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Take the closed form of this Hadamard/Weierstrass product:

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

Easy to see that the complex zeros occur in conjugated pairs at $ni$ and $-ni$ and, similar to the prime-counting function, these can be plugged into an infinite series:

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

Note that this series only converges for $x>0, x \in \mathbb{R}$. It is easy to see that $F(x)=-F(\frac{1}{x})$.

I then put $F(x)$ into the following 'counting' function:

$$\displaystyle N(x) := x + F(x)$$

and I like to conjecture that: $\displaystyle\frac{N(x)}{x} \sim 1$, but how could I prove this?

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Writing $x = e^\xi$, we see that

$$f(\xi) = F(e^\xi) = \sum_{n=1}^\infty \left(\frac{e^{in\xi}-e^{-in\xi}}{ni} \right) = 2\sum_{n=1}^\infty \frac{\sin (n\xi)}{n}$$

is periodic, with period $2\pi$. It is the Fourier series of a bounded, piecewise continuous function (the $2\pi$-periodic extension of $\pi-\xi$ on $[0,2\pi)$; thanks to Marko Riedel for remembering that), so the sum is bounded too.

Hence $F$ is bounded, and $\dfrac{N(x)}{x} \sim 1$.

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  • $\begingroup$ Many thanks, Daniel. Transforming it into a sine via $x=e$ is a very nice trick! $\endgroup$ – Agno Oct 27 '13 at 23:38
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    $\begingroup$ We can identify the series in question, it is $\pi-\xi$ with period $2\pi$. This is not difficult to prove. $\endgroup$ – Marko Riedel Oct 27 '13 at 23:40
  • $\begingroup$ Thanks, @MarkoRiedel. I knew it was something like that, but couldn't remember what exactly. $\endgroup$ – Daniel Fischer Oct 27 '13 at 23:42

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