bibkey: bianepitmanyor1999xiprice
authors: Philippe Biane, Jim Pitman, Marc Yor
year: 1999
title: Probability laws related to the Jacobi theta and Riemann zeta functions, and Brownian excursions
doi: null
url: https://arxiv.org/abs/math/9912170v1
claim: The classical positive xi law admits positive inverse-moment calibration for the Robin price observer. Occupied layers have nonnegative cost, and absent-layer responses have a unique asymptotic sign boundary determined by an exact theta budget; the full signed Robin comparison remains unpaid.
strata_touched: []
license: citation-only
triage: anchor

Positive inverse-moment xi observation and actual occupied layers

The primary is Biane–Pitman–Yor, arXiv:math/9912170v1, §2.2, equation (20), printed p.6, and Proposition 1(i), equation (21), p.7. The inspected 40-page PDF has SHA-256 04a444275e5522cef9a1ba9f7d1b9f20a752764d3548f9c48be6dbc055bb12ea. The probability law and its reciprocal identity are classical; the Nicolas card’s XLR1 already owns their mean-normalized Robin use. They are not reconstructed here. The application below uses inverse normalization instead, with KTO’s existing observation kernel. It supplies a sign for the actual occupied layers, rather than positivity of a signed measure or a new prime-distribution estimate. No literature priority or Lean certification is claimed, and RH and full strict Robin remain unresolved.

A positive source with exact inverse moment

Reuse the primary’s positive variable $Y$ with $\mathbb{E}{Y}^s=2\xi (s)$ for every complex $s$. Set

$$C=Y^{-1},h(s)=\mathbb{E}{C}^{s-1}=2\xi (s),m=\pi /3,b_{+}=h(1/2),a_{+}=\mathbb{E}\log C.\text{\hspace{1em}(IPX1)}$$

All fixed real moments are finite, $\mathbb{E}{C}^{-1}=1$ and $\mathbb{E}C=m>1$. Nondegeneracy and strict Jensen give $0<b_{+}=\mathbb{E}\sqrt{Y}<1$ and $a_{+}=-\mathbb{E}\log Y>0$. The zeros $h(\rho )=0$ include every actual nontrivial zeta zero, without a premise about its real part. This source differs from XLR’s $1/(mY)$; its ordinary mean is not one.

For $A=\log n$ sufficiently large, put $L=\log A$, $T_A=\sqrt{A}L$ and define

$${\tau }_A=2e/A,U_A=\max \{C,{\tau }_A\},z_A=\mathbb{E}{U}_A^{-1},{\eta }_A=\mathrm{Law}(z_A{U}_A).\text{\hspace{1em}(IPX2)}$$

With $M_{-4}=\mathbb{E}{C}^{-4}<\infty$,

$$0\le 1-z_A\le \mathbb{E}[C^{-1};C<{\tau }_A]\le {\tau }_A^3{M}_{-4}=O(A^{-3}).$$

Hence $z_A\ge 1/2$ eventually. The source ${\eta }_A$ is positive, has mass one, support $c\ge e/A$, and inverse moment exactly one. All fixed real and weighted logarithmic moments are uniformly bounded. Its ordinary mean is $m+O(A^{-3})$, retaining a different normalization from the earlier mean-one source. In particular

$$\int c^{-1/2}d{\eta }_A=b_{+}+O(A^{-3}),\int \log cd{\eta }_A=a_{+}+O(A^{-3}).\text{\hspace{1em}(IPX3)}$$

For the first error use $\mathbb{E}[C^{-1/2};C<{\tau }_A]\le {\tau }_A^{7/2}{M}_{-4}$. For the second, integrate $\mathbb{P}(C<u)\le M_{-4}{u}^4$ over $0<u<{\tau }_A$ to bound $\mathbb{E}[\log ({\tau }_A/C);C<{\tau }_A]$ by $M_{-4}{\tau }_A^4/4$. The scale change $\log z_A=O(A^{-3})$ accounts for the remaining error. No negative correction atom is used.

Keep the actual price observer and its full tail

Retain exactly KTO5’s kernel ${\kappa }_{A,c}$, price $\lambda (x)=1/(x\log x)$, weight $K=-{\lambda }^{\prime }$ and benchmark $B(x)=\gamma +\log \log x-1/\log x$. Define

$${\omega }_A(E)=\int {\kappa }_{A,c}(E)d{\eta }_A(c).\text{\hspace{1em}(IPX4)}$$

KTO6–KTO8 and the uniform moments above give absolute kernel Fubini, finite variation, and the exact identities

$$\int d{\omega }_A=1,\int \lambda (Av)d{\omega }_A=\lambda (A),\int B(Av)d{\omega }_A=B(A).\text{\hspace{1em}(IPX5)}$$

These use the inverse moment, not $\mathbb{E}C=1$. The positive source does not make ${\omega }_A$ positive: KTO’s $c>1$ kernel still has its negative infinite density tail.

