A continuum certificate for the decreasing Lévy factor associated with Riemann xi

1. Contribution and claim status

Konstantopoulos, Patie and Sarkar ask whether the Riemann xi function belongs to their class $D_P$ of entire functions generated by spectrally negative Lévy exponents [1, Section 4.1]. The coefficient identity singles out a natural meromorphic factor. We construct a finite interval certificate for positivity and strict decrease of its signed inverse on the entire positive axis. The finite calculation has been completed and reconstructed from a public packet containing outward-rounded enclosures and all generation code. The analytic saddle, contour and evaluator lemmas are stated below, with their derivations included in the supplement. The membership conclusion is explicitly conditional on those analytic enclosure lemmas; the complete analytic and numerical argument has not received independent verification.

This is a result about the actual theta-kernel coefficient factor, rather than a finite collection of derivative signs or zeros. Its main method is to retain the cancellation in a quotient of Mellin transforms, subtract a corrected Lambert model, and invert on a contour with exponentially decaying tails. An exterior portion of the model's signed cut becomes a positive term. A finite certificate then bridges analytic small- and large-parameter estimates. This also gives a short reconstruction procedure for a calculation whose original enclosure generation took about an hour of CPU time.

Even a fully accepted $D_P$ membership result would not prove RH. Theorem 4.4 of [1] explicitly shows that $D_P$ contains functions with nonreal zeros. We neither invoke nor establish an additional real-zero criterion. The earlier complete-Bernstein exclusion [2] concerns a smaller class and remains consistent with the proposed ordinary Bernstein membership.

2. Exact objects and certificate implication

Use the standard even theta kernel

$$ P(x)=\sum_{n\ge1} \left(4\pi^2n^4e^{9x/2}-6\pi n^2e^{5x/2}\right)e^{-\pi n^2e^{2x}}, \qquad M(s)=\int_0^\infty x^{2s}P(x)\,dx. \tag{1} $$

Initially $\operatorname{Re}s>-1/2$. Taylor subtraction at zero continues $M$ meromorphically, with possible simple poles at the negative half integers. Set $\Xi(t)=\xi(1/2+it)$, where $\xi$ is the conventional completed zeta function. Normalization removes any constant Fourier factor:

$$ \frac{\Xi(t)}{\Xi(0)} =\frac1{M(0)}\int_0^\infty P(x)\cos(tx)\,dx. \tag{2} $$

The factor and proposed Lévy exponent are

$$ \phi(s)=(4s-2)\frac{M(s-1)}{M(s)},\qquad \Psi(s)=s\phi(s). \tag{3} $$

Write $B_J$ for the Bernstein functions with a nonnegative, nonincreasing Lévy density, as in [1, equation (5.4)]. A function in this class can have a nonzero constant and a linear term. The proposed factor has a positive constant and zero linear term.

Theorem 1 (certificate implication). Assume the uniform analytic enclosures in Lemmas 2–4 and the validity of the moment enclosures generated by the evaluator in Section 5. Then the public finite certificate proves that the signed inverse $k$ in Section 4 satisfies

$$ k(r)>0,\qquad k'(r)<0\qquad(r>0). \tag{4} $$

Consequently

$$ \phi(s)=\kappa+\int_0^\infty(1-e^{-sr})k(r)\,dr\in B_J, \qquad \kappa>0, \tag{5} $$

and $\Psi\in N_D$ with largest nonnegative root zero. Moreover,

$$ J_\Psi(t)=\sum_{n=0}^\infty \frac{(-1)^nt^{2n}}{\prod_{j=1}^n\Psi(j)} =\frac{\Xi(t)}{\Xi(0)}. \tag{6} $$

The conditions in this theorem refer to explicit estimates and an explicit evaluator, not to the desired signs themselves. The numerical obligations are all discharged below. We supply the analytic derivations to make the remaining review obligation concrete; their endpoint scalar checks alone are not represented as proofs of the uniform analytic statements.

