[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*}[/step]

[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: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]