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\title{Infinite divisibility of hitting times\\for symmetric stable processes}
\author{Min Wang\\[3pt]\small Wuhan University of Technology}
\date{October 27, 2026}
\hypersetup{pdftitle={Infinite divisibility of hitting times for symmetric stable processes},
 pdfauthor={Min Wang},pdfsubject={Symmetric stable processes and additive infinite divisibility}}
\begin{document}
\maketitle
\begin{abstract}
Let \(T_a\) be the first hitting time of a nonzero point \(a\) by a
symmetric strictly \(\alpha\)-stable L\'evy process, where \(1<\alpha<2\).
We give a computer-assisted classification of its ordinary additive
infinite divisibility. There is a unique threshold \(\alpha_*\), with
\(1.6670659903<\alpha_*<1.6670659915\), such that \(T_a\) is infinitely
divisible if and only if \(1<\alpha\le\alpha_*\). The threshold is
characterized by a unique zero contact of the canonical inverse
logarithmic-derivative kernel. The proof combines explicit analytic
estimates, rigorous interval covers, Gaussian necessary conditions and
Rouch\'e's theorem. In particular, a fully analytic argument proves
failure of infinite divisibility for every \(15/8\le\alpha<2\).
Scalar bounds at the two time ends reduce the compact interval
computations. The finite computations and their acceptance
conditions are supplied as ancillary material.
\end{abstract}
\noindent\textbf{Keywords:} stable process; first hitting time; infinite
divisibility; Laplace transform; computer-assisted proof.\par
\medskip
\noindent\textbf{MSC 2020:} 60G51; 60E07; 60G52.
\bigskip

\section{Introduction}
Let $X=(X_t)_{t\ge0}$ be a symmetric strictly $\alpha$-stable L\'evy
process, with $X_0=0$ and
\[
 \mathbb E[e^{i\xi X_t}]=e^{-t|\xi|^\alpha},\qquad 1<\alpha<2.
\]
For a nonzero real number $a$, let
\[
 T_a=\inf\{t>0:X_t=a\}
\]
be its first hitting time of $a$. A probability distribution is called
\emph{infinitely divisible} if, for every integer $n\ge1$, it is the
distribution of a sum of $n$ independent identically distributed random
variables. Throughout this paper the term refers to addition, and is
applied to $T_a$ itself.

For background on stable processes and their potential theory we refer
to \cite{Bertoin1996,Zolotarev1986}. Explicit distributional descriptions
of stable hitting times are also developed in \cite{KKPW2014}.


The distribution of a point-hitting time of a stable process has attracted
considerable attention. Yano, Yano and Yor~\cite{YYY2009} obtained several
distributional identities, including a relation with the $\alpha$-Cauchy
distribution. Letemplier and Simon~\cite{LS2014} proved unimodality for the
hitting times of general strictly stable processes. In
\cite[Proposition~4.1(b)]{LS2019}, they also proved infinite divisibility of
the logarithm of the hitting time, and conjectured, immediately after that
proposition, that the hitting time itself is infinitely divisible.

For the symmetric process, this conjecture is related to a question about
the random variable $C_\alpha$ with density
\[
 \frac{\alpha\sin(\pi/\alpha)}{2\pi(1+|x|^\alpha)},
 \qquad x\in\mathbb R.
\]
Indeed, if $\widehat X$ is an independent copy of the normalized symmetric
$\alpha$-stable process, then
\[
 \widehat X_{T_a}\ \stackrel{\mathrm d}{=}\ |a|C_\alpha;
\]
see~\cite[Theorem~5.3, formula~(5.12)]{YYY2009}. Consequently, infinite
divisibility of $T_a$ implies infinite divisibility of $C_\alpha$ by
subordination. Wang~\cite[Theorem~1.1]{Wang2026} proved that $C_\alpha$ is
infinitely divisible precisely when $1<\alpha\leq2$. As explained on the
second page of that paper, this result is consistent with the hitting-time
conjecture. The implication used there is one-way, however, and does not
settle infinite divisibility of $T_a$.



Here is our main result. The function $K_\alpha$ appearing in its
statement is defined canonically in Lemma~\ref{lem:canonical} by inversion
of $-L_\alpha'/L_\alpha$, where $L_\alpha(q)=\mathbb E[e^{-qT_1}]$.
The quantitative parts of the proof use rigorous ball arithmetic;
they are stated separately from the analytic lemmas.

\begin{theorem}[Computer-assisted classification]\label{thm:main}
There is a unique number $\alpha_*\in(1,2)$ such that, for every
$a\ne0$,
\begin{equation}\label{eq:classification}
 T_a\text{ is infinitely divisible}
       \quad\Longleftrightarrow\quad 1<\alpha\le\alpha_*.
\end{equation}
More precisely,
\begin{equation}\label{eq:alphastar}
 1.6670659903<\alpha_*<1.6670659915.
\end{equation}
It is the parameter coordinate of the unique solution of
\begin{equation}\label{eq:contactdefinition}
 K_\alpha(t)=0,\qquad \partial_tK_\alpha(t)=0
\end{equation}
in
\[
 [1.66706598,1.667066]\times[.162799,.162801].
\]
For $1<\alpha<\alpha_*$, $K_\alpha(t)>0$ for every $t>0$.
For $\alpha=\alpha_*$, the kernel is nonnegative and has precisely
one zero on $(0,\infty)$. In the infinitely divisible range the
L\'evy measure of $T_1$ is $K_\alpha(t)dt/t$, with zero drift.
\end{theorem}

The endpoint $\alpha=2$ is distinct: its hitting time has Laplace
transform $e^{-|a|\sqrt q}$ and is infinitely divisible. Thus
infinite divisibility fails on a whole interval approaching $2$ from
the left and holds again at the Brownian endpoint. There is no
contradiction with closure of infinitely divisible laws under weak
limits: that closure theorem imposes no such restriction on limits
of laws which are not infinitely divisible.

The proof has three parts. First, a finite large-frequency expansion
constructs $K_\alpha$ without presupposing infinite divisibility, and
an exponential-tilting argument proves that infinite divisibility is
equivalent to $K_\alpha\ge0$. Second, analytic estimates at zero and
infinity, together with a finite cover by interval enclosures, prove
strict positivity below a small critical parameter interval. Inside
that interval a strictly convex time section has a minimum that
decreases strictly through zero; positivity outside the section
identifies this contact with the global boundary. Third, a negative
kernel value, Gaussian necessary conditions and zeros of the Laplace
transform cover the entire range above the contact. A separate
analytic Rouch\'e argument treats the whole interval $[15/8,2)$.

The finite enclosures concern whole parameter--time rectangles or
whole parameter intervals, rather than samples. Appendix~\ref{sec:certificates}
specifies the certificate format, the finite acceptance conditions
and the accompanying reproducibility files. The computational proof
uses the guarantees of Arb/FLINT ball arithmetic and the explicit
analytic truncation bounds established below. It is not a formal
proof-assistant verification. No assertion about the classification
of nonsymmetric stable processes is made here.

\section{Normalization and the Laplace transform}
\label{sec:source-preliminaries}

Let $X=(X_t)_{t\geq0}$ be the symmetric strictly $\alpha$-stable process
starting from zero, normalized by
\begin{equation}\label{eq:source-normalization}
 \mathbb E[e^{i\xi X_t}]=e^{-t|\xi|^\alpha},
 \qquad \xi\in\mathbb R,\quad t\geq0,
\end{equation}
where $1<\alpha<2$. For $a\ne0$, put
\[
 T_a=\inf\{t>0:X_t=a\},\qquad T=T_1,
 \qquad L_\alpha(q)=\mathbb E[e^{-qT}].
\]
Write $\beta=1/\alpha$ and $\theta=\pi\alpha/2$. We use the principal
branch for all complex powers.

\begin{lemma}\label{lem:source-resolvent}
The random variable $T_a$ is finite almost surely and
\begin{equation}\label{eq:source-scaling}
 T_a\stackrel{\mathrm d}{=}|a|^\alpha T.
\end{equation}
For $q>0$, define
\[
 u_q(x)=\frac1\pi\int_0^\infty\frac{\cos(x\xi)}{q+\xi^\alpha}\,d\xi.
\]
Then
\begin{equation}\label{eq:source-resolvent-ratio}
 \mathbb E[e^{-qT_a}]=\frac{u_q(a)}{u_q(0)},
 \qquad
 u_q(0)=\frac{q^{\beta-1}}{\alpha\sin(\pi\beta)}.
\end{equation}
\end{lemma}

\begin{proof}
The resolvent identity is the standard hitting-time formula; in this
normalization it is stated in~\cite[formula~(5.10)]{YYY2009}. The value of
$u_q(0)$ follows by substituting $\xi=q^\beta v$ and using
\(\int_0^\infty(1+v^\alpha)^{-1}\,dv
=\pi/(\alpha\sin(\pi/\alpha))\).
Moreover,
\[
 0\leq u_q(0)-u_q(a)
 \leq\frac1\pi\int_0^\infty
       \frac{1-\cos(a\xi)}{\xi^\alpha}\,d\xi<\infty.
\]
Since $u_q(0)\to\infty$ as $q\downarrow0$, the ratio in
\eqref{eq:source-resolvent-ratio} tends to one. This proves almost sure
finiteness. Self-similarity and symmetry give~\eqref{eq:source-scaling};
see also~\cite[formula~(5.16)]{YYY2009}.
\end{proof}

The normalization entails no loss of generality. If the characteristic
exponent is $c|\xi|^\alpha$ with $c>0$, the corresponding hitting time
has law $|a|^\alpha T/c$. Multiplication by a positive constant preserves
infinite divisibility in both directions.

\begin{lemma}\label{lem:transform}
For $\operatorname{Re}q>0$,
\begin{equation}\label{eq:source-laplace-integral}
 L_\alpha(q)=
 \frac{\alpha\sin\theta\sin(\pi\beta)}{\pi}
 \int_0^\infty
 \frac{e^{-q^\beta r}r^\alpha}
 {1+2\cos\theta\,r^\alpha+r^{2\alpha}}\,dr.
\end{equation}
\end{lemma}

\begin{proof}
First suppose that $q>0$. Rotation through the first quadrant gives
\[
 \int_0^\infty\frac{e^{ix}}{q+x^\alpha}\,dx
 =i\int_0^\infty\frac{e^{-y}}{q+e^{i\theta}y^\alpha}\,dy.
\]
There is no pole in this quadrant because $\theta<\pi$. The small arc
vanishes as its radius tends to zero, and the large arc vanishes because
$\alpha>1$. Taking real parts yields
\[
 u_q(1)=\frac{\sin\theta}{\pi}
 \int_0^\infty
 \frac{e^{-y}y^\alpha}
 {q^2+2q\cos\theta\,y^\alpha+y^{2\alpha}}\,dy.
\]
Substitute $y=q^\beta r$ and divide by $u_q(0)$. Both sides of
\eqref{eq:source-laplace-integral} are holomorphic in
$\operatorname{Re}q>0$, so the identity extends there by analytic
continuation.
\end{proof}

Formula~\eqref{eq:source-laplace-integral} is also obtained from the
independent-product representation in~\cite[Section~2.4 and final
remark~(a)]{LS2014}. The contour argument above fixes all normalizing
constants directly. The formula is for the transform $L_\alpha$;
it does not assert that this transform is zero-free in the right
half-plane.

\section{The canonical logarithmic-derivative kernel}\label{sec:kernel}

Throughout the paper, powers and logarithms of complex numbers have their
principal values. Put
\begin{equation}\label{eq:notation}
 \beta=\alpha^{-1},\quad \delta=\pi(1-\alpha/2),\quad m=\alpha+1,
 \quad D_\delta(z)=1-2\cos(\delta)z+z^2,
 \quad \kappa_\alpha=\frac{\alpha\sin\delta\sin(\pi/\alpha)}{\pi}.
\end{equation}
Thus $0<\delta<\pi/2$. The transform obtained in
Lemma~\ref{lem:transform} is
\begin{equation}\label{eq:L}
 L_\alpha(q)=\kappa_\alpha\int_0^\infty
       \frac{e^{-q^\beta r}r^\alpha}{D_\delta(r^\alpha)}\,dr,
       \qquad \Re q>0.
\end{equation}
The integral converges locally uniformly in this half-plane. In particular,
it defines the analytic continuation of the real Laplace transform.
Write $F_\alpha=-L_\alpha'/L_\alpha$ wherever the denominator is nonzero.
We first define its inverse on a sufficiently far right vertical line;
no global assertion that $L_\alpha$ is zero-free is implicit in this definition.

