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.?
Let me carry out @Ninad_Munshi's suggestion in the first case, so that we can compare it with what I did yesterday.
First of all, $$\iint_U \frac{\log x \log y}{x^\alpha y^\alpha \max(x,y)} dx dy = 2 \frac{\partial}{\partial a} \frac{\partial}{\partial b} \mathop{\iint_U}_{x>y} \frac{dx dy}{x^{a+1} y^b} \bigg|_{(a,b)=(\alpha,\alpha)},$$ where $U=\{[(x,y)\in [1,\infty)^2: (x y)^\alpha \max(x,y)>R\}$ as before. Since $x^{\alpha+1} y^\alpha>R \iff x>(R/y^{\alpha})^{1/(\alpha+1)}$ and $(R/y^\alpha)^{1/(\alpha+1)}>y \iff R>y^{2\alpha+1}$, $$\begin{aligned}\mathop{\iint_U}_{x>y} \frac{dx dy}{x^{a+1} y^b} &= \int_{1}^{R^{\frac{1}{2\alpha+1}}} \frac{1}{y^b}\int_{(R/y^{\alpha})^{1/(\alpha+1)}}^\infty \frac{dx}{x^{a+1}} dy+ \int_{R^{\frac{1}{2\alpha+1}}}^\infty \frac{1}{y^{b}} \int_y^\infty \frac{dx}{x^{a+1}} dy\\ &= \int_{1}^{R^{\frac{1}{2\alpha+1}}} \frac{1}{y^b}\frac{1}{a (R/y^{\alpha})^{a/(\alpha+1)}} dy + \int_{R^{\frac{1}{2\alpha+1}}}^\infty \frac{1}{y^{b}} \frac{1}{a y^a} dy \\ &= \frac{1}{a R^{a/(\alpha+1)}} \int_{1}^{R^{\frac{1}{2\alpha+1}}} \frac{dy}{y^{b-\alpha a/(\alpha+1)}} + \frac{1}{a} \int_{R^{\frac{1}{2\alpha+1}}}^\infty \frac{dy}{y^{a+b}}\\ &= \frac{1}{a R^{a/(\alpha+1)}} \cdot \frac{-\frac{1}{b-1-\alpha a/(\alpha+1)}}{y^{b-1-\alpha a/(\alpha+1)}}\bigg|_1^{R^{\frac{1}{2\alpha+1}}} + \frac{1}{a} \frac{1}{(a+b-1)R^{\frac{a+b-1}{2\alpha+1}}}\\ &= \left(\frac{1}{a (a+b-1)} - \frac{\alpha+1}{a \left((\alpha+1)(b-1)-\alpha a\right)}\right) \frac{1}{R^{\frac{a+b-1}{2\alpha+1}}} +\frac{\alpha+1}{a \left((\alpha+1)(b-1)-\alpha a\right) R^{\frac{a}{\alpha+1}}} \end{aligned}$$ Here the coefficient of $\frac{1}{R^{\frac{a+b-1}{2\alpha+1}}}$ simplifies to $\frac{2\alpha+1}{(a+b-1) (\alpha a - (\alpha+1) (b-1))}$. Since $\frac{\partial}{\partial a} \frac{\partial}{\partial b} \frac{1}{R^{\frac{a+b-1}{2\alpha+1}}} = \frac{(\log R)^2}{(2\alpha+1)^2} \frac{1}{R^{\frac{a+b-1}{2\alpha+1}}}$, it is clear that the main term will be $\frac{1}{(2\alpha+1) (2\alpha-1)} \frac{(\log R)^2}{R^{\frac{2\alpha-1}{2\alpha+1}}}$ (multiplied by $2$, in the end; let's leave that part out for all terms). It is also clear that there will be no term proportional to $\frac{(\log R)^2}{R^{\frac{\alpha}{\alpha+1}}}$, as $\frac{\partial}{\partial b} \frac{1}{R^{\frac{a}{\alpha+1}}}=0$. The coefficient of $\frac{\log R}{R^{\frac{2\alpha-1}{2\alpha+1}}}$ will be $$\begin{aligned}- \left(\frac{\partial}{\partial a} + \frac{\partial}{\partial b}\right) \frac{1}{(a+b-1) (\alpha a - (\alpha+1)(b-1))} \bigg|_{(a,b)=(\alpha,\alpha)} &= \frac{2}{(a+b-1)^2 (\alpha a - (\alpha+1)(b-1))}\bigg|_{(a,b)=(\alpha,\alpha)} + \frac{\alpha-(\alpha+1)}{(a+b-1) (\alpha a - (\alpha+1)(b-1))^2} \bigg|_{(a,b)=(\alpha,\alpha)} \\ &= \frac{3-2\alpha}{(2\alpha-1)^2}. \end{aligned}$$ The coefficient of $\frac{1}{R^{\frac{2\alpha-1}{2\alpha+1}}}$ will be $2\alpha+1$ times $$\begin{aligned}\frac{\partial}{\partial a} \frac{\partial}{\partial b} \frac{1}{(a+b-1) (\alpha a - (\alpha+1)(b-1))} &= \frac{2}{(a+b-1)^3 (\alpha a - (\alpha+1)(b-1))} - \frac{2 \alpha (\alpha+1)}{(a+b-1) (\alpha a - (\alpha+1)(b-1))^3} + \frac{\alpha-(\alpha+1)}{(a+b-1)^2 (\alpha a - (\alpha+1)(b-1))^2} \end{aligned}$$ evaluated at $(a,b)=(\alpha,\alpha)$, and that is $$\begin{aligned}(2\alpha+1) \left(\frac{2}{(2\alpha-1)^3}-\frac{2 \alpha(\alpha+1)}{2\alpha-1}- \frac{1}{(2\alpha-1)^2}\right) &= \frac{2\alpha+1}{(2\alpha-1)^3} (-8 \alpha^4+6\alpha^2-4\alpha+3 )\\&= - (2\alpha^2+4\alpha+3) - \frac{16(\alpha^2-\alpha)}{(2\alpha-1)^3}, \end{aligned}$$ which agrees with the coefficient we had before. The coefficient of $\frac{\log R}{R^{\frac{a}{a+1}}}$ is $$-\frac{\partial}{\partial b} \frac{1}{a((a+1)(b-1)-\alpha a)}\bigg|_{(a,b)=(\alpha,\alpha)} = \frac{a+1}{a ((a+1)(b-1)-\alpha a)^2}\bigg|_{(a,b)=(\alpha,\alpha)} = \frac{\alpha+1}{\alpha},$$ which agrees with