[proofplan]
We compute the entropy by applying the countable full-branch Rokhlin formula to the Gauss map with its continued-fraction digit partition. The main point is to verify the measure-theoretic hypotheses rather than merely quote the formula: the Gauss density is invariant under the inverse branches, the digit partition has finite entropy, and the logarithmic derivative is integrable. The formula then identifies the entropy with the $\mu_G$-integral of $\log |G'|$. The remaining calculation reduces this integral to the classical alternating zeta value $\sum_{n=1}^{\infty}(-1)^{n-1}n^{-2}=\pi^2/12$.
[/proofplan]
[step:Apply the full-branch Rokhlin formula to the continued-fraction partition]
For each $k\in\mathbb N$, define the interval
\begin{align*}
I_k:=\left(\frac{1}{k+1},\frac{1}{k}\right].
\end{align*}
The family $\mathcal P:=\{I_k:k\in\mathbb N\}$ is a countable measurable partition of $(0,1]$ modulo the point $0$, which is outside the domain under the present convention. On $I_k$, the map $G$ is given by
\begin{align*}
G|_{I_k}(x)=\frac{1}{x}-k,
\end{align*}
and maps $I_k$ bijectively onto $[0,1)$ modulo endpoints. Its inverse branch is the map
\begin{align*}
\psi_k:[0,1)\to I_k,\quad y\mapsto \frac{1}{k+y}.
\end{align*}
Let $\rho:(0,1]\to(0,\infty)$ denote the Gauss density, defined by
\begin{align*}
\rho(x):=\frac{1}{\log 2}\frac{1}{1+x}.
\end{align*}
For every non-negative Borel function $\varphi:(0,1]\to[0,\infty]$, the change of variables $x=\psi_k(y)=1/(k+y)$ on each branch gives
\begin{align*}
\int_{(0,1]} \varphi(G(x))\rho(x)\,d\mathcal L^1(x)=\sum_{k=1}^{\infty}\int_0^1 \varphi(y)\rho(\psi_k(y))|\psi_k'(y)|\,d\mathcal L^1(y).
\end{align*}
Here $|\psi_k'(y)|=(k+y)^{-2}$, and the branch weights satisfy
\begin{align*}
\sum_{k=1}^{\infty}\rho(\psi_k(y))|\psi_k'(y)|=\frac{1}{\log 2}\sum_{k=1}^{\infty}\frac{1}{(k+y)(k+y+1)}=\frac{1}{\log 2}\frac{1}{1+y}=\rho(y),
\end{align*}
where the series telescopes because
\begin{align*}
\frac{1}{(k+y)(k+y+1)}=\frac{1}{k+y}-\frac{1}{k+y+1}.
\end{align*}
Therefore $\mu_G$ is $G$-invariant.
The digit partition has finite entropy. Indeed,
\begin{align*}
\mu_G(I_k)=\frac{1}{\log 2}\int_{1/(k+1)}^{1/k}\frac{1}{1+x}\,d\mathcal L^1(x)=\frac{1}{\log 2}\log\left(\frac{(k+1)^2}{k(k+2)}\right).
\end{align*}
Since $\log(1+t)\leq t$ for $t>-1$, this gives $\mu_G(I_k)\leq 1/(k^2\log 2)$. Hence the series $-\sum_{k=1}^{\infty}\mu_G(I_k)\log\mu_G(I_k)$ converges by comparison with $\sum_{k=1}^{\infty}k^{-2}\log k$.
Moreover,
\begin{align*}
G'(x)=-\frac{1}{x^2}
\end{align*}
for every $x\in I_k$, and $\log |G'|=2\log(1/x)$ is $\mu_G$-integrable because
\begin{align*}
\int_0^1 \log\left(\frac{1}{x}\right)\frac{1}{1+x}\,d\mathcal L^1(x)\leq\int_0^1 \log\left(\frac{1}{x}\right)\,d\mathcal L^1(x)=1.
\end{align*}
The partition $\mathcal P$ is generating modulo the countable endpoint orbit because two non-endpoint points with the same itinerary have the same continued-fraction expansion; the exceptional endpoint orbit has $\mu_G$-measure zero. Thus the hypotheses of the countable full-branch Rokhlin entropy formula are satisfied: $G$ is nonsingular and invariant for $\mu_G$, the branch partition is countable, generating, and finite-entropy, and $\log |G'|\in L^1(\mu_G)$. Therefore
\begin{align*}
h_{\mu_G}(G)=\int_{(0,1]} \log |G'(x)|\,d\mu_G(x).
