Confirming that an integral equals 1 I was looking at Wolfram Alpha's interpretation of the integral $\int_{-\infty}^{\infty} e^{-e^{-x} - x}dx = 1$ and I'm trying to explicitly compute that now. My first attempt was to substitute with $u = -e^{-x}$, but I ended up getting a $u^u$ in the resulting integral, which I don't think can be derived in elementary functions. After several more unsuccessful attempts at solving this integral, I think that either I'm missing something obvious or that this integral can't be evaluated with elementary functions. I am curious to read your htoughts.

( http://www.wolframalpha.com/input/?i=integrate+-infinity+to+infinity+e%5E%28-x-e%5E%28-x%29%29 )
Set $y:=e^{-x}$ then (since $-dy=e^{-x}dx$) :
$$\int_{-\infty}^{\infty} e^{-e^{-x} - x}dx=\int_{-\infty}^{\infty} e^{-e^{-x}}e^{-x}dx=\int_{\infty}^0-e^{-y}dy=\int_0^\infty e^{-y}dy=1$$
This integral cannot be solved in terms of elementary functions. Try using Maclaurin series!