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let's define $p_\alpha(n) = \displaystyle\int_1^n x^\alpha dx$ so that $\left\{\begin{array}{lll} p_0(n) &=& n-1 \\ p_{-1}(n) &=& \ln n \\ p_\alpha(n) &=& \frac{\textstyle n^{\alpha+1} - 1}{\alpha+1} \quad \text{for} \ \ \ \alpha \ne -1\end{array}\right.$

Then, take $p_\alpha(n)$ to make a Dirac comb and look at its Laplace transform : $$P_\alpha(s) = LT\left\{ \sum_{n=1}^\infty \delta\left(t - p_\alpha(n)\right)\right\} = \sum_{n=1}^\infty e^{\textstyle-s \,p_\alpha(n)}$$

We get that $\displaystyle P_0(s) = \sum_{n=1}^\infty e^{-s (n-1)} = \frac{1}{1-e^{-s}}$ and $\displaystyle P_{-1}(s) = \sum_{n=1}^\infty n^{-s} = \zeta(s)$.

Question : what do we know about the zeros and poles localization of the analytic continuation of $P_\alpha(s)$ for other values of $\alpha$ in $[-1;\infty[$ ?

Are there numerical evidences that they are always aligned on a few lines of the complex plane ?

Are these localization at least continuous with respect to slow variations of $\alpha$ ?

reuns
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1 Answers1

1

You ask for localization of the poles and zeroes for the analytic continuation of $P_{\alpha}(s)$ for $\alpha \in [-1,\infty]$.

Below, I prove that for $\alpha \in (0,\infty]$, there is no analytic continuation to start with.

$$ \begin{align} P_{\alpha}(s) &= \sum\limits_{n=1}^{\infty} e^{-s \frac{n^{\alpha+1}-1}{\alpha+1}} \\ &= \underbrace{e^{\frac{s}{\alpha+1}}}_{{f(s)}} \overbrace{\sum\limits_{n=1}^{\infty} \left(e^{\frac{-s}{\alpha+1}}\right)^{n^{\alpha+1}}}^{g(s)} \\ \end{align} $$

$f(s)$ is an entire function.

To analyze $g(s)$,

  1. lets substitute $e^{\frac{-s}{\alpha+1}} \to z$.
  2. This maps the right half of the $s$-plane to interior of unit circle in $z$-plane.
  3. $g(z)$ has no singularities inside the unit circle.

$$g(z) = \sum\limits_{n=1}^{\infty} z^{n^{\alpha+1}}$$

  1. $\lim\limits_{n \to \infty} \frac{n^{\alpha+1}}{n} = \lim\limits_{n \to \infty} n^{\alpha} = \infty$

So unit circle is a natural boundary$^{\dagger}$. Analytical extension from inside the unit circle to outside the unit circle is not possible for $g(z)$. Translating this to $s$-plane, we can't analytically extend $g(s)$ to left half of plane.

Note$^{\dagger}$: Fabry gap theorem is clear for $\alpha \in \mathbb{Z}^{+}$. I see some lacunary gap theorems upon search that apply to $\alpha \in \mathbb{R}^+$ (which loosely translate to $\left(1+\frac1n\right)^{\alpha+1} \gt 1+\delta$ lacunacy condition for $g(s)$). But I don't have a reliable reference.

Since $g(s)$ has a natural boundary at $\mathfrak{R}(s) = 0$, $P_{\alpha}(s) = f(s) g(s)$ also cannot be analytically extended to left half of $s$-plane.

Edit 1:

For $\alpha \in (-1,0)$, Let $\delta = \alpha + 1$ , where $\delta \in (0,1)$

$$P_{\alpha}(s) = e^{\frac{s}{\delta}} \sum\limits_{n=1}^{\infty} e^{-\frac{s n^{\delta}}{\delta}}$$

By Mellin inversion,

$$e^{-\frac{s n^{\delta}}{\delta}} = \frac{1}{2\pi i} \int\limits_{c-i \infty}^{c + i \infty} \Gamma(z) \left(\frac{s n^{\delta}}{\delta}\right)^{-z} dz$$

Therefore $$P_{\alpha}(s) = e^{\frac{s}{\delta}} \frac{1}{2\pi i} \int\limits_{c-i \infty}^{c + i \infty} \Gamma(z) \left(\frac{1}{\delta^z}\right) \zeta(\delta z) s^{-z}$$

The poles of $\Gamma(z)$ are at $z = 0, -1, -2, ....$. The pole of $\zeta(\delta z)$ is at $z=\frac{1}{\delta} \gt 0$. So the dominant pole is from $s^{-z}$ at $s=0$ when combined with the Zeta residue at $\frac{1}{\delta}$.

The pole analysis is complete for $\alpha \in (-1,0)$. I am not sure about the zeroes though. That is like another Riemann Hypothesis, no? :)

Srini
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