\begin{lemma}[Finite expansion with an explicit error]\label{lem:expansion}
Let $N\ge2$, $Y>0$, and set
\begin{align}
 c_n&=\frac{\sin((n+1)\delta)}{\sin\delta},&
 A_n&=c_n\frac{\Gamma(m+n\alpha)}{\Gamma(m)},\label{eq:An}\\
 d_0&=1+\beta,&d_n&=nA_n-\sum_{j=1}^{n-1}A_jd_{n-j}\quad(1\le n<N).
 \label{eq:dn}
\end{align}
Put $\mu=2^{-1/2}$, $\eta=\sin(\delta/4)\sin(3\delta/4)$ and
\begin{align}
 C_A&=\frac{\max(|c_N|,|c_{N-1}|)}{\eta}
       \frac{\Gamma(m+N\alpha)}{\Gamma(m)\mu^{m+N\alpha}},\\
 C_H&=\frac{\max(|c_N|,|c_{N-1}|)}{\eta}
       \frac{\Gamma(m+N\alpha+1)}{\Gamma(m)\mu^{m+N\alpha+1}},\\
 \lambda_Y&=1-\sum_{n=1}^{N-1}|A_n|Y^{-n}-C_AY^{-N}.
 \label{eq:lambda}
\end{align}
If $\lambda_Y>0$, then $L_\alpha(q)\ne0$ for
$\Re q\ge0$, $|q|\ge Y$, and
\begin{equation}\label{eq:finiteF}
 \left|F_\alpha(q)-\sum_{n=0}^{N-1}d_nq^{-n-1}\right|
       \le C_F|q|^{-N-1},
\end{equation}
where a valid constant is
\begin{equation}\label{eq:CF}
 C_F=\frac{1}{\lambda_Y}\left\{
 \sum_{n=N}^{2N-2}\left|\sum_{\substack{0\le j<N\\0\le n-j<N}}
                A_jd_{n-j}\right|Y^{N-n}
 +\beta C_H+C_A\sum_{n=0}^{N-1}|d_n|Y^{-n}\right\}.
\end{equation}
All these bounds are uniform when their right-hand sides are evaluated by
interval arithmetic on a closed parameter interval in $(1,2)$.
\end{lemma}

\begin{proof}
The recurrence $c_0=1$, $c_1=2\cos\delta$,
$c_n=2\cos\delta\,c_{n-1}-c_{n-2}$ gives the finite identity
\begin{equation}\label{eq:geometric}
 \frac1{D_\delta(z)}=\sum_{n=0}^{N-1}c_nz^n+
          \frac{z^N(c_N-c_{N-1}z)}{D_\delta(z)}.
\end{equation}
In \eqref{eq:L}, make the substitution $w=q^\beta r$ and rotate the
$w$ ray to $\arg w=(\arg q)/2$. This rotation crosses no pole: along it
the argument of $w^\alpha/q$ stays between zero and
$(\alpha/2-1)\arg q$, whose modulus is at most $\delta/2$.
On the final ray $\Re w\ge\mu|w|$, and, for $z=w^\alpha/q$,
\begin{equation}\label{eq:sectorD}
 |D_\delta(z)|=|1-e^{i\delta}z|\,|1-e^{-i\delta}z|
       \ge\eta(1+|z|)^2.
\end{equation}
Indeed $|1-xe^{i\theta}|\ge(1+x)|\sin(\theta/2)|$ for $x\ge0$;
the product of the two sine factors is minimized at
$|\arg z|=\delta/2$. The small circular arc vanishes because $m>0$;
the large arc vanishes by exponential decay. These observations justify
the rotation, including its limiting values on $\Re q=0$.

Normalize the transform and its first moment as follows:
\begin{align}
 A(q)&=\frac{L_\alpha(q)}{\kappa_\alpha\Gamma(m)q^{-1-\beta}}
     =\frac1{\Gamma(m)}\int_{\arg w=(\arg q)/2}
              \frac{e^{-w}w^\alpha}{D_\delta(w^\alpha/q)}\,dw,\\
 H(q)&=\frac1{\Gamma(m)}\int_{\arg w=(\arg q)/2}
              \frac{e^{-w}w^{\alpha+1}}{D_\delta(w^\alpha/q)}\,dw.
\end{align}
Then $F_\alpha(q)=\beta H(q)/(qA(q))$. Equation~\eqref{eq:geometric}
and \eqref{eq:sectorD} yield
\begin{equation}\label{eq:AH}
 A(q)=\sum_{n=0}^{N-1}A_nq^{-n}+E_A(q),\qquad
 H(q)=\sum_{n=0}^{N-1}(m+n\alpha)A_nq^{-n}+E_H(q),
\end{equation}
with $|E_A(q)|\le C_A|q|^{-N}$ and
$|E_H(q)|\le C_H|q|^{-N}$. For example, the first error follows by
integrating
\[
 \frac{\max(|c_N|,|c_{N-1}|)}{\eta\Gamma(m)}
       e^{-\mu u}u^{\alpha+N\alpha}|q|^{-N}
\]
over $u>0$. Thus $|A(q)|\ge\lambda_Y>0$.
Multiplying the polynomial in \eqref{eq:dn} by the polynomial for $A$
cancels all powers of $q^{-1}$ of degree less than $N$ in
$\beta H-A\sum_{n<N}d_nq^{-n}$. Retaining every higher-degree product
and then dividing by $qA$ gives \eqref{eq:CF}.
This proof uses only a finite algebraic identity. In particular, no
convergence of the infinite asymptotic expansion is asserted.
\end{proof}

\begin{lemma}[Existence and uniqueness of the canonical kernel]\label{lem:canonical}
For each $1<\alpha<2$ there is a unique real continuous function
$K_\alpha:[0,\infty)\to\mathbb R$, of exponential order, such that
\begin{equation}\label{eq:canonical}
 F_\alpha(q)=\int_0^\infty e^{-qt}K_\alpha(t)\,dt
\end{equation}
for all sufficiently large real $q$. It is smooth on $[0,\infty)$ and
\begin{equation}\label{eq:initial}
 K_\alpha^{(n)}(0)=d_n,\qquad n\ge0.
\end{equation}
Its time derivatives and its first real parameter derivative are
continuous locally in $(\alpha,t)\in(1,2)\times[0,\infty)$.
\end{lemma}

\begin{proof}
Fix a compact parameter interval $I\subset(1,2)$. Lemma~\ref{lem:expansion},
with $N=2$ and $c$ sufficiently large, gives $\lambda_c>0$ uniformly
on $I$. Hence $F_\alpha$ is holomorphic on $\Re q\ge c$ there.
The same lemma supplies its finite expansion to every order in this
half-plane, uniformly on $I$.
For real parameter differentiation on $q=c\pm iy$, use the fixed rays
$w=ue^{\pm i\pi/4}$ when $|y|$ is sufficiently large. Then
$|\arg(w^\alpha/q)|<\delta/2$, uniformly on $I$, and
\eqref{eq:sectorD} still applies. Differentiating the finite remainder
introduces only $\log w$. The bound
$|\log w|\le u+u^{-1}+\pi/4$ gives an integrable majorant against
$e^{-u/\sqrt2}$, so the finite expansion can also be differentiated
uniformly on $I$. Explicit constants are given in
Lemma~\ref{lem:derivexpansion}. On bounded parts of the vertical line,
differentiation under \eqref{eq:L} is immediate.

Fix $h>0$. For $n\ge1$ define
\begin{equation}\label{eq:model}
 R_n(q)=\sum_{k=1}^n\frac{b_k}{(q+h)^k},\qquad
 b_k=\sum_{j=0}^{k-1}\binom{k-1}{j}h^{k-1-j}d_j,
 \quad r_n(t)=e^{-ht}\sum_{k=1}^n\frac{b_kt^{k-1}}{(k-1)!}.
\end{equation}
These coefficients give $F_\alpha-R_n=O(|q|^{-n-1})$ and
$r_n^{(j)}(0)=d_j$ for $j<n$. Set
\begin{equation}\label{eq:bromwich}
 K_\alpha(t)=r_n(t)+\frac1{2\pi}
       \int_{-\infty}^{\infty}e^{(c+iy)t}
             (F_\alpha(c+iy)-R_n(c+iy))\,dy.
\end{equation}
Absolute convergence, the elementary inverse transform of $R_n-R_1$,
and contour deformation show that this expression is independent of
$n$ and of the sufficiently large $c$. Conjugation makes it real.
For $n>j$, differentiation $j$ times in $t$ is justified by an
integrable bound, uniformly on compact time intervals. At $t=0$,
close the integral of $q^j(F_\alpha-R_n)$ to the right: there is no
pole, and the arc integral tends to zero. Thus
$K_\alpha^{(j)}(0)=r_n^{(j)}(0)=d_j$.
The same construction gives an exponential bound for each derivative.

For real $q>c$, Fubini's theorem followed by Cauchy's formula shows that
the Laplace transform of the integral in \eqref{eq:bromwich} is
$F_\alpha(q)-R_n(q)$. Adding the transform of $r_n$ proves
\eqref{eq:canonical}. Uniqueness follows from uniqueness of Laplace
transforms after exponential weighting. Finally, the differentiated
remainder bounds above permit parameter differentiation of
\eqref{eq:bromwich} and prove the asserted local continuity.
\end{proof}

\begin{lemma}[A sign criterion for additive infinite divisibility]\label{lem:ID}
The distribution of $T_1$ is infinitely divisible if and only if
$K_\alpha(t)\ge0$ for every $t>0$. In that event its drift is zero and
its L\'evy measure is
\begin{equation}\label{eq:levy}
 \nu_\alpha(dt)=\frac{K_\alpha(t)}t\,dt.
\end{equation}
\end{lemma}

\begin{proof}
For a nonnegative infinitely divisible random variable, the
L\'evy--Khintchine representation (see \cite{Sato1999,SSV2012}) and its
derivative are
\begin{equation}\label{eq:LK}
 -\log L(q)=dq+\int_0^\infty(1-e^{-qt})\nu(dt),\qquad
 -\frac{L'(q)}{L(q)}=d+\int_0^\infty te^{-qt}\nu(dt),
\end{equation}
where $d\ge0$ and $\int(1\wedge t)\nu(dt)<\infty$.
Since $F_\alpha(q)\to0$ as $q\to\infty$, $d=0$. Uniqueness of
Laplace transforms after exponential weighting gives
$t\nu(dt)=K_\alpha(t)dt$. Continuity implies $K_\alpha\ge0$.

Conversely, suppose $K_\alpha\ge0$ and choose $c>0$ larger than its
exponential growth rate and the threshold in \eqref{eq:canonical}.
Then $\nu_c(dt)=e^{-ct}K_\alpha(t)dt/t$ is a L\'evy measure.
Integrating \eqref{eq:canonical} from $c$ to $c+s$ gives
\begin{equation}\label{eq:tilt}
 \frac{L_\alpha(c+s)}{L_\alpha(c)}
 =\exp\left\{-\int_0^\infty(1-e^{-st})\nu_c(dt)\right\}.
\end{equation}
Thus the actual tilted probability law
$P_c(dt)=e^{-ct}\mathbb P(T_1\in dt)/L_\alpha(c)$ is infinitely divisible.
For each $n\ge1$, let $Q_{n,c}$ be its convolution root with L\'evy
measure $\nu_c/n$ and zero drift, so $Q_{n,c}^{*n}=P_c$.
Tonelli's theorem gives
\begin{equation}\label{eq:momenttilt}
 \left(\int_0^\infty e^{ct}Q_{n,c}(dt)\right)^n
 =\int_0^\infty e^{ct}P_c(dt)=L_\alpha(c)^{-1}<\infty.
\end{equation}
Consequently $Q_n(dt)=L_\alpha(c)^{1/n}e^{ct}Q_{n,c}(dt)$ is a
probability law, and multiplication of the exponential weights yields
\[
 Q_n^{*n}(dt)=L_\alpha(c)e^{ct}P_c(dt)=\mathbb P(T_1\in dt).
\]
This proves infinite divisibility directly from its definition.
The already established necessary direction then identifies its L\'evy
measure as \eqref{eq:levy}, including the required integrability.
\end{proof}

\begin{remark}\label{rem:notSD}
The first two coefficients are
\[
 K_\alpha(0)=1+\alpha^{-1},\qquad
 K_\alpha'(0)=2\cos\delta\,
          \frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}>0.
\]
A law is self-decomposable if, for every $b\in(0,1)$, it admits a
representation $Z\stackrel{\mathrm d}=bZ_1+Y_b$, with $Z_1$ distributed
as $Z$ and independent of $Y_b$. For such a law on $[0,\infty)$ the L\'evy density has
the form $k(t)/t$ with $k$ nonincreasing
\cite[Theorem~15.10]{Sato1999}. Uniqueness in
Lemma~\ref{lem:ID} rules out self-decomposability whenever the law
is infinitely divisible; outside that range it is ruled out a fortiori.
Thus this argument also recovers the result of
Wang and Yin~\cite[Corollary~1.2]{WY2026}: non-self-decomposability for every
$1<\alpha<2$. It does not apply at $\alpha=2$, where the limiting
normalization in \eqref{eq:L} degenerates.
\end{remark}

\section{Positivity up to the critical index}
\label{sec:positive-proof}

We retain the notation
\[
 \beta=\alpha^{-1},\qquad \delta=\pi(1-\alpha/2),\qquad
 D_\delta(z)=1-2\cos(\delta)z+z^2,
 \qquad d_0=1+\beta.
\]
In this section \(K_\alpha\) is the canonical, initially signed inverse
Laplace transform of \(F_\alpha=-L_\alpha'/L_\alpha\).  Its definition does
not presuppose infinite divisibility.  We prove
\begin{equation}\label{eq:positive-goal}
 K_\alpha(t)>0\quad(1<\alpha<\alpha_*,\ t>0),\qquad
 K_{\alpha_*}(t)\geq0\quad(t>0),
\end{equation}
and show that the last kernel has precisely one zero.  Every decimal
endpoint of a parameter or time interval below is an exact rational number.
The finite inequalities used in the proof, together with their exact
interval records and verification programs, are indexed in Section~\ref{sec:certificates}.  The analytic reductions and the quantities tested by those
programs are given explicitly here.