The unclamped law of $z_{A}C$ has Mellin transform $z_A^{s-1}2\xi (s)$, so it retains all actual zero cancellations. Apply the full-response proof GPC5–GPC6 to this law; its needed moments are uniformly bounded for $1/2\le z_A\le 1$. On $c\ge A^{-1/2}$ the clamp is inactive, and the integrated explicit formula therefore has only the omitted small-source zero remainder, of normalized size $O(A^{-1})$. The entire $c<A^{-1/2}$ response costs $O(L{A}^{-3/2})$ by GPC6, for both the unclamped law and ${\eta }_A$. The absolute constant and trivial-zero cost is $O(A^{-1/2})$. Consequently the unchanged full response $J_c^{\psi }$ satisfies

$$T_A\left|\int J_c^{\psi }(A)d{\eta }_A(c)\right|=O(A^{-1/2}),T_A\left|\int I_{\psi }(Av)d{\omega }_A(v)\right|=O(A^{-1/2}).\text{\hspace{1em}(IPX6)}$$

The second equality uses KTO7’s exact transported response. Every zero, small clock, elementary correction and infinite kernel tail is retained. Mellin cancellation is assigned to the unclamped source, not to ${\omega }_A$.

Every occupied layer has an exact nonnegative observation

For any actual exponent layer, write $r=r_{p,j}=\log (1+1/(p+\cdots +p^j))/\log p>0$, and let $t>1$ be the unique solution of $\lambda (t)=r$. The actual occupied cost is $\log p(\lambda (Av)-r{)}_{+}$. For every $c\ge e/A$, finite Stieltjes integration by parts in KTO5 gives

$$\int (\lambda (Av)-r{)}_{+}d{\kappa }_{A,c}(v)=\frac{\lambda (A)-\lambda (\max \{A,t/c\})}{c}\ge 0.\text{\hspace{1em}(IPX7)}$$

If $t\le Ac$, the integrand is zero on the whole kernel support and both sides are zero. Otherwise its support ends at $v=t/A$; the upper boundary value is zero, while the cumulative kernel is $c^{-2}K(Av/c)/K(Av)$. Multiplication by $-d\lambda (Av)=AK(Av)dv$ leaves $c^{-1}{\int }_A^{t/c}K(u)du$, which proves the formula including its lower atom. This also covers $t<e$, when the first zero case applies.

Positivity of ${\eta }_A$ transports this sign exactly. If $a_p=v_p(n)$, summing all occupied layers of the same original integer yields

$$0\le O_A(n):=\sum _p\sum _{j\le a_p}\log p\int (\lambda (Av)-r_{p,j}{)}_{+}d{\omega }_A(v)\le \lambda (A)\sum _p{a}_p\log p=\frac{1}{L}.\text{\hspace{1em}(IPX8)}$$

The upper bound uses IPX7 and the exact inverse moment. There is no factor counting divisors or exponent layers, no CA replacement, and no discarded correction error. The statement is uniform for every original integer at its own $A$. Negative observed contributions can only enter through unoccupied layers. Dropping positive occupied terms is a valid lower-bound step, but need not preserve the cancellation needed at the root scale.

Retain the complete higher layers and the original integer

Use HLD1–HLD4’s actual higher pressure and tail $F_2$, and KTO1’s actual reserve $R$, without changing any exponent cutoff. Both have the owned quantitative form $c/(\sqrt{x}\log x)+O(1/(\sqrt{x}{\log }^{2}x))$ and are bounded on $[e,\infty )$. KTO9–KTO10’s root-tail argument, now with IPX3’s bounded fixed moments, gives

$$\begin{array}{rlrl}{T}_A\int R(Av)d{\omega }_A(v)& =c_R{b}_{+}+O(L^{-1}),& c_R& =2(\sqrt{2}-1),\\ T_A\int F_2(Av)d{\omega }_A(v)& =2\sqrt{2}{b}_{+}+O(L^{-1}).& & \end{array}\text{\hspace{1em}(IPX9)}$$

Its small-clock bound and the complete later kernel support are the same as KTO’s. No positive-power moment is assigned to ${\omega }_A$. Let $G_{2,A}(n)=d_2(n,\lambda (A))\ge 0$ be the unchanged original higher-layer baseline, and $D_{2,A}(n)$ its observed complete defect. Exact mass and price, as in HLD7–HLD8, therefore give

$$T_A{D}_{2,A}(n)=T_A{G}_{2,A}(n)+2\sqrt{2}(1-b_{+})+O(L^{-1}).\text{\hspace{1em}(IPX10)}$$

Thus the actual complete higher-layer cost is eventually positive, uniformly for all original integers. On the actual strict CA family, HCB3 makes its normalized limit exactly $2\sqrt{2}(1-b_{+})>0$. This differs from HCB4’s negative limit under its other, unchanged source.

Let $D_{1,A}(n)$ be the complete observed first-layer defect for this ${\omega }_A$, with both occupied and absent primes retained. The same exact pressure account KTO12, IPX6 and IPX9–IPX10 yield

$$\boxed{T_A\Delta (n)=T_A[D_{1,A}(n)+G_{2,A}(n)]+c_R+2(1-b_{+})+O(L^{-1}).}\text{\hspace{1em}(IPX11)}$$

Here $\Delta (n)=\gamma +\log \log \log n-\log (\sigma (n)/n)$. The new constant exceeds $c_R$, but it is paired with a different first-layer account. IPX11 is not a stronger proved Robin margin.

