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This question aims to extend this question to (automorphic) Dirichlet L-functions.

Take: $$f(s,q,j) = (s-1)\,L(s, \chi_{q,j})$$

with $q$ the modulus and $j$ the index of a character $\chi$. A fast way to compute the Riemann-Liouville fractional differintegral for $f(s,q,j)$ is to use the following MacLaurin series for some finite $M$ (with the tail small):

$${}_0D^{-\alpha}_s \bigl(f(s,q,j)\bigr) = s^{\alpha}\,\sum^M_{m \ge 0}\frac{f^{(m)}(0, q,j)}{\Gamma(m + \alpha +1)}\,s^m$$

The numerator of the coefficients is independent of $\alpha$, hence it can be precomputed once. The function is valid for $\alpha, s \in \mathbb{C}, \alpha \ne -1,-2,-3,...$ and $s \ne 0$ for $\Re(\alpha) < 0$.

For $\alpha = 0, -1,-2,-3,...$ the precomputed numerators can be reused as follows:

$${}_0D^{-\alpha}_s \bigl(f(s,q,j)\bigr) = \sum^M_{m \ge 0}\frac{f^{(m-\alpha)}(0,q,j)}{\Gamma(m +1)}\,s^m$$

These series enable fast computation of the complex roots ($\mu$) of the fractional integrals $(\alpha > 0)$ and derivatives $(\alpha < 0)$. A value $M=1500$ proved to be sufficient to "trace" the "trajectories" of the first $25$ known complex zeros of $L(s, \chi_{q,j})$ when $\alpha$ is varied in steps $0.1$ over the range $\alpha = -30, \cdots, 30$. Computations were done for first $72$ Dirichlet characters (i.c. modulus $q = 1, \cdots ,15$).

Here are the graphs up to modulus $q=10$. Also included is a link to the data and its format.

BEGIN EDIT:

Based on the comment to possibly exclude the redundant factor $(s-1)$ for non-principal characters, here are some updated graphs up to modulus $q=10$. The patterns largely remain the same, except that for non-principal characters the specific trajectories now cannot exceed the first derivative instead of the second (bullet a. below).

END EDIT:

A couple of observations about these graphs:

  • For all primitive characters, subsequent fractional integration seems to always converge the trajectories into a (vertical) arithmetic progressions.

  • For most imprimitive characters, subsequent fractional integration seems to increasingly make the distribution of the zeros more "pronounced arithmetically", i.e. the trajectories tend to team up in groups of various integer sizes.

  • The zero trajectories of the fractional derivatives appear more complex, however patterns do emerge. On a high level, there appear to be two types of trajectories, i.e. those on which the zeros:

    a. rapidly accelerate in speed, however they'll never exceed $\alpha = 2$ (see end of deck).

    b. slow down or reach a constant speed but continue to have zeros for all $\alpha$'s.

Q1: Is there a logical explanation for the observation in the last bullet? What is so special about the second derivative that zeros cease to exist at higher (fractional) derivatives? And why do these zeros increase in speed under repeated differentiation and also have a tendency to organise themselves on clear curves (see end of deck)?

Q2: Any other observations about these graphs? (I am aware of the work by Farr-Pauli-Saidak on zeros of fractional derivatives of $\zeta(s)$, e.g. here, however have not seen any extensions to L-functions).

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    $\begingroup$ I understand the factor of $(s-1)$ in front of $\zeta(s)$ in the previous question - it kills off the pole at $s=1$. Dirichlet $L$-functions for nontrivial characters don't have a pole, so why include the $(s-1)$? $\endgroup$
    – Stopple
    Commented Sep 12, 2023 at 21:03
  • $\begingroup$ @Stopple, great question. Before landing on the finite MacLaurin series to compute the zeros, I had used the more complicated differintegral-expression outlined in this question: mathoverflow.net/questions/451646/… An expression like that could only be derived when the factor $(s-1)$ was included (to neutralise the $s-1$ division in the second term of the Hurwitz integral representation DLMF 25.11.29). It then stuck, but you are right that it is only required to remove the poles of the trivial characters. $\endgroup$
    – Agno
    Commented Sep 12, 2023 at 23:03

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