**A hint.**
Your integral is:
$$I=\int_0^\infty \frac{dx}{\Gamma(x)}$$
By reflection formula for the Gamma function:
$$\frac{1}{\Gamma(x)}=\frac{1}{\pi} \Gamma(1-x) \sin \pi x$$
$$I=\frac{1}{\pi} \int_0^\infty \Gamma(1-x) \sin \pi x~ dx$$
Now by the integral definition of the Gamma function we get:
$$\Gamma(1-x)= \int_0^\infty \frac{e^{-t}}{t^x}dt$$
So now the integral becomes:
$$I=\frac{1}{\pi} \int_0^\infty \int_0^\infty \sin (\pi x)~ \frac{e^{-t}}{t^x}~dt ~ dx$$
$$I=\frac{1}{\pi} \int_0^\infty \int_0^\infty \sin (\pi x)~ e^{- \ln(t) x} e^{-t}~dt ~ dx$$
See @nospoon's comment. No closed form is available, apparently, but maybe this hint might still help.