The same positive source produces a negative actual far tail

Keep $P=A^{16}$, $Z=\log P=16L$, and define the exact actual-prime tail

$${\mathcal{U}}_A=\int \sum _{p>P}\log p(r_{p,1}-\lambda (Av){)}_{+}d{\omega }_A(v).\text{\hspace{1em}(IPX12)}$$

The full-pressure majorant in JPT3 applies because ${\eta }_A$ has uniform fixed moments. It proves absolute kernel/source Fubini and absolute interchange of the entire prime sum. Restricting the source to $A^{-1/2}\le c\le A^8$ costs $O(A^{-1}(1+\log L))$, by JPT4. For these clocks every far-prime atom sees zero hinge. Reuse JPT5’s exact response, with $\lambda (t_p)=r_{p,1}$:

$$\int (r_{p,1}-\lambda (Av){)}_{+}d{\kappa }_{A,c}(v)=-\frac{\log c}{t_p\log t_p\log (t_p/c)}.$$

JPT6’s complete prime-tail calculation, including $t_p=p+O(1)$ and the effective unconditional PNT error, gives on that source interval

$$\int \sum _{p>P}\log p(r_{p,1}-\lambda (Av){)}_{+}d{\kappa }_{A,c}(v)=\log (1-\log c/Z)+O\left(|\log c|e^{-d\sqrt{Z}}+\frac{|\log c|}{P\log P}\right)$$

for a fixed $d>0$. Since $|\log c|/Z\le 1/2$, the Taylor remainder is bounded by $C(\log c{)}^2/Z^2$. IPX3 and the uniform second logarithmic moment, with the same source-tail bounds, consequently give

$$\boxed{{\mathcal{U}}_A=-\frac{a_{+}}{16L}+O(L^{-2})<0\text{eventually}.}\text{\hspace{1em}(IPX13)}$$

The comparison error here is not root-small; it determines the tail’s sign and size, rather than replacing its exact value in IPX11. Exact mass and price keep the same original integer’s entire far account:

$$D_{\mathrm{far},A}(n)={\mathcal{U}}_A+\sum _{\begin{array}{c}p>P\\ p\mid n\end{array}}(\lambda (A)-r_{p,1})\log p.\text{\hspace{1em}(IPX14)}$$

Each occupied-prime correction is nonnegative. For an actual strict CA integer $N_x$, $A\sim x$ and every prime factor is less than $x$; therefore this correction is zero eventually. IPX13 is then a negative observed far defect on actual integers, although its integrand and its source ${\eta }_A$ are nonnegative. The sign is introduced by the transport kernel, not by an invented arithmetic model.

At any actual own-price zero-defect source, $G_{2,A}=0$. Combining the original exact $\Delta (n)=I_{\psi }(A)+R(A)$ with IPX11 gives

$$T_A{D}_{1,A}(n)=T_A{I}_{\psi }(A)-2(1-b_{+})+O(L^{-1}).\text{\hspace{1em}(IPX15)}$$

Thus the apparent constant gain is compensated in the same first-layer account. The sign interface in IPX7–IPX8 locates negative observations in unoccupied layers, and IPX10 pays the complete higher sector, but neither supplies a sufficient lower bound for the remaining first-prime account. Its center and exact negative far tail still need a joint estimate. No improved finite verification range, unbounded Robin bound, RH proof or new Lean theorem follows from these applications.

An exact theta budget for the absent-layer sign

The same positive source admits a more precise absent-layer analysis. Reuse GP’s already owned weight $w_c(t)=c^{-1}\lambda (\max \{A,t/c\})$ in the Nicolas card. Its complete response, the mass/price identities and IPX7 are inputs, not new formulas to reprove. The additional classical input is BPY, Proposition 1(ii), equation (23), printed p.7. The estimates below control the cap residual for IPX2’s broad positive source; GP2’s ordinary-mean-one hypothesis does not hold for this source.

Fix $A$ sufficiently large. Write $C_A$ for a variable of law ${\eta }_A$, $X=\log C_A$, $a_A=\mathbb{E}X=a_{+}+O(A^{-3})>0$, and $M_{2,A}=\mathbb{E}{X}^2=O(1)$. For an actual absent layer with $\lambda (t)=r_{p,j}$, its full observed hinge is

$${\mathcal{H}}_A(t)=\int (r_{p,j}-\lambda (Av){)}_{+}d{\omega }_A(v)=\lambda (t)-\mathbb{E}\frac{\lambda (\max \{A,t/C_A\})}{C_A}.\text{\hspace{1em}(IPX16)}$$

The second equality follows from IPX7 and the exact mass and price identities, using $(r-q{)}_{+}-(q-r{)}_{+}=r-q$. It retains the lower atom and the entire signed kernel tail. It is also the negative of GP’s existing weight difference for $t\ge A$, rather than a new smoothing kernel.

For $t=AB\ge A$ set $s=\log B$, $v=L+s$, and define

$${\mathcal{S}}_A(B)=\mathbb{E}(1-B/C_A{)}_{+},{\mathcal{P}}_A(s)=\mathbb{P}(X\ge s),{\mathcal{T}}_A(s)=\mathbb{E}(X-s{)}_{+}.\text{\hspace{1em}(IPX17)}$$

Split the expectation in IPX16 at $C_A=B$. On $X<s$ use the exact identity $(v-X{)}^{-1}=v^{-1}+X{v}^{-2}+X^2/[v^2(v-X)]$. The resulting cap account is

$$\begin{array}{rl}{v}^{2}t{\mathcal{H}}_A(t)=& \frac{v^2}{L}{\mathcal{S}}_A(B)-a_A+{\mathcal{T}}_A(s)\\ & -\frac{s^2}{L}{\mathcal{P}}_A(s)-\mathbb{E}\left[\frac{X^2}{v-X};X<s\right].\end{array}\text{\hspace{1em}(IPX18)}$$

