Orbit spectral zeta: continuation, trace zeros, and regularized product
This chapter extends the previously defined orbit family with a self-contained continuation proof. The Mittag–Leffler seed and Mellin continuation mechanism are classical.
The continuation theorem
Let \(N\ge2\) be an integer, and put \(\alpha=1/N\). Define
List its nonzero zeros \(w_j\) with multiplicity, and use the principal logarithm, with \(-\pi<\operatorname{Arg}w_j<\pi\). No zero lies on the omitted negative ray. The orbit zeta function
converges absolutely exactly when \(\Re s>1/N\). It has a meromorphic continuation to the whole plane, with exactly one pole: a simple pole at \(s=1/N\), of residue
At every nonpositive integer,
At the origin,
The last equality is the zeta regularized product of the orbit zeros under this branch convention. It is a regularized product, not convergence of the ordinary product of the zeros. Positive integer values retain the exact rational coefficient law established in the preceding construction.
The exponential packet and its exact asymptotic
Let \(\omega=e^{2\pi i/N}\). The finite root-of-unity filter gives, for every \(x\in\mathbb C\),
This equality follows by expanding the exponentials and using \(N^{-1}\sum_j\omega^{jm}=1\) when \(N\mid m\), and zero otherwise. For real \(x\ge0\), the power series makes \(F_N(x)>0\). Put
Then
Since \(1+h_N(x)=Ne^{-x}F_N(x)>0\), the real logarithm is well defined for all \(x\ge0\). At infinity,
At the origin \(g_N(x)=O(x^{N-1})\). Its only Taylor powers are \(x^{Nm-1}\), with \(m\ge1\).
The branch and absolute-interchange issue
The pre-existing genus-zero product is
For completeness, the series bound \(|d_N(w)|\le\exp(|w|^{1/N})\) gives order at most \(1/N<1\). Hadamard factorization therefore has genus zero and a constant exponential factor, which equals one because \(d_N(0)=1\). The usual zero-count consequence gives \(\sum_j|w_j|^{-\sigma}<\infty\) for every \(\sigma>1/N\). Along the negative real axis (S3) also shows the order is exactly \(1/N\).
It is not enough simply to say that no zero lies on the integration contour. We need a uniform angular gap from the branch cut to interchange the Mellin integral with the zero sum. Here is that gap.
For \(|\arg(-w)|\le\pi/2\), choose \(x=(-w)^{1/N}\) with \(|\arg x|\le\pi/(2N)\). For every \(j\ne0\),
Indeed, the nearest competing root is separated from the positive direction by at least \(3\pi/(2N)\), while the distinguished root is within \(\pi/(2N)\). Cosine decreases on \([0,\pi]\). Thus, once \((N-1)e^{-\gamma_N|x|}<1\), the distinguished term \(e^x\) in (S1) is larger in modulus than the sum of all competitors, and \(d_N(w)\ne0\). Consequently, sufficiently large zeros avoid the entire closed left half-plane. There are only finitely many remaining zeros in a bounded disk. None lies on the negative ray, since \(d_N(-t)>0\) for \(t\ge0\). It follows that some \(\eta_N>0\) satisfies
With \(c=\sin(\eta_N/2)>0\), this implies
To check the elementary estimate, write \(w=re^{i\theta}\): \(|t+w|^2=(t+r)^2-4tr\sin^2(\theta/2)\ge(t+r)^2\cos^2(\theta/2)\).
Mellin representation in the original strip
Let \(H_N(t)=d_N(-t)=E_{N,1}(t)\), and \(G_N(t)=H_N'(t)/H_N(t)\). Logarithmic differentiation of (S4), locally uniformly off the zeros, gives
For \(1/N<\sigma<1\), (S7) yields
This is the absolute interchange needed for complex \(s\) with \(\Re s=\sigma\). Euler's beta integral and the reflection identity give
First prove this for positive \(w\) by substitution and the beta integral, then extend to the slit \(w\)-plane by holomorphic dependence. On compact subsets of that slit plane the denominator estimate supplies the needed local domination.
Combining (S8)–(S10), and substituting \(t=x^N\), proves
For \(x>0\), \(x^{-Ns}=e^{-Ns\log x}\) uses the real logarithm. Multiplicities are already incorporated in (S4), so no simple-zero assumption entered the proof.
