TITLE: Showing that the integral of an implicit function is $\pi^2/6 -1 = \zeta(2) -1$
QUESTION [5 upvotes]: For $x \ge 1$, define $f(x) \in [0,1]$ implicitly by the equation $1 - f(x) = e^{-x f(x)}$.
Numerically, it seems that
$\int_1^\infty (1-f(x)) dx = \pi^2/6 - 1$.
How can I show this analytically?
This calculation arose in a problem I am studying on Erdős–Rényi random graphs -- thanks for the help!

REPLY [5 votes]: Let $y = f(x)$.  Then $$1-y = e^{-xy}$$ implies $$x = - \frac{\log (1-y)}{y},$$ which we can see is strictly increasing for $0 < y < 1$. Hence $$\int_{x=1}^\infty (1 - f(x)) \, dx = \int_{y=0}^1 - \frac{\log (1-y)}{y} - 1 \, dy = - 1 + \frac{\pi^2}{6}.$$
If the above is not quite clear, we insert some intermediate steps via Fubini's theorem:
$$\begin{align}
\int_{x=1}^\infty (1 - f(x)) \, dx &= \int_{x=1}^\infty \int_{y=f(x)}^1 1 \, dy \, dx \\
&= \int_{y=0}^1 \int_{x=1}^{-\log(1-y)/y} 1 \, dx \, dy \\
&= \int_{y=0}^1 - \frac{\log (1-y)}{y} - 1 \, dy.
\end{align}$$