I'm interested to find the value of the following integral involving the reprocal gamma function
$$\int_0^{\infty}\frac{(u+\beta)^n}{\Gamma(u+\alpha)}du$$
where $\alpha, \beta>0$ (can be the same) and $n=0,1,2...$.
If we use the binomial formula for $(u+\beta)^n$ we can reduce to calculate integrals of the type $$\int_0^{\infty}\frac{u^n}{\Gamma(u+\alpha)}du$$
I've checked books of finite integrals but a havent't found anything.
Not an answer but some things I have noticed
The first thing I encountered right now is the special case $\alpha=0$ and $n=0$ for which the integral reduces to
$$\int_0^\infty \frac{\mathrm dx}{\Gamma(x)}=F=2.807~770...$$
The constant $F$ is known as Fransén–Robinson constant. So it seems like integrals of this type, at least this particular integral, has been studied already but, as Wikipedia states, "It is however unknown whether $F$ can be expressed in closed form in terms of other known constants".
By considering positive integer $\alpha$ we can always reduce the integral via the functional relation of the Gamma Function combined with partial fraction decomposition to something of the form
$$I_1=\int_0^\infty \frac{\mathrm dx}{(x+t)\Gamma(x)}$$
whereas for negative integer $\alpha$ we will arrive at something of the form
$$I_2=\int_0^\infty \frac{x^t}{\Gamma(x)}\mathrm dx$$
So the real question is how to evaluate $I_1$ and $I_2$ hence we can reduce at least integer $\alpha$ back to these two integrals. Considering real values for $\alpha$ I have no idea where to get started
Overall I have to admit that I think it is highly improbable that there are known closed-form expressions for your integral hence even the simplest case $($i.e. $\alpha=n=0$ $)$ is not expressable in terms of known constants yet; or will never be.
According to http://mathworld.wolfram.com/MuFunction.html,
$\int_0^\infty\dfrac{u^n}{\Gamma(u+\alpha)}~du=n!\mu(1,n,\alpha-1)$