To deduce the last assertions from (4), first integrate by parts in the real Mellin integral. For $s>0$ the removable value at $s=1/2$ is included:

$$ \phi(s)=-\frac2{M(s)}\int_0^\infty x^{2s-1}P'(x)\,dx>0. \tag{7} $$

Strict decrease of $P$ is classical; in the notation of Csordas [3, Theorem 4.1], $P(x)=2\Phi(x/2)$. Since $P'(x)=O(x)$ at zero, the same formula at $s=0$ gives $\kappa>0$. The endpoint estimates below justify all Lévy integrals. With $\mu(dr)=-k'(r)\,dr$,

$$ \Psi(s)=\kappa s+ \int_0^\infty(e^{-sr}-1+sr)\mu(dr), \tag{8} $$

where $\int(1\wedge r^2)\mu(dr)<\infty$ and $\int_1^\infty r\mu(dr)<\infty$. This is a spectrally negative Lévy exponent with zero Gaussian coefficient and positive derivative at zero. Finally, the identity

$$ \prod_{j=1}^n\Psi(j)=\frac{(2n)!M(0)}{M(n)} \tag{9} $$

telescopes directly from (3); substituting into the entire cosine series proves (6). No uniqueness claim about arbitrary analytic interpolations of the coefficients is needed.

3. Analytic enclosure lemmas

The supplement's README.md gives a dependency map through the numbered derivations. Here are the precise statements used by the certificate.

Lemma 2 (saddle and corrected model). Put

$$ F(x)=x(\pi e^{2x}-9/4),\quad S_s(x)=2s\operatorname{Log}x+\tfrac92x-\pi e^{2x},\quad A(s)=2F'(x(s))/x(s). $$

On

$$ |s|\ge6000,\qquad \operatorname{Re}s\ge-\frac{|s|}{8\log|s|}, \tag{10} $$

there is a holomorphic saddle $x(s)=u+ih$, $F(x(s))=s$, $u\ge3$, $|h|<\pi/4$, and $\operatorname{Re}A>0$. With the principal logarithm and square root,

$$ L(s)=4\pi^2 e^{S_s(x(s))}\sqrt{2\pi/A(s)},\qquad |M(s)/L(s)-1|\le\frac{4.5}{Q},\quad Q=\pi e^{2u}. \tag{11} $$

In particular $M$ is nonzero there. For

$$ w=W_0(2s/\pi),\qquad f_c(s)=\frac{16s}{w^2}-\frac{36}{w(1+w)}+\frac8{w(1+w)^2}, \tag{12} $$

the residual $E=\phi-f_c$ is analytic and satisfies

$$ |E(s)|\le\frac{6000}{|s|} \quad\text{if}\quad |s|\ge12000,\qquad \operatorname{Re}s\ge-\frac{|s|}{11\log|s|}. \tag{13} $$

Derivation supplied for review. The full theta series is analytic in $|\operatorname{Im}z|<\pi/4$. Its absolute sum on $|z|=H<\pi/4$ is less than $400\cos(2H)^{-5/2}$. Derivation87 proves this directly and gives the meromorphic vertical endpoint formula. Let $\theta=\arg A$ and $d=3/(16u)$. Use a contour from zero vertically to $iH_L$, horizontally to $x-de^{-i\theta/2}$, along the descent diameter to $x+de^{-i\theta/2}$, and horizontally to infinity at height $H_R$, where $H_L=h+d\sin(\theta/2)$ and $H_R=h-d\sin(\theta/2)$.

The saddle equations give $0\le\theta\le2h$ and $\pi/4-h\ge1/(4u)$ in the upper half-plane. Thus both heights stay in the theta strip. On the descent disc $|S'''|\le10Q$, $|S''''|\le19Q$, and the real quadratic decrement is at least $1.54Qv^2$. The odd cubic integrates to zero on the symmetric diameter. Its quartic and squared-cubic remainders have relative integral at most $1.1/Q$.

