:4923333 �1:9346896 �4:7295503 �7:5952520 �10:05845074:0 1:3333333 �1:2123398 �4:2610060 �7:5435362 �10:59828234:1 2:2943333 �0:3425167 �3:6224581 �7:3117162 �10:97049444:2 3:3813333 0:6845856 �2:7988889 �6:8781425 �11:14573154:3 4:6003333 1:8788594 �1:7749958 �6:2204963 �11:09330044:4 5:9573333 3:2502815 �0:5351981 �5:3158000 �10:78118184:5 7:4583333 4:8089097 0:9363567 �4:1404290 �10:17604234:6 9:1093333 6:5648813 2:6557865 �2:6701210 �9:24324534:7 10:9163333 8:5284105 4:6394688 �0:8799864 �7:94686124:8 12:8853333 10:7097863 6:9040356 1:2554829 �6:24967784:9 15:0223333 13:1193709 9:4663676 3:7624043 �4:11320965:0 17:3333333 15:7675971 12:3435899 6:6674961 �1:49770755:1 19:8243333 18:6649676 15:5530666 9:9980699 1:63783285:2 22:5013333 21:8220522 19:1123970 13:7820221 5:33566195:3 25:3703333 25:2494871 23:0394107 18:0478272 9:63926855:4 28:4373333 28:9579729 27:3521639 22:8245302 14:59337165:5 31:7083333 32:9582735 32:0689354 28:1417400 20:24391285:6 35:1893333 37:2612142 37:2082227 34:0296230 26:63804875:7 38:8863333 41:8776811 42:7887388 40:5188963 33:82414405:8 42:8053333 46:8186194 48:8294083 47:6408222 41:85176465:9 46:9523333 52:0950320 55:3493642 55:4272017 50:7716704Table 1. Values of q�(u) (continued)

A SIEVE AUXILIARY FUNCTION 35un� 2:5 2:6 2:7 2:8 2:91:0 �2:7500000 �4:3749277 �5:4031157 �5:0933651 �2:67423231:1 �2:1292333 �4:0168146 �5:6002049 �6:1323890 �4:72738991:2 �1:4230667 �3:4677596 �5:5018414 �6:8127828 �6:45501111:3 �0:6639000 �2:7630670 �5:1339481 �7:1330788 �7:80731861:4 0:1182667 �1:9381278 �4:5271187 �7:1028117 �8:75215051:5 0:8958333 �1:0278948 �3:7153951 �6:7406573 �9:27292691:6 1:6436000 �0:0664595 �2:7352620 �6:0728617 �9:36685961:7 2:3387667 0:9132980 �1:6248025 �5:1318880 �9:04334751:8 2:9609333 1:8800020 �0:4229747 �3:9552308 �8:32251661:9 3:4921000 2:8040314 0:8310157 �2:5843615 �7:23387412:0 3:9166667 3:6577369 2:0982680 �1:0637805 �5:81505542:1 4:2214333 4:4156297 3:3406780 0:5598468 �4:11064752:2 4:3956000 5:0545501 4:5213805 2:2384820 �2:17107602:3 4:4307667 5:5538166 5:6051508 3:9234466 �0:05154592:4 4:3209333 5:8953616 6:5587711 5:5661059 2:18897182:5 4:0625000 6:0638529 7:3513685 7:1185086 4:48871362:6 3:6542667 6:0468063 7:9547276 8:5339910 6:78406232:7 3:0974333 5:8346880 8:3435819 9:7677480 9:01045612:8 2:3956000 5:4210094 8:4958864 10:7773759 11:10326982:9 1:5547667 4:8024151 8:3930734 11:5233914 12:99867273:0 0:5833333 3:9787654 8:0202946 11:9697269 14:63446493:1 �0:5079000 2:9532125 7:3666497 12:0842073 15:95089373:2 �1:7057333 1:7322733 6:4254038 11:8390077 16:89145263:3 �2:9945667 0:3258968 5:1941944 11:2110963 17:40366473:4 �4:3564000 �1:2524712 3:6752293 10:1826614 17:4398505Table 1. Values of q�(u) (continued)

