# Analytic Function with positive integers as zeros?

Do you know any nontrivial analytic function f(z) with zeros only at positive integer values of the argument z = 1, 2, 3, 4, ... ? If yes, please give some example.

PS: I already thought of $f(x)=\frac{1}{\Gamma(-x+1)}$. Any other nice options?

EDIT: To avoid trivial solutions due to restriction of definition range, please consider the required function to be defined in the whole complex plane.

• What about $\frac{\sin(\pi z)}{z}$? – Pragabhava Oct 12 '12 at 14:21
• @Pragabhava : your function has zeros at $z=\pm 1,~\pm 2,~\pm 3,~...$ – Kagaratsch Oct 12 '12 at 14:29
• Right, Didn't thought about the negative ones. As I did with the zero: $\frac{\sin(\pi z)}{\Pi_0^\infty (z + n)}$, but now I'm not sure about the analiticity (I think it is). If not, how about removing the negative ones using the $\Gamma$ function? They are of order one. – Pragabhava Oct 12 '12 at 14:34
• @Pragabhava: nice idea to use $\Gamma$ to remove zeros, but somehow it does not really enrich the pool of functions with that property... – Kagaratsch Oct 12 '12 at 14:37
• If you already have an example, then what is it that you're looking for in additional examples that isn't already provided by the one you have? – Ben Crowell Oct 12 '12 at 14:47

All other entire functions $f$ with simple zeros exactly at positive integers differ from your $f_o(z)=1/\Gamma(-z+1)$ by a function of the form $e^{g(z)}$ for entire $g$, and vice-versa. That is, $f(z)=f_o(z) \cdot e^{g(z)}$ for entire $g$, and vice-versa. Indeed: $f(z)/f_o(z)$ has no zeros and is entire, so is of the form $e^{g(z)}$, since we can define its logarithm.

• Very nice proof! (+1) – Pragabhava Oct 12 '12 at 15:54

$$\zeta(-2x)+\sin(\pi x),$$ but $x \neq 0.109737160708907...$

Denote $(p \in \mathbb{N}\cup \{0\})$ $$E\left({z},\, p\right)=\begin{cases}1-z, & p=0, \\ (1-z)e^{z+\frac{z^2}{2}+\ldots+\frac{z^p}{p}}, & p\geqslant{1}. \end{cases}$$ Function $E\left({z},\, p\right)$ is called by primary or elementary Weierstrass factor of genus $p$. Let $\{a_\nu \} \subset \mathbb{C}$ be a sequence of non-zero complex numbers such that $|{a_\nu}| \rightarrow \infty.$ Then function $$f(z)=\prod\limits_{\nu=1}^{\infty}{E\left(\dfrac{z}{a_\nu},\, p\right)}$$ called as Weierstrass canonical product, is entire function that has zeroes at points $a_\nu$ and only at them.