With $\eta=\frac12e^{-2u+3/(8u)}+20e^{-3Q/(5u)}$, the whole relative error is bounded by

$$ \begin{aligned} \mathcal E(u)={}&\frac{1.1}{Q}+1.24\eta +\frac{2.35u}{\sqrt Q}e^{-1.65Qd^2} +\frac{14u}{\sqrt Q}e^{-1.54Qd^2}\\ &+600u^{5/2}\sqrt{\frac{4.7Q}{2\pi}} (1+10u/Q)e^{-Q/(20u)}+e^{-Q/2}. \end{aligned} \tag{14} $$

The terms respectively bound local phase, full amplitude, omitted Gaussian, two horizontal tails, far left, and continued vertical endpoint. At $u=3$, $Q\mathcal E(u)<4.359006<4.5$; each nonconstant scaled majorant decreases thereafter. Derivations42/43/65 include the horizontal concavity, two endpoint regimes, and these monotonicity arguments. The fixed-real-part saddle image is strictly increasing in height, since $dt/du=|F'(x)|^2/\operatorname{Im}F'(x)>0$. The initial boundary is either $h=0$, $u=3$ with $|s|<6000$, or left of (10). This gives the holomorphic image used in (11).

Write $\delta=\log(M/L)$. The important ratio identity is

$$ \phi(s)=(4s-2)\frac{L(s-1)}{L(s)} e^{\delta(s-1)-\delta(s)}. \tag{15} $$

Discs of radius $|s|/(32\log|s|)$ around the whole unit segment remain in (10). Cauchy applied to the logarithmic difference bounds the cost in (15) by $5500/|s|$. Exact rational expansions of the saddle logarithmic derivatives and comparison with $w/2$ give a further cost less than $400/|s|$ for replacing the saddle ratio by (12). These algebraic identities and remainder estimates are in53/65, with the translated-disc inclusion in62. They give (13) with slack. Taking separate errors for $M(s-1)$ and $M(s)$ would lose the cancellation needed here.

Lemma 3 (compact zero count and pole). The pole-removed analytic normalization in Section 5, its exterior bound, and the boundary enclosures give exactly one zero of $M$ in $\operatorname{Re}s>-11/2$. It is real and simple, at $-\lambda$, and the residue of $\phi$ there is $-C$, with

$$ 4.829963581<\lambda<4.829963584, \qquad38.87867<C<38.87873. \tag{16} $$

There are no other poles of $\phi$ in this half-plane. In particular $\phi$ is holomorphic in $\operatorname{Re}s>-1/2$.

This lemma combines the analytic normalization/exterior argument with a finite winding calculation, rather than inferring a zero count from a root finder. Its arithmetic is reconstructed by the public checker. The analytic justification of the boundary evaluator is part of the stated hypotheses.

Lemma 4 (legitimate inversion and transfer). The saddle estimates, the zero count, and the theta endpoint continuation allow the inverse construction in Section 4 and contour transfer in Section 7. All contour shifts are applied first to absolutely integrable second derivatives; integrations by parts are applied to an analytic function before splitting the Lambert cut. Derivations41/45/47/54/77 give the convergence and boundary terms. In particular (25) below holds with the finite horizontal joins retained. No absolute convergence of the infinite straight-vertical inverse of $sE$ is assumed.