\subsection{The cut and a common zero-free region}

Put
\[
 \kappa_\alpha=\frac{\alpha\sin\delta\sin(\pi\beta)}{\pi},\qquad
 \eta_\alpha=\frac{\alpha\sin\delta}{\pi},\qquad
 M_\alpha(q)=\int_0^\infty e^{-q^\beta r}
                  \frac{r^\alpha}{D_\delta(r^\alpha)}\,dr.
\]
For \(\alpha>1\), \(L_\alpha=\kappa_\alpha M_\alpha\), so the logarithmic
derivative can be computed using \(M_\alpha\).  The latter family, unlike
its vanishing normalization factor, remains meaningful at \(\alpha=1\):
\begin{equation}\label{eq:auxiliary-one}
 M_1(q)=\int_0^\infty e^{-qr}\frac{r}{1+r^2}\,dr.
\end{equation}
The endpoint \(\alpha=1\) used in the estimates below always refers to
this auxiliary analytic family, and never to a Cauchy point-hitting law.
Its kernel $K_1$ is defined by the same regularized Bromwich construction
as in Lemma~\ref{lem:canonical}, with $M_1$ in place of $L_\alpha$.

\begin{lemma}[Cut phase]\label{lem:positive-cut}
For \(1\leq\alpha<2\), the upper-bank continuation of
\(M_\alpha\) has no zeros on \(-1\leq q<0\).  There is a continuous
phase \(\theta_\alpha(v)\), normalized by
\(\theta_\alpha(0+)=0\), such that
\[
 -\pi\beta<\theta_\alpha(v)<0\quad(0<v\leq1).
\]
Consequently
\[
 w_\alpha(v)=-\frac1\pi\operatorname{Im}F_\alpha(-v+i0)>0,
 \qquad
 \int_0^b w_\alpha(v)\,dv=-\frac{\theta_\alpha(b)}\pi<\beta
 \quad(0<b\leq1).
\]
\end{lemma}
\begin{proof}
Let \(x=v^\beta\) and
\[
 H_\alpha(x)=\int_0^\infty e^{-xu}
 \frac{u^\alpha}{1+2\cos\delta\,u^\alpha+u^{2\alpha}}\,du.
\]
Rotating the integration ray by \(-\pi\beta\) crosses the lower simple
pole \(p=\exp[-i\pi(\beta-1/2)]\).  The integral on the new ray and
that residue give the exact identity
\begin{equation}\label{eq:positive-cut-identity}
 M_\alpha(-v+i0)=\eta_\alpha^{-1}e^{-i\pi\beta}N_\alpha(x),
 \qquad N_\alpha(x)=i e^{-ix}-\eta_\alpha H_\alpha(x).
\end{equation}
In particular \(\operatorname{Im}N_\alpha(x)=\cos x>0\) for
\(0<x\leq1\).  The density under the integral defining \(H_\alpha\)
is at most \(1/2\), because \(\cos\delta\geq0\) and
\(1+u^{2\alpha}\geq2u^\alpha\).  Thus \(xH_\alpha(x)\leq1/2\),
whereas \(H_\alpha'(x)<0\).  Direct differentiation gives
\begin{equation}\label{eq:positive-cut-density}
 w_\alpha(v)=\frac{\beta v^{\beta-1}}\pi
 \frac{1-\eta_\alpha\{H_\alpha(x)\sin x+
                      H_\alpha'(x)\cos x\}}
      {|N_\alpha(x)|^2}>0,
\end{equation}
since \(\eta_\alpha<1\) and the numerator is greater than \(1/2\).

The phase \(\arg N_\alpha(v^\beta)-\pi\beta\) lies in
\((-\pi\beta,\pi(1-\beta))\), and its derivative is \(-\pi w_\alpha\).
For \(\alpha>1\), its limit at zero is zero because
\(M_\alpha(0)=1/\kappa_\alpha>0\).  At \(\alpha=1\), the same limit follows
from \(M_1(q)=-\log q+O(1)\) near the origin.  Its strict decrease proves
the asserted phase interval and, by integration, the mass identity.
\end{proof}

\begin{lemma}[Zero-free slit region]\label{lem:positive-strip}
For \(1\leq\alpha\leq1.6671\), \(M_\alpha\) has no zeros in
\[
 \{q:\operatorname{Re}q\geq-1\}\setminus(-\infty,0],
\]
and its boundary values on the cut segment \([-1,0)\) are nonzero.
\end{lemma}
\begin{proof}
We describe separately the origin, the exterior and the finite vertical
boundary; their argument changes, rather than boundary nonvanishing alone,
are needed to exclude interior zeros.

For \(1\leq\alpha\leq6/5\), write \(\varepsilon=\alpha-1\),
\(s=q^\beta\), and \(r=|s|\).  Subtraction of the leading
\(u^{-\alpha}\) tail gives, for \(\varepsilon>0\),
\[
 M_\alpha(q)=M_\alpha(0)
       -\frac{\Gamma(1-\varepsilon)}{\varepsilon}s^\varepsilon
       +R_\alpha(s),
 \qquad |R_\alpha(s)|\leq60|s|,\quad |R_\alpha'(s)|\leq60.
\]
The two divergent terms have their removable logarithmic limit when
\(\varepsilon=0\).  To obtain these bounds on both banks, use the rays
\(-\pi/4\) and \(-3\pi/4\).  Their denominators exceed
\((1+u^\alpha)^2/10\); the subtracted first moment is less than
\(10(5/4+1/3+1+\pi/2)<42\), and the crossed-pole contribution and its
derivative are less than \(12\).  The exact inequality
\(\varepsilon M_\alpha(0)\geq1\) and
\(\Gamma(1-\varepsilon)\leq e^{2\varepsilon}\) now imply, with
\(\ell=\log(1/r)-2\),
\[
 \operatorname{Re}M_\alpha(q)
 \geq\frac{1-e^{-\varepsilon\ell}}{\varepsilon}-60r>2
 \quad (|q|\leq10^{-4}).
\]
Here the quotient has its continuous value at \(\varepsilon=0\);
\(\ell>4\), and its lower bound \(20/9\) suffices.  These same
estimates yield \(qF_\alpha(q)\to0\) uniformly at this auxiliary endpoint.
For \(6/5\leq\alpha\leq13/10\), the corresponding origin estimate is
\[
 |L_\alpha(q)-1|<20|q|^{1/6}<1\quad (|q|\leq10^{-8}).
\]
For \(13/10\leq\alpha\leq1.6671\), the finite parameter records give
\(C_\alpha^{(0)}(10^{-12})^{1-\beta}<1/2\), where
\( |L_\alpha(q)-1|\leq C_\alpha^{(0)} |q|^{1-\beta}\).
Here the constant is completely explicit. With
\[
 \gamma=\frac{3\pi\beta}{4},\quad
 \Delta=\sin\frac{\alpha\gamma-\delta}{2}
          \sin\frac{\alpha\gamma+\delta}{2},\quad
 V_\alpha=2^{2-\alpha}
          \left(\frac1{2-\alpha}+\frac1{\alpha-1}\right),
\]
one may take
\[
 C_\alpha^{(0)}=\max\left\{\sin(\pi\beta)
            +\frac{\kappa_\alpha V_\alpha}{\Delta},
             \frac{\kappa_\alpha V_\alpha}{\sin^2(\delta/2)}\right\}.
\]
These last two estimates follow by integrating
\(\min(|s|u,2)u^{-\alpha}\) on a decaying ray and adding the explicit
pole difference.  Thus an origin circle has its image in a right half-plane.

For \(|q|\geq400\) and \(|\arg q|\leq\pi/2+1/20\), normalize by
\(A_\alpha(q)=q^{d_0}M_\alpha(q)/\Gamma(\alpha+1)\).
A finite geometric expansion of \(D_\delta^{-1}\), with its integrated
remainder, gives
\[
 |A_\alpha(q)-1|<1/4\quad(1\leq\alpha\leq13/10),\qquad
 |A_\alpha(q)-1|<1/2\quad(13/10\leq\alpha\leq1.6671),
\]
and \(F_\alpha(q)-d_0/q=O(|q|^{-2})\), uniformly on each parameter
range.  The first range has the explicit bound \(1000/|q|^2\).
For the last range the normalized ray \(\arg w=\arg q/2\),
\(w=q^\beta r\), has exponential cosine at least \(2/3\) and denominator
constant \(\sin(\delta/5)\sin(3\delta/5)\).  No pole is crossed.
The finite expansion, including its full quotient remainder, supplies the
stated exterior inequalities.

On \(q=-1+iy\), \(0\leq y\leq400\), the verified boundary conditions are
\[
\begin{array}{c|c|c}
 \text{parameter range}&\text{parameter boxes}&\text{boundary condition}\\ \hline
 {}[1,6/5]&20&\operatorname{Re}M_\alpha<0\\
 {}[6/5,13/10]&100&\operatorname{Re}L_\alpha<0\\
 {}[13/10,1.6671]&74&
       \operatorname{Re}L_\alpha<0\ \text{or}\ \operatorname{Im}L_\alpha<0.
\end{array}
\]
For clarity, these are conditions on entire closed parameter--\(y\)
rectangles.  If \(b_j\) is the lower bound for the negative component at
\(y_j\), and \(B_j\) bounds \(|L_\alpha'|\) on
\([y_j,y_{j+1}]\), the checked inequality is
\(b_j-(y_{j+1}-y_j)B_j>0\); use \(M\) in the first row.
The bounds \(B_j\) include the residue derivative and the full ray-integral
derivative.  Exact rational adjacency gives a cover of all of \([0,400]\).
For example, for the third row, put \(r=(1+y^2)^{1/2}\),
\(p=e^{-i\delta/\alpha}\), and use the \(\Delta\) above. A bound at the
left endpoint, valid on the whole subsequent segment, is
\begin{align*}
 B(y)={}&\beta\sin(\pi\beta)\,r^{\beta-1}
            e^{-\Re(p(-1+iy)^\beta)}\\
 &+\frac{\beta\kappa_\alpha}{\Delta}
 \min\left\{\frac{\Gamma(2-\alpha)}{(3/4)^{2-\alpha}}r^{-\beta},
       \frac{\Gamma(\alpha+2)}{(3/4)^{\alpha+2}}r^{-2-\beta}\right\}.
\end{align*}
This follows by differentiating the residue and the rotated integral,
using \(\Re(se^{-i\gamma})\ge(3/4)|s|\). The two powers are the bounds
obtained respectively from the large- and small-radial-power estimates
for the denominator. Both decrease with \(y\). The residue bound also
decreases: \(r^{\beta-1}\) decreases, whereas
\[
 \frac{d}{dy}\Re(p(-1+iy)^\beta)
 =\beta r^{\beta-1}\cos((1-\beta)(\pi-\arg(-1+iy)))>0.
\]
Thus this formula supplies the required whole-segment \(B_j\), including
parameter interval rounding.

All three rows place the vertical image in
$\{z:\Re z<0\text{ or }\Im z<0\}$, which admits a continuous
argument in $(-3\pi/2,0)$. At the cut endpoint this argument agrees
with the phase in Lemma~\ref{lem:positive-cut}, which lies in $(-\pi,0)$.
At $q=-1+400i$ it agrees with
$-d_0\arg q+\arg A_\alpha(q)$: indeed $3/2<d_0\le2$,
$\pi/2<\arg q<\pi/2+1/20$ and $|\arg A_\alpha|<\pi/6$ give
\[
 -\frac{3\pi}{2}<-\frac{7\pi}{6}-\frac1{10}
 <-d_0\arg q+\arg A_\alpha(q)<-\frac{7\pi}{12}<0.
\]
The exterior arc continues this phase to zero on the positive axis.
The positive axis and the small origin arc, whose image lies in a
right half-plane, complete a contour with zero total argument change.
The argument principle excludes zeros in the bounded upper slit region.
Conjugation and the exterior estimate cover the remaining parts of the
stated domain.
\end{proof}

\subsection{Control of the two time ends}

\begin{lemma}[A signed cut representation]\label{lem:positive-cut-inverse}
Throughout the range of Lemma~\ref{lem:positive-strip},
\begin{equation}\label{eq:positive-cut-inverse}
 K_\alpha(t)=d_0e^{-2t}+\int_0^1 e^{-vt}w_\alpha(v)\,dv+E_\alpha(t),
 \qquad |E_\alpha(t)|\leq B_\alpha e^{-t}.
\end{equation}
One can take \(B_\alpha<94,37,26\) on the respective three parameter
ranges in the preceding table.  In particular the signed kernel is bounded
on all positive times, before any positivity conclusion is made.
\end{lemma}
\begin{proof}
Subtract \(d_0/(q+2)\) from \(F_\alpha\), and deform its inverse contour
to \(\operatorname{Re}q=-1\), detouring around the cut.  The exterior
bound is integrable and eliminates the horizontal ends.  The origin bound
\(qF_\alpha(q)\to0\) eliminates the small circle, also for the logarithmic
auxiliary endpoint.  Lemma~\ref{lem:positive-cut} gives the jump and its
finite mass.  The pole of the subtracted model is at \(-2\), outside
the deformed domain.