Here $v-X\ge L$ on the last event, and $s^2{\mathcal{P}}_A(s)\le M_{2,A}$. Since $s\ge 0$ and $a_A>0$, multiplying by $L^2/v^2$ gives the uniform lower bound; dropping the last two nonpositive terms gives the upper bound:

$$\begin{array}{rl}{L}^{2}t{\mathcal{H}}_A(t)& \ge L{\mathcal{S}}_A(B)-a_A-\frac{2M_{2,A}}{L},\\ v^{2}t{\mathcal{H}}_A(t)& \le \frac{v^2}{L}{\mathcal{S}}_A(B)-a_A+{\mathcal{T}}_A(s).\end{array}\text{\hspace{1em}(IPX19)}$$

These bounds apply to every $B\ge 1$. No term involving the cap or source tail has been treated as a root-small prime error.

Put $C_0=1/Y$. BPY’s cited equation gives for every $u>0$

$$\mathbb{P}(C_0>u)=4\pi u^3\sum _{n\ge 1}{n}^2{e}^{-\pi n^2{u}^2}.$$

Tonelli’s theorem therefore evaluates its positive cap budget exactly:

$${\mathcal{S}}_0(B)=B{\int }_B^{\infty }\frac{\mathbb{P}(C_0>u)}{u^2}du=2B\sum _{n\ge 1}{e}^{-\pi n^2{B}^2},{\mathcal{S}}_A(B)={\mathcal{S}}_0(B/z_A)(B\ge 1).\text{\hspace{1em}(IPX20)}$$

The last equality is exact: for sufficiently large $A$, $z_A{\tau }_A<1$, so the positive part in IPX17 vanishes on the entire clamped part of the source. Scaling the surviving upper source by $z_A$ is the only change. No estimate for a newly truncated law is being substituted for ${\eta }_A$.

For $u=B/z_A\ge 1$, differentiation of the absolutely convergent positive series yields

$$\begin{array}{rl}\frac{d}{ds}{\mathcal{S}}_A(e^s)& ={\mathcal{S}}_A(e^s)-{\mathcal{P}}_A(s)\le (1-2\pi u^2){\mathcal{S}}_A(e^s)<0,\\ {\mathcal{T}}_A(s)& ={\mathcal{S}}_A(e^s)+{\int }_s^{\infty }{\mathcal{S}}_A(e^w)dw\le \left(1+\frac{1}{2\pi -1}\right){\mathcal{S}}_A(e^s).\end{array}\text{\hspace{1em}(IPX21)}$$

Thus IPX19 has an explicit decreasing theta budget, with a uniformly controlled logarithmic tail. The statement concerns the full absent response of the fixed source, rather than positivity of ${\omega }_A$.

A unique absent-layer sign boundary and its scale

There is, for every sufficiently large $A$, exactly one $t_{\ast }(A)>A$ where IPX16 vanishes. Its position is determined by the actual calibrated source:

$$\boxed{\pi {\left(\frac{t_{\ast }(A)}{A{z}_A}\right)}^2=\log L+\frac{1}{2}\log \log L+\log \frac{2}{a_{+}\sqrt{\pi }}+o(1),t_{\ast }(A)\sim A\sqrt{\frac{\log \log A}{\pi }}.}\text{\hspace{1em}(IPX22)}$$

The statement is asymptotic: no certified numerical starting $A$ is specified. All source parameters remain those of IPX2; a new source is not chosen as $t$ varies.

To prove existence and locate every possible zero, define $B_{-},B_{+}>1$ by

$$L{\mathcal{S}}_A(B_{-})=2a_A,L{\mathcal{S}}_A(B_{+})=a_A/2.$$

For large $A$ these unique numbers exist by IPX20–IPX21. The first term of the positive theta series, with its exponentially smaller remaining terms, gives for both endpoints

$$\pi (B_{\pm }/z_A{)}^2=\log L+\frac{1}{2}\log \log L+O(1).\text{\hspace{1em}(IPX23)}$$

IPX19’s lower bound proves ${\mathcal{H}}_A(AB)>0$ for every $1\le B\le B_{-}$. For $B\ge 1$ and $L>2$, the upper-bound supplier

$$\frac{(L+\log B{)}^2}{L}{\mathcal{S}}_A(B)+{\mathcal{T}}_A(\log B)$$

strictly decreases: IPX21 bounds the logarithmic derivative of its first positive term by $2/(L+\log B)+1-2\pi (B/z_A{)}^2<0$, and the second term decreases. At $B_{+}$ this supplier is $a_A/2+o(1)<a_A$. IPX19 therefore proves ${\mathcal{H}}_A(AB)<0$ for every $B\ge B_{+}$, including arbitrarily large $B$. Every zero is confined to $[A{B}_{-},A{B}_{+}]$; no uncontrolled outer region remains.

