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.

  • $\begingroup$ You didn't mention the integral representation $\varphi(u) = e^{{iu}+e^{iu}}+\int_0^\infty \frac{e^{-t}\ dt}{(\ln t-iu)^2+\pi^2}$ (the corresponding formula for the Laplace transform, $\int_0^\infty \frac{e^{-ux}\ dx}{\Gamma(x)}=e^{-u+e^{-u}}+\int_0^\infty \frac{e^{-t}\ dt}{(\ln t+u)^2 + \pi^2}$ is proved in a paper by Fransén & Wrigge, in Math. Computation, 1984). Observe that a complex-valued function of a real variable "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

This is not an answer. It's just a comment about the above integral representation.

If $H$ is a Hankel contour to the left ( a path that comes from $\infty$ with argument $-\pi$, turns around 0 - always at a distance from 0 bigger than 1) and then comes back to $\infty$ with argument $\pi,$ we have $${1\over{\Gamma (z)}}={1\over {2\pi i}}\int_H {{ e^t}\over {t^z}}dt, \ \ z\in C,$$ where $t^z$ uses a branch of the logarithm $L,$ with $-\pi <Im L <\pi.$

When we compute now its Fourier transform, we can interchange integrals and write $$\varphi (u)=\int_0^{\infty} {1\over{\Gamma (z)}}e^{iz u}dz={1\over {2\pi i}}\int_H \int_0^\infty e^te^{izu}e^{-z L t}dzdt={1\over{2\pi i}}\int_H {{ e^t}\over{ Lt-iu} }dt.$$ For $u\in (-\pi,\pi),$ we can deform the path to the negative real line plus a circle surrounding the singularity $e^{iu},$ and so we get $$\varphi(u)= e^{iu}e^{e^{iu}}+\int_0^\infty{{e^{-t}}\over{(\log t-iu)^2+\pi^2}}dt, \ \ \ \ u\in (-\pi,\pi).$$ Diferently, for $ u\in R-[-\pi,\pi],$ we write: $$\varphi(u)= \int_0^\infty{{e^{-t}}\over{(\log t-iu)^2+\pi^2}}dt, \ \ \ \ u\not\in [-\pi,\pi].$$

  • $\begingroup$ That's right, and it leads to a potentially useful alternative expression, $\varphi(u) = \lim_{\lambda\to\infty}\frac{\lambda}{2\pi}\int_0^{2\pi}\frac{\exp(-\lambda e^{i\theta})e^{i\theta}\ d\theta}{\log\lambda+i(\theta-u-\pi)}$ (using a contour integral $\oint_\mathcal{C}\frac{f(s)\ ds}{s-iu-i\pi}$ with $f(s):=\exp(s-e^s)$, $f(s+2\pi i)= f(s)$, and $\mathcal{C}=$ the rectangle $[-\log\lambda,+\log\lambda]$, $[\log\lambda,\log\lambda+2\pi i]$, $[\log\lambda+2\pi i, -\log\lambda+2\pi i]$ and $[-\log\lambda+2\pi i,-\log\lambda]$). Some asyptotic analysis might give something, I guess... $\endgroup$ – Jean Duchon Aug 3 '15 at 9:28
  • $\begingroup$ Sorry for the sign mistake: $\varphi(u)=-\lim$ etc $\endgroup$ – Jean Duchon Aug 4 '15 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.