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Some sums over the non-trivial zeros of the Riemann zeta function

We prove some identities, which involve the non-trivial zeros of the Riemann zeta function. From them we derive some convergent asymptotic expansions related to the work by Cram\'er, and also new representations for some arithmetical functions in terms of the non-trivial zeros.

Riemann also knew this product and arrived at it using, in Edwards' words [6, p. 18], an obscure argument. As the product shows that the zeros of zeta determine the whole function, Riemann thought that there had to be a connection among the prime numbers and the zeros of zeta. He showed directly this relation by proving an explicit formula for the number of primes π(x) less or equal than x in terms of the complex zeros of ζ(s). A simpler variant of his formula is valid for x not a prime power, where Λ(n) is the Mangoldt function, which is defined by Λ(n) = log p if n is a power of p, and 0 otherwise. The asymptotic approximation ψ(x) ∼ x is equivalent to the Prime Number Theorem conjectured by Gauss, namely From the functional equation of zeta, Riemann proved that all the non-trivial zeros of zeta are in the band β ∈ [0, 1] and in 1896 Hadamard and de La Vallée Poussin achieved to prove the Prime Number Theorem by showing that ζ(s) has no zeros of the form ρ = 1+iγ. The Riemann's famous conjecture stating that the real part of all the non-trivial zeros was equal to 1/2 remains unproved. Riemann introduced the notation ρ = 1/2 + iτ for the zeros of zeta. In this way his conjecture is the statement that all the τ ′ s are real. In addition, Riemann conjectured a convergent asymptotic behavior, as T → +∞, for the number of complex zeros with positive imaginary part less or equal than T : which was proved by von Mangoldt, who improved this result, by showing in 1905 that Around 1912 Backlund derived an exact formula for counting the zeros [6]: where θ(T ) is the Riemann-Siegel theta function, which is defined by In 1911 E. Landau proved the following formula for the Mangoldt function [12], [11]: which implies It has the surprising property that neglecting a finite number of zeros of zeta we still recover the Mangoldt function. Related to it is the self-replicating property of the zeros of zeta observed recently in the statistics of [15], and later proved in [8].
In this paper we prove a formula which resembles in a certain way to that of Landau: which implies sinh xτ sinh πτ cos(τ log t) .
We see that it shares with that given by Landau the property of invariance when we neglect a finite number of zeros. However our representation is smooth. In addition we give representations, of the same nature, for the functions of Moebius and Euler-Phi.
Other results in this paper, are some asymptotic expansions related to Crámer's work but which are more explicit, and other kind of limit evaluations, one of which writes if we assume the Riemann Hypothesis.

Series involving the Riemann zeta function
The formulas that we will prove involve a sum over the non-trivial zeros of zeta. We use the notation ρ = β + iγ for these non-trivial zeros. Following Riemann, we define τ = −i(ρ − 1 2 ). Hence ρ = 1/2 + iτ (Riemann used the notation α instead of τ ). The Riemann Hypothesis is the statement that all the τ ′ s are real. We use C for Euler's constant as Euler did. As usual in papers of Number Theory, log denotes the naperian logarithm.
The proof for t < −1 is similar.
In a similar way for t < 0, we get z s sin πs ≤ 2 e σ log |z| e t(π−arg(z)) .

Lemma 2.3. Let us consider the vertices
for |z| < 1 and H being each of the segments AB, BC, CD, and for |z| > 1 and H being each of the segments AB ′ , B ′ C ′ , C ′ D.
Proof. By Lemma 2.1 the integrals along the segments AB, CD for |z| < 1, and AB ′ , C ′ D for |z| > 1 are equal to 0, and by Lemma 2.2 the integrals along the segments BC for |z| < 1 and B ′ C ′ for |z| > 1 are also 0.
We will choose numbers T which satisfy the above condition. In addition, suitable bounds for σ ≥ 1 and T ≥ 8 are known [7, Theorem 2.4-(d)]: For σ ≥ 2, we have and for σ < −1, we know that supposing that circles of radius 1/4 around the trivial zeros s = −2k of ζ(s) are excluded [14,Lemma 12.4]. The lemma follows from these bounds. Observe that the circles we need to exclude justify the choice S = 1/2 − 2⌊T ⌋ in Lemma 2.3.
Lemma 2.5. If we assume the Riemann hypothesis, then we can prove that the function φ(s) = 1/ζ(s + 1/2) satisfies the hypothesis of Lemma 2.3, for suitable numbers T .

