The equation, $x^y = y^x ~(x≠y)$ does not seem to have an elegant method to separate the variables. It is intuitive from the graph shown below (Desmos output) that it should have a convergent area bounded by the lines x = 1, y = 1. If the area does converge, are there any numerical methods to compute the same? Is there a closed form value to it? In general, what is the approach to identify whether or not the area bounded by a curve with its asymptotic curve converges or not?
If $x^y=y^x$ for $x,y>0$, then we have by taking logarithms $$\frac{\ln(x)}x=\frac{\ln(y)}y.$$
Let $K\overset{\text{Def.}}=\frac{\ln(x)}x$. Then we have the equation $$K = \frac{\ln(y)}y.$$
Introducing $\tilde y \overset{\text{Def.}}=-\ln(y)$, we get $y=\exp(-\tilde y)$ and therefore $$\tilde y \exp(\tilde y)=-K.$$
The solution to the latter equation is usually denoted by $\tilde y = W(-K)$, where $W$ is called the Lambert $W$ function.
Ignoring details with the multiple branches of the Lambert W function, if $x>e$, then a Taylor expansion gives that $$W(-K)=-K+O(K^2)$$ as $K\to0$.
Therefore, keeping in mind that $y=\exp(-\tilde y)$, the integral for the area for $x>e$ looks like $$\int_{e}^\infty \exp(-W(-K))-1\,\mathrm dx=\int_e^\infty \exp\left(\frac{\ln(x)}x+O\left(\frac{\ln(x)^2}{x^2}\right)\right)-1\,\mathrm dx,$$ which diverges since $\exp(1+y)\ge y$ for all $y\in\mathbb R$.
More explicitly, the mentioned integral is bounded from below by $\int_e^\infty \ln(x)/x+O(\ln(x)^2/x^2)\,\mathrm dx$. Now, the $O$ term converges absolutely, since $\ln(x)$ is locally bounded for $x\ge e$ and therefore $\ln(x)\le C \sqrt[3] x$ for some $C>0$ and all $x\ge e$, which means that $\frac{\ln(x)^2}{x^2}\le C x^{-\frac 43}$. The latter has an anti-derivative that is globally bounded for $x\ge e$. Furthermore, since $\frac{\ln(x)}x\ge\frac 1x$ for all $x\ge e$, the integral of the first term diverges, so the whole thing diverges.