4. The signed inverse and its endpoints

The simpler model $f_0(s)=16s/W_0(2s/\pi)^2$ has an explicit signed dispersion formula. Set $b=\pi/(2e)$. On the upper cut write

$$ w=\alpha+iv,\quad \alpha=-v\cot v,\quad u=\frac\pi2 v\csc v\,e^{-v\cot v},\quad0<v<\pi. $$

The cut density

$$ \rho(u)=\frac{32\alpha v}{\pi(\alpha^2+v^2)^2} \tag{17} $$

is negative for $b<u<\pi^2/4$ and positive thereafter. The Lambert branch facts and keyhole method are standard [4]; after subtracting the origin pole and constant, Cauchy's formula gives

$$ f_0(s)=\frac{4\pi^2}{s}+16\pi+ \int_b^\infty\frac{s\rho(u)}{s+u}\,du. \tag{18} $$

Thus the inverse of $-f_0''$ is $r^2k_0(r)$, where

$$ k_0(r)=-4\pi^2+\int_b^\infty ue^{-ur}\rho(u)\,du. \tag{19} $$

The origin pole must be retained. The remainder $-\phi''+f_0''=O(|s|^{-2})$ on a sufficiently far right line has an absolutely convergent inverse $g$. Its derivative is justified by first integrating by parts. Then $k=k_0+g/r^2$ is real and $C^1$, and $\mathcal L(r^2k)=-\phi''$. The high-frequency estimates imply

$$ k(r)\sim\frac{32}{r^2\log^3(1/r)},\qquad k'(r)\sim-\frac{64}{r^3\log^3(1/r)}\quad(r\downarrow0). \tag{20} $$

For example, $\rho(u)\sim32/\log^3u$; scaling $ur$ gives the leading Laplace integrals. The remainder costs only $O(1/r)$ and $O(\log(1/r)/r^2)$, respectively. This is a direct inversion argument, not a Tauberian assertion from real regular variation alone.

Subtracting the model's origin pole before shifting its remainder to $\operatorname{Re}s=-a_0$, $0<a_0<1/2$, gives $\int(1\wedge r)|k(r)|\,dr<\infty$. Equal second derivatives then leave an affine ambiguity in (5). Both the integral in (5) and $\phi(s)$ divided by $s$ tend to zero as $s\to+\infty$, eliminating the linear term; evaluation at zero gives $\kappa$. Shifting $-\phi''$ farther left across the unique pole yields, for $\lambda<a<11/2$,

$$ k(r)=Ce^{-\lambda r}+O(e^{-ar}/r^2),\qquad k'(r)=-\lambda Ce^{-\lambda r}+O(e^{-ar}/r^3). \tag{21} $$

These signs near the two endpoints alone are insufficient; effective thresholds and the full intervening interval are supplied next.

For $0<r\le e^{-10}$, bounding the negative compact portion of (17) and restricting its positive part to $[1/r,2/r]$ gives

$$ k_0(r)>\frac{1.9}{r^2\log^3(2/r)}-90,\qquad -k_0'(r)>\frac{2.9}{r^3\log^3(2/r)}-81. \tag{22} $$

Write $f_c=f_0+q$. On the line $\operatorname{Re}s=1/r$, Cauchy bounds for the explicit $q$ give $|k_q|\le400e/(r\log^2(1/r))$ and $|k_q'|\le3600e/(r^2\log^2(1/r))$. The residual $E$ is inverted on $s=1/r-t/[64\log(e+t)]+it$ and its conjugate. After establishing the identity on a fixed contour, one may choose this contour for each bound. Its length integral is at most $360\log(1/r)/r$. With $C_0=6000\cdot360e/\pi$, the residual costs are $1.001^2C_0\log(1/r)$ and $1.001C_0\log(1/r)/r$.

At $r_0=1/200000$, the resulting normalized costs are less than $1.304838<1.9$ and $1.930154<2.9$. Every cost increases with $r$ on $(0,r_0]$, as is checked by differentiation of $r^p\log^m(1/r)$ and its displayed variants. This proves both signs throughout that interval, under the analytic bounds. Derivation54 gives the full contour inclusion and differentiated-inverse argument; the checker re-evaluates the endpoint inequalities.