2.2.
Series with the Mangoldt function.
Let Ω = C−(−∞, 0] (the plane with a cut along the real negative axis). We shall denote by log z the main branch of the log function defined on Ω taking | arg(z)| < π.
We also denote by z s = exp(s log(z)), the usual branch of z s defined also on Ω. For all Remark 1. This formula is a specialization of [10, Lemma 1] taking It needs to evaluate 1 2π However our proof of this particular case is direct and probably simpler.
Proof. Let I(z) be the analytic continuation of the function along the vertical axis x = −1/2. For evaluating I, we first consider rectangles of vertices If we choose suitable real numbers T and use Lemma 2.4, we see that for |z| < 1 the integrals, as T → ∞, along the sides AB, BC and CD are equal to zero. By applying the residues theorem, we obtain .
By analytic continuation, we have that for all z ∈ Ω In a similar way, for |z| > 1 we integrate along the rectangles of vertices A = −1/2 + iT , By Lemma 2.4 we see that when T → +∞, the integrals along the sides AB ′ , B ′ C ′ and C ′ D are equal to zero. Hence, by the residues method, we deduce that where we understand the expression inside the first sum of (9) as a limit based on the identity We use the functional equation (which comes easily from the functional equation of ζ(s)): to simplify the sums in (9). For the first sum in (9), we obtain and for the last sum in (9), we have where ψ is the digamma function, which satisfies the property Using the identity, due to Hongwei Chen [2, p.299, exercise 34] we get that for |z| > 1 Then, by analytic continuation, we obtain that for all z ∈ Ω: By identifying (8) and (11), and observing that the pole at z = 1 is removable, we can complete the proof.
Corollary 2.7. For x > 0, the function h(z) in (7) becomes Proof. For x > 0 we have and the formula follows.
Corollary 2.8. Replacing z with the real number x in (6), multiplying by √ x and taking the limit as x → 0 + , we see that Corollary 2.9. For x real sufficiently large the inequality holds in case that the Riemann Hypothesis is true.
Proof. The function h(x) and the second sum, replacing z with x, of the left side of (7) tend to 0 as x → +∞. On the other hand, if we assume the RH then the sum over the zeros of zeta in (7) is of order less than 10 −18 for all x > 0.
Theorem 2.10. The following identity where holds for |Re(z)| < π. If in addition |Im(z)| < log 2, then we have Proof. Let That is From (6), we see that the function H(z) has the property H(z) = −H(z −1 ). Hence When |Re(z)| < π we have e iz ∈ Ω so that, we may put e iz instead of z in Theorem 2.6. If in addition we multiply by −i/2, we get From (16), we have which we can write as Using elementary trigonometric formulas we arrive at (14). Finally, as we see that the expression in (15) is convergent for |Im z| < log 2.
It is interesting to notice the known expression: In addition, from the functional equation of the Riemann zeta function, one can prove that we can check however that it is not possible to use the functional equation to derive an identity for the constant ζ ′′ (1/2)/ζ(1/2).

Series with the Moebius function.
Theorem 2.14. The following identity holds for |Re(z)| < π and Im(z) < 0 assuming the Riemann Hypothesis and that all the zeros of zeta are simple. Proof.
Let Ω = C − (−∞, 0] (the plane with a cut along the real negative axis). We shall denote by log z the main branch of the log function defined on Ω taking | arg(z)| < π. We also denote by z s = exp(s log(z)), the usual branch of z s defined also on Ω. Then, let I(z) be the analytic continuation of the function π sin πs z s ds.
along the vertical axis x = −1/2. To evaluate I(z), we first consider rectangles of vertices If we choose suitable real numbers T then, by Lemma 2.5 (which assumes the RH), we see that for |z| < 1 the integrals, as T → ∞, along the sides AB, BC and CD are equal to zero. Hence we can apply the residues theorem, and we obtain By analytic continuation, we deduce that for z ∈ Ω we have Integrating along the rectangles of vertices using Lemma 2.5 (which assumes the RH), we see that the integrals as T → ∞ along the sides AB ′ , B ′ C ′ and C ′ D are equal to zero. By the residues method, we obtain where we understand the expression inside the first sum of (24) as a limit based on the identity Identifying (23) and (24), multiplying by i and replacing z with e iz , we complete the proof of the theorem.

