A class of Dirichlet series integrals

Transcription

1 A class of Dirichlet series integrals J.M. Borwein July 3, 4 Abstract. We extend a recent Monthly problem to analyse a broad class of Dirichlet series, and illustrate the result in action. In [5] the following integral evaluation is obtained. ( ) 3 cos (t ln ()) ( ζ (/ + it) ) t dt = π log. () + /4 This somewhat recondite looking result transpires to be a case of a rather pretty class of evaluations given in Theorem, and worth making explicit. We let := n> λ n n s represent a formal Dirichlet series, with real coefficients λ n and we set s := + iτ with R() >. We refer to [,, 4] for other, largely standard, details. We shall consider the following integral: ι λ () := s dτ, () as a function of λ. The question of evaluating () when λ is the alternating zeta function was posed and solved in [5]. It is convenient to know that for u,a > cos (at) t + u dt = π u e au, (3) as may be established by contour integration, by consulting a computer algebra system, or otherwise. We write n,m> ι λ () = τ + dτ = λ n λ m n +iτ m iτ τ + dτ = λ n λ m (n/m) iτ (nm) τ + dτ n,m> = = π τ n,m> λ n λ m (nm) n,m> λ n λ m (nm) (n/m) τ cos (τ log(n/m)) τ + Faculty of Computer Science, Dalhousie University, Halifax NS, B3H W5. Research supported by NSERC, the Canada Foundation for Innovation and the Canada Research Chair Program. jborwein@cs.dal.ca dτ

2 on applying (3). Thus, ι λ () = π n> λ n n + π n> λ n n m= λ m n = π n> λ n Λ n, (4) n where Λ n := n m= λ m + λ n / = n m= λ m λ n /, on taking care to legitimate the steps. Examples. (Primitive L-series) First, we note that with λ := we recover the arctangent formula + t dt = π. R Next, we let L µ (s) := ( µ n= n) n s denote the primitive L-function corresponding to the Kronecker symbol ( µ n), [3]. (All other real character sums can be reduced to such primitive ones.) Then:. For the Riemann zeta function (ζ = L ): as λ n = and Λ n = n /. π ι ζ() = ζ( ) ζ(), There are similar formulae for s ζ(s k) with k integral.. For the alternating zeta function, α := s ( s )ζ(s), we recover, [5]: π ι α() = α(), as λ n = ( ) n+,λ n = / and λ n Λ n = ( ) n+ /. Set := /. Then α(/ + it) /4 + t = / i t ζ(/ + it) ( /4 + t = 3 ) ζ(/ + it) cos (t ln ()) /4 + t is precisely the integrand in (). Thus, since α() = log we see that is the evaluation in [5]. 3. For the Catalan zeta function (β := L 4 ): ι α (/) = π log, π ι β() = β(), as λ n =,λ n+ = ( ) n and again λ n Λ n = λ n /.

3 4. For L 3, the same pattern holds, in that π ι L 3 () = L 3(), but not for L 5,L 7, and so on. In general L ±d does not lead to output which is again a primitive L-series mod d. For example, π ι L 5 () = 5 n ( ) n mod 5 n, and π ι L 8 () = L 8 () L 4(). These are not character sums, though always the coefficients repeat modulo d. 5. Let ϑ := s ( + s ). Then again π ι ϑ() = ϑ(). These results are nicely explained by the following result. Theorem. The unique Dirichlet series with all non-vanishing coefficients satisfying the functional equation π ι ϑ() = ϑ() is the alternating zeta function α. Moreover, all solutions are alternating of the form ( ) k ϑ(s) = k= n s k (5) for increasing integers < n < n < < n k. Proof. Comparing coefficients we see, using (4), that, in the ring of formal Dirichlet series π ι ψ() = ψ() if and only if ( n ) ψn + ψ n = ψ n m= for all n. Written in terms of Ψ n := n m= ψ m, we have Ψ n = Ψ n or Ψ n + Ψ n =, for all n. Noting that Ψ =, we deduce that Ψ n = or for each n. This leads to (5) when reexpressed in terms of ψ n. Note that λ and s 3 s have the requisite form, as do L 3, L 4 and ϑ in Examples, while the other functions therein do not. The intermediate form of (4) can be rewritten as s dτ + π n> λ n n = ι λ() + π where Θ n := n n m= λ m, is the average order of λ n, [4]. n> λ n n = π n> λ n Θ n n (6) 3