Global continuation and the only pole
One can see directly that \(d_N\) has no zero on \(|w|\le1\):
The strict inequality also gives a slightly larger zero-free disk by continuity. Thus \(G_N(t)\) has a Taylor series converging uniformly on \([0,1]\),
For any integer \(K\ge0\), define
The first integral is holomorphic for \(\Re s<K+1\). This includes local uniform domination, since the Taylor remainder is \(O(x^{N(K+1)-1})\). The last integral is entire in \(s\), because (S3) gives exponential decay against every power of \(x\), uniformly on compact \(s\)-sets, including after any number of \(s\)-derivatives. The rational terms exhibit the only possible poles: positive integers \(1,\ldots,K\) and \(1/N\).
On their overlapping domains the functions (S13) agree: adding the next Taylor term subtracts its integral and adds precisely the same rational function. They also agree with (S11) in its original strip. Therefore
provides compatible meromorphic continuations covering the whole plane as \(K\to\infty\). At each positive integer \(m\), the apparent simple pole \(a_{N,m-1}/(m-s)\) cancels with the simple zero of \(\sin(\pi s)\). The term \(1/(Ns-1)\) has residue \(1/N\), and its multiplier at \(s=1/N\) is nonzero. This proves the stated unique simple pole and its residue.
It also proves that the original series cannot converge absolutely at \(s=1/N\). If it did, its absolute terms would dominate the series for real \(s\downarrow1/N\), except for finitely many zeros of modulus below one. Dominated convergence would then keep that real limit bounded, contradicting the nonzero pole. Absolute convergence at any smaller real part would imply convergence at \(1/N\). The abscissa is consequently exactly \(1/N\).
Trace zeros, rational values, and the origin
For \(\Re s<1\) we may use \(K=0\):
This expression is holomorphic at every nonpositive integer. Hence (S14) proves \(\mathcal Z_N(-m)=0\). These statements do not assert that every such zero is simple.
At a positive integer \(m\), taking the removable limit in (S14) gives
If \(c_n=(-1)^n/(Nn)!\), the previous logarithmic-product law therefore remains
In particular all these values are rational. This is the finite coefficient dictionary, independent of numerical zero location.
At \(s=0\), the elementary antiderivatives in (S15) give
Since \(\sin(\pi s)/\pi=s+O(s^3)\), (S14) implies \(\mathcal Z_N(0)=0\) and \(\mathcal Z_N'(0)=-\log N\).
Classical benchmarks and numerical examples
The inherited identities
now hold as meromorphic identities, by equality in the original half-plane and uniqueness of continuation. Their pole residues are respectively \(1/(2\pi)\) and \(1/(4\sqrt2\pi)\), agreeing with the general theorem. Their zero values at nonpositive integers follow from the Hurwitz formula, and their regularized products are respectively 2 and 4.
The companion script performs 411 exact rational assertions: independent formal-logarithm and quotient calculations for \(2\le N\le10\) through degree 40; three explicit integer values for each \(N\); and 12 Bernoulli/Hurwitz anchors each for \(N=2,4\). It then evaluates (S13) using a truncated local Taylor sum and high-precision exponential-tail quadrature. Sixteen \(N=2,4\) benchmarks sample both sides of the original convergence boundary, negative real values, and complex values. Values for \(N=3,5\) and checks of \(\mathcal Z_N'(0)\) are recorded in the JSON evidence.
Those quadrature results are approximations. Agreement at two precisions and with known functions tests normalization and implementation; it does not replace the proof, certify intervals, or settle historical novelty.
A worked Hurwitz deformation
There is a concrete shifted family. For a real shift \(a\ge0\), define
again with the principal logarithm and multiplicities. A rightward shift preserves the angular gap from the negative ray: for each nonreal \(w\), \(|\arg(w+a)|\) is nonincreasing in \(a\), and positive real zeros remain positive. The zero sequence is unchanged at infinity up to a bounded translation, so the previous absolute convergence argument applies. The normalized determinant is exactly
This follows by taking the ratio of the two convergent products in (S4), not by introducing an unverified exponential factor. Its logarithmic derivative is \(G_{N,a}(t)=H_N'(a+t)/H_N(a+t)\). The same dominated beta-integral argument gives
At infinity, (S2)–(S3), with \(x=(a+t)^{1/N}\), imply
For fixed \(a\) and any integer \(K\ge0\), the binomial expansion therefore gives
To be explicit about continuation, choose a finite splitting point \(T>2a+1\). Subtract the first \(K+1\) terms of (T5) inside the integral from \(T\) to infinity, and add their integrals back as the rational terms
The remainder integral is holomorphic for \(\Re s>\alpha-K-1\). The integral on any compact interval away from zero is entire. At zero, \(G_{N,a}\) is analytic since \(H_N(a)>0\); subtracting any finite Taylor polynomial there extends the other side, with apparent poles only at positive integers. Such Taylor poles cancel against \(\sin(\pi s)\), just as in Section Global continuation and the only pole. Choosing arbitrarily large subtractions at both ends covers every \(s\in\mathbb C\). The resulting continuations agree on overlaps because all added terms are exactly the integrals of the terms removed.
