REVIEW
Infinitely Many Off-Critical-Line Zeros of the Tempered Xi Function: A Disproof of Yang's Conjecture
Number Theory · Submitted 24 Sep 2026 · 5 references
Pith's verdict: A short, correct disproof of Yang's conjecture on the tempered xi function — elementary, self-contained, and publishable as a note.
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Referee report
The paper
The paper claims
A sine-based cousin of the Riemann xi function, obtained by replacing the hyperbolic cosine with a hyperbolic sine in a classical integral representation, is proved to have infinitely many zeros off the critical line, refuting Yang's…
The reading
What carries the argument
The central object is the Fourier sine transform $S(z) = \int_0^\infty K(t)\sin(zt)\,dt$ obtained from the tempered xi function by the change of variables $x = e^{2t}$, where $K(t) = 8e^{2t}M(e^{2t})$ is a positive, smooth, doubly exponentially decaying weight. Two mechanisms carry the proof: (i) the endpoint term in the two-integration-by-parts identity $S(y) = K(0)/y - y^{-2}\int_0^\infty K''(t)\sin(yt)\,dt$, which with $K(0) > 0$ and $K'' \in L^1$ fixes the sign of $S$ on the far real axis and confines all real zeros to a compact interval; and (ii) Hadamard factorization for entire functions of order at most one, which forbids a polynomial-times-exponential shape from decaying along both ends of the real axis and therefore forces $S$ to have infinitely many zeros in the plane. The contrast with the classical xi function is that its cosine transform has no analogous nonzero endpoint term.
Extended reading notes
Core claim
In the paper's own terms: with $x = e^{2t}$, $\widehat\xi(\tfrac12 + iz) = iS(z)$, where $S(z) = \int_0^\infty K(t)\sin(zt)\,dt$ and $K$ is positive, smooth, doubly exponentially decaying, with $K(0) > 0$. Two integrations by parts give the endpoint asymptotic $S(y) = K(0)/y + O(y^{-2})$, so $S$ keeps one sign on each end of the real axis and has only finitely many real zeros (Corollary 1). Because $S$ is an odd entire function of order at most one, if it had only finitely many zeros overall, Hadamard factorization would force $S(z) = P(z)e^{az+b}$, contradicting $S(y) \to 0$ as $y \to \pm\infty$ (Lemma 4). Hence $S$ has infinitely many nonreal zeros, and $\widehat\xi$ has infinitely many zeros off the critical line — Theorem 1, which refutes Yang's Conjecture 1.
Load-bearing premise
The load-bearing premise is that the sine transform $S$ is an odd entire function of order at most one whose real-axis values obey $S(y) = K(0)/y + O(y^{-2})$ with $K(0) > 0$, so that $S$ keeps a fixed sign out on the real axis; if the endpoint term were zero or of the wrong sign, the finitely-many-real-zeros step would fail and the contradiction would collapse.
Editorial extensions
If this is right
- Yang's Conjecture 1 is false: the tempered sine xi function $\widehat\xi$ has infinitely many distinct zeros with $\Re s \neq 1/2$ and only finitely many on the critical line, with a simple zero at $s = 1/2$.
- The disproof is unconditional — it uses neither the Riemann Hypothesis nor numerical zero finding — so the real-rootedness question for this sine-tempered family is settled in the negative on purely analytic grounds.
- The mechanism is specific to the sine transform: the nonzero endpoint term $K(0)/y$ has no analogue in the cosine transform of the classical xi function, and the argument does not transfer to the Riemann Hypothesis.
- From any off-line zero, the symmetries $\widehat\xi(s) = -\widehat\xi(1-s)$ and $\widehat\xi(s) = \widehat\xi(\bar s)$ generate the corresponding reflected and conjugate zeros.
Reading between the lines
- The same contradiction — finitely many real zeros alongside infinitely many total zeros — applies to any sine transform $\int_0^\infty K(t)\sin(zt)\,dt$ with a smooth, rapidly decaying kernel and $K(0) \neq 0$, so Yang-type conjectures for other sine-based tempered variants would fail by the same mechanism unless the kernel vanishes at $t = 0$.
- The paper establishes infinitude but not location: it says nothing about where the off-line zeros sit relative to the critical line, and the incomplete-gamma series (14)–(15) that produced the displayed zero offers a practical route to map their distribution numerically.