4 For an interesting variety of arithmetic functions, each of λ() and λ () := n> λ n n has a closed form (see [4, 3]) and so (6) can be viewed as giving information about the more recondite average order series on the far right. Thus, an application of the arithmetic-geometric mean inequality for λ n /n and nθ n /n yields Thus, s π n> dτ = π λ n Θ n n n> λ n Θ n n π s Examples 3. (Some other number theoretic cases) n> dτ π λ n n π n> n> Θ n n. (7) Θ n n. (8) Clearly, (L ±d ) is simple. In particular, for p prime (L ±p ) (s) = ( p s )ζ(s). By contrast, for the möbius function we have µ(s) := n> µ(n)n s = /ζ(s) directly from the Möbius inversion formula while µ (s) = n SF n s, where SF represents the squarefree numbers, has no explicit formula. Recall also, n> φ(n)n s = ζ(s )/ζ(s). We illustrate further with a few more interesting evaluations detailed in [3, ]:. Let k denote the kth power sum of the divisors of n. There is a beautiful formula given as Theorem 9 of [4]: n= k (n) n s = ζ(s)ζ(s k), R(s) > max{,k + } (9) which is in terms of only ζ(s). In Hardy and Wright, one also finds the following remarkable identity due to Ramanujan (Theorem 35 of [4]) n= for R(s) > max{,a +,b +,a + b + }. a (n) b (n) ζ(s)ζ(s a)ζ(s b)ζ(s a b) n s = ζ(s a b). Let r N (n) be the number of solutions to x + x + + x N = n (counting permutations and signs). Define r N (n) rn L N (s) := n s and R N (s) := (n) n s. n= Simple closed forms for L N (s) are known for N =,4,6,8, via the explicit formulae for r N (n) for these N. Indeed the corresponding q series were known to Jacobi. We have L (s) = 4ζ(s)L 4 (s), L 4 (s) = 8( 4 s )ζ(s)ζ(s ), n= L 6 (s) = ζ(s )L 4 (s) 4ζ(s)L 4 (s ), L 8 (s) = 6( s + 4 s )ζ(s)ζ(s 3). 4 ()

5 t t Figure : The integrands for α and L 3 respectively with = /,, Correspondingly, we have R (s) = (4ζ(s)L 4(s)) ( + s, R(s) >, )ζ(s) R 4 (s) = 64 (8 3 3s s + s + )ζ(s )ζ (s )ζ(s) ( + s, R(s) > 3, )ζ(s ) R 6 (s) = 6 (7 3 s ) ζ(s 4)L 4 (s )ζ(s) ( 6 s ) ζ(s 4) 8 ( + 4 s ) L 4 (s 4)ζ (s )L 4 (s), R(s) > 5, ζ(s 4) and R 8 (s) = 56 (3 6 s 3 3 s + )ζ(s 6)ζ (s 3)ζ(s) ( + 3 s, R(s) > 7. )ζ(s 6) Final Remarks. While (3) gives an effective way of evaluating the integral, directly evaluating the integral numerically to high precision presents a more interesting challenge. This is in part because of the severe oscillations of the integrand as illustrated in Figure with = /,, for α and L 3 respectively. In large part, the issue appears to lie estimating the integrand well and so is intrinsically non-trivial. References [] Jonathan Borwein and David Bailey, Mathematics by Experiment. Plausible Reasoning in the st Century, AK Peters, 3. 5

6 [] Jonathan Borwein, David Bailey and Roland Girgensohn, Experimentation in Mathematics. Computational Paths to Discovery, AK Peters, 4. [3] Jonathan Borwein and Kwok-Kwong (Stephen) Choi, On Dirichlet Series for Sums of Squares, Ramanujan Journal, 7 (3), [4] Godfrey H. Hardy and Edward M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, Oxford, 985. [5] Solution to MAA Problem #939, proposed by Ivić, with his solution in MAA Monthly, November (3),