For \(a>0\), the complete pole set is consequently
and every pole is simple. Its residue is
Here \((b)_0=1\) and \((b)_k=b(b+1)\cdots(b+k-1)\). The sign calculation uses both \(\sin(\pi(\alpha-k))=(-1)^k\sin(\pi\alpha)\) and \(\binom{\alpha-1}{k}=(-1)^k(1-\alpha)_k/k!\); their signs cancel. All residues are positive when \(a>0\). At \(a=0\), the secondary coefficients vanish, so the only remaining pole is the unshifted pole at \(1/N\). As before, its nonzero residue makes the absolute convergence abscissa exactly \(1/N\).
None of the pole locations \(\alpha-k\) is an integer. Thus the sine factor vanishes at every nonpositive integer without an opposing pole:
Finally, the one-term subtraction is already convergent at \(s=0\), since \(G_{N,a}(t)-\alpha t^{\alpha-1}=O(t^{\alpha-2})\) and \(\alpha<1\). Let
Integrating exactly and using \(\log H_N(a+t)-t^\alpha\to-\log N\) gives
The multiplier \(\sin(\pi s)/\pi\) then gives
For \(N=2\), the product reduces to \(2\cosh\sqrt a\). Four direct quadrature checks at \(a=0,1/4,2,9\) agree with \(-\log(2\cosh\sqrt a)\) at 80-digit working precision. They use a local integral, an exactly integrated algebraic leading tail, and an exponentially decaying remainder; the quadrature does not sample numerical zero locations. These are four additional numerical benchmarks, with the original 411 exact assertions unchanged. Results are in evidence/spectral_continuation_checks.json under shifted_N2_product_benchmarks.
This is a classical Hurwitz-type spectral deformation, with a packet-specific explicit regularized determinant. It is useful because a single shift reveals an entire pole ladder that collapses at the unshifted boundary. No historical-priority claim is made for the deformation or determinant identity.
Source comparison and bounded search
The continuation proof uses the finite-order Hadamard product and the beta and reflection identities. DLMF [DLMF], §§5.5, 5.12 and 10.46 provides the special-function identities and identifies the seed as a Mittag–Leffler function. The finite-order factorization theorem is also summarized in Elkies's notes [Elkies]. The normalization, zero counts, complex branch gap and domination estimates needed here were proved in Sections The exponential packet and its exact asymptotic–Global continuation and the only pole.
Mellin subtraction and spectral regularization are established methods. Voros [Voros2003], §§2.1–2.2, with its published erratum [Voros2004], provides a related framework. The present proof checks the hypotheses for this Mittag–Leffler seed directly and does not import a theorem about Riemann zeros. The coefficient approach also has antecedents in the hypergeometric zero spectra of Byrnes, Jiu, Moll and Vignat [Byrnes2013].
This comparison establishes methodological context, not priority for the particular specialization. The archived research notes preserve the broader search record. A complete historical review of Mittag–Leffler and hyper-Bessel zero-zeta functions remains outside the scope of this volume.
Scope and remaining questions
The local proof establishes global continuation for the unweighted orbit family and its nonnegative real shifts. A functional equation is not supplied or presumed. The only-pole theorem concerns the unshifted case; positive shifts have the full pole ladder in (T7). The pole and trace-zero theorems follow from the exact positive-axis exponential asymptotic, and do not imply an RH-type critical-line property. Residue-weighted sums still require separate multiplicity and convergence analysis. A historical comparison against all Mittag-Leffler and hyper-Bessel zero-zeta literature remains open.
The evidence archive distinguishes exact identities, numerical checks and previously corrected implementation failures. None of the open questions listed above is assumed in the continuation proof.