For uniqueness, differentiate at fixed $A$ and put $Q_A(s)=L^{2}A{e}^s{\mathcal{H}}_A(A{e}^s)$. The only source atom is below $C_A=1$, so the BPY density makes this differentiation valid for $s\ge 0$. Differentiating the two cap pieces cancels their boundary terms and gives

$$\begin{array}{rl}{Q}_A^{\prime }(s)=& L\frac{d}{ds}{\mathcal{S}}_A(e^s)\\ & +L^2\left[-\frac{1}{v^2}+\mathbb{E}\left[\frac{1}{(v-X{)}^2};X<s\right]\right],\\ \left|L^2\left[-\frac{1}{v^2}+\mathbb{E}\left[\frac{1}{(v-X{)}^2};X<s\right]\right]\right|& \le {\mathcal{P}}_A(s)+\frac{2\mathbb{E}|X|}{L}.\end{array}\text{\hspace{1em}(IPX24)}$$

For the last bound, subtract $v^{-2}$ inside the expectation on $X<s$ and apply the mean value theorem; both denominators are at least $L$. The omitted probability contributes at most ${\mathcal{P}}_A(s)$. On $B_{-}\le e^s\le B_{+}$, IPX20 gives ${\mathcal{P}}_A(s)=O((\log L)/L)$, while IPX21 and IPX23 give $Ld{\mathcal{S}}_A(e^s)/ds\le -c\log L$ for some fixed $c>0$. Thus $Q_A^{\prime }<0$ throughout this whole band for sufficiently large $A$. Continuity and the endpoint signs yield exactly one zero.

At the zero, IPX18, IPX21, bounded $M_{2,A}$ and IPX23 imply $L{\mathcal{S}}_A(t_{\ast }/A)=a_A+o(1)$. Write $u=t_{\ast }/(A{z}_A)$. Using $\mathcal S_A(t_*/A)=2u e^{-\pi u^2} (1+O(e^{-3\pi u^2}))$and$a_A\to a_+>0$, taking logarithms first gives $u^2\sim (\log L)/\pi$ and then IPX22, including its constant $\log (2/(a_{+}\sqrt{\pi }))$.

For $1<t<A$, IPX7 gives ${\mathcal{H}}_A(t)\ge \lambda (t)-\lambda (A)>0$. At $t=A$, every source value except $C_A=1$ has $C_A^{-1}\lambda (\max \{A,A/C_A\})<\lambda (A)$, so nondegeneracy gives strict positivity there as well. The complete single-layer classification is consequently

$${\mathcal{H}}_A(t)>0(1<t<t_{\ast }),{\mathcal{H}}_A(t_{\ast })=0,{\mathcal{H}}_A(t)<0(t>t_{\ast }).\text{\hspace{1em}(IPX25)}$$

This locates the actual absent-layer sign; it does not assert that all actual primes are present below that boundary. For first layers, $\lambda (t_p)=r_{p,1}$ satisfies $p<t_p<p+1$: use $1/(p+1)<\log (1+1/p)<1/p$ and monotonicity of $\lambda$. Thus an absent first prime $p\le t_{\ast }-1$ has positive response and an absent first prime $p\ge t_{\ast }$ has negative response. IPX7 separately keeps every occupied response nonnegative. Negative observed terms can therefore occur only in absent layers whose actual activation threshold exceeds $t_{\ast }$.

The unique boundary improves the sign description, rather than the proved complete Robin margin. The positive and negative prime sums still use the same source and actual integer; neither may be omitted from IPX11. In particular the negative far contribution in IPX13 is still present, and no root-small replacement for its exact prime sum has been supplied. A sign boundary at $A\sqrt{\log \log A/\pi }$ does not make the full comparison error root-small. The quantitative joint arithmetic estimate, strict Robin above 5040 and RH remain unproved. These are paper-level applications of the cited BPY tail and the existing full kernel response, with no Lean certification or literature-priority claim.

The complete negative first-prime mass at the natural sign boundary

IPX22–IPX25 locate the negative responses. Their complete size can also be estimated on the actual prime set. The inputs are GP’s owned absolute-weight Fubini and primitive, the unconditional Chebyshev and effective PNT bounds already used in JPT and ACC, and the unchanged calibrated source. No new prime-counting theorem or qualitative cancellation mechanism is used.

Keep $A$ sufficiently large and put $B_{\ast }=t_{\ast }/A$, $s_{\ast }=\log B_{\ast }$, $v_{\ast }=\log t_{\ast }$ and $u_{\ast }=B_{\ast }/z_A$. In addition to $a_A$ and $M_{2,A}$, use the uniform third absolute logarithmic moment $M_{3,A}=\mathbb{E}|X{|}^3=O(1)$. For every actual first-prime activation $\lambda (t_p)=r_{p,1}$ define

$$g_A(t)=(-{\mathcal{H}}_A(t){)}_{+},{\mathfrak{N}}_A=\sum _p\log p{g}_A(t_p),{\mathfrak{n}}_A={\int }_{t_{\ast }}^{\infty }{g}_A(t)dt.\text{\hspace{1em}(IPX26)}$$

The sum uses every actual prime, not a prime-density model. For an integer $n$, its actual negative first-layer part includes only $p\nmid n$, so it is bounded by ${\mathfrak{N}}_A$. Equality requires that all primes with $t_p>t_{\ast }$ be absent; it is not assumed for an arbitrary integer. The continuous quantity ${\mathfrak{n}}_A$ is only an intermediate comparison.

