What is $\int_0^\infty e^{-e^x}dx$?

Question

Can we compute the following integral? $$\int_0^\infty e^{-e^x}dx\tag{1}$$ I think it converges because $e^x\geq 1+x$ for $x\geq0$, so we have $$0\leq\int_0^\infty e^{-e^x}dx\leq\int_0^\infty e^{-1-x}dx=-\frac{1}{e}e^{-x}\bigg|_0^\infty=\frac{1}{e}.$$ But which specific number in $[0,1/e]$ is $(1)$ equal to?

Answer 1

Let $u = e^x$. Then $x = \ln u$, which implies $dx = (1/u)du$. It follows that

$$\int_0^\infty e^{-e^x}\, dx = \int_1^\infty e^{-u} \frac{du}{u} = -\operatorname{Ei}(-1),$$

where $\operatorname{Ei}(x)$ is the exponential integral function. The value is about $0.219$.