Double integral: elegant way? I need to evaluate (or, if that is not feasible, bound well) some integrals of the type
$$\mathop{\int \int}_{(x,y)\in U} \frac{\log x \log y}{F(x,y)} dx dy,$$
where $U = \{(x,y)\in [1,\infty)^2: F(x,y)>R\}$, $R$ is positive, and $F(x,y)$ is one of the following:

*

*$F(x,y) = x^{5/3} y^{5/3} \max(x,y)$

*$F(x,y) = x^{5/3} y^{5/3} (x+y)$

*$F(x,y) = x^{5/3} y^{5/3} |x-y|$ or $F(x,y) = x^{5/3} y^{5/3} \max(|x-y|,1)$.

My main concern is: how to do this without inflicting a painful mess on myself and the reader?
In case 1, (a) a direct approach is feasible but results in a mess, (b) the substitution $u = x y$ helps a great deal (I'll show how in an answer below). Can something like that (or better) be done in cases 2. and 3.?
 A: We can rewrite a few things to make the computations a bit nicer. First denote $G$ as
$$F(x,y) = x^{\frac{5}{3}}y^{\frac{5}{3}}G(x,y)$$
which means $G$ now encompasses all of the change between the functions. Next denote
$$I(a,b) = \iint\limits_U \frac{\log x \log y}{x^ay^bG(x,y)}dxdy = \frac{\partial}{\partial a}\frac{\partial}{\partial b}\iint\limits_U \frac{1}{x^ay^bG(x,y)}dxdy$$
In all cases cutting the region of integration in half at the line $x=y$ by symmetry is useful.
$\textbf{Case 1}$
$$x^{\frac{8}{3}}y^{\frac{5}{3}} > R \implies y > \left(\frac{R^3}{x^{8}}\right)^{\frac{1}{5}}  $$
$$I = 2\frac{\partial}{\partial a}\frac{\partial}{\partial b}\int_{R^{\frac{3}{13}}}^\infty \int_{\max\left\{\left(\frac{R^3}{x^{8}}\right)^{\frac{1}{5}},1\right\}}^x\frac{1}{x^{a+1}y^b}dydx$$
$\textbf{Case 2}$
$$\begin{cases}u = xy \\ v = \frac{y}{x}\end{cases} \to J^{-1} = 2\frac{y}{x} \implies J = \frac{1}{2v}$$
$$x+y = \left(\frac{u}{v}\right)^{\frac{1}{2}}(v+1)$$
$$\frac{u^{\frac{13}{6}}}{v^{\frac{1}{2}}}(v+1) > R \implies u > \left(\frac{vR^2}{(v+1)^2}\right)^\frac{3}{13}$$
$$I = 2\frac{\partial}{\partial a}\frac{\partial}{\partial b}\int_0^1 \int_{\max\left\{\left(\frac{vR^2}{(v+1)^2}\right)^\frac{3}{13},\frac{1}{v}\right\}}^{\infty}\frac{dudv}{2u^{a+b+\frac{1}{2}}\sqrt{v}(1+v)}$$
And the same coordinate change can be used for cases 3 and 4, but now the boundary term would read
$$x-y = \left(\frac{u}{v}\right)^{\frac{1}{2}}(v-1)$$
