Consider the open set $E=\{(x,y)\in(0,1)^2:x+y<1\}$ and define $\text{Cant}:E\to(0,1)$ such that $x^{\text{Cant}(x,y)}+y^{\text{Cant}(x,y)}=1$ which we may call the Cantor dimension function. My question then is: is there a closed form of the integral $$\bbox[#EDDDFF,2pt,border: 2px solid #B8A5FF]{\mathcal{I}:=\int_{(x,y)\in E}\text{Cant}(x,y)dxdy=\int_0^1\int_0^{1-y}\text{Cant}(x,y)dxdy}$$
Clearly the Cantor dimension function is well defined since $x^t+y^t-1$ is continuous, monotonically decreasing as, for $x,y\in E$, we have $\frac{d}{dt}[x^t+y^t-1]=\log(x)x^t+\log(y)y^t<0$ and $x^0+y^0-1=1>0, x^1+y^1-1<0$. It is also easy to see that, if $x_0\in(0,1)$ is such that $x_0^a+x_0^b=1$, then $\text{Cant}(x^a,x^b)=\frac{\log x_0}{\log x}$ which implies that $\text{Cant}(x,x)=-\frac{\log 2}{\log x}$ and $\text{Cant}(x^2,x)=-\frac{\log \varphi}{\log x}$. Similarly, it is easy to see that $\text{Cant}(x^a,y^a)=\frac{1}{a}\text{Cant}(x,y)$ which means that it is exponentially inverse-linear.
$\color{#0072FF}{\bf Motivation.}$ Let us consider the generalized Cantor set of ratios $\alpha,\beta\in(0,1)$ such that $\alpha+\beta<1$ defined inductively. First, set $C_0:=[0,1]$. Then, assuming $C_n$ has been defined, set the mappings $f_\alpha:t\mapsto\alpha t$ and $f_\beta:t\mapsto 1-\beta t$ and define $C_{n+1}:=f_\alpha[C_n]\cup f_\beta[C_n]$. Now, the said Cantor set is defined as the intersection $C(\alpha,\beta):=\bigcap_{n=1}^\infty C_n$. It is not hard to see (although somewhat trickier to formalize) that it has fractal dimension (say box counting dimension) $D=\text{Cant}(\alpha,\beta)$ since it must satisfy $\alpha^D+\beta^D=1$. Then, a natural question that arises is what's the average dimension of the Cantor sets? And, it is not hard to see that, since $E$ has area $1/2$, it should be $\mathcal{I}/(1/2)=2\mathcal{I}$ (uniformly distributed).
$\color{#5A59FF}{\bf My Attempt.}$ By the fact that Cant is exponentially inverse-linear, we can define $C(u):=\text{Cant}(e^{-u},e^{-1})$ for values of $u>u_0:=-\log(1-e^{-1})\approx 0.458675$ in such a way that $$\text{Cant}(x,y)=\text{Cant}\left(x^{\frac{-\log y}{-\log y}},e^{\log y}\right)=-\frac{1}{\log y}\text{Cant}\left(x^{-\frac{1}{\log y}},e^{-1}\right)=-\frac{1}{\log y}C\left(\frac{\log x}{\log y}\right)$$ Then, the desired integral becomes $$\mathcal{I}=\int_0^1\int_0^{1-y}\text{Cant}(x,y)dxdy=-\int_0^1\frac{1}{\log y}\int_0^{1-y}C\left(\frac{\log x}{\log y}\right)dxdy$$ and then, by substituting on the inner integral $u=\frac{\log x}{\log y}\Rightarrow x=y^u\Rightarrow dx=y^u \log (y) du$, we get $$\mathcal{I}=-\int_0^1\frac{1}{\log y}\int_{+\infty}^{\frac{\log(1-y)}{\log(y)}}C(u)y^u\log( y)dudy=\int_0^1\int_\frac{\log(1-y)}{\log y}^{+\infty} y^u C(u)dudy$$ and then I tried to obtain the indefinite integral $\int y^u C(u) du$ by exploiting the fact that we know $$1=e^{-u C(u)}+e^{-C(u)}\Rightarrow 0=e^{-uC(u)}(-C(u)-uC'(u))-C'(u)e^{-C(u)}$$ $$\Rightarrow C'(u)=-\frac{C(u)e^{-uC(u)}}{ue^{-uC(u)}+e^{-C(u)}}=\frac{C^2(u)(1-e^{-C(u)})}{(-\log(1-e^{-C(u)}))(1-e^{-C(u)})+e^{-C(u)}}$$ which means that we know how to substitute $C=C(u)$ on the integral since $dC=C'(u)du\Rightarrow du=\frac{dC}{C'(u)}$ and we have just expressed $C'(u)$ completely in terms of $C=C(u)$. And so we get (setting $y=e^{-k}$ for simplicity) $$\int e^{-ku} C(u)du=\int (1-e^{-C})^{-\frac{k}{C}}\frac{(-\log(1-e^{-C}))(1-e^{-C})+e^C}{C^2(1-e^{-C})}dC$$ $$=\int \frac{1}{C^2}\left(-\log(1-e^{-C})+(e^C-1)^{-1}\right)(1-e^{-C})^{-\frac{k}{C}}dC$$ But, sadly, this leads to nothing...