A CLASS OF DIRICHLET SERIES INTEGRALS JONATHAN M. BORWEIN 1. Preliminaries. In this note we extend the solution to a recent Monthly problem to analyze a broad class of Dirichlet series and illustrate the result in action in various ways. More precisely, in [6] the following integral evaluation is obtained: Z © p ª 1 3 ¡ 2 2 cos (t log 2) j³ (1=2 + it)j2 2 dt = ¼ log 2: (1) 0 t + 1=4 This somewhat recondite-looking result transpires to be a case of a rather pretty class of evaluations given in Theorem 1, and worth making explicit. P ¡s We let ¸(s) := n>0 ¸n n represent a formal Dirichlet series, with real coe±cients ¸n and we set s := σ + i ¿ with σ = <(s) > 0. We refer to [2], [3], and [7] for other, largely standard, notation. We consider the following integral Z ¯ ¯ 1 ¯ ¯2 ¯¸(s)¯ ¶¸(σ) := ¯ ¯ d¿; (2) 0 s as a function of the Dirichlet series ¸ (i.e., of the coe±cients ¸n). The question of evaluating (2) when ¸ is the alternating zeta-function was posed and solved (at least formally) in [6]. It is convenient to know that for strictly positive u and a Z 1 cos (at) ¼ ¡au 2 2 dt = e ; (3) 0 t + u 2u as may be established by contour integration, by consulting a computer algebra system, or otherwise. 2. The Main Result. P1 ¡s Theorem 1. For ¸(s) = n=1 ¸n n and s = σ + i ¿ with ¯xed σ = <(s) > 0 such that the Dirichlet series is absolutely convergent it is true that Z ¯ ¯ 1 ¯¸(s)¯2 ¼ X1 ¤2 ¡ ¤2 ¶ (σ) = ¯ ¯ d¿ = n n¡1 ; (4) ¸ ¯ s ¯ 2σ n2σ 0 n=1 Pn where ¤n := k=1 ¸k and ¤0 := 0. 1 2 P1 ¡s More generally, for given absolutely convergent Dirichlet series ®(s) := n=1 ®n n P1 ¡s and ¯(s) := n=1 ¯n n Z 1 ®(s) ¯(s) ¼ X1 A B ¡ A B < d¿ = n n n¡1 n¡1 ; (5) σ2 + ¿ 2 2σ n2σ 0 n=1 Pn Pn in which An = k=1 ®k and Bn = k=1 ¯k: PN ¡s Proof. Let ¸N (s) = n=1 ¸n n : Then P Z 1 2 Z 1 ¡σ+i¿ ¡σ¡i¿ j¸N (s)j N¸n;m>0 ¸n ¸mn m 2 2 d¿ = 2 2 d¿ 0 ¿ + σ 0 ¿ + σ Z Z X ¸ ¸ 1 (n=m)i¿ XN ¸ ¸ 1 (n=m)i¿ = 2 n m d¿ + n m d¿ (nm)σ ¿ 2 + σ2 (nm)σ ¿ 2 + σ2 N¸n>m>0 0 n=m=1 0 Z ¡ ¢ Z X ¸ ¸ 1 cos ¿ log(n=m) XN ¸2 1 1 = 2 n m d¿ + n d¿ (nm)σ ¿ 2 + σ2 n2σ ¿ 2 + σ2 N¸n>m>0 0 n=1 0 ¼ X ¸ ¸ ¼ XN ¸2 = n m + n σ (nm)σ (n=m)σ 2σ n2 σ N¸n>m>0 n=1 ¼ XN ¸ ¤ ¼ XN ¸2 ¼ XN ¤2 ¡ ¤2 = n n ¡ n = n n¡1 : σ n2σ 2σ n2σ 2σ n2σ n=1 n=1 n=1 2 P1 ¡σ 2 Next, we observe that j¸N (s)j · ( n=1 j¸njn ) = M < 1; where M is independent of ¿: Hence ¯ ¯ ¯¸ (s)¯2 M ¯ N ¯ · ; ¯ s ¯ ¿ 2 + σ2 and (4) follows from Lebesgue's theorem on dominated convergence when N ! 1. We can establish (8) in a similar fashion. 2 By Dirichlet's test, the ¯nal series in (4) is convergent when σ > 0 provided that f¤ng is bounded, but we cannot automatically guarantee that it is equal to the integral in (4) in this case, or even that the integral is ¯nite. Simple continuation arguments just will not work. It is easy to check that Z 1 ®(s) ¯(s) h®; ¯i¿ := 2 2 d¿ 0 σ + ¿ de¯nes an extended-value inner product on the space of Dirichlet series with h®; ®iσ = ¶®(σ); which is typically ¯nite for σ large enough. 3 ¡ ¢ P1 ¹ ¡s Next, let L¹(s) := n=1 n n denote the primitive L-function corresponding to the ¡ ¹ ¢ Legendre-Jacobi-Kronecker symbol n [5]. An imprimitive character sum modulo m is induced by another character sum modulo m0 for which mjm0 and Â(m) = Â0 (n mod m0). ³We´ remind the reader that, for a prime p and ¹ relatively prime to p, the Legendre symbol ¹ p is 1 when ¹ is a quadratic residue modulo p and is ¡1 otherwise. The Jacobi symbol extends this multiplicatively for ¹ the product of distinct primes, while the Kronecker symbol provides the ¯nal extension to all relatively prime pairs ¹ and n. Example 1. First, we note that with ¸ ´ 1 we recover the arctangent formula Z 1 ¼ 2 2 dt = : R σ + t σ ¡s ¡1 σ 1 Similarly, let # := s 7! (1 + 2 ) . Here again ¼ ¶#(σ) = 2 #(2σ): 2 Example 2. For the Riemann