36 DAVID BRADLEYun� 2:5 2:6 2:7 2:8 2:93:5 �5:7708333 �2:9858275 1:8754758 8:7415261 16:95788273:6 �7:2150667 �4:8535534 �0:1931574 6:8815496 15:92192853:7 �8:6639000 �6:8313594 �2:5136458 4:6030180 14:30317993:8 �10:0897333 �8:8912332 �5:0636803 1:9130250 12:08057363:9 �11:4625667 �11:0013886 �7:8155053 �1:1741585 9:24150064:0 �12:7500000 �13:1262176 �10:7357614 �4:6367217 5:78250664:1 �13:9172333 �15:2262444 �13:7853338 �8:4449618 1:70998354:2 �14:9270667 �17:2580801 �16:9192046 �12:5609376 �2:95914764:3 �15:7399000 �19:1743805 �20:0863101 �16:9381317 �8:19676094:4 �16:3137333 �20:9238044 �23:2294023 �21:5211192 �13:96285804:5 �16:6041667 �22:4509739 �26:2849130 �26:2452429 �20:20490684:6 �16:5644000 �23:6964356 �29:1828233 �31:0362942 �26:85718734:7 �16:1452333 �24:5966230 �31:8465346 �35:8102002 �33:84014354:8 �15:2950667 �25:0838211 �34:1927441 �40:4727159 �41:05974104:9 �13:9599000 �25:0861302 �36:1313225 �44:9191218 �48:40683135:0 �12:0833333 �24:5274328 �37:5651946 �49:0339264 �55:75652015:1 �9:6065667 �23:3273598 �38:3902232 �52:6905724 �62:96754215:2 �6:4684000 �21:4012585 �38:4950944 �55:7511493 �69:88164015:3 �2:6052333 �18:6601612 �37:7612061 �58:0661078 �76:32294905:4 2:0489333 �15:0107546 �36:0625587 �59:4739802 �82:09738465:5 7:5625000 �10:3553503 �33:2656472 �59:8011036 �86:99203625:6 14:0062667 �4:5918551 �29:2293566 �58:8613468 �90:77456455:7 21:4534333 2:3862568 �23:8048582 �56:4558415 �93:19260255:8 29:9796000 10:6899718 �16:8355081 �52:3727157 �93:97316095:9 39:6627667 20:4347639 �8:1567475 �46:3868313 �92:8220372Table 1. Values of q�(u) (continued)References[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions,Dover, New York, 1972.[2] D. Bradley, A Sieve Auxiliary Function, Ph.D. Thesis, University of Illi-nois, Urbana, 1995.[3] H. Diamond, H. Halberstam, and H.-E. Richert, Combinatorial sieves ofdimension exceeding one, J. Number Theory 28 (1988), 306{346.[4] H. Diamond, H. Halberstam, and H.-E. Richert, Sieve auxiliary func-

A SIEVE AUXILIARY FUNCTION 37tions I, in \Number Theory, Proceedings of the First Conference of theCanadian Number Theory Association" (R. Mollin, ed.), W. de Gruyter,Berlin, 1990, pp. 99{113.[5] H. Diamond, H. Halberstam, and H.-E. Richert, A boundary value prob-lem for a pair of di�erential delay equations related to sieve theory, I,in \Analytic Number Theory: Proceedings of a Conference in honourof P. T. Bateman" (B. Berndt, et al. eds.), Birkh�auser, Boston, 1990,pp. 133{157.[6] H. Diamond, H. Halberstam, and H.-E. Richert, A boundary value problemfor a pair of di�erential delay equations related to sieve theory, II, J.Number Theory 45 (1993), 129{185.[7] H. Diamond, H. Halberstam, and H.-E. Richert, Sieve auxiliary functionsII, in \A tribute to Emil Grosswald: Number theory and related analysis"(M. Knopp and M. Sheingorn, eds.), Contemporary Math., Vol. 143,Amer. Math. Soc., Providence, RI, 1993, pp. 247{253.[8] H. Diamond, H. Halberstam, and H.-E. Richert, A boundary value problemfor a pair of di�erential delay equations related to sieve theory, III, J.Number Theory 47 (1994), 300{328.[9] H. Diamond, H. Halberstam, and H.-E. Richert, Estimation of the sieveauxiliary functions q� in the range 1 < � < 2, Analysis 14 (1994), 75{102.[10] H. Diamond, H. Halberstam, and H.-E. Richert, Combinatorial sieves ofdimension exceeding one II, This Proceedings, XX{YY.[11] H. Iwaniec, Rosser's sieve, Acta Arith. 36 (1980), 171{202.[12] H. Iwaniec, J. van de Lune, and H. J. J. te Riele, The limits of Buchstab'siteration sieve, Nederl.-Akad.-Wetensch.-Indag.-Math. 42 (1980), no. 4,409-417.[13] A. Odlyzko, e-mail communication, July 9, 1993.[14] F. W. J. Olver, Asymptotics and Special Functions, Academic Press, NewYork, 1974.[15] H. J. J. te Riele, Numerical Solution of two coupled nonlinear equationsrelated to the limits of Buchstab's iteration sieve, Afdeling NumeriekeWiskunde [Department of Numerical Mathematics], 86, MathematischCentrum, Amsterdam, 1980.[16] H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, New York,1967.[17] F. Wheeler, On Two Di�erential-Di�erence Equations Arising in AnalyticNumber Theory, Ph.D. Thesis, University of Illinois, Urbana, 1988.[18] F. Wheeler, Two di�erential-di�erence equations, Trans. Amer. Math.Soc. 318 (1990), 491{523.

38 DAVID BRADLEYDavid BradleyCentre for Experimentaland Constructive MathematicsSimon Fraser UniversityBurnaby B.C. V5A 1S6CANADAdbradley@cecm.sfu.ca