Closed form for $\int_1^\infty\frac{\operatorname dx}{\operatorname \Gamma(x)}$

Solution 1:

As @Lucian said in comment this integral is related to Fransén–Robinson constant. That is, $$F = \int_{0}^\infty \frac{1}{\Gamma(x)}\, dx.$$

It is unknown whether $F$ can be expressed in closed form. I think because of this fact, your problem also doesn't have a known closed form.

I can tell you just some results about this problem. We could write your integral into the form:

$$\int_{1}^\infty \frac{1}{\Gamma(x)}\, dx = \int_{0}^\infty \frac{1}{\Gamma(1+x)}\, dx.$$

A numerical approximation of the integral is (A247377)

$$2.2665345076998488350719638576782209184088297...$$

Ramanujan observed that

$$ \int_0^{\infty} \frac{w^x}{\Gamma(1+x)} \, dx = e^w - \int_{-\infty}^{\infty} \frac{\exp(-we^y)}{y^2+\pi^2} \, dy.$$

For $w=1$ Ramanujan's formula gives an alternate form of you problem.

$$ \int_0^{\infty} \frac{1}{\Gamma(1+x)} \, dx = e - \int_{-\infty}^{\infty} \frac{\exp(-e^y)}{y^2+\pi^2} \, dy.$$

And at last the last page of this paper is also related.

I could get a connection between Fransén–Robinson constant and your integral. If we denote the Fransén–Robinson constant with $F$, then

$$\int_{1}^\infty \frac{1}{\Gamma(x)}\, dx = F - \int_{-\infty}^{\infty} \frac{\exp(-e^y) \cdot (1+e^y)}{y^2 + \pi^2} \, dy.$$

But I couldn't find a closed form for this integral. Of course it is true, that

$$\int_{0}^1 \frac{1}{\Gamma(x)}\, dx = \int_{-\infty}^{\infty} \frac{\exp(-e^y) \cdot (1+e^y)}{y^2 + \pi^2} \, dy.$$