zeta-function (³ = L1) and for σ > 1, σ 1 ¶ (σ) = ³(2σ ¡ 1) ¡ ³(2σ); ¼ ³ 2 as ¸n = 1 and ¤n = n. By contrast it is known|see [4, eqn. (69)] or [1]|that on the critical line µ ¶ 1=2 1 p 1 ¶ = log( 2 ¼) ¡ γ: ¼ ³ 2 2 More broadly, Crandall (in unpublished notes) has shown that when 0 < σ < 1 Z Z 1 X θ2 1 ¶ (σ) = ¼ dθ = ¼ θ2³(1 + 2σ; θ) dθ; ³ (n + θ)1+2σ 0 n¸0 0 where ³(s; a) is the Hurwitz zeta-function, which is easy to compute. Working a little harder [1], we obtain 8 ¡ 1 [³(2σ) + 2 ³(2σ ¡ 1)=(2σ ¡ 1)] if 0 < σ < 1; σ < 2 ¶³ (σ) = (6) ¼ : 1 ¡ 2 [³(2σ) ¡ 2 ³(2σ ¡ 1)] ; if 1 < σ < 1. 2 Example 3. There are similar formulas for s 7! ³(s ¡ k) with k integral. For instance, applying the result in (4) with ³1 the function sending t 7! ³(t + 1) yields Z µ ¶ 1 1 j³(3=2 + i¿)j2 1 1 2 d¿ = ¶³1 = 2 ³(2; 1) + ³(3) = 3 ³(3); ¼ 0 1=4 + ¿ ¼ 2 P 2 Pn¡1 on using Euler's result [3] that ³(2; 1) := n 1=n k=1 1=k = ³(3): 2 4 Example 4. For the classical alternating zeta-function, ®(s) := (1¡21¡s)³(s), we recover (see [6]): σ 1 ¶ (σ) = ®(2σ); ¼ ® 2 n+1 2 2 as ¸n = (¡1) = ¤n ¡ ¤n¡1: Set σ = 1=2. Then j®(1=2 + it)j2 j1 ¡ 21=2¡i tj2 j³(1=2 + i t)j2 ³ p ´ j³(1=2 + i t)j2 = = 3 ¡ 2 2 cos (t ln (2)) 1=4 + t2 1=4 + t2 1=4 + t2 is precisely the integrand in (1). Thus, since ®(2σ) = log 2, we see that ¶®(1=2) = ¼ log 2 is the evaluation in [6]. Note that to justify the exchange of sum and integral implicit in (4) we have to group the terms more carefully as 1=2 is below the abscissa of absolute convergence of the series|or use the approach in [1] on the Hurwitz zeta-function. 2 Example 5. For the Catalan zeta-function, ¯ := L¡4: σ 1 ¶ (σ) = ¯(2σ); ¼ ¯ 2 n 2 2 as ¸2n = 0; ¸2n+1 = (¡1) , and again ¤n ¡ ¤n¡1 = ¸n: σ 1 For L¡3, the same pattern holds, in that ¼ ¶L¡3 (σ) = 2 L¡3(2σ); but it fails for L5; L¡7, and so on. In general L§d does not lead to output that is again a primitive L-series modulo d. For example, σ X (¡1)(n mod 5) ¶ (σ) = ¡ ; ¼ L5 n2σ 5-n and σ 1 ¶ (σ) = L (2σ) ¡ L (2σ): ¼ L¡8 ¡8 2 ¡4 Unlike the case of primitive L-series (meaning that the coe±cients are not induced by restriction of another L-series), these right-hand side series have coe±cients that are not multiplicative, though always the coe±cients repeat modulo d. 2 While (4) gives an e®ective way of evaluating the integral, direct numerical computation to high precision presents a more interesting challenge. This is in part because of the severe oscillations of the integrand| shown in Figure 1 with σ = 1=2; 1; and 2 for ® and L¡3, respectively. In large part, the issue lies in estimating the integrand well, and so is intrinsically no-trivial. 3. A Characterization and Some Reworkings. The preceding results are nicely explained by the following theorem: 5 0.35 2 0.3 1.5 0.25 0.2 1 0.15 0.1 0.5 0.05 0 5 10 15 20 25 30 0 5 10 15 20 25 30 t t Figure 1. The integrands for ® and L¡3 respectively with σ = 1=2; 1 and 2 Theorem 2. The unique Dirichlet series with all nonvanishing coe±cients satisfying the functional equation ¼ 1 ¶ (σ) = #(2σ) σ # 2 is the alternating zeta-function ®. Moreover, each solutions is alternating of the form X1 (¡1)k #(s) = (7) ns k=1 k for an increasing sequence of integers 0 < n1 < n2 < ¢ ¢ ¢ < nk ¢ ¢ ¢ . Proof. Comparing coe±cients and invoking (4), we see that, in the ring of formal Dirichlet σ 1 series ¼ ¶Ã(σ) = 2 Ã(2σ) if and only if 1 ª2 ¡ ª2 = à (ª + ª ) = à n n¡1 n n¡1 n 2 n Pn for all n, where as before ªn := m=1 Ãm. Equivalently, we have either ªn = ªn¡1 or ªn + ªn¡1 = 1 for all n. Observing that ª0 = 0, we deduce that ªn = 0 or 1 for each n. This leads to (7) when reexpressed in terms of Ãn. 2 ¡s Note that ¸ ´ 1 and s 7! 1 ¡ 3 have the requisite form, as do L¡3,L¡4; and # in the previous examples, whereas the other functions therein do not.