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I've been looking at the zeros of the incomplete zeta function $\zeta_{lower}(s, z)$ recently. $$ \zeta_{\mathrm{lower}}(s,z)=-\frac{{\Gamma(1-s)}}{2\pi i}\int_{z}^{\infty}\frac{{(-t)^{s-1}}}{e^{t}-1}dt $$ The subscript "lower" indicates that the contour of integration passes below the branch cut on the positive real axis. By zeros, I mean I mean the zeros in $z$, for fixed $s$. I find that this function can have an infinite number of complex zeros $z_k$, each of which is a function of the parameter $s$. The zeros (for fixed $s$) live on a smooth curve and have an accumulation point at the origin.

Let $s_0$ be any zero of the (original, complete) Riemann zeta function. A remarkable thing happens when $s$ is near $s_0$. The kth zero, $z_k(s)$ depends logarithmically on $(s_0-s)$. $$ \log(z_k(s)) \approx \frac{1}{s-1} \log(s_0-s) + C_k(s_0) $$ Assuming $|\operatorname{Im}(s)| >> |\operatorname{Re}(s)|$, the entire set $z_k$ rotate about the origin as $s \rightarrow s_0$. Imagine a (slightly bent) barn door rotating about its hinge.

For added fun, if you move $s$ around $s_0$ and return to the original point, $z_k$ will become $z_{k+1}$ or $z_{k-1}$ depending on the sense of the rotation in $s$.

I've verified these prediction numerically down to a distance of $10^{-20}$ between $s$ and $s_0$.

So my question is whether anyone has seen this before, for the zeta function or for other incomplete functions.

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  • $\begingroup$ You displayed equation is not precise enough to say anything. The left side depends on $k$, the right does not (or only implicitly $C=C_k$). The right side depends on $s_0$, the left does not. What happens if $s_0$ is replaced with another zeta zero $s_0^\prime$? How can the left be $\approx$ both? $\endgroup$
    – Stopple
    Commented 2 days ago
  • $\begingroup$ Fair catch. $C_k$ does depend on $s_0$ and the relation is only valid in the $s$ neighborhood of $s_0$. I've edited the question to make that clear. $\endgroup$
    – Zhobbyist
    Commented 2 days ago

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