# Integral involving reciprocal gamma function

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.

• – cgiovanardi Jan 30 '19 at 3:12

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

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

• Also called "Volterra function" – popi Nov 18 '19 at 16:28