On every finite vertical segment the preceding margin \(\ell_j>0\)
gives \(|L|\geq\ell_j\), hence
\[
 B_\alpha\leq\frac1\pi\left\{
 \sum_j h_j\left(\frac{B_j}{\ell_j}
              +\frac{d_0}{\sqrt{1+y_j^2}}\right)
       +\text{the complete exterior tail}\right\}.
\]
For the first two ranges the last tail is at most \(1004/400\).
Evaluating this finite inequality on every parameter box gives the constants
stated above; the first sharper bound is \(93.0745928287506103515625\).
The integral on the left line has absolute value at most \(B_\alpha e^{-t}\),
which proves the representation.  Its cut mass is less than \(\beta\),
so it also proves \(|K_\alpha(t)|\leq1+2\beta+B_\alpha\).
\end{proof}

\begin{lemma}[Time-end positivity]\label{lem:positive-time-ends}
The following inequalities hold uniformly on the indicated parameter sets:
\[
\begin{array}{c|c|c}
 \alpha\text{-range}&\text{small time}&\text{large time}\\ \hline
 {}[1,6/5]&K_\alpha(t)>1/2\ (0<t\leq1/4)&K_\alpha(t)>0\ (t\geq33/5)\\
 {}[6/5,13/10]&K_\alpha(t)>1/2\ (0<t\leq1/5)&K_\alpha(t)>0\ (t\geq6)\\
 {}[13/10,3/2]&K_\alpha(t)>1/2\ (0<t\leq9/100)&K_\alpha(t)>0\ (t\geq6)\\
 {}[3/2,1.6671]&K_\alpha(t)>1/2\ (0<t\leq9/200)&K_\alpha(t)>0\ (t\geq6).
\end{array}
\]
In particular \(K_\alpha(t)>0\) for \(0<t\leq9/200\) on the parameter
ranges used in the contact and bridge arguments below.
\end{lemma}
\begin{proof}
For small time we sharpen the finite expansion by splitting its inversion
line. The proof of Lemma~\ref{lem:expansion} permits any lower bound
$\mu$ for the exponential cosine; use
$\max(|c_N|,|c_{N-1}|)\le1/\sin\delta$ in its remainder constants.
On $q=c+iy$, $c=N/t$, the part $|\arg q|\le\pi/3$ permits
$\mu=\sqrt3/2$, while the remaining part has $|q|\ge2c$ and permits
$\mu=1/\sqrt2$. For $0<t\le T$, let $C_F^{\rm in}$ and
$C_F^{\rm out}$ be the corresponding constants computed at
$Y=N/T$ and $Y=2N/T$. Their checked positive denominator bounds also
exclude zeros on $\Re q\ge N/T$, so the canonical inversion line may
be moved there. The inner integral is bounded by the whole-line integral;
on the outer part,
\[
 \int_{|y|\ge\sqrt3c}|c+iy|^{-N-1}\,dy
 \le (2c)^{1-N}\int_{|y|\ge\sqrt3c}\frac{dy}{c^2+y^2}
 =c^{-N}\frac\pi3\,2^{1-N}.
\]
Thus the entire inversion error is at most
\[
 E_N(T)=e^N(T/N)^N\left\{
 \frac{C_F^{\rm in}\Gamma(N/2)}{2\sqrt\pi\Gamma((N+1)/2)}
       +\frac{C_F^{\rm out}2^{1-N}}6\right\}.
\]
To retain the positive polynomial terms, let $\underline d_n$ be lower
bounds for $d_n$ on the parameter box and use the Bernstein coefficients
\[
 \mathcal B_k=\sum_{n=0}^k\frac{\underline d_n T^n}{n!}
                 \frac{\binom{k}{n}}{\binom{N-1}{n}},
 \qquad 0\le k<N.
\]
The Bernstein basis is nonnegative and sums to one on $[0,1]$, giving
$\sum_{n<N}d_nt^n/n!\ge\min_k\mathcal B_k$ for $0\le t\le T$.
The finite checks are therefore
\begin{equation}\label{eq:positive-small-time-test}
 \min_{k<N}\mathcal B_k-E_N(T)>\frac12.
\end{equation}
They use $N=40,32,24,16$ on the successive parameter ranges
$[1,6/5]$, $[6/5,13/10]$, $[13/10,3/2]$ and $[3/2,1.6671]$,
with four adjacent parameter boxes in each range, of maximum widths
$1/20,3/100,1/20,1/20$, respectively; the last box is shortened to
the range endpoint. This is a finite expansion
bound on the entire time interval, with no convergence assertion about
an infinite series.

For large time we use one cut-mass estimate on all three parameter ranges.
Put $b=1/20$, $\chi=\cos\delta$ and
$\lambda=(1-2\chi^2)/(2(1+\chi))$. The elementary inequality
\[
 \frac1{1+2\chi v+v^2}\le1-\chi v-\lambda v^2,
 \qquad 0\le v\le1,
\]
follows by multiplying through: the difference is
$v(1-v)\{\chi+(\chi+\lambda+2\chi\lambda)v+\lambda v^2\}$.
For $\lambda\ge0$ all coefficients in braces are nonnegative; for
$\lambda<0$ the quadratic is concave with positive endpoint values
$\chi$ and $1+2\chi-2\chi^2$.
Applying this inequality with $v=\min(u^\alpha,u^{-\alpha})$ gives
\begin{align*}
 H_\alpha(x)&\le\widehat H_\alpha(x)
       :=I_\alpha(x)-\chi I_{2\alpha}(x)-\lambda I_{3\alpha}(x),\\
 I_r(x)&:=\int_0^1e^{-xu}u^r\,du+\int_1^\infty e^{-xu}u^{-r}\,du\\
 &=x^{-r-1}\gamma(r+1,x)+x^{r-1}\Gamma(1-r,x).
\end{align*}
The lower and upper incomplete Gamma functions here represent convergent
real integrals, including when $1-r\le0$.

The function $H_\alpha(x)$ decreases with $x$ and with $\alpha$:
its radial denominator is $2\cosh(\alpha\log u)+2\cos\delta$.
For a parameter box $A=[a_0,a_1]$ set
\[
 x_0=b^{1/a_0},\quad x_1=b^{1/a_1},\quad
 \bar\eta_A=\frac{a_1\sin\delta(a_0)}\pi,\quad
 n_A=\bar\eta_A\widehat H_{a_0}(x_0)-\sin x_0.
\]
On the boxes below $n_A>0$. Since
$x_0\le x\le x_1$ and $\eta_\alpha\le\bar\eta_A$, the exact cut phase yields
\[
 W_\alpha(b):=\int_0^b w_\alpha(v)\,dv
 \ge\frac1{a_1}-\frac12-
          \frac1\pi\arctan\!\left(\frac{n_A}{\cos x_1}\right).
\]
Only the following scalar inequalities are required:
\[
\begin{array}{c|c|c|c|c}
 \alpha\text{-range}&\text{boxes}&\text{maximum width}&W_\alpha(1/20)>&T\\ \hline
 {}[1,6/5]&5&1/10&899/5000&33/5\\
 {}[6/5,13/10]&2&1/20&1351/10000&6\\
 {}[13/10,1.6671]&10&1/10&441/5000&6
\end{array}
\]
The exact endpoints of these nonuniform partitions are listed in the scalar
certificates. These whole-box bounds use real incomplete Gamma and elementary functions,
with no transform quadrature. The final time comparisons follow from
$e^{627/100}>523$, $e^{57/10}>298$ and
\[
 \frac{899}{5000}\,523>94,\qquad
 \frac{1351}{10000}\,298>37,\qquad
 \frac{441}{5000}\,298>26.
\]
The two exponential bounds already follow from their degree-$16$ Taylor
sums. Lemma~\ref{lem:positive-cut-inverse} now gives
$K_\alpha(t)\ge W_\alpha(b)e^{-bt}-B_\alpha e^{-t}>0$ for all $t\ge T$,
since $W_\alpha(b)e^{(1-b)t}$ increases with $t$.
\end{proof}


\subsection{Validated inversion on compact rectangles}

We use the rational model $R_p$ and its inverse $r_p$ from
\eqref{eq:model}, and write $R=R_p$, $r=r_p$, $G=F-R$ and $g=K-r$.
Thus $G=O(q^{-p-1})$. Choose $P>t$, $\omega=2\pi/P$ and
$J\omega\geq Y>h$. The following bound controls both omitted
frequencies and the error caused by periodization.

\begin{lemma}[Complete Fourier error]\label{lem:positive-fourier}
If \(|g(t)|\leq M\) for all \(t>0\), the finite expression
\[
 \mathcal H_J(t)=r(t)+\frac{e^{\sigma t}}P
 \left\{G(\sigma)+2\operatorname{Re}\sum_{j=1}^J
          e^{ij\omega t}G(\sigma+ij\omega)\right\}
\]
satisfies, for \(0<t<P\),
\begin{align*}
 |K(t)-\mathcal H_J(t)|
 &\leq\frac{M e^{-\sigma P}}{1-e^{-\sigma P}}\\
 &\quad+\frac{e^{\sigma t}}\pi
 \left\{\sum_{n=p}^{N-1}\frac{|d_n-r_n|}{nY^n}
                    +\frac{C_F+C_R}{NY^N}\right\},
\end{align*}
where \(r_n\) are the finite inverse-power coefficients of \(R\), and
\[
 C_R=\sum_{k=1}^p |b_k|\binom N{k-1}
                 \frac{h^{N+1-k}}{(1-h/Y)^k}.
\]
The same assertion holds for \(|g(t)|\leq M e^{\lambda t}\),
\(\sigma>\lambda\), if the alias term is replaced by
\(M e^{\lambda t}e^{-(\sigma-\lambda)P}/
 (1-e^{-(\sigma-\lambda)P})\).
\end{lemma}
\begin{proof}
Extend \(g\) by zero to negative times and periodize \(e^{-\sigma t}g(t)\).
Its Fourier coefficients are \(G(\sigma+ij\omega)/P\).  The matched
coefficients give an absolutely summable Fourier series, hence, at
\(0<t<P\),
\[
 g(t)=\frac{e^{\sigma t}}P\sum_{j\in\mathbb Z}
                e^{ij\omega t}G(\sigma+ij\omega)
       -\sum_{\ell\geq1}e^{-\sigma\ell P}g(t+\ell P).
\]
The last sum has exactly the stated alias bound.  Expand the rational
model to the same finite order as \(F\); its convergent binomial remainder
is bounded by \(C_R|q|^{-N-1}\).  For each remaining power use
\(\sum_{j>J}j^{-n-1}\leq J^{-n}/n\) and \(J\omega\geq Y\).
Including both frequency signs and the factor \(1/P\) gives the stated
\(1/\pi\) bound.  This accounts for every omitted frequency.
\end{proof}

The bound in Lemma~\ref{lem:positive-cut-inverse} supplies \(M\), after
adding
\[
 \sup_{t\geq0}|r(t)|\leq |b_1|+
 \sum_{k=2}^p\frac{|b_k|}{(k-1)!}
                  \left(\frac{k-1}{eh}\right)^{k-1}.
\]
A real parameter mean-value enclosure is applied to the finite function
\(\mathcal H_J\), and the uniform inversion error is added afterwards.
No derivative of an unknown alias error is used.  On a time interval
centered at \(t_m\), finite Taylor enclosures through degree four use the
whole interval for the fifth derivative.  A complex coefficient rectangle
can instead be split into its exact binary center and its radius: the
uncertainty is then added once using \(|e^{ij\omega t}|=1\), rather than
being differentiated repeatedly with the Taylor polynomial.

For the earlier ranges one uses directly the absolutely convergent
Bromwich integral after subtracting \(d_0/q\), or a two-term shifted
model.  The finite inverse-power tail at a time midpoint is evaluated
exactly using the upper incomplete gamma function.  For its time variation,
the \(q^{-2}\) term has the bound \(2e^{ct}/(\pi tY)\) obtained by one
integration by parts; all subsequent terms have absolutely integrable
derivatives.  The remaining expansion error is
\(e^{ct}C_F/(\pi NY^N)\).  Parameter derivatives are obtained on fixed
rays before integration, with separate bounds on the omitted origins and
tails.  For example, putting \(w=e^{i\phi}u\), \(z=w^\alpha/q\),
\[
 A=\frac{e^{i(\alpha+1)\phi}}{\Gamma(\alpha+1)}
       \int_0^\infty\frac{e^{-e^{i\phi}u}u^\alpha}{D_\delta(z)}\,du,
 \qquad F=\frac\beta q\frac{N}{A},
\]
where \(N\) has one extra factor \(w\).  The fixed-ray integrand derivative
is its value times
\[
 \log u+i\phi-\psi(\alpha+1)
                 -\frac{\partial_\alpha D_\delta(z)}{D_\delta(z)},
\quad
 \partial_\alpha D_\delta(z)
  =-\pi\sin\delta\,z-2\cos\delta\,z\log w+2z^2\log w.
\]
Thus \(\partial_\alpha F=-\beta F+
 (\beta/q)(N_\alpha/A-NA_\alpha/A^2)\).
The ray is fixed on the entire parameter box.  Its angular distance from
both poles and its exponential decay are checked before it is used; a
representation crossing a pole includes its residue explicitly.

\begin{lemma}[The lower parameter ranges]\label{prop:positive-lower}
For \(1<\alpha\leq33/20\), \(K_\alpha(t)>0\) for every \(t>0\).
\end{lemma}
\begin{proof}
The compact parts not covered by Lemma~\ref{lem:positive-time-ends} have
the following finite, closed-rectangle covers.  The bounds in the last
column are conservative rational lower bounds, restricted to the displayed
compact time domains.
\[
\begin{array}{c|c|r|r|c}
 \alpha\text{-range}&t\text{-range}&\alpha\text{-boxes}&
       \text{rectangles}&K\text{ lower bound}\\ \hline
 {}[1,6/5]&[1/4,33/5]&20&6360&1/50000\\
 {}[6/5,13/10]&[1/5,6]&20&5800&7/100000\\
 {}[13/10,3/2]&[9/100,6]&40&11840&63/1000\\
 {}[3/2,33/20]&[9/200,6]&30&8941&2/125
\end{array}
\]
The first two rows use \(c=1/8,Y=40,N=12\).  The third uses
\(c=1/32,Y=120,N=8\), a two-term shifted model with \(h=1\), and the
whole-kernel enclosure
\[
 K_\alpha(t)\in K_{\alpha_m}(t_m)
       +(t-t_m)\,\partial_tK_{\alpha_m}(T)
       +(\alpha-\alpha_m)\,\partial_\alpha K_A(T).
\]
The last row uses Lemma~\ref{lem:positive-fourier} with
\(\sigma=1/8,P=64,J=4096,Y=400,N=18,p=4,h=20\).
Every accepted output is an enclosure of the full parameter and time
rectangle. Exact rational adjacency joins the rectangles; a failed
rectangle is subdivided, never skipped. The saved enclosure lower endpoints
are strictly above the displayed bounds.  In the first two rows, input
containment is justified by the exact rational construction in the frozen
producer; later rows also retain their binary input balls for a separate
containment check. Every retained saved output has a strictly positive
lower endpoint. The finite subset is selected by exact interval endpoints:
a rectangle crossing a new time cutoff is retained and restricted to the
required subinterval. The parameter intervals and clipped time intervals
meet exactly.
Combining these compact covers with both analytic time ends proves the
proposition.  The auxiliary row endpoint \(\alpha=1\) makes no assertion
about a point-hitting distribution at that index.
\end{proof}

