Integrating a Bivariate Piecewise Function

Question

Given $$ f_{ab}(u,v)= \begin{cases} (1-a)u^{-a}v,\quad u^a>v^b\\ v^{1-b}, \quad\quad\quad\quad\: u^a<v^b\\ \end{cases} $$ where the domain is on $(u,v)\in[0,1]\times[0,1].$

How would I split the integral: $$\int_0^1\int_0^1f_{ab}(u,v)dudv.$$

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Answer 1

Well, the natural thing is $$ \int_0^1\int_0^1f_{ab}(u,v)~du~dv = \int_0^1\int_0^{v^\frac{b}{a}}f_{ab}(u,v)~du~dv + \int_0^1\int_{v^\frac{b}{a}}^1 f_{ab}(u,v)~du~dv, $$ where the formulas used for $f_{ab}$ in the two integrals are the two "parts" of the $f$ formula in your question.

This assumes that $a,b > 0$, so that $v^{b/a}$ makes sense; in general, if it's ambiguous (e.g., $v^\frac{1}{2}$), my intended meaning is "the positive value".