# 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.

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Okay. By the integral, I meant negative infinity at the bottom, sorry. – user16647 Oct 14 '12 at 22:11
Although the question has apparently been settled by Raymond below, try changing $x\to ix$. Perhaps this integral could also be expressed in terms of Hankel functions – Valentin Oct 14 '12 at 23:02

## 2 Answers

Set $y:=e^{-x}$ then (since $-dy=e^{-x}dx$) : $$\int_{-\infty}^{\infty} e^{-e^{-x} - x}dx=\int_{-\infty}^{\infty} e^{-e^{-x}}e^{-x}dx=\int_{\infty}^0-e^{-y}dy=\int_0^\infty e^{-y}dy=1$$

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This integral cannot be solved in terms of elementary functions. Try using Maclaurin series!

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Or not.   – Did Oct 14 '12 at 22:39
you are wrong about elementary functions ... $\frac{d}{dx} \exp(-e^{-x}) = \exp(-e^{-x}-x)$ ... but maybe Maclaurin series will work, too, who knows? – GEdgar Oct 14 '12 at 23:20