\subsection{The contact and completion of the positive range}

Put
\begin{gather*}
 \alpha_0=1.66706599,\quad u=10^{-8},\quad
 t_0=.1628,\quad v=10^{-6},\\
 A_*=[\alpha_0-u,\alpha_0+u],\qquad I_*=[t_0-v,t_0+v].
\end{gather*}
\begin{lemma}[One local contact]\label{lem:contact}
There is exactly one pair \((\alpha_*,t_*)\in A_*\times I_*\) satisfying
\(K_{\alpha_*}(t_*)=\partial_tK_{\alpha_*}(t_*)=0\), and
\[
 1.6670659903<\alpha_*<1.6670659915,\qquad .162799<t_*<.162801.
\]
Each \(K_\alpha\), \(\alpha\in A_*\), has a unique interior minimum
on \(I_*\), and its minimum value decreases strictly with \(\alpha\).
\end{lemma}
\begin{proof}
The interval inversion gives the following rational inequalities:
\begin{align*}
 3\cdot10^{-9}&<K_{\alpha_0}(t_0)<13\cdot10^{-9},&
 |\partial_tK_{\alpha_0}(t_0)|&<57\cdot10^{-6},\\
 1402&<\partial_t^2K_\alpha(t)<1421,&
 -9.068&<\partial_\alpha K_\alpha(t)<-9.059
       \quad((\alpha,t)\in A_*\times I_*).
\end{align*}
For these local derivative evaluations, the line is \(\operatorname{Re}q=20\),
with \(Y=1000,N=18\); the regularized inverse of \(q^jF\) subtracts
its polynomial terms and its \(d_j/q\) term and adds back \(d_j\).
Appendix~\ref{app:inversion} gives the complete finite remainder and its
parameter derivative. Lemma~\ref{lem:positive-strip} identifies the
inverse with the canonical kernel, independently of positivity.

At \(\alpha_0\), strong convexity raises either time endpoint above
the center by at least \(-57\cdot10^{-6}v+701v^2\). The variation
of this difference when \(|\alpha-\alpha_0|\leq u\) is at most
\(u(9.068-9.059)\). Thus both endpoints exceed the center by more than
\[
 -57\cdot10^{-12}+701\cdot10^{-12}-90\cdot10^{-12}
       =554\cdot10^{-12}>0.
\]
Strict convexity therefore gives a unique interior minimizer for every
\(\alpha\in A_*\). Write \(m(\alpha)=\min_{t\in I_*}K_\alpha(t)\).
For \(a<b\) in \(A_*\), integrate the parameter bounds and then take
minima to obtain
\begin{equation}\label{eq:minimum-comparison}
 -9.068(b-a)\leq m(b)-m(a)\leq-9.059(b-a).
\end{equation}
In particular \(m\) is continuous and strictly decreasing; no
differentiation of its minimizer is required. Strong convexity also gives
\[
 m(\alpha_0)>3\cdot10^{-9}
       -\frac{(57\cdot10^{-6})^2}{2\cdot1402}>0,
 \qquad
 m(\alpha_0+u)<13\cdot10^{-9}-9.059\cdot10^{-8}<0.
\]
Consequently \(m\) has exactly one zero, and \eqref{eq:minimum-comparison}
implies
\[
 \frac{3\cdot10^{-9}-(57\cdot10^{-6})^2/(2\cdot1402)}{9.068}
 <\alpha_*-\alpha_0<\frac{13\cdot10^{-9}}{9.059}.
\]
These bounds give the stated enclosure. The unique interior minimizer
at \(\alpha_*\) is \(t_*\); strict convexity proves uniqueness of the
pair in \(A_*\times I_*\).
\end{proof}

\begin{lemma}[Complete positive side]\label{lem:positivecomplete}
\label{prop:positive-global-contact}\label{prop:positive-bridge}
For every \(1<\alpha<\alpha_*\), \(K_\alpha(t)>0\) for all \(t>0\).
The kernel at \(\alpha_*\) is nonnegative and has exactly one zero,
at the time \(t_*\) of Lemma~\ref{lem:contact}. The corresponding
hitting-time laws are infinitely divisible under ordinary addition.
\end{lemma}
\begin{proof}
By Lemmas~\ref{prop:positive-lower} and~\ref{lem:positive-time-ends},
it remains to consider \(33/20\leq\alpha\leq\alpha_*\) and
\(.045\leq t\leq6\). Put \(a_-=\alpha_0-u=1.66706598\).

First let \(a_-\leq\alpha\leq\alpha_*\). On
\(A_*\times[.145,.175]\), thirty adjacent time intervals give
\(\partial_t^2K_\alpha>242\). Hence the minimum on \(I_*\) is also
the unique minimum on this larger interval. Its value is nonnegative
by Lemma~\ref{lem:contact}, and is zero only when
\((\alpha,t)=(\alpha_*,t_*)\). On the complement
\[
 A_*\times\bigl([.045,.145]\cup[.175,6]\bigr),
\]
891 adjacent time rectangles give \(K_\alpha>10^{-5}\).
These use \(\sigma=1/2,P=32,J=2048,Y=400,N=18,p=4,h=20\) in
Lemma~\ref{lem:positive-fourier}. The convexity calculation uses the
filtered second derivative with its full inversion remainder. The finite
sum enclosures and error constants are verified independently at higher
precision while preserving the exact original parameter centers.

Now let \(33/20\leq\alpha\leq a_-\). On
\([33/20,a_-]\times[.15,.17]\), ninety-eight parameter boxes and
980 time rectangles give \(\partial_\alpha K_\alpha<-1\).
Its derivative alias bound is independent of infinite divisibility.
Specifically, put
\(Z(q)=L_\alpha(q)(q+2)^{d_0}/(\kappa_\alpha\Gamma(\alpha+1))\).
The finite vertical boundary margins, the cut bound
\(|L_\alpha(-v+i0)|\geq\sin(\pi\beta)\cos1\), and the exterior
estimate give \(|Z|>0.1476\) by the maximum principle applied to \(1/Z\).
For \(\operatorname{Re}q\geq1\), this implies \(|A_\alpha|>0.0252\).
Bounding the fixed-ray moment derivatives by
\(|\log u|\leq u+u^{-1}\) yields
\(|\partial_\alpha F_\alpha(q)|<4.178\cdot10^9/|q|\).
An integrable inversion after subtraction of
\(-\beta^2/(q+2)\), with the finite derivative remainder on \(|y|\geq400\),
then gives
\[
 |\partial_\alpha K_\alpha(t)|<5.319\cdot10^{11}e^t.
\]
The exponential-growth version of Lemma~\ref{lem:positive-fourier},
with \(\sigma=5,P=8,J=512\), supplies the complete alias and
omitted-frequency errors in the derivative cover. Integration in
\(\alpha\) now gives the direct comparison
\[
 K_\alpha(t)\geq K_{a_-}(t)+(a_--\alpha)>0,
 \qquad .15\leq t\leq.17,
\]
where positivity at \(a_-\) was proved in the first case. On the
remaining compact times \([.045,.15]\cup[.17,6]\), eighteen parameter
boxes and 3096 rectangles give \(K_\alpha>2\cdot10^{-4}\).
All parameter and time endpoints meet exactly, so the two cases and
the time-end bounds prove the kernel assertions. Lemma~\ref{lem:ID}
gives infinite divisibility and the L\'evy representation.
\end{proof}

The negative-side argument is needed separately to establish that this
locally unique contact is the unique classification threshold. No claim
that other contacts cannot occur at larger indices is used or implied.

\section{The non-infinitely-divisible range}\label{sec:negative}

The local minimum constructed below the critical index already gives a
negative value of $K_\alpha$ immediately to the right of that index.
For larger parameters two necessary conditions for infinite divisibility
are more efficient than a global inversion of $F_\alpha$. We give both
conditions, their validated evaluation, and an analytic argument that
closes the interval at $2$.

\begin{lemma}[Gaussian necessary condition]\label{lem:gaussian}
Suppose a nonnegative random variable has an infinitely divisible law
with Laplace transform $L$. Put $F=-L'/L$. For $\sigma,h>0$,
$t_0\in\mathbb R$ and $\Omega>0$, define
\begin{equation}\label{eq:gaussianfinite}
 J(\sigma,t_0,h,\Omega)=\sqrt{\frac2\pi}\,h\,
 \Re\int_0^\Omega e^{-h^2y^2/2-iyt_0}F(\sigma-iy)\,dy.
\end{equation}
Then
\begin{equation}\label{eq:gaussiannecessary}
 J(\sigma,t_0,h,\Omega)+
 \sqrt{\frac2\pi}F(\sigma)
       \frac{e^{-h^2\Omega^2/2}}{h\Omega}\ge0.
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{eq:LK}, $F(q)=\int_{[0,\infty)}e^{-qt}\mu(dt)$, where
$\mu=d\delta_0+t\nu(dt)$ is nonnegative and
$\int e^{-\sigma t}\mu(dt)=F(\sigma)<\infty$. Consequently
$|F(\sigma-iy)|\le F(\sigma)$. The Fourier formula for a Gaussian
and Fubini's theorem give
\[
 0\le\int_{[0,\infty)}e^{-\sigma t-(t-t_0)^2/(2h^2)}\mu(dt)
 =\sqrt{\frac2\pi}h\,
      \Re\int_0^\infty e^{-h^2y^2/2-iyt_0}F(\sigma-iy)\,dy.
\]
The absolute value of the omitted part is at most
\[
 \sqrt{\frac2\pi}h F(\sigma)\int_\Omega^\infty e^{-h^2y^2/2}dy
 \le \sqrt{\frac2\pi}F(\sigma)
          \frac{e^{-h^2\Omega^2/2}}{h\Omega}.
\]
This proves the assertion. Notice that the estimate is used under the
infinite-divisibility hypothesis; no signed inverse kernel is needed.
\end{proof}

\begin{lemma}[Rotated moments and their error bounds]\label{lem:moments}
Let $k$ be a nonnegative integer and let $s=q^\beta$, with $\Re q>0$.
Set
\[
 J_k(s)=\int_0^\infty\frac{v^{\alpha+k}e^{-sv}}
                              {D_\delta(v^\alpha)}\,dv.
\]
On a ray $v=e^{i\gamma}u$, chosen so that $\Re(se^{i\gamma})=\lambda>0$,
the same integral equals
\begin{equation}\label{eq:rotmom}
 J_k(s)=P_k(s)+e^{i(\alpha+k+1)\gamma}
 \int_0^\infty\frac{u^{\alpha+k}e^{-se^{i\gamma}u}}
                         {D_\delta(e^{i\alpha\gamma}u^\alpha)}\,du.
\end{equation}
Here $P_k=0$ if $0\le\alpha\gamma<\delta$, whereas, if
$\delta<\alpha\gamma$ and $\alpha\gamma+\delta<\pi$,
\begin{equation}\label{eq:residue}
 P_k(s)=\frac{\pi p^{k+1}e^{-ps}}{\alpha\sin\delta},
       \qquad p=e^{i\delta/\alpha}.
\end{equation}
In either case a denominator lower bound is
\begin{equation}\label{eq:Dray}
 |D_\delta(e^{i\alpha\gamma}u^\alpha)|\ge
 D_*:=\sin|\alpha\gamma-\delta|\sin(\alpha\gamma+\delta)>0.
\end{equation}
For $0<u_0<U$, the errors from deleting $(0,u_0)$ and $(U,\infty)$
from \eqref{eq:rotmom} are bounded respectively by
\begin{equation}\label{eq:radialtails}
 \frac{u_0^{\alpha+k+1}}{(\alpha+k+1)D_*},\qquad
 \frac{U^{\alpha+k}e^{-\lambda U}}
          {D_*[\lambda-(\alpha+k)/U]},
\end{equation}
provided the denominator in the latter bound is positive.
Moreover $F_\alpha(q)=\beta sJ_1(s)/(qJ_0(s))$.
\end{lemma}

\begin{proof}
Rotation of the original contour proves \eqref{eq:rotmom}. In the
second case only $v=p$ is crossed. At that pole the residue is
$p^{k+1}e^{-ps}/(2i\alpha\sin\delta)$, giving \eqref{eq:residue}.
The factorization of $D_\delta$ and
$|1-xe^{i\theta}|\ge|\sin\theta|$ for $x\ge0$ give
\eqref{eq:Dray}. The first estimate in \eqref{eq:radialtails} follows
by dropping the exponential. For the second, write $u=U+v$ and use
$u^p\le U^p e^{pv/U}$, $p=\alpha+k$. Differentiating
\eqref{eq:L} under the integral gives the stated formula for $F$.
The rotations are valid by exponential decay on the outer arcs and
the positive power at zero. Conjugation gives the corresponding
formulas in the opposite half-plane.
\end{proof}