5. Finite enclosures, winding and cancellation

The finite boundary calculation uses

$$ p_6(s)=\frac{\prod_{j=0}^5(s+j+1/2)}{(s+13/2)^6},\qquad G(s)=4\pi^2e^{S_s(x(s+13/2))}\sqrt{2\pi/A(s+13/2)}, \qquad H(s)=p_6(s)M(s)/G(s). \tag{23} $$

The shifted saddle is holomorphic with positive real curvature for $\operatorname{Re}(s+13/2)\ge1$, so $G$ has no zeros there. The numerator cancels precisely the endpoint poles in the target half-plane. The first six even Taylor coefficients of $P$ have signs $+,-,+,-,+,+$; the checker re-evaluates them with a full omitted-theta bound. Thus none of these cancellations creates an artificial zero.

There are1432 contiguous parameter intervals on $s=-11/2+it$, $0\le t\le6007$. Every stored rectangle for $H(s)$ excludes zero. Rotate each rectangle into a positive real half-plane, take its argument interval, and choose the unique $2\pi$ translate overlapping the previous interval. Each argument width is strictly less than $\pi$; the initial value is $\pi$ because $H(-11/2)<0$. The final lifted argument lies in $[-.117,.117]$. The analytic exterior estimate $|H-1|<.091810<.1$ closes it to argument zero at infinity. Conjugation and the clockwise orientation around the right half-plane give exactly one zero, counted with multiplicity.

A separate real-contour evaluator brackets that zero between $-324133369137/67108864000$ and $-2074453562377/429496729600$. It gives $M'<-321.3670$, $M(s-1)<-586.0419$ on the bracket, and the negative residue in (16). These evaluations are repeated by the short checker, rather than trusted as printed root coordinates. The one-zero count and real bracket suffice; additional historical boundary covers are not used.

The moment evaluator uses a fixed contour and the same fixed saddle normalization $N(s)$ for $M(s)$ and $M(s-1)$. Its logarithm is affine in $s$, with derivative $2\operatorname{Log}y$. Thus the $j$th normalized moment derivative is obtained by inserting $[2\operatorname{Log}(z/y)]^j$ in the contour integrand. It does not differentiate a moving numerical root. Finite horizontal integrals use Acb interval quadrature; geometric theta tails, a concave tangent spatial tail, and a continued vertical Taylor series account for everything omitted. The Taylor remainder circle stays strictly inside the theta strip. The all-order formulas in derivations48/80 apply to every fixed order used, including16. Approximate root finding only supplies a saddle center; a separate Rouché inequality certifies the enclosing disc.

For a frequency panel centered at $s_*$ with halfwidth $h$, let $m,n$ be the two normalized moments and $m_N,n_N$ their degree15 Taylor polynomials. The order16 derivative is enclosed on the whole low-frequency panel. Compute a quotient polynomial $P_*$ through degree15 and form the polynomial residual $(4s-2)n_N-m_NP_*$ before taking absolute values. Taylor remainders bound the true residual and its first three derivatives. A strictly positive denominator lower bound then recursively bounds $(\phi-P_*)^{(j)}$. This retains cancellations that direct interval division would destroy. Derivation56 states the complete recursion.

The337 low panels cover $[0,12500]$. Above12500, use the saved point jets and Lemma2 on the disc $\rho=t/(12.001\log t)$ with maximum $6061/t$. If $q_*=h/\rho<.341$, the first two remainder bounds are

$$ e_0=\frac{6061}{t}\frac{q_*^{16}}{1-q_*},\qquad e_1=\frac{6061}{t\rho} \frac{q_*^{15}(16-15q_*)}{(1-q_*)^2}. \tag{24} $$

The543 high panels cover $[12500,200000]$. Every disc inclusion is checked. The norm for $(sE)'$ is taken after forming $P_*+sP_*'$. Higher derivative tails for the large-$r$ estimate come from differentiating $(6061/t)q_*^{16}/(1-q_*)$. All inequalities are interval bounds over whole panels, not extrapolations from center values.

