Let us call a measure $\mu$ on the Borel $\sigma$-algebra $\mathfrak{B}_{(0,\infty)}$ of subsets of $(0,\infty)$ a *reciprocal gamma measure* if it is absolutely continuous with respect to the Lebesgue measure $m_L$ on $((0,\infty),\mathfrak{B}_{(0,\infty)})$ with the Radon-Nikodym derivative given by $$\frac{d\mu}{dm_L}(x) = \frac{1}{\Gamma(x)}\quad\text{for } x\in (0,\infty).$$
Let $\varphi$ be the characteristic transformation of the reciprocal gamma measure, that is, the complex valued function on $\mathbb{R}$ defined by
$$\varphi(u)=\int_{0}^{\infty}\!e^{ixu}\, \mu(dx)\quad\text{for } u \in \mathbb{R}.$$

**Question:** Is the collection of all zeros $\{u\in\mathbb{R}\colon \varphi(u)=0\}$ an empty set?

**Remarks I:** From numerical inspections, I am tempted to believe that the answer is in the affirmative. But I do not think there are any standard methods of deciding whether or not there are any zeros.

If we set $c=\varphi(0)$ then $c^{-1}\frac{d\mu}{dm_L}$ is a density of a probability measure $\lambda$ on $\mathfrak{B}_{(0,\infty)}$ having moments of every order. It is well known that if $\lambda$ is infinitely divisible, then $\varphi(u)$ is non-zero for every real number $u\in\mathbb{R}$. I have not been able to verify this infinite divisibility for $\lambda$. I have scrolled through the pages of Steutel & van Harn's book on infinitely divisible probability distributions, but I have not found anything on a probability distribution where the density is given by the reciprocal gamma function normalized. I therefore do not think it is known whether or not the probability distribution $\lambda$ is infinitely divisible. The density is not likely to belong to the class *Bondesson* so it might be very difficult to determine whether or not it is infinitely divisible.

Now the moments of the probability distribution $\lambda$ satisfy Carleman's condition so that $\lambda$ is "determined", that is, there are no other probability distributions on $((0,\infty),\mathfrak{B}_{(0,\infty)})$ having the same sequence of moments. As far as I can tell, the characteristic transformation can be extended to an entire holomorphic function and this too gives that the distribution is determined.

**Remarks II:** It is immediate, from the continuity of $\varphi$, that the set $\Lambda=\{u\in \mathbb{R}\colon \varphi(u)\neq 0\}$ is an open set in the Euclidean metric topology of $\mathbb{R}$. Moreover, since 0 is an element of $\Lambda$, it is a nonempty set. To prove that the collection of zeros of $\varphi$ is an empty set, according to the connectedness of $\mathbb{R}$, it suffices to show that $\Lambda$ is also a *closed* subset of $\mathbb{R}$. I have not been able to prove that $\Lambda$ is closed, but perhaps my patience was too short.

inMath. Computation, 1984). Observe that acomplex-valued function of arealvariable "generally" has no zeroes, so it would be surprising if $\varphi$ had zeroes. On the other hand, if $dx/\Gamma(x)$ was infinitely divisible it should be well known... $\endgroup$ – Jean Duchon Jul 23 '15 at 14:21