Asymptotic behaviors
The formulas here are related to the work by Cramér [5] Example 3.1. If we let in (15) z = π − 1/T , then we have the following convergent asymptotic expansion as T → ∞: where A is the constant as T → +∞.

Representation of arithmetical functions
In this section we give representations of the functions: Mangoldt Λ(t), Moebius µ(t) and Euler phi ϕ(t), in terms of the non-trivial zeros of the Riemann zeta function. We consider that these functions are defined in R + rather than in Z + , and we will see that it is natural to define them as 0 when t is a non-integer.
Proof. Replace z with x − i log t in (14) for t > 1, take real parts and observe that We have to prove that But this identity is evident if t is not the power p k of a prime p. When t = p k the formula comes observing that the only term that contributes to the sum is n = p k . Using the elementary trigonometric identity tan we complete the proof of the theorem.
sinh xτ sinh πτ cos(τ log t) If we assume the Riemann Hypothesis, then we can replace τ and Re τ with γ.
We have used Sage [17] to write the code below. In it we have taken x = 3.14 and the first 10000 zeros of zeta.
Proof. Substitute t = 2 in formula (28), then use this particular formula together with the general formula (28) to eliminate the term involving tan(x/2).
Observe that when t is not a prime nor a power of prime, we have and that the limit is infinite otherwise.
Corollary 4.4. The following formula: holds whenever t is not a prime nor a power of prime. Otherwise the limit is infinite.
and observe that we can commute the limits.
is true whenever t is not a prime nor a power of prime. Otherwise the limit is divergent.
In these formulas γ 1 , γ 2 , · · · , denote the imaginary positive parts of the zeros of zeta (a countable set).

4.2.
A known sum over the zeros of zeta. In next theorem, we reprove in a different way a known formula which is proposed in the book [7,Exercise 5.11]. Its representation has the aspect of a staircase which jumps in the primes, and in the powers of primes. See also the staircase of ψ(x) in [13].
Theorem 4.6. The identity (which assumes the Riemann Hypothesis) where holds for t > 1.
Observe again that although we have written the formula assuming RH, we can write a version of it which does not assume it summing over Re τ > 0 and replacing γ with τ in the other places.
Proof. Replace z with π − i log t in (18). Then take imaginary parts and observe that each logarithm of a negative real number contributes to the imaginary part with iπ. Finally, observe that for powers of prime numbers we have in addition a logarithm of type log(e ix + 1) with x → π − . As the argument of 1 + e ix = (1 + cos x) + i sin x is given by we get that ω = π/2. Hence, in case that t be a power of a prime, one of the logarithms gives the extra contribution iπ/2 to the imaginary part.
We have used Sage [17] to represent the graphic of the function in the left side of (33).

4.3.
Formula for the function of Moebius.
Theorem 4.7. For the Moebius function we have the following representation as x → π − for t > 1: if we assume the Riemann Hypothesis and that all the zeros of zeta are simple.
Proof. Replace z with x − i log t in Theorem 2.14 and take real parts. Simplify and prove that similarly as we have done in Theorem 4.1. On the other hand observe that the real parts of ζ ′ (1/2 + iγ) and ζ ′ (1/2 − iγ) are equal, and that their imaginary parts are opposite. Then, continue as in Corollary 4.2.
Here is a Sage code which represents the Moebius function using the zeros of zeta. In it we have taken x = 3.14 and the first 10000 zeros of zeta.
Proof. The left hand side of (35) comes for the analytic continuation of i 2π e iz ∞ n=0 (−1) n ζ( 1 2 + n) ζ( 3 2 + n) e izn , and the proof of the formula (35) is similar to that of Theorem 2.14.
Proof. Replace z with x − i log t in (35) and continue as in the proof of Theorem 4.7.
Here is a Sage code for the Euler's phi function. In it we have taken x = 3.14 and the first 10000 zeros of zeta.