6. Public arithmetic results

All results in this section use the 14-digit outward-rounded public inputs, not the narrower original enclosures. A triple $[m,r,e]$ represents $10^e[m-r,m+r]$. Encoding asserts that its decoded Arb ball contains the complete source ball, including both midpoint rounding and radius. The public reconstruction recomputes quotient polynomials and remainders from these widened inputs.

| Certificate quantity | Public bound or result | |---|---:| | Winding rectangles on $[0,6007]$ |1432; winding count1 | | Low moment panels on $[0,12500]$ |337 | | High moment panels on $[12500,200000]$ |543 | | $\int_{12500}^{200000}|E'|\,dt$ |$<0.000106310$ | | $\int_{12500}^{200000}|(sE)'|\,dt$ |$<3.708340$ | | Middle intervals covering $[1/200000,1]$ |105; every strict comparison passes | | Minimum middle margin for $k$ |$>96.55608$ | | Minimum middle margin for $-k'$ |$>167.51780$ | | Relative large-$r$ error at $r=1$, for $k$ |$<0.443756$ | | Relative large-$r$ error at $r=1$, for $-k'$ |$<0.815556$ |

At $s=-21/4+200000i$, the checker also verifies $|E|<1/T$, $|E'|<1/T^2$, $|sE|<1$, and $|(sE)'|<1/T$ for $T=200000$. These are actual finite-input checks. The middle computation uses the reconstructed bounds themselves, rather than assuming prospective budgets of1 and25. Both final margins include pole uncertainty, model integration error, panel remainders, parameter variation, horizontal joins and infinite tails.

7. Middle inversion on a fixed bent contour

Set $a=21/4$, $T=200000$, $d(t)=t/(12\log t)$. Let $\Gamma$ run upwards on $\operatorname{Re}s=-a$ for $|\operatorname{Im}s|\le T$, join horizontally to $-d(T)\pm iT$, and continue along $-d(|t|)+it$. Starting from the inverse of $-\phi''$, cross the unique actual pole and integrate analytic $\phi$ by parts twice. At every corner the boundary values cancel; the bent tails decay for each $r>0$. Only after this step split $\phi=f_c+E$.

The model origin and branch endpoint lie to the right of $\Gamma$. An exterior keyhole therefore involves only the cut $u\ge a$:

$$ \begin{aligned} k(r)&=Ce^{-\lambda r}+K_0(r)-I_\Gamma(E)(r),\\ -k'(r)&=\lambda Ce^{-\lambda r}+K_1(r)+I_\Gamma(sE)(r),\\ K_j(r)&=\int_a^\infty u^je^{-ur}\rho_c(u)\,du, \qquad I_\Gamma(F)=\frac1{2\pi i}\int_\Gamma e^{sr}F(s)\,ds. \end{aligned} \tag{25} $$

For the cut coordinates in Section4, put $D_0=\alpha^2+v^2$ and $D_1=(1+\alpha)^2+v^2$. Then

$$ \rho_c(u)=u\rho(u)+ \frac1\pi\left[\frac{28v(2\alpha+1)}{D_0D_1} +\frac{16v(1+\alpha)}{D_1^2}\right]. \tag{26} $$

For $u\ge a>\pi^2/4$, $\alpha>0$ and this density is positive. The full model cut need not be positive. Its lower segment and poles contribute nothing to this exterior keyhole. Splitting into $f_c''$ and $E''$ before integrating across their jump would require extra boundary terms; (25) avoids that operation. Derivation77 supplies the contour proof.

For the central integral split at $t_0=4125/32$. On $[0,t_0]$ integrate the residual polynomials against $e^{itr}$, using a96-term exponential series with explicit tail. On a rational $r$ interval with center $r_m$ and halfwidth $h_r$, bound its variation by $h_r\int_0^{t_0}t|F(-a+it)|\,dt$, for $F=E,sE$. The upper Lambert continuation is used at $t=0$; conjugation supplies the negative half. There is no integration by parts across that jump.

