调和Xi函数的无穷多个临界线外零点：对Yang猜想的反驳
Infinitely Many Off-Critical-Line Zeros of the Tempered Xi Function: A Disproof of Yang's Conjecture
- SDU University（SDU大学）
- Ajman University（阿治曼大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
本文反驳Yang关于调和xi函数零点全在临界线上的猜想，通过积分变换和Hadamard分解证明其存在无穷多个非实零点，即临界线外零点，证明无条件且不依赖Riemann假设。
AI中文摘要：
Yang通过将Riemann xi函数经典积分表示中的双曲余弦替换为双曲正弦，引入了调和xi函数$\widehat\xi(s)$，并猜想$\widehat\xi$的每个零点都位于临界线$\Re s=1/2$上。我们反驳了这一猜想。在变量替换$x=e^{2t}$之后，有\\[ \widehat\xi\\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\\,dt, \\] 其中$K$是正的、光滑的且双重指数衰减的。两次分部积分给出$S(y)=K(0)/y+O(y^{-2})$，对于实数$|y|\to\infty$，且$K(0)>0$；因此$S$只有有限多个实零点。另一方面，$S$是一个阶数至多为一的整函数。如果它在复平面中只有有限多个零点，Hadamard分解将迫使$S(z)=P(z)e^{az+b}$，这与$S(y)\to0$（当$y\to\pm\infty$时）不相容。因此$S$有无限多个非实零点，从而$\widehat\xi$有无限多个临界线外的零点。该证明是无条件的，不使用Riemann假设或数值零点查找。文中包含一个简短的数值示例。
英文摘要：
Yang introduced a tempered xi function $\widehatξ(s)$ 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 $\widehatξ$ lies on the critical line $\Re s=1/2$. We disprove this conjecture. After the change of variables $x=e^{2t}$, one has \[ \widehatξ\!\left(\tfrac12+iz\right)=iS(z), \qquad S(z)=\int_0^\infty K(t)\sin(zt)\,dt, \] 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 $\widehatξ$ 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.