\end{align*}
Substituting the density of $\mu_G$ and the derivative of $G$ yields
\begin{align*}
h_{\mu_G}(G)
=
\frac{1}{\log 2}
\int_0^1
\log\left(\frac{1}{x^2}\right)\frac{1}{1+x}\,d\mathcal L^1(x).
\end{align*}
[guided]
We use the partition by first continued-fraction digit. For each $k\in\mathbb N$, define
\begin{align*}
I_k:=\left(\frac{1}{k+1},\frac{1}{k}\right].
\end{align*}
If $x\in I_k$, then $k\leq 1/x<k+1$, so $\lfloor 1/x\rfloor=k$. Hence on this interval the Gauss map has the explicit branch formula
\begin{align*}
G|_{I_k}(x)=\frac{1}{x}-k.
\end{align*}
This branch maps $I_k$ onto $[0,1)$ up to endpoints, and the corresponding inverse branch is
\begin{align*}
\psi_k:[0,1)\to I_k,\quad y\mapsto \frac{1}{k+y}.
\end{align*}
The relevant entropy theorem is the countable full-branch version of Rokhlin's entropy formula. It requires an invariant nonsingular probability measure, a countable generating branch partition of finite entropy, and integrability of the logarithmic derivative. We verify these requirements explicitly.
First, define the Gauss density $\rho:(0,1]\to(0,\infty)$ by
\begin{align*}
\rho(x):=\frac{1}{\log 2}\frac{1}{1+x}.
\end{align*}
To prove invariance, take a non-negative Borel function $\varphi:(0,1]\to[0,\infty]$. Changing variables on the branch $I_k$ with $x=\psi_k(y)=1/(k+y)$ and $|\psi_k'(y)|=(k+y)^{-2}$ gives
\begin{align*}
\int_{(0,1]}\varphi(G(x))\rho(x)\,d\mathcal L^1(x)=\sum_{k=1}^{\infty}\int_0^1\varphi(y)\rho(\psi_k(y))|\psi_k'(y)|\,d\mathcal L^1(y).
\end{align*}
The branch densities add up to the original density:
\begin{align*}
\sum_{k=1}^{\infty}\rho(\psi_k(y))|\psi_k'(y)|=\frac{1}{\log 2}\sum_{k=1}^{\infty}\frac{1}{(k+y)(k+y+1)}=\frac{1}{\log 2}\frac{1}{1+y}=\rho(y).
\end{align*}
The middle equality is a telescoping sum, since
\begin{align*}
\frac{1}{(k+y)(k+y+1)}=\frac{1}{k+y}-\frac{1}{k+y+1}.
\end{align*}
Hence
\begin{align*}
\int_{(0,1]}\varphi\circ G\,d\mu_G=\int_{(0,1]}\varphi\,d\mu_G,
\end{align*}
so $\mu_G$ is $G$-invariant.
Second, the partition has finite entropy. We compute
\begin{align*}
\mu_G(I_k)=\frac{1}{\log 2}\int_{1/(k+1)}^{1/k}\frac{1}{1+x}\,d\mathcal L^1(x)=\frac{1}{\log 2}\log\left(\frac{(k+1)^2}{k(k+2)}\right).
\end{align*}
Since $\log(1+t)\leq t$ for $t>-1$, this implies $\mu_G(I_k)\leq 1/(k^2\log 2)$. Therefore $-\sum_{k=1}^{\infty}\mu_G(I_k)\log\mu_G(I_k)$ converges by comparison with $\sum_{k=1}^{\infty}k^{-2}\log k$. Define the partition entropy $H_{\mu_G}(\mathcal P)$ by
\begin{align*}
H_{\mu_G}(\mathcal P):=-\sum_{k=1}^{\infty}\mu_G(I_k)\log\mu_G(I_k).
\end{align*}
The convergence just proved gives $H_{\mu_G}(\mathcal P)<\infty$.
Third, the partition is generating modulo a null set. If two points outside the countable endpoint orbit have the same itinerary with respect to the intervals $I_k$, then they have the same continued-fraction digits and hence the same continued-fraction expansion. The endpoint orbit is countable, so it has $\mu_G$-measure zero because $\mu_G$ is absolutely continuous with respect to $\mathcal L^1$.
Finally, we verify integrability of the logarithmic derivative. On every branch,
\begin{align*}
G'(x)=-\frac{1}{x^2},
\end{align*}
so $\log |G'(x)|=2\log(1/x)$. Using the density of $\mu_G$ and the bound $1/(1+x)\leq 1$ for $x\in(0,1]$, we get
\begin{align*}
\int_{(0,1]}\log |G'(x)|\,d\mu_G(x)=\frac{2}{\log 2}\int_0^1\log\left(\frac{1}{x}\right)\frac{1}{1+x}\,d\mathcal L^1(x)\leq\frac{2}{\log 2}\int_0^1\log\left(\frac{1}{x}\right)\,d\mathcal L^1(x)=\frac{2}{\log 2}.
