Find the integral $\int_0^\infty \mu^x / \Gamma(x + 1) dx$ 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!
 A: $$I(\mu)=\int_0^\infty \frac{\mu^x}{\Gamma(x + 1)}dx$$  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 $I(2)\sim 6.998$
A: 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 :)
