Integral involving reciprocal gamma function

Question

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.

Answer 1

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.

Answer 2

According to http://mathworld.wolfram.com/MuFunction.html,

$\int_0^\infty\dfrac{u^n}{\Gamma(u+\alpha)}~du=n!\mu(1,n,\alpha-1)$