- Because every ingredient — positivity of $K$, $K(0) > 0$, the decay and order bounds — is verified directly from the definition of $K$, modest perturbations of Yang's construction (for instance replacing $e^{2t}$ with $e^{ct}$ for other $c$) would very likely fail to reinstate the conjecture, since the structural source of the failure persists.
Original abstract
Yang introduced a tempered xi function $\widehat\xi(s)$ by replacing the hyperbolic cosine in a classical integral representation of the Riemann xi function by a hyperbolic sine, and conjectured that every zero of $\widehat\xi$ lies on the critical line $\Re s=1/2$. We disprove this conjecture. After the change of variables $x=e^{2t}$, one has \[ \widehat\xi\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\,dt, \] where $K$ is positive, smooth, and doubly exponentially decaying. Two integrations by parts give $S(y)=K(0)/y+O(y^{-2})$ for real $|y|\to\infty$, with $K(0)>0$; hence $S$ has only finitely many real zeros. On the other hand $S$ is an entire function of order at most one. If it had only finitely many zeros in the complex plane, Hadamard factorization would force $S(z)=P(z)e^{az+b}$, which is incompatible with $S(y)\to0$ as $y\to\pm\infty$. Thus $S$ has infinitely many nonreal zeros, and consequently $\widehat\xi$ has infinitely many zeros off the critical line. The proof is unconditional and does not use the Riemann Hypothesis or numerical zero finding. A short numerical illustration is included.
The compiled record
The paper's own sentences, arranged, kept beside the review and separate from it. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.
The abstract, sentence by sentence
Yang introduced a tempered xi function $\widehat\xi(s)$ by replacing the hyperbolic cosine in a classical integral representation of the Riemann xi function by a hyperbolic sine, and conjectured that every zero of $\widehat\xi$ lies on the critical line $\Re s=1/2$. We disprove this conjecture. After the change of variables $x=e^{2t}$, one has \[ \widehat\xi\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\,dt, \] where $K$ is positive, smooth, and doubly exponentially decaying. Two integrations by parts give $S(y)=K(0)/y+O(y^{-2})$ for real $|y|\to\infty$, with $K(0)>0$; hence $S$ has only finitely many real zeros. On the other hand $S$ is an entire function of order at most one. If it had only finitely many zeros in the complex plane, Hadamard factorization would force $S(z)=P(z)e^{az+b}$, which is incompatible with $S(y)\to0$ as $y\to\pm\infty$. Thus $S$ has infinitely many nonreal zeros, and consequently $\widehat\xi$ has infinitely many zeros off the critical line. The proof is unconditional and does not use the Riemann Hypothesis or numerical zero finding. A short numerical illustration is included.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Yang introduced a tempered xi function $\widehat\xi(s)$ by replacing the hyperbolic cosine in a classical integral representation of the Riemann xi function by a hyperbolic sine, and conjectured that every zero of $\widehat\xi$ lies on the critical line $\Re s=1/2$. | background 1.00 | 0.03 | 0.03 | asserted 0.81 |
| s1 | We disprove this conjecture. | result 0.75 | 0.96 | 0.95 | asserted 1.00 |
| s2 | After the change of variables $x=e^{2t}$, one has \[ \widehat\xi\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\,dt, \] where $K$ is positive, smooth, and doubly exponentially decaying. | method 0.94 | 0.56 | 0.02 | asserted 1.00 |
| s3 | Two integrations by parts give $S(y)=K(0)/y+O(y^{-2})$ for real $|y|\to\infty$, with $K(0)>0$; hence $S$ has only finitely many real zeros. | result 0.81 | 0.86 | 0.03 | asserted 1.00 |
| s4 | On the other hand $S$ is an entire function of order at most one. | result 0.67 | 0.52 | 0.02 | asserted 0.95 |
| s5 | If it had only finitely many zeros in the complex plane, Hadamard factorization would force $S(z)=P(z)e^{az+b}$, which is incompatible with $S(y)\to0$ as $y\to\pm\infty$. | method 0.65 | 0.52 | 0.08 | asserted 0.96 |
| s6 | Thus $S$ has infinitely many nonreal zeros, and consequently $\widehat\xi$ has infinitely many zeros off the critical line. | result 0.96 | 0.93 | 0.86 | asserted 1.00 |
| s7 | The proof is unconditional and does not use the Riemann Hypothesis or numerical zero finding. | method 0.50 | 0.88 | 0.05 | asserted 1.00 |
| s8 | A short numerical illustration is included. | contribution 0.63 | 0.86 | 0.01 | asserted 0.99 |
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References
Discussion
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