Find the integral $\int_0^\infty \mu^x / \Gamma(x + 1) dx$

Question

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!

Answer 1

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 :)

Answer 2

$$I(\mu)=\int_0^\infty \frac{\mu^x}{\Gamma(x + 1)}dx$$ 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 $I(2)\sim 6.998$