For $B\ge 1$, write $s=\log B$, $v=L+s$. GP’s absolute convergence permits the full-source primitive

$$J_A(B):={\int }_{AB}^{\infty }{\mathcal{H}}_A(t)dt=\mathbb{E}\log \left(1-\frac{\min (X,s)}{v}\right)-\frac{{\mathcal{S}}_A(B)}{L}.\text{\hspace{1em}(IPX27)}$$

For a source value $X<s$, the primitive is $\log (1-X/v)$. For $X\ge s$, it is $\log (L/v)-(1-e^{s-X})/L$. Thus IPX27 includes the entire capped source, rather than extending the reciprocal logarithm across its pole. It uses absolutely convergent differences, with no subtraction of two divergent $\lambda$ integrals. By IPX25, ${\mathfrak{n}}_A=-J_A(B_{\ast })$.

Write $S={\mathcal{S}}_A(B_{\ast })$ and

$$R_1={\int }_{s_{\ast }}^{\infty }{\mathcal{S}}_A(e^w)dw,R_2={\int }_{s_{\ast }}^{\infty }{\mathcal{T}}_A(w)dw,k_{\ast }=2\pi u_{\ast }^2-1\asymp \log L.$$

IPX21 gives $R_1\le S/k_{\ast }$, ${\mathcal{T}}_A(s_{\ast })=S+R_1$ and $R_2\le (1+1/k_{\ast })S/k_{\ast }$. For $Y=\min (X,s_{\ast })$ the first two moments are exactly

$$\mathbb{E}Y=a_A-{\mathcal{T}}_A(s_{\ast }),\mathbb{E}{Y}^2=M_{2,A}-2s_{\ast }{\mathcal{T}}_A(s_{\ast })-2R_2.$$

Taylor expansion in IPX27 therefore yields

$$\begin{array}{rl}{J}_A(B_{\ast })=& -\frac{a_A}{v_{\ast }}-\frac{M_{2,A}}{2v_{\ast }^2}+\frac{R_1}{v_{\ast }}+\frac{s_{\ast }{R}_1}{v_{\ast }^2}+\frac{R_2}{v_{\ast }^2}-\frac{s_{\ast }^{2}S}{L{v}_{\ast }^2}+O(L^{-3}),\\ {\mathfrak{n}}_A=& \frac{a_A}{v_{\ast }}+\frac{M_{2,A}}{2v_{\ast }^2}+O\left(\frac{1}{L^2\log L}\right).\end{array}\text{\hspace{1em}(IPX28)}$$

The expansion does not require $|X|<v_{\ast }$. For the unbounded negative part use $|\log (1+z)-z+z^2/2|\le z^3/3$ for $z\ge 0$; for the positive part, $Y\le s_{\ast }=o(v_{\ast })$ gives the same third-moment allowance. The second-order cap terms are retained: their cancellation uses $S/v_{\ast }+s_{\ast }S/v_{\ast }^2-S/L=-s_{\ast }^{2}S/(L{v}_{\ast }^2)$. Finally $S=O(1/L)$, $s_{\ast }=O(\log \log L)$ and the bounds on $R_1,R_2$ make all remaining boundary terms $O(1/(L^2\log L))$.

For the literal activations, differentiate IPX16 at fixed $A$:

$${\mathcal{H}}_A^{\prime }(t)=-K(t)+\mathbb{E}[C_A^{-2}K(t/C_A);C_A<t/A].$$

On $t\ge t_{\ast }$ the last kernel has $\log (t/C_A)\ge L$, so $|g_A^{\prime }(t)|\le C/(t^{2}L)$ almost everywhere, with $g_A(t_{\ast })=0$. Below $t_{\ast }$, $g_A$ is zero. The already owned bracket $p<t_p<p+1$ and the Chebyshev bound consequently give

$$\begin{array}{rl}\left|\sum _p\log p[g_A(t_p)-g_A(p)]\right|& \le \frac{C}{L}\sum _{p>t_{\ast }-1}\frac{\log p}{p^2}=O\left(\frac{1}{t_{\ast }L}\right),\\ T_A\left|\sum _p\log p[g_A(t_p)-g_A(p)]\right|& =o(1).\end{array}\text{\hspace{1em}(IPX29)}$$

There is no factor counting primes, divisors or exponent layers. This activation shift is controlled at the root scale.

Retain the exact primary-prime quadrature error

$${\mathfrak{E}}_A={\int }_{t_{\ast }}^{\infty }{g}_A(t)d(\vartheta (t)-t)=-{\int }_{t_{\ast }}^{\infty }(\vartheta (t)-t)g_A^{\prime }(t)dt.\text{\hspace{1em}(IPX30)}$$

The boundary terms vanish. The same effective PNT supplier used in ACC gives $|\vartheta (t)-t|\le Ct{e}^{-d\sqrt{\log t}}$ for a fixed $d>0$ and all sufficiently large $t$. Hence

$$|{\mathfrak{E}}_A|\le \frac{C}{L}{\int }_{\log t_{\ast }}^{\infty }{e}^{-d\sqrt{u}}du=O\left(\frac{e^{-d\sqrt{L}}}{\sqrt{L}}\right).\text{\hspace{1em}(IPX31)}$$

This allowance is smaller than every fixed negative power of $L$. It is insufficient to guarantee a root-small actual quadrature error; the exact ${\mathfrak{E}}_A$ remains available when that precision is required.