\end{align*}
Thus $\log |G'|\in L^1(\mu_G)$. All hypotheses of the countable full-branch Rokhlin entropy formula are now verified, and the formula gives
\begin{align*}
h_{\mu_G}(G)=\int_{(0,1]} \log |G'(x)|\,d\mu_G(x).
\end{align*}
Substituting $|G'(x)|=x^{-2}$ and the Gauss density gives
\begin{align*}
h_{\mu_G}(G)
=
\frac{1}{\log 2}
\int_0^1
\log\left(\frac{1}{x^2}\right)\frac{1}{1+x}\,d\mathcal L^1(x).
\end{align*}
[/guided]
[/step]
[step:Evaluate the logarithmic integral by Abel limiting of the power series]
For $0<r<1$, define
\begin{align*}
S_r:(0,1)\to\mathbb R,\qquad x\mapsto \frac{-\log x}{1+rx}.
\end{align*}
Since
\begin{align*}
\frac{1}{1+rx}=\sum_{n=0}^{\infty}(-1)^n r^n x^n
\end{align*}
with [uniform convergence](/page/Uniform%20Convergence) on $[0,1]$, termwise integration gives
\begin{align*}
\int_0^1 \frac{-\log x}{1+rx}\,d\mathcal L^1(x)
=
\sum_{n=0}^{\infty}(-1)^n r^n
\int_0^1 x^n(-\log x)\,d\mathcal L^1(x).
\end{align*}
For each $n\in\mathbb N\cup\{0\}$, [integration by parts](/theorems/210) with $u(x)=-\log x$ and $dv=x^n\,d\mathcal L^1(x)$ gives
\begin{align*}
\int_0^1 x^n(-\log x)\,d\mathcal L^1(x)=\frac{1}{(n+1)^2}.
\end{align*}
Thus
\begin{align*}
\int_0^1 \frac{-\log x}{1+rx}\,d\mathcal L^1(x)
=
\sum_{n=0}^{\infty}\frac{(-1)^n r^n}{(n+1)^2}.
\end{align*}
Letting $r\uparrow 1$, the functions $S_r$ converge pointwise to
\begin{align*}
S_1:(0,1)\to\mathbb R,\qquad x\mapsto \frac{-\log x}{1+x}.
\end{align*}
They are dominated by the integrable function $x\mapsto -\log x$ on $(0,1)$. Hence the [dominated convergence theorem](/theorems/4) gives
\begin{align*}
\int_0^1 \frac{-\log x}{1+x}\,d\mathcal L^1(x)
=
\lim_{r\uparrow 1}
\sum_{n=0}^{\infty}\frac{(-1)^n r^n}{(n+1)^2}.
\end{align*}
Since the series $\sum_{n=0}^{\infty}(-1)^n(n+1)^{-2}$ converges, [Abel's theorem for power series](/theorems/3833) yields
\begin{align*}
\lim_{r\uparrow 1}
\sum_{n=0}^{\infty}\frac{(-1)^n r^n}{(n+1)^2}
=
\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}
=
\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m^2}.
\end{align*}
Therefore
\begin{align*}
\int_0^1 \frac{-\log x}{1+x}\,d\mathcal L^1(x)
=
\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m^2}.
\end{align*}
[/step]
[step:Insert the alternating zeta value and conclude the entropy formula]
From the entropy integral obtained above,
\begin{align*}
h_{\mu_G}(G)
=
\frac{1}{\log 2}
\int_0^1
\log\left(\frac{1}{x^2}\right)\frac{1}{1+x}\,d\mathcal L^1(x).
\end{align*}
Since $\log(1/x^2)=2\log(1/x)=-2\log x$, the previous step gives
\begin{align*}
h_{\mu_G}(G)
=
\frac{2}{\log 2}
\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m^2}.
\end{align*}
Using the classical alternating zeta value
\begin{align*}
\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m^2}=\frac{\pi^2}{12},
\end{align*}
we obtain
\begin{align*}
h_{\mu_G}(G)
=
\frac{2}{\log 2}\cdot \frac{\pi^2}{12}
=
\frac{\pi^2}{6\log 2}.
\end{align*}
This is the asserted entropy of the Gauss map with respect to Gauss measure.
[/step]
