I'm sure that's a coincidence, but the Laplace transform of $1/\Gamma(x)$ at $s=1$ turns out to be pretty close to the inverse of the Golden ratio:

$$F(1)=\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx=0.61985841414477344973$$

Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx>\frac{1}{\phi}$$

The continued fraction of this number also starts very beautifully:

$$F(1)=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\dots}}}}}}}}}}=$$

$$=[0; 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 7, 2, 1, 1, 1, 2, 6, 1, 4, 7, 1, 3, 1, 1, 6, 1,\dots]$$

The question has no practical application, but everyone here loves Golden ratio questions, right?

I'm sure there is no closed form for this integral, but if there is some useful transformation, I would like to see it as well.

If we replace every partial quotient in the CF after the first 5 2's by $2$, we obtain a great approximation to the integral:

$$F(1) \approx \frac{1}{41} \left(24+\sqrt{2}\right)$$

So, the more complicated question would be:

Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx<\frac{1}{41} \left(24+\sqrt{2}\right)$$

The difference is about $4 \cdot 10^{-7}$.