Combining IPX28–IPX31 gives the full literal-prime expansion

$$\boxed{{\mathfrak{N}}_A=\frac{a_A}{v_{\ast }}+\frac{M_{2,A}}{2v_{\ast }^2}+O\left(\frac{1}{L^2\log L}\right)\sim \frac{a_{+}}{L}.}\text{\hspace{1em}(IPX32)}$$

In particular this includes all negative first-prime responses from the natural boundary onward, rather than just IPX13’s artificial $p>A^{16}$ tail. The continuous integral is used only with its paid literal activation and prime-quadrature errors. No improvement of the underlying PNT is asserted.

For the same original integer $n$, let $h_{p,A}(n)$ be the observed first-layer response before multiplication by $\log p$: it is IPX7’s nonnegative expectation when $p\mid n$ and ${\mathcal{H}}_A(t_p)$ when $p\nmid n$. Define its complete positive and negative first-layer parts by

$${\mathfrak{P}}_A(n)=\sum _p\log p(h_{p,A}(n){)}_{+},{\mathfrak{N}}_A(n)=\sum _{p\nmid n}\log p{g}_A(t_p).$$

The owned absolute convergence gives $D_{1,A}(n)={\mathfrak{P}}_A(n)-{\mathfrak{N}}_A(n)$ and ${\mathfrak{N}}_A(n)\le {\mathfrak{N}}_A$. Reuse ACC3–ACC5’s actual strict products $N_x$, with $A=\log N_x\sim x$. Every prime factor lies below $x$, whereas $t_{\ast }/x\to \infty$. Thus eventually all negative first primes are absent from this same $N_x$, and ${\mathfrak{N}}_A(N_x)={\mathfrak{N}}_A$ exactly.

ACC5 also gives $0\le G_{2,A}(N_x)\le d_A(N_x)$ and $d_A(N_x)=O(e^{-d\sqrt{L}}/L)$ for some $d>0$. Use the unchanged exact account $\Delta (N_x)=d_A(N_x)+I_{\psi }(A)+R(A)$, IPX11 and the owned $R(A)=c_R/T_A+O(1/(T_{A}L))$. They give

$$\begin{array}{rl}{D}_{1,A}(N_x)& =d_A(N_x)-G_{2,A}(N_x)+I_{\psi }(A)-\frac{2(1-b_{+})}{T_A}+O\left(\frac{1}{T_{A}L}\right)\\ & =O(e^{-d_1\sqrt{L}})+O(T_A^{-1})=o(L^{-m})\text{for every fixed }m>0,\end{array}$$

for a fixed $d_1>0$, by the same effective PNT input as ACC2. Consequently on these actual integers,

$${\mathfrak{P}}_A(N_x)=\frac{a_A}{v_{\ast }}+\frac{M_{2,A}}{2v_{\ast }^2}+O\left(\frac{1}{L^2\log L}\right),\frac{{\mathfrak{P}}_A(N_x)}{{\mathfrak{N}}_A(N_x)}⟶1.\text{\hspace{1em}(IPX33)}$$

Both parts use the same integer, source and literal prime set. This quantifies the full sign split at the natural boundary; ACC already owns the actual-integer qualitative compensation mechanism.

The surviving sufficient estimate for IPX11 is, for some fixed $ϵ>0$ and every sufficiently large original integer $n$,

$$T_A[{\mathfrak{P}}_A(n)-{\mathfrak{N}}_A(n)+G_{2,A}(n)]\ge -c_{+}+ϵ,c_{+}=c_R+2(1-b_{+}).\text{\hspace{1em}(IPX34)}$$

It would imply eventual strict Robin; a complete proof above 5040 would also have to cover the remaining integers. On the actual family in IPX33 the residual scale $T_A^{-1}$ is of relative order $A^{-1/2}$ compared with either sign part $\sim a_{+}/L$. The logarithmic remainder in IPX32–IPX33 and the unconditional prime-quadrature allowance do not supply that one-sided precision. A larger positive residual is favorable: no two-sided smallness of the actual difference is required or claimed. The joint bound IPX34, a certified numerical starting scale, full strict Robin and RH remain unproved. IPX26–IPX33 are paper-level applications of the cited source and existing arithmetic estimates, without Lean certification or a literature-priority claim.

The complete first-layer response has a sharp conditional root benchmark

IPX33’s actual first-layer account also has a root-scale consumer. The inputs are the already owned complete signed explicit formula, NPC1’s two-price benefit bound, and classical simultaneous phase recurrence. No new Ramanujan theorem, zero-distribution estimate or recurrence principle is introduced. Keep exactly the IPX source and observer, and the actual strict products $N_x$ of ACC, with $A=A_x=\log N_x$, $L=\log A$ and $T_A=\sqrt{A}L$.

The classical xi logarithmic-derivative identity, already used in GE1, and differentiation of IPX1’s Mellin identity give

$$c_0=2+\gamma -\log (4\pi )=2a_{+},{\kappa }_{\xi }=c_0+2(1-b_{+})>0.\text{\hspace{1em}(IPX35)}$$

Indeed $a_{+}=-2{\xi }^{\prime }(0)=2{\xi }^{\prime }(1)$ and $\xi (1)=1/2$; the classical zero weight is $2{\xi }^{\prime }(1)/\xi (1)$. The positivity also follows directly from IPX1’s strict Jensen inequalities. This connects existing constants; it is not a new identity for the zeta zeros.