For completeness, the parameter differentiation used in the Gaussian
cover is specified here. The ray is held fixed while $\alpha$ varies.
Writing $\theta=\pi\alpha/2$, $c=\cos\theta$, $z=v^\alpha$, one has
\begin{align}
 s_\alpha&=-s\log q/\alpha^2, & c'&=-(\pi/2)\sin\theta,\\
 (D_\delta(z))_\alpha&=2z[(z+c)\log v+c'],\label{eq:Dalpha}\\
 \partial_\alpha\!\left(\frac{v^{\alpha+k}e^{-sv}}{D_\delta(z)}\right)
 &=\frac{v^{\alpha+k}e^{-sv}}{D_\delta(z)}
     \left(\log v-s_\alpha v-
                      \frac{(D_\delta(z))_\alpha}{D_\delta(z)}\right).
 \label{eq:momentderivative}
\end{align}
If $Q_k$ denotes this differentiated moment, including the derivative
of the pole term when present, then
\begin{equation}\label{eq:Falpha}
 \partial_\alpha F_\alpha(q)=
 F_\alpha(q)\left(-\frac1\alpha-\frac{\log q}{\alpha^2}\right)
 +\frac{\beta s}{q}
      \left(\frac{Q_1}{J_0}-\frac{J_1Q_0}{J_0^2}\right).
\end{equation}
For the pole term, $p_\alpha/p=-i\pi/\alpha^2$, and
\[
 \partial_\alpha P_k=P_k\left\{-\alpha^{-1}
 +\frac\pi2\cot\delta+(k+1)\frac{p_\alpha}p
                  -p_\alpha s-ps_\alpha\right\}.
\]
These identities include the dependence of $q^{1/\alpha}$ on $\alpha$.
Endpoint errors for \eqref{eq:momentderivative} are obtained from
\begin{align}
 \int_0^{u_0}u^p(|\log u|+\gamma)du
 &=u_0^{p+1}\left(\frac{-\log u_0+\gamma}{p+1}
                               +\frac1{(p+1)^2}\right),\quad u_0\le1,\\
 |\log(ue^{i\gamma})|&\le2u\quad(u\ge100, 0\le\gamma<1),
\end{align}
together with \eqref{eq:radialtails}, applied also to the powers
$3\alpha+k$, $2\alpha+k$ and their successors. Thus these derivatives
are enclosed on whole parameter intervals.

\begin{lemma}[The finite Gaussian cover; computer-assisted]\label{lem:gaussiancover}
For every $\alpha\in[1.6671,1.7386]$, the distribution of $T_1$ is
not infinitely divisible.
\end{lemma}

\begin{proof}
All decimals in this proof are exact rationals. For the first part take
$\sigma=1$, $t_0=.163$, $h=1/3000$ and $\Omega=7/h=21000$.
The evaluated test at $\alpha=1.6671$, including the conditional tail,
is enclosed by
\[
 [-1.403446562329232\cdot10^{-7} \pm\ 3.90\cdot10^{-10}].
\]
For the same finite function \eqref{eq:gaussianfinite}, the derivative
intervals on the four successive parameter intervals of radius $.0001$
and centers $1.6672,1.6674,1.6676,1.6678$ are contained respectively in
\[
 [-.00963,-.00304],\quad[-.00962,-.00301],\quad
 [-.00960,-.00299],\quad[-.00956,-.00299].
\]
Their exact unions cover $[1.6671,1.6679]$. Each corresponding uniform
tail bound, when added to the upper bound for the finite test at the
leftmost endpoint, gives a value less than $-1.39\cdot10^{-7}$.
Monotonicity and Lemma~\ref{lem:gaussian} prove non-infinite-divisibility
throughout that interval.

For the rest of the cover, 33 overlapping parameter intervals cover
$[1.6679,1.7386]$. Each uses either direct interval evaluation or the
following mean-value enclosure. If $\alpha\in[a-r,a+r]$, $J(a)$ is
enclosed by $B$, and $|\partial_\alpha J|\le M$ on the interval, then
\begin{equation}\label{eq:gaussianaccept}
 \sup B+rM+R<0
\end{equation}
certifies the entire interval, where $R$ is the uniform conditional
Gaussian tail in \eqref{eq:gaussiannecessary}. The parameters, input
balls, integral enclosures, derivative balls and tails of every interval
are supplied with the ancillary certificate. The exact coverage test
is to sort the rational endpoints, check that the first left endpoint
is at most $1.6679$, each subsequent left endpoint is at most the
current right endpoint, and the final right endpoint is at least $1.7386$.
All 33 inequalities \eqref{eq:gaussianaccept} are strict. Two of the
intervals use more of the parameter range already covered by their
derivative enclosures:
\[
\begin{array}{c|c|c}
 a& r&\text{available derivative radius}\\\hline
 1.685422064&.0011&.0012842363\\
 1.687466206&.00095&.0013556707
\end{array}
\]
Recombining the same point, derivative and tail balls on these radii
still gives strict negativity. Their overlap makes one former bridging
interval unnecessary; two further intervals are redundant. The active
list and its exact endpoint checks are supplied in the ancillary material.
For example,
the last interval $[1.7356,1.7386]$ has its entire test enclosed by
$[-.00515256635094\pm.00126000000001]$.

Here are the evaluation details needed to reproduce the inequalities.
The radial integral is regularized by $u=x^4$, truncated at
$u_0=2^{-64}$ and $U=100$, and evaluated by complex ball integration.
The deleted parts are bounded by \eqref{eq:radialtails} and its
differentiated versions. The noncrossing ray is
$\gamma=\pi(1/\alpha_0-1/2)/2$, fixed at the interval center;
where used, the crossing ray is $\gamma=\pi/4$ with the residue
\eqref{eq:residue}. Every ray inequality is checked on the whole
parameter interval. High frequencies in the point tests use the finite
identity \eqref{eq:geometric}, with $N=18$, retaining the integrated
remainder for each moment. Direct differentiated moments are used for
the derivative intervals. The last Fourier tail is always included
with $h\Omega=7$. No sign claim is inferred from a floating-point
inverse transform.
\end{proof}

\begin{lemma}[A uniform zero certificate]\label{lem:rouche}
Let $I\subset(1,2)$ be a parameter interval. Set
$G_\alpha(s)=\sin\delta\,J_0(s)$, using analytic continuation by
\eqref{eq:rotmom}. Suppose the disk $|s-z_0|\le r$ lies in $\Re s>0$ and in its
domain of analyticity for every $\alpha\in I$. If a nonzero complex
number $D_0$ satisfies
\begin{equation}\label{eq:rouchetest}
 \sup_{\alpha\in I}|G_\alpha(z_0)|
 +r\sup_{\alpha\in I}|G_\alpha'(z_0)-D_0|
 +\frac{r^2}{2}\sup_{\substack{\alpha\in I\\|s-z_0|\le r}}
                         |G_\alpha''(s)|<r|D_0|,
\end{equation}
and $\Re(s^\alpha)>0$ on the whole disk for every $\alpha\in I$,
then $T_1$ is not infinitely divisible for any $\alpha\in I$.
\end{lemma}

\begin{proof}
Taylor's formula with integral remainder gives
\[
 G_\alpha(z_0+z)=G_\alpha(z_0)+zG_\alpha'(z_0)
     +z^2\int_0^1(1-u)G_\alpha''(z_0+uz)du.
\]
On $|z|=r$, \eqref{eq:rouchetest} compares this expression with
$D_0z$. Rouch\'e's theorem gives exactly one zero in the disk,
counting multiplicity. Since $|\arg s|<\pi/2$ and $\alpha<2$,
the condition $\Re(s^\alpha)>0$ forces
$|\alpha\arg s|<\pi/2$. The principal mapping $q=s^\alpha$ therefore
sends that zero into $\Re q>0$ and satisfies $q^{1/\alpha}=s$ there. It
gives a zero of $L_\alpha$. The representation \eqref{eq:LK} shows
that a nonnegative infinitely divisible law has a zero-free Laplace
transform throughout that half-plane, a contradiction.
\end{proof}

\begin{lemma}[The finite zero cover; computer-assisted]\label{lem:zerocover}
The distribution of $T_1$ is not infinitely divisible for
$1.7386\le\alpha\le15/8$.
\end{lemma}

\begin{proof}
The active certificate contains 102 parameter intervals and disks
to which Lemma~\ref{lem:rouche} applies. Their rational endpoints
cover the stated closed interval with overlaps. The moment enclosures
in Lemma~\ref{lem:moments}, on the ray $\gamma=\pi/4$, give
\[
 G_\alpha=\sin\delta J_0,\quad G_\alpha'=-\sin\delta J_1,
                 \quad G_\alpha''=\sin\delta J_2.
\]
For the last expression the complex rectangle
$[\Re z_0-r,\Re z_0+r]+i[\Im z_0-r,\Im z_0+r]$
contains the whole disk. Exact rational comparisons also give
$\Re z_0-r>5.63$ for every disk. The same rectangle is used to enclose
$\exp(\alpha\log s)$, and its real part has a strictly positive lower
bound in every record. All 102 inequalities \eqref{eq:rouchetest}
hold strictly. Recombining the saved balls gives a maximum left/right
ratio below $.999668$ and a minimum lower bound for $\Re(s^\alpha)$
above $.0266059$. These margins concern entire parameter intervals
and entire disks. Approximate root finding is used only to choose the
centers $z_0$; it is not an existence argument.
\end{proof}

\begin{lemma}[An analytic zero for every $15/8\le\alpha<2$]\label{lem:near2}
For $15/8\le\alpha<2$, the transform $L_\alpha$ has a zero in
$\Re q>0$. Consequently the law of $T_1$ is not infinitely divisible
throughout this interval.
\end{lemma}

\begin{proof}
Put $\varepsilon=2-\alpha$, so $0<\varepsilon\le1/8$ and
$\delta=\pi\varepsilon/2\le\pi/16$. Write
\[
 B=\kappa_\alpha\Gamma(m),\quad p=e^{-i\delta/\alpha},\quad
 d=\frac{\pi\kappa_\alpha}{\alpha\sin\delta},\quad A=dp,
 \quad \ell=\log(d/B),\quad\nu=\pi+\delta.
\]
We shall use the elementary digamma bound
\begin{equation}\label{eq:digamma-lower}
 \psi(t)>\log(t-1/2),\qquad t>1/2.
\end{equation}
Indeed, subtracting the integral representations of its two sides gives
\[
 \psi(t)-\log(t-1/2)
 =\int_0^\infty e^{-(t-1/2)u}
       \left(\frac1u-\frac1{2\sinh(u/2)}\right)du>0.
\]
Since $\sin\delta\le\pi\varepsilon/2$, integration of
\eqref{eq:digamma-lower} gives
\[
 \ell\ge-\log\bigl(\varepsilon(2-\varepsilon)\bigr)
                  +\int_{3-\varepsilon}^{3}\log(t-1/2)\,dt.
\]
This lower bound decreases with $\varepsilon\in(0,1/8]$; its derivative
is $-1/\varepsilon+1/(2-\varepsilon)+\log(5/2-\varepsilon)<0$.
Consequently
\begin{equation}\label{eq:ell19}
 \ell\ge\log\frac{64}{15}+\frac52\log\frac52
                     -\frac{19}{8}\log\frac{19}{8}-\frac18>\frac{31}{20}.
\end{equation}
The duplication formula and \eqref{eq:digamma-lower} also give
\[
 \frac{d}{d\alpha}\log\frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}
 =\log4+\psi(\alpha+1/2)>\log(4\alpha)\ge\log\frac{15}{2}>2.
\]
Here $e<87/32$ and $(87/32)^2<15/2$. Thus
\begin{equation}\label{eq:gamma19}
 \frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}\le12e^{-2\varepsilon}.
\end{equation}

We first locate an exact zero of the exponential model. For $x>m$ let
$y(x)>0$ be the unique solution of
\[
 y-m\arctan(y/x)=\nu.
\]
Existence and uniqueness follow because the left side increases from
zero to infinity, with derivative at least $1-m/x>0$. Implicit
differentiation gives
\[
 y'(x)=-\frac{my}{x^2+y^2-mx},\qquad
 \frac{d}{dx}\left(x-\frac m2\log(x^2+y^2)\right)
 =\frac{(x-m)^2+y^2}{x^2+y^2-mx}>0.
\]
The expression differentiated on the left is negative as $x\downarrow m$
and tends to infinity as $x\to\infty$. By \eqref{eq:ell19}, there is
therefore exactly one $x>m$ for which
\begin{equation}\label{eq:exact-model19}
 w_0=x+iy(x),\qquad w_0-m\log w_0=\ell+i\nu.
\end{equation}

For useful uniform bounds, put $m_0=23/8$ and let $h(x)$ solve
$h-m_0\arctan(h/x)=\pi$, $x>m_0$. Then $y(x)\ge h(x)$.
At $x_0=79/10$ the elementary comparisons
\begin{equation}\label{eq:model-scalars19}
 \frac{14}{3}-\frac{23}{8}\arctan\frac{140}{237}<\pi,
 \qquad
 \frac{79}{10}-\frac{23}{16}\log\frac{75769}{900}<\frac{31}{20}
\end{equation}
show that $h(x_0)>14/3$ and that the real equation in
\eqref{eq:exact-model19} cannot hold for $x\le x_0$. Indeed its left
side is at most $x-m_0\log\sqrt{x^2+h(x)^2}$, an increasing function
which is below $31/20$ at $x_0$. Also $\sqrt{x^2+h(x)^2}$ increases with
$x$. Finally, $\arctan(51/79)<7/12$ and
$\nu\le17\pi/16<67/20=51/10-3(7/12)$ imply $y(x)<51/10$.
We have obtained
\begin{equation}\label{eq:model-location19}
 x>\frac{79}{10},\qquad |w_0|>\frac{55}{6},\qquad \pi<y<\frac{51}{10}.
\end{equation}

