Here is a heuristic.
Observe that $f$ is decreasing on any interval of the form $I_n:=(1/(n+1),1/n)$. If $xin I_n$ and $f(x)=a$, then $x=frac{1}{a+n}$, and so $f^{-1}(a,1)$ is the union of the intervals $(1/(n+1),1/(n+a))$. Supposing there were an $f$-invariant measure $mu=gdx$, you can see that
$displaystyle{sum_{n=1}^infty int_{1/n+1}^{1/n+a}g(x)dx=int_a^1g(x)dx}$.
Supposing that $g$ is continuous, take the derivative of both sides:
(**) $displaystyle{sum_{n=1}^infty gleft(1/(n+a)right)/(n+a)^2=g(a)}$.
Approximating the sum by an integral, we get
$displaystyle{g(a)approxint_1^infty gleft(frac{1}{x+a}right)frac{1}{(x+a)^2}dx}$.
Supposing that $g$ is never too big or too small, this gives
$g(a)approx frac{C}{1+a}$.
One can then plug this kind of function into (**) to see that such a function works for any C. Then just pick C to normalize.
Probably someone from the era of Gauss (especially with his acuity at mathematics) did not need to do anything past (**) since people back then seem like magicians when it comes to expressions with infinite sums.
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