Assume RH throughout the following conditional calculation. The classical estimate $\vartheta (x)-x=O(\sqrt{x}{\log }^{2}x)$, ACC’s literal first-layer bracket and its complete higher inventory give $A_x-x=O(\sqrt{x}{\log }^{2}x)$. In NPC1, $h=O({\log }^{2}x/\sqrt{x})$, so $T_A{d}_A(N_x)\to 0$. Since $0\le G_{2,A}(N_x)\le d_A(N_x)$, the same holds for its higher baseline. These facts apply to every large real price $x$, including strict endpoint conventions; no own-price equality $A_x=x$ is assumed.

Reuse the complete signed formula, Broadbent–Fiori–Kadiri–Ng–Wilk, Proposition 13(i), as recorded there. Under RH its normalized form and IPX33’s existing account give

$$\begin{array}{rl}{T}_A{I}_{\psi }(A)& =-\mathcal{B}(\log A)+O(L^{-1}),\\ \mathcal{B}(y)& =2\sum _{\tau >0}\frac{\cos (\tau y)}{1/4+{\tau }^2},\\ T_A{D}_{1,A}(N_x)& =-\mathcal{B}(\log A_x)-2(1-b_{+})+o(1).\end{array}\text{\hspace{1em}(IPX36)}$$

Here $\tau$ runs over all positive nontrivial-zero ordinates with their multiplicities. No infinite-height tail is omitted. The leading factor of the exact formula is $1+1/L$; its complete integral remainder is $O(L^{-1})$ after normalization, by the convergent weight ${\sum }_{\rho }1/(|\rho ||1-\rho |)$. The constant and trivial-zero corrections are root-small. On RH this weight is $c_0$, and the cosine series is absolutely and uniformly convergent.

Its coefficients are positive, so $\mathcal{B}(y)\le \mathcal{B}(0)=c_0$. For any positive error tolerance, retain a finite head with smaller absolute tail. The classical simultaneous approximation theorem, as in the existing Kronecker source, returns arbitrarily large $y$ for which every head phase is arbitrarily close to one. This uses recurrence toward the identity, not density in the whole phase torus or rational independence of zero ordinates. The paid tail then gives ${\mathrm{limsup}}_{y\to \infty }\mathcal{B}(y)=c_0$.

Set $x=e^y$ and retain its same strict integer $N_x$. Since $\log A_x-\log x\to 0$, uniform continuity of this series transports the recurrence to $\log A_x$. The step-valued inventory therefore does not change the extremum. IPX36 yields the sharp conditional benchmark

$$\boxed{\mathrm{RH}⟹\underset{x\to \infty }{\mathrm{liminf}}{T}_{A_x}{D}_{1,A_x}(N_x)=-{\kappa }_{\xi }<0.}\text{\hspace{1em}(IPX37)}$$

This is a paper-level application at the specified observed first layer. The RH premise remains essential to its lower bound and sharp value; it is not an unconditional estimate for IPX34 or a Lean certification.

Actual negative first-layer responses persist without an RH premise

For the complementary case reuse the existing actual regular-source family S5. If RH fails, it supplies an unbounded family of actual event-free own-price integers $N$, $A=\log N$, with $G(N)>e^{\gamma }(1+c_{\eta }{A}^{-\eta })$ for fixed $c_{\eta }>0$ and $0<\eta <1/2$. The source’s already retained prime-local bridge gives the strict identity $N=N_A$; there is no new replacement of tied endpoints or separately chosen integer. On this same family,

$$T_A\Delta (N)\le -\frac{c_{\eta }}{2}{A}^{1/2-\eta }L⟶-\infty .$$

IPX11 and $G_{2,A}(N)\ge 0$ then give $T_A{D}_{1,A}(N)\le T_A\Delta (N)-c_{+}+O(L^{-1})\to -\infty$. Together with IPX37 this establishes the paper-level dichotomy

$$\underset{x\to \infty }{\mathrm{liminf}}{T}_{A_x}{D}_{1,A_x}(N_x)=\{\begin{array}{ll}-{\kappa }_{\xi },& \text{if RH holds},\\ -\infty ,& \text{if RH fails}.\end{array}\text{\hspace{1em}(IPX38)}$$

In particular, without assuming either answer to RH, ${\mathrm{liminf}}_{x\to \infty }{T}_{A_x}{D}_{1,A_x}(N_x)\le -{\kappa }_{\xi }<0$. Thus on an unbounded family of these actual strict CA integers,

$$T_A[{\mathfrak{N}}_A(N_x)-{\mathfrak{P}}_A(N_x)]>{\kappa }_{\xi }/2.\text{\hspace{1em}(IPX39)}$$

This rules out eventual nonnegativity of the complete observed first layer on the actual CA family. It does not follow merely from IPX13’s negative far tail: here the occupied and positive absent responses have also been retained. In particular IPX33’s ratio tending to one coexists with an actual negative difference at the root scale.

The sufficient target IPX34 remains a lower bound for the complete joint account, and has not been proved. It allows the negative values above: the constant gap $c_{+}-{\kappa }_{\xi }=c_R-c_0>0$ is the classical RH-conditional Ramanujan margin in this logarithmic normalization, as in the existing Ramanujan source, rather than an extra unconditional reserve created by this observer. IPX38 is a quantitative consumer for the specified source, not a new RH criterion or a proof of either branch’s premise. No numerical starting threshold, stronger unconditional Robin bound, mathematical priority or Lean certification is claimed.