Put $r_0=13/10$ and consider the closed disk $|w-w_0|\le r_0$.
Write $U=\Re w$, $V=\Im w$ and $\lambda=7/8$. The center satisfies
$\lambda x-y>29/16$, while
\[
 r_0\sqrt{1+\lambda^2}<\frac{29}{16}.
\]
Cauchy--Schwarz therefore gives $\lambda U-V>0$ throughout the disk.
Also $V>\pi-r_0>0$. Set
\[
 \theta=\frac{\pi}{2\alpha},\qquad
 \gamma=\theta-\frac{\delta}{\alpha}
        =\frac{\pi(\alpha-1)}{2\alpha}.
\]
Since
\[
 0<\arg w<\arctan\frac78<\frac{73}{100}<\frac{7\pi}{30}\le\gamma,
\]
the point $s=w/p$ satisfies $0<\arg s<\theta$, and
$q=s^\alpha$ belongs to $\Re q>0$ with $q^{1/\alpha}=s$.
All powers and logarithms hence have their principal values.
Moreover, if $a=\pi/4-\gamma=\delta/(2\alpha)\le\pi/60$, then
\[
 \frac{U-V}{U+V}>\frac1{15}>\frac1{30}>\tan(a/2).
\]
Expanding $\cos(\pi/4-a)$ and $\sin(\pi/4-a)$ proves
\begin{equation}\label{eq:decay19}
 \Re(se^{-i\theta})=\Re(we^{-i\gamma})\ge\frac{U+V}{\sqrt2}.
\end{equation}

Rotate the $r$ contour in \eqref{eq:L} downward through angle $\theta$.
The sole crossed pole is $p$, and the clockwise residue contribution is
$Ae^{-ps}$. The small arc vanishes; the large arc decays because
$0<\arg s<\theta<\pi/2$. Consequently
\begin{align}
 \mathcal G(s):=L_\alpha(s^\alpha)&=Ae^{-ps}+J(s),\label{eq:twoterm}\\
 J(s)&=\kappa_\alpha e^{-im\theta}\int_0^\infty
       \frac{e^{-se^{-i\theta}u}u^\alpha}{D_\delta(-iu^\alpha)}\,du.
\end{align}
For $z=-ib$, $b\ge0$, direct squaring gives
\[
 4|D_\delta(-ib)|^2-|2\cos\delta+ib|^2
 =4\sin^2\delta+(16\cos^2\delta-9)b^2+4b^4\ge0.
\]
Here $\cos\delta>3/4$. Applying
$D_\delta(z)^{-1}=1+z(2\cos\delta-z)/D_\delta(z)$ therefore gives
\begin{equation}\label{eq:Jnear2}
 J(s)=Bs^{-m}(1+R_0(s)),\qquad
 |R_0(w/p)|\le
 2\frac{\Gamma(2\alpha+1)}{\Gamma(\alpha+1)}
 \frac{|w|^m}{\Re(we^{-i\gamma})^{2\alpha+1}}.
\end{equation}
The remainder is analytic on a neighborhood of the disk.

To bound \eqref{eq:Jnear2}, define
\[
 Q_\alpha(U,V)=\frac{(U^2+V^2)^{(\alpha+1)/2}}{(U+V)^{2\alpha+1}}.
\]
This function decreases in each coordinate on $(0,\infty)^2$.
It also decreases along $(U,V)=(x-r_0,h(x)-r_0)$ for $x\ge79/10$.
For completeness,
\[
 |h'(x)|\le\frac{m_0}{2\sqrt{x(x-m_0)}}<\frac14,
 \qquad
 |\partial_U\log Q_\alpha|>\frac14|\partial_V\log Q_\alpha|.
\]
The second inequality follows from positivity of
$(2\alpha-1)U^2-3(\alpha+1)UV+(7\alpha+4)V^2$; its discriminant is
$-47\alpha^2+14\alpha+25<0$. Both coordinate derivatives are negative,
as follows similarly from $1-2\alpha-7\alpha^2<0$. Hence, with
$u_0=33/5$ and $v_0=101/30$,
\[
 Q_\alpha(U,V)\le Q_\alpha(x-r_0,y-r_0)
 \le Q_\alpha(x-r_0,h(x)-r_0)\le Q_\alpha(u_0,v_0).
\]
Put $H=(u_0+v_0)^2/\sqrt{u_0^2+v_0^2}$. Direct rational comparisons,
using $e>163/60$, give
\[
 H<\frac{59}{4}<2e^2,\qquad
 24\sqrt{32}\,Q_2(u_0,v_0)<\frac9{16}.
\]
Combining \eqref{eq:gamma19}, \eqref{eq:decay19} and
\eqref{eq:Jnear2}, and writing $R(w)=R_0(w/p)$, we conclude throughout the disk that
\begin{equation}\label{eq:near2-remainder}
 |R(w)|
 \le24\sqrt{32}\,Q_2(u_0,v_0)
             \left(\frac{H}{2e^2}\right)^\varepsilon<\frac9{16}.
\end{equation}

Now let $\Psi(w)=w-m\log w-\ell-i\nu$. Equation
\eqref{eq:exact-model19} implies, on $|w-w_0|=r_0$,
\begin{equation}\label{eq:model-zero-bound}
 |\Psi(w)-(w-w_0)|
 \le-m\log(1-r_0|w_0|^{-1})<3\log\frac{275}{236}<\frac{47}{100}.
\end{equation}
Since $|R|<9/16$, the function
$\Xi(w)=\Psi(w)+\log(1+R(w))$ is analytic on a neighborhood of the disk.
Furthermore,
\[
 |\log(1+R(w))|<\log\frac{16}{7}<\frac{83}{100}.
\]
Thus $|\Xi(w)-(w-w_0)|<13/10=r_0$ on the boundary. A single application of
Rouch\'e's theorem gives a zero $w_*$ of $\Xi$ inside it. Finally,
$Bp^m/A=e^{-\ell-i\delta}$ yields the exact identity
\[
 \frac{\mathcal G(w/p)}{Bp^mw^{-m}}=1+R(w)-e^{-\Psi(w)}.
\]
At $w_*$ this is zero, so $L_\alpha$ has a zero at
$q=(w_*/p)^\alpha\in\{\Re q>0\}$. Equation~\eqref{eq:LK} excludes
infinite divisibility.

All fixed scalar comparisons in this proof reduce to rational arithmetic.
One may use $157/50<\pi<22/7$, $163/60<e<87/32$, and the alternating
series for $\arctan z$. The logarithmic comparisons in \eqref{eq:ell19}
and \eqref{eq:model-scalars19} follow respectively from
\[
 \left(\frac{64}{15}\right)^{40}
 \left(\frac52\right)^{100}
 \left(\frac8{19}\right)^{95}>\left(\frac{87}{32}\right)^{67},
 \qquad
 \left(\frac{75769}{900}\right)^{115}
                       >\left(\frac{87}{32}\right)^{508}.
\]
The remainder bound is rational after squaring. The two last logarithmic
bounds follow respectively from
$-\log(1-t)\le t+t^2/(2(1-t))$ at $t=39/275$ and the first five terms of
$e^{83/100}$. No interval enclosure of a transform or parameter subdivision
is used in this lemma.
\end{proof}

\begin{lemma}[Completion of the negative side]\label{lem:negativecomplete}
For every $\alpha_*<\alpha<2$, the distribution of $T_1$ is not
infinitely divisible.
\end{lemma}

\begin{proof}
The contact argument gives a negative minimum for
$\alpha_*<\alpha\le1.667066$. At $t_0=.1628$ its point and
derivative bounds imply
\[
 K_{1.667066}(t_0)<13\cdot10^{-9}-9.059\cdot10^{-8}<0.
\]
On $\alpha\in[1.667066,1.6671]$ the separate uniform derivative
enclosure is
\[
 \partial_\alpha K_\alpha(t_0)
 \in[-9.06079735696894\pm2.78000000000001]\subset(-\infty,-6).
\]
It is evaluated by the same regularized inversion and differentiated
finite-tail estimates as the contact enclosure. Canonical identification
is supplied by the zero-free strip for $1.3\le\alpha\le1.6671$.
Thus a negative value persists through $1.6671$, and
Lemma~\ref{lem:ID} applies. Lemmas~\ref{lem:gaussiancover}
and~\ref{lem:zerocover} then cover $[1.6671,15/8]$.
Lemma~\ref{lem:near2} covers $[15/8,2)$ analytically, completing
the range without an extrapolation from finitely many zero disks.
\end{proof}

\section{Proof of the main theorem}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Lemma~\ref{lem:source-resolvent} reduces the assertion to $a=1$.
Lemma~\ref{lem:contact} constructs the unique contact in the stated
rectangle and gives \eqref{eq:alphastar}. Lemma~\ref{lem:positivecomplete}
gives $K_\alpha>0$ on $(0,\infty)$ for $1<\alpha<\alpha_*$,
and $K_{\alpha_*}\ge0$ with exactly one zero. By Lemma~\ref{lem:ID}
these laws are infinitely divisible, with the stated L\'evy measure
and drift. Lemma~\ref{lem:negativecomplete} excludes every
$\alpha\in(\alpha_*,2)$. The positive and negative parameter intervals
meet at the contact with no unclassified parameter. This also proves
the uniqueness of the threshold in \eqref{eq:classification}.
\end{proof}

\begin{corollary}\label{cor:notSD}
For every $1<\alpha<2$ and every $a\ne0$, the law of $T_a$ is not
self-decomposable.
\end{corollary}
\begin{proof}
This follows from Remark~\ref{rem:notSD} and scaling.
\end{proof}

The definition \eqref{eq:contactdefinition} specifies the critical
parameter independently of its decimal approximation. The argument
uses uniqueness of the contact only in its stated rectangle. It does
not assert that the kernel has no other critical points for larger
values of $\alpha$; a single negative value, or either necessary-condition
violation used above, is sufficient there.

