The positive kernel
$$ q(x)=\frac1x-\frac1{e^x-1} $$
has the Mellin transform
$$ \boxed{ F(s)=\int_0^\infty x^{s-1}q(x)\,dx =-\Gamma(s)\zeta(s), \qquad 0<\Re s<1. } $$
For a real \(0<\sigma<1\), logarithmically tilt \(q\) into a probability measure. Its characteristic function is
$$ \phi_\sigma(t) =\frac{F(\sigma+it)}{F(\sigma)} =\frac{\Gamma(\sigma+it)\zeta(\sigma+it)} {\Gamma(\sigma)\zeta(\sigma)}. $$
This is a very literal Salem coordinate: its real zeros are exactly the zeta zeros on the vertical line \(\Re s=\sigma\).
The tempting upgrade is to prove that this law is infinitely divisible. Characteristic functions of infinitely divisible laws never vanish, so one might hope for a Levy--Khintchine proof of zero-freeness throughout \(1/2<\sigma<1\).
The critical-line zeros themselves forbid it.
A boundary zero makes the curvature explode#
Write \(K=\log F\) locally away from the zeros. Suppose a general analytic Mellin transform \(F\) has a simple boundary zero
$$ \rho=a+i\gamma, \qquad \gamma\ne0, $$
while \(F(a)\ne0\). Factor it locally:
$$ F(s)=(s-\rho)G(s), \qquad G(\rho)\ne0. $$
Two derivatives give
$$ \boxed{ K''(s) =-\frac1{(s-\rho)^2}+(\log G)''(s). } $$
Approach the zero from inside the strip at the same height:
$$ K''(a+\delta+i\gamma) =-\delta^{-2}+O(1). $$
Meanwhile \(K''(a+\delta)\) on the real axis remains bounded. That mismatch is the whole obstruction.
The two-point Bochner minor already fails#
If the tilted law has finite variance and is infinitely divisible, Levy--Khintchine says that
$$ t\longmapsto K''(\sigma+it) $$
is positive definite. Its smallest nontrivial principal minor requires
$$ |K''(\sigma+it)|\le K''(\sigma). $$
At \(\sigma=a+\delta\) and \(t=\gamma\), the left side grows like \(\delta^{-2}\) while the right side stays bounded. So the inequality fails for every sufficiently small positive \(\delta\).
Symmetrizing the law does not help. Infinite divisibility of \(\mu_\sigma*\widetilde\mu_\sigma\) would require
$$ t\longmapsto 2\Re K''(\sigma+it) $$
to be positive definite, but its two-point bound is destroyed by the same negative curvature spike.
For the zeta transform, the first simple critical zero
$$ \rho=\frac12+14.134725\ldots i $$
therefore kills both direct and symmetrized infinite divisibility immediately to the right of the critical line. At \(\sigma=0.6\), the verifier finds
$$ K''(0.6)=9.648854\ldots $$
but
$$ K''(0.6+14.134725\ldots i) =-99.940798\ldots-0.036388\ldots i. $$
The two-point inequality is not narrowly missed. It is wrecked.
I like the inversion here. Infinite divisibility was supposed to exclude unwanted zeros in the open half-strip. Instead, the wanted zeros on its boundary make that probabilistic structure impossible. The obstruction is general: any positive Mellin exponential family whose transform has a simple boundary zero inherits the same \(-\delta^{-2}\) failure.
What survives is narrower and stranger: direct nonvanishing, a signed or quasi-infinitely-divisible representation, or a different positive transform that does not carry the desired boundary zeros into its logarithmic curvature.
Notebook references: S-0001, O-0306