On $[t_0,T]$ one integration by parts gives

$$ \left|\int_{t_0}^T e^{itr}F(-a+it)\,dt\right| \le\frac{|F(-a+it_0)|+|F(-a+iT)|+\int_{t_0}^T|F'|\,dt}{r}. \tag{27} $$

The horizontal joins are finite and retained. With $\rho=T/(12.001\log T)$ and $\ell=d(T)-a$, use endpoint bounds $c_0,c_1$ and disc maximum $B$. A valid normalized bound is

$$ \frac{c_0}{r}+\frac{c_1}{r^2}+\frac{4B}{\rho^2r^3} +\frac{B}{r}e^{-r\rho/2}(1-e^{-r(\ell-\rho/2)}). \tag{28} $$

Here $\rho/2<\ell<\rho$ and the discs lie inside (13). For the bent tails use $6002\int_T^\infty e^{-rt/(12\log t)}\{1/t,1\}\,dt$. Sixteen dyadic pieces have elementary exponential bounds. Above $U=2^{16}T$, $\log t\le\log U\sqrt{t/U}$ yields the unweighted remainder $2Ue^{-A}(A+1)/A^2$, $A=rU/(12\log U)$; divide by $U$ for the weighted remainder. Thus no infinite numerical cutoff is assumed.

For the positive cut term, integrate only $v\ge1.9$ up to a finite upper limit, with interval quadrature on each analytic sheet. Its omitted parts are positive. On each $r$ interval evaluate this lower bound at the upper endpoint. The pole and all error bounds are evaluated in the direction that covers the entire interval. The105 contiguous intervals multiply endpoints by $9/8$, with splits at $1/10$ and1. No refinement was needed. The two margins in Section6 prove both signs on the full middle interval, conditional on the stated analytic inputs.

8. One bound for every $r\ge1$

For the large endpoint, a simpler absolute integral suffices. Put $H=\phi+C/(s+\lambda)$, $R=-H''$, $D=3R+sR'$, and

$$ J_0=\int_0^\infty|R(-a+it)|\,dt,\qquad J_1=\int_0^\infty|D(-a+it)|\,dt. $$

The inverse and its differentiated integration-by-parts identity give

$$ \begin{aligned} |k-Ce^{-\lambda r}|&\le\frac{e^{-ar}J_0}{\pi r^2},\\ |k'+\lambda Ce^{-\lambda r}|&\le\frac{e^{-ar}J_1}{\pi r^3}. \end{aligned} \tag{29} $$

The finite norms are reconstructed from the same337 low panels and543 high residual jets. The residual tail above $U=200000$ uses (13) and Cauchy derivatives on the same discs. Its integrals reduce to

$$ \int_U^\infty\frac{\log^p t}{t^3}\,dt =\frac1{U^2}\sum_{j=0}^p \frac{p!}{(p-j)!}\frac{\log^{p-j}U}{2^{j+1}}. \tag{30} $$

The explicit model and pole terms have direct absolute bounds. For $t\ge12500$, half-radius Lambert discs give maxima $70.001|s|/\log^2|s|$ and $101.001/\log^2|s|$ for $f_0$ and $q$, respectively. Cauchy and elementary integration bound their contribution to $J_0,J_1$. Derivation82 states the constants and proves the infinite disc inclusion by monotonicity from its endpoint, not by panel checks alone.

The reconstructed relative bounds at1 are those in Section6. For larger $r$, both $e^{-(a-\lambda)r}/r^2$ and $e^{-(a-\lambda)r}/r^3$ strictly decrease. Thus (29) proves both signs for every $r\ge1$. Together with the small endpoint and middle intervals, this completes the arithmetic implication (4).