\appendix

\section{Differentiated remainders and finite inversion}\label{app:inversion}
This appendix specifies the remainder bounds used in the critical enclosure.
A dot denotes differentiation with respect to the real parameter $\alpha$,
with $q$ fixed. The notation is that of Lemma~\ref{lem:expansion}.
\begin{lemma}[Differentiated finite expansion]\label{lem:derivexpansion}
Fix $c,Y>0$ and $N\ge2$, and suppose
$\arctan(Y/c)>\alpha\pi/4$ and $\lambda_Y>0$.
For $q=c\pm iy$, $y\ge Y$, the remainders in \eqref{eq:AH} satisfy
\[
 |\dot E_A(q)|\le C_A'|q|^{-N},\qquad
 |\dot E_H(q)|\le C_H'|q|^{-N}.
\]
Here primes on error constants denote bounds for differentiated remainders,
not derivatives of previously chosen upper bounds. Explicit choices are
\begin{align*}
 G_0&=\frac{\max(|c_N|,|c_{N-1}|)}\eta,&
 G_1&=\frac{\max(|\dot c_N|,|\dot c_{N-1}|)}\eta,\\
 D_0&=\frac{\pi\sin\delta}{4\eta},&
 D_1&=\frac{2(1+|\cos\delta|)}\eta,\\
 B_0&=G_0(|\psi(m)|+D_0)+G_1,&
 B_1&=G_0(N+1)+\frac{|c_{N-1}|}{4\eta}+G_0D_1,
\end{align*}
where $\psi=\Gamma'/\Gamma$. For $v=0,1$, set
\[
 p_v=m+N\alpha+v-1,\quad M_v=\frac{\Gamma(p_v+1)}{\mu^{p_v+1}},\quad
 W_v=\frac{\Gamma(p_v+2)}{\mu^{p_v+2}}+
       \frac{\Gamma(p_v)}{\mu^{p_v}}+\frac\pi4M_v.
\]
Then $C_A'=(B_0M_0+B_1W_0)/\Gamma(m)$ and
$C_H'=(B_0M_1+B_1W_1)/\Gamma(m)$.
The coefficient derivatives are computed without division by $c_n$:
\begin{align}
 \dot c_n&=-\frac\pi{2\sin\delta}
       \{(n+1)\cos((n+1)\delta)-c_n\cos\delta\},\label{eq:dotcn}\\
 \dot A_n&=\frac{\Gamma(m+n\alpha)}{\Gamma(m)}
       \{\dot c_n+c_n[(n+1)\psi(m+n\alpha)-\psi(m)]\},\\
 \dot d_0&=-\beta^2,\qquad
 \dot d_n=n\dot A_n-\sum_{j=1}^{n-1}
                  (\dot A_jd_{n-j}+A_j\dot d_{n-j}).
\end{align}
For $N\le n\le2N-2$ put
$V_n=\sum_{0\le j<N,\,0\le n-j<N}A_jd_{n-j}$, and define
\begin{align*}
 U_0&=\sum_{n=N}^{2N-2}|V_n|Y^{N-n},&
 U_1&=\sum_{n=N}^{2N-2}|\dot V_n|Y^{N-n},\\
 P_0&=\sum_{n=0}^{N-1}|d_n|Y^{-n},&
 P_1&=\sum_{n=0}^{N-1}|\dot d_n|Y^{-n},\\
 Z_1&=\sum_{n=0}^{N-1}|\dot A_n|Y^{-n}+C_A'Y^{-N},\\
 C_R&=U_0+\beta C_H+C_AP_0,&
 C_R'&=U_1+\beta^2C_H+\beta C_H'+C_A'P_0+C_AP_1.
\end{align*}
An explicit differentiated quotient bound is
\begin{equation}\label{eq:derivative-tail}
 \left|\dot F_\alpha(q)-\sum_{n=0}^{N-1}\dot d_nq^{-n-1}\right|
 \le C_F'|q|^{-N-1},\qquad
 C_F'=\frac{C_R'}{\lambda_Y}+\frac{C_RZ_1}{\lambda_Y^2}.
\end{equation}
\end{lemma}
\begin{proof}
For $q=c-iy$, use the fixed ray $w=ue^{-i\pi/4}$ in the normalized
integrals for $A,H$. Then $z=w^\alpha/q$ has $0<\arg z<\delta/2$;
the rotation is pole-free and \eqref{eq:sectorD} applies. On this fixed
ray $\dot z=z\log w$, and hence
\[
 \left|\frac{\partial_\alpha D_\delta(z)}{D_\delta(z)}\right|
 \le D_0+D_1|\log w|.
\]
Differentiate the exact remainder
$w^\alpha z^N(c_N-c_{N-1}z)/(\Gamma(m)D_\delta(z))$.
Its modulus is bounded by
$u^{\alpha+N\alpha}|q|^{-N}(B_0+B_1|\log w|)/\Gamma(m)$.
Use $|\log w|\le|\log u|+\pi/4\le u+u^{-1}+\pi/4$ and
$|e^{-w}|\le e^{-\mu u}$ to obtain the two gamma bounds. Conjugation
handles the upper ray. In particular no $\log q$ is introduced by this
normalized, fixed-ray differentiation; the factor $\kappa_\alpha$
cancels from $F_\alpha=\beta H/(qA)$.
If $S=\sum_{n<N}d_nq^{-n}$, the exact residual is
$\beta H-AS=-\sum_{n=N}^{2N-2}V_nq^{-n}+\beta E_H-E_AS$.
Its bounds are $C_R|q|^{-N}$ and $C_R'|q|^{-N}$ after differentiation.
Use $|A|\ge\lambda_Y$, $|\dot A|\le Z_1$ and differentiate its quotient
by $qA$. This proves \eqref{eq:derivative-tail}. All estimates remain
valid on closed real parameter intervals by outward interval evaluation.
\end{proof}
\begin{lemma}[Filtered finite Bromwich formula]\label{lem:filteredinversion}
Suppose $F_\alpha$ is holomorphic on $\Re q\ge c$ and $\lambda_Y>0$.
For the dotted formula also impose the ray condition of
Lemma~\ref{lem:derivexpansion}. Let $0\le j<p<N$ be integers, $h>0$,
and use $R_p,r_p,b_k$ from \eqref{eq:model}. Set $q_y=c+iy$ and
\[
 I_k(t;x,Y)=\frac{(-t)^{k-1}}\pi
       \Im\Gamma(1-k,-t(x+iY)),\qquad k\ge1,\quad t>0,
\]
where the upper incomplete gamma function has its principal continuation.
Then
\begin{align}
 K_\alpha^{(j)}(t)&=r_p^{(j)}(t)+\frac{e^{ct}}\pi\Re
   \int_0^Y e^{iyt}q_y^j(F_\alpha(q_y)-R_p(q_y))\,dy
   +T_j(t)+\varepsilon_j(t),\label{eq:filteredfinite}\\
 T_j(t)&=\sum_{n=j}^{N-1}d_n I_{n+1-j}(t;c,Y)
 -e^{-ht}\sum_{k=1}^p\sum_{l=0}^{\min(j,k-1)}
   b_k\binom jl(-h)^{j-l}I_{k-l}(t;c+h,Y),\\
 |\varepsilon_j(t)|&\le\frac{e^{ct}C_F}{\pi(N-j)Y^{N-j}},\qquad
 r_p^{(j)}(t)=e^{-ht}\sum_{k=1}^p\sum_{l=0}^{\min(j,k-1)}
  \binom jl(-h)^{j-l}\frac{b_kt^{k-1-l}}{(k-1-l)!}.
\end{align}
The alpha derivative is given by the same formulas with
$F_\alpha,d_n,b_k,r_p$ replaced by their dotted versions and $C_F$ by $C_F'$.
\end{lemma}
\begin{proof}
The coefficients of $R_p$ agree with those of $F_\alpha$ through degree
$p$ at infinity. Thus $q^j(F_\alpha-R_p)$ is absolutely integrable on
the vertical line. Split its Bromwich integral at $Y$ and use
Lemma~\ref{lem:expansion}. The identity
$I_k=\pi^{-1}\Re\int_Y^\infty e^{(x+iy)t}(x+iy)^{-k}\,dy$
follows by substitution into the incomplete gamma integral; for $k=1$
the integral is interpreted as its ordinary oscillatory limit.
In $q^j(\sum_{n<N}d_nq^{-n-1}-R_p)$ the polynomial parts cancel
because $p>j$. The remaining proper fractions give $T_j$ exactly.
Finally $|q_y|\ge y$ gives the displayed absolute remainder. The
differentiated statement follows from Lemma~\ref{lem:derivexpansion}.
\end{proof}
For whole time intervals one may evaluate $T_j$ at their midpoint $t_*$.
Here is an explicit continuous-time error, when $Y>h$ and $p\ge j+2$.
Define
\[
 a_n^{R}=\sum_{k=1}^{\min(p,n+1)}b_k(-h)^{n+1-k}\binom n{k-1},\qquad
 Q_N=\sum_{k=1}^p\frac{|b_k|\binom N{k-1}h^{N+1-k}}{(1-h/Y)^k}.
\]
On $[t_-,t_+]$ with $t_->0$ and radius $r$, the additional error is at most
\begin{equation}\label{eq:filtertimevariation}
 \frac{re^{ct_+}}\pi\left\{
 \sum_{n=p}^{N-1}\frac{|d_n-a_n^R|}{(n-j-1)Y^{n-j-1}}
 +\frac{Q_N}{(N-j-1)Y^{N-j-1}}\right\}.
\end{equation}
Indeed the binomial tail of $R_p$ is bounded by $Q_N|q|^{-N-1}$;
differentiate the combined finite tail once in $t$ and integrate its
absolute value. The same bound with dotted coefficients handles its
alpha derivative. The filtered convexity enclosures use
$c=h=20$, $Y=1000$, $N=18$, $p=4$ and $j=2$.
The local contact records also use the following unshifted version,
for $j=0,1,2$:
\begin{align*}
 K_\alpha^{(j)}(t)={}&d_j+\frac{e^{ct}}\pi\Re\int_0^Y e^{iyt}
 \left(q_y^jF_\alpha(q_y)-\sum_{n=0}^j d_nq_y^{j-n-1}\right)dy\\
 &+\sum_{n=j+1}^{N-1}d_n I_{n+1-j}(t;c,Y)+\varepsilon_j(t).
\end{align*}
The error $\varepsilon_j$ has the same bound as in
\eqref{eq:filteredfinite}. The polynomial terms that are subtracted
have no ordinary inverse on $t>0$, and the term $d_j/q$ contributes
the displayed constant $d_j$. At a midpoint $t_*$ in $[t_-,t_+]$,
the extra error for the finite tail is at most
\[
 \frac{re^{ct_+}}\pi\left\{
 \frac{2|d_{j+1}|}{t_-Y}
 +\sum_{n=j+2}^{N-1}\frac{|d_n|}{(n-j-1)Y^{n-j-1}}\right\}.
\]
Indeed the derivative of the first tail term contains the oscillatory
integral of $e^{iyt}/(c+iy)$; integration by parts bounds it by
$2/(tY)$. All remaining terms are absolutely integrable.
The first alpha derivative uses the $j=0$ formula with dotted
coefficients, dotted $F_\alpha$, and $C_F'$ in its absolute remainder.
These formulas require the independent zero-free-strip argument to
identify the chosen line with the canonical kernel.

\section{Computational certificates and reproduction}\label{sec:certificates}

The numerical part of the proof consists of enclosures of entire closed
parameter--time rectangles, together with analytic estimates on the two
unbounded time ranges.  No inference from a grid of point evaluations is
used.  The ancillary material contains the actual interval outputs, the
programs that produced them, the independent readers, and the analytic
error notes. The current entries are \texttt{PROOF\_INDEX.txt},
\texttt{select\_positive\_records.json} and
\texttt{reduced\_active\_cover.json}. The original complete snapshot
is preserved; these lists specify the smaller subset required here.

\subsection{Finite certificates}
Put $a_-=1.66706598$ and $a_+=1.667066$.  The finite positive-side
certificates are summarized below.  Every displayed bound has been rounded
in the conservative direction.  Bounds in this table apply only to the
listed compact time domains; the small- and large-time arguments in the
proof are indispensable.
\begin{center}
\small
\begin{tabular}{llll}
\hline
Parameter range & Time domain & Rectangles & Certified bound\\
\hline
$[1,6/5]$ & $[1/4,33/5]$ & $6360$ & $K>2\cdot10^{-5}$\\
$[6/5,13/10]$ & $[1/5,6]$ & $5800$ & $K>7\cdot10^{-5}$\\
$[13/10,3/2]$ & $[9/100,6]$ & $11840$ & $K>0.063$\\
$[3/2,33/20]$ & $[9/200,6]$ & $8941$ & $K>0.016$\\
$[33/20,a_-]$ & $[.15,.17]$ & $980$ & $\partial_\alpha K<-1$\\
$[33/20,a_-]$ & $[.045,.15]\cup[.17,6]$ & $3096$ & $K>2\cdot10^{-4}$\\
$[a_-,a_+]$ & $[.045,.145]\cup[.175,6]$ & $891$ & $K>10^{-5}$\\
$[a_-,a_+]$ & $[.145,.175]$ & $30$ & $\partial_t^2K>242$\\
\hline
\end{tabular}
\end{center}
The endpoint $\alpha=1$ in the first row belongs to the auxiliary normalized
analytic family.  It is not interpreted as a point-hitting probability law.
There are $37\,938$ rectangles in the table. The fifth row propagates an
already established positive right endpoint to smaller $\alpha$; its
negative derivative is not itself a positivity statement.

The time-end estimates require $16$ finite-expansion parameter boxes and
$17$ real special-function bounds, without numerical transform integration.
They replace $11\,164$ of the $49\,102$ compact rectangles in the complete
archive. All $115$ rectangles crossing a new cutoff are retained and
restricted to the required time range; none is discarded by its midpoint.

For the remaining negative side, the saved-data readers verify four
Gaussian monotonicity segments covering $[1.6671,1.6679]$, thirty-three
Gaussian certificates covering $[1.6679,1.7386]$, and $102$ Rouch\'e discs
covering $[1.7386,15/8]$. The local negative
minimum and its fixed-time derivative bridge connect these intervals to
$\alpha_\ast$.  All joins are comparisons of exact rational endpoints.
The entire interval $[15/8,2)$ is supplied by Lemma~\ref{lem:near2}.
Thus $244$ of the $346$ archived zero discs and three of the $36$
archived Gaussian intervals are unnecessary. Two Gaussian radii are
enlarged within their already validated derivative domains as described
in Lemma~\ref{lem:gaussiancover}; their strict inequalities are recomputed.

\subsection{What an enclosure certifies}
Most real balls are recorded as a binary midpoint and a nonnegative binary
radius. Their endpoints are therefore exact rational numbers. Early
$[6/5,13/10]$ records instead preserve outward decimal enclosures;
the active reader checks their displayed lower endpoints directly. The readers
check strict signs at these endpoints and verify that the input balls
contain their claimed rational rectangles.  They also check that the
rectangles meet without gaps and that a saved finite approximation is
expanded by every stated error term before the sign is tested.
In the early slabs where input balls were not saved, input containment
instead follows from the exact rational construction in the archived
producer. The six archived display-rounding repair records concern only
discarded large-time rectangles; every active saved kernel enclosure
has a strictly positive lower endpoint without those repairs.

The integration programs use Arb/FLINT ball arithmetic
\cite{Johansson2016Arb} and explicit
analytic bounds for the omitted integration ranges.  For Fourier inversion,
the error includes both the entire omitted frequency sum and the
periodization error.  Time and parameter variations are enclosed over the
whole rectangle.  Independent finite-sum readers recompute from the
persisted transform balls, preserving their original radii and parameter
midpoints.  Increasing the precision of that last sum does not improve the
accuracy of the earlier transform integrations.  The Rouch\'e readers
recompute the strict boundary inequality from the saved function and
first- and second-derivative balls, and check that the entire certified
zero disc maps into the right half-plane under the principal power.

The analytic identification of these inversions with the canonical kernel,
and the analytic bounds on the unbounded time intervals, are proved in the
preceding sections.  They are separate obligations from the finite
rectangle checks.  Likewise, a digest establishes which source and data
were used; digest agreement alone does not establish an inequality.

\subsection{Portable reproduction}
The unchanged ancillary snapshot comprises $1262$ files and $3450$ explicit
keyed source/data digest bindings. The current scalar producers and active
subset readers are supplied beside it. The launcher
\texttt{reproduce\_reduced.py} recomputes the time-end scalar bounds,
checks the selected saved signs and exact interval geometry, and replays
the reduced negative cover. It also checks the fixed rational comparisons
in Lemma~\ref{lem:near2}. It does not repeat original transform integrations.

The original relative-path launcher, \texttt{reproduce.py}, retains three
verification levels for the complete archived computations:
\begin{enumerate}
\item \texttt{verify} checks archive integrity and the declared reference
closure, without numerical computation.
\item \texttt{saved} reads the actual saved interval balls, recombines the
stated budgets, verifies the finite signs and exact interval joins, and
rechecks the local contact scalar inequalities.  It does not repeat the
original transform integrations.
\item \texttt{finite-sums} independently resums the critical Fourier
certificates at the verification precision, including their analytic
alias and frequency-tail bounds, using the saved integral balls.
\end{enumerate}
The separate \texttt{script} mode runs an original producer and can perform
fresh validated integration.  The README gives example commands and the
batch specifications needed for a full reproduction.  All replay writes
are isolated in a temporary copy; the snapshot is unchanged.

The recorded production environment is Python~3.12.14,
\texttt{python-flint}~0.9.0, and FLINT~3.6.0.  The software version used by an
actual replay is recorded in its output.  Numerical records and program
sources are preserved exactly, apart from relative-path normalization of
two metadata strings in each of $33$ JSON files and one historical
executable path.  A change ledger records original and public digests and
verifies that no numerical field was modified.  The public package requires
no access to the original machine or its directory structure.

These certificates constitute a computer-assisted proof with explicit
analytic error bounds.  They are not a formal proof-kernel verification,
and the internal independent reviews should not be confused with external
peer review.  The distinction between integrity checks, saved-data checks,
finite-sum recomputation, and fresh integration is retained in every replay
log.

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