# A sum of zeros of an infinite Hadamard product

I was experimenting with pairs of zeros of the following function ($i$ = imaginary unit):

$\displaystyle \xi_{int}(s) := \xi_{int}(0) \prod_{n=1}^\infty \left(1- \frac{s}{+ ni} \right) \left(1- \frac{s}{{- ni}} \right) = \frac{\sinh(\pi s)}{s}$

and plugged these zeros into the following (paired) sum:

$f(s):=\displaystyle \sum_{n=1}^\infty \left( \frac{s^{ni}}{ni} +\frac{s^{-ni}}{-ni} \right)$

which can be nicely transformed into this closed form:

$f(s) := i( \ln(1-s^i) - \ln(1-s^{-i}))$

I then found that numerically:

$f(e) = \pi-1$

and

$f(\pi) = \pi-ln(\pi)$

but also that:

$\displaystyle \lim_{x \to 0} f(1 \pm xi) = \pi$

however I struggle to properly mathematically derive these outcomes.

Grateful for any help. Thanks.

## 1 Answer

If $s\ne e^{2n\pi},\quad n\in \mathbb{Z}$, you have $$f(s)=i\ln\left(\frac{1-s^i}{1-s^{-i}}\right)=i\ln\left(\frac{1-s^i}{1-s^{-i}}\right)=i\ln(-s^i)\\ =i\ln((se^{(2k+1)\pi})^i)=i\ln\left(\left(r^ie^{(-\theta+i(2k+1)\pi)}\right)\right),\quad (k\in \mathbb{Z})\\= -((2k+1)\pi+\ln r)-i\theta$$where $s=re^{i\theta}$. So, $f(s)$ is actually multi-valued. One value of $f(e)$ is thus $\pi -1$ which is obtained by putting $k=-1$ in the above equation. Similarly, putting $k=-1$ in the equation for $f(\pi)$, you get the result you've got numerically.

Also, $$f(1\pm xi)=f(\sqrt{1+x^2}e^{i\theta})=\pi-\frac{1}{2}\ln(1+x^2)-i\theta\quad (k=-1)$$ where $$\theta=\pm \tan^{-1}x$$So, as $x\rightarrow 0$, $\theta\rightarrow 0$ and $f(1\pm xi)\rightarrow \pi$