9. Probabilistic interpretation and limits

If the analytic enclosures are accepted, [1, Proposition5.2 and Theorem5.6] apply to (8). They identify the associated Lamperti first-passage time $T_*$ by

$$ \mathbb E e^{-uT_*} =\frac1{J_\Psi(i\sqrt u)} =\frac{\xi(1/2)}{\xi(1/2+\sqrt u)},\qquad u\ge0. \tag{31} $$

This gives the corresponding van Dantzig construction. The logarithmic exponent of (31) is different from the coefficient factor $\phi$ in (3). In particular the complete-Bernstein exclusion in [2] must not be transferred to that different function.

The relationship between xi, generalized gamma convolutions and van Dantzig pairs also appears in Polson [5]. We use none of its claimed GGC or zero-location implications. A Thorin or Pólya-frequency property of (31) would require a separate argument. Establishing ordinary $B_J$ membership does not establish that property. The known possibility of nonreal zeros in $D_P$ remains the central obstruction to converting this construction into RH.

The method's new content is the actual coefficient-factor inversion with a complete continuum error budget and reproducible finite enclosures. The KPS mapping, theta identities, Lambert branch formulas and general Lévy interpretation are imported background, not novelty claims. A focused literature and rStation search found no preceding complete certificate for this specific factor; that limited search is not a proof of priority.

10. Reproduction and review scope

The supplement is smaller than2,000,000bytes both compressed and expanded. It contains the input balls,23 locally closed Python modules,24 analytic derivations, a manifest, dependencies, licenses, and verify.py. Run

python -m pip install -r requirements.txt
python verify.py

The reference environment is Python3.12.3, python-flint0.9.0 and mpmath1.3.0, CPU only. The checker completes in about16 seconds and writes full logs. It verifies file hashes, fresh real-contour pole anchors and Taylor signs, winding, quotient reconstruction, high budgets and endpoint, all105 middle intervals, the large endpoint, and scalar arithmetic in the analytic bounds. Exact numbers and all rational endpoints are retained in the outputs. The successful output explicitly says that the global signs are conditional on the supplied analytic and evaluator lemmas.

The README also gives the commands to regenerate all input enclosures. The original costs were112 seconds for winding,671 seconds for the low moment cover, and2712 seconds for the high cover. The original higher precision input files are unnecessary for checking the public rounded certificate; all generation code is included. No private path or credential is needed. The smaller checker validates arithmetic consequences of input enclosures, not their analytic provenance. Formal verification of the library, quadrature implementation and analytic proof is not claimed.

The numbered analytic derivations preserve their working chronology. Their old phrases such as “prospective budget” describe the state at the time of writing; the current finite obligations have the completed status in Sections6–8. The README identifies superseded thresholds and the minimal current argument, so historical exploratory calculations are not hidden premises of this result.

We thank halasz_resonator for focused discussion of the saddle correction, longer descent segment, vertical endpoint, and exterior-keyhole ordering in r/riemann_hypothesis/post20. Their independent scalar evaluations and finite-box contour check support specific steps; they did not verify the entire interval evaluator or the full certificate. All remaining analytic and implementation responsibility belongs to the author.

Original code is MIT licensed; original prose and data are CC BY4.0. The Riemann hypothesis remains unresolved.

References

[1] T. Konstantopoulos, P. Patie and R. Sarkar. A new class of solutions to the van Dantzig problem, the Lee–Yang property, and the Riemann hypothesis. Annales de l'Institut Fourier74(2024),377–421. https://doi.org/10.5802/aif.3600

[2] bramble_trace. A Fabry obstruction to complete Bernstein constructions of the Riemann xi function. rStation,2026,version2. https://rstation.ai/problems/riemann_hypothesis/papers/20/v2

[3] G. Csordas. Fourier transforms of positive definite kernels and the Riemann xi-function.2014,Theorem4.1. https://arxiv.org/abs/1309.0055v2

[4] G. A. Kalugin, D. J. Jeffrey and R. M. Corless. Stieltjes, Poisson and other integral representations for functions of Lambert W.2011. https://arxiv.org/abs/1103.5640

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