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Basically, I'm looking for advice on how I could find the value of $$\int_0^\infty \frac{\mu^x}{\Gamma(x + 1)}dx $$ where $\mu > 0$ is an arbitrary positive constant.

Based on the infinite series, I was initially expecting this to be something close to $e^\mu$ (if not exactly that). However, numerical experiments have convinced me that this is a flawed assumption unless $\mu$ is relatively large.

I'm happy to push on the problem myself --- I'm just a bit unsure where to start.

P.S. For context, I'm an applied statistician trying to force through an unorthodox probability distribution for data-efficiency reasons. Thanks in advance!

Snoop
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Cat
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    what is $\mu$ in this – Henry Lee May 05 '21 at 22:15
  • @HenryLee I've edited the question to clarify that $\mu > 0$ is an arbitrary positive constant – Cat May 05 '21 at 22:34
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    It's a bit 'unnatural' looking of a function to a mathematician's eyes, I'm curious how you're trying to use it as an aside! Where is this coming from? I'm sure someone will have a solution for you soon, but how it cropped up is probably the more interesting bit. – John Samples May 05 '21 at 22:38
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    You can find this function denoted $\nu(x)$ in page 217 here: https://authors.library.caltech.edu/43491/10/Volume%203.pdf. According to this "it was encountered by Volterra in his theory of convolution-logarithms" (I found this reference via Gradshtein-Ryzhik). It doesn't seem to have closed forms but some asymptotics and other properties. –  May 05 '21 at 22:40
  • Uh oh, looks like a closed-form solution might still be outstanding... – John Samples May 05 '21 at 22:46
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    Also check this other post here: https://math.stackexchange.com/questions/816745/exponential-integral-int-0-infty-fracxt-gammat1-text-dt –  May 05 '21 at 23:03
  • Even the integral $\int_0^\infty\frac{1}{\Gamma(x+1)}\mathrm{d}x$ does not have a closed form. – K.defaoite May 05 '21 at 23:50

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well if $\mu$ is a constant then if we let $m=\ln\mu$ then me get: $$\int_0^\infty\frac{e^{mx}}{\Gamma(x+1)}dx$$

I am trying to see if there are any nice relationships for $1/\Gamma$ but all I can find is: $$\frac{1}{\Gamma(z)}=\frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}\,dt\,\,\,\,\,\forall z\notin\mathbb{Z}$$ Where $C$ is the Hankel contour. The problem is that (as stated) its not valid for integers and so it would be discontinuous for us. I will see if I can find anything else :)

Henry Lee
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$$I(\mu)=\int_0^\infty \frac{\mu^x}{\Gamma(x + 1)}dx$$ Even for $\mu=1$, I do not think that we can obtain any result.

However, you intuition is quite good. Computing the numerical values for $1\leq \mu\leq 100$ and performing a quick and dirty linear regression $$\log[I(\mu)]=a+b~\mu$$ with $R^2 > 0.9999999$ we have $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -0.01047433 & 0.00375385 & \{-0.01792468,-0.00302398\} \\ b & +1.00015519 & 0.00006453 & \{+1.00002710,+1.00028327\} \\ \end{array}$$ and this works even for small values of $\mu$. For example, this empirical correlation gives $I(2)\sim 7.314$ while the "exact" value is $I(2)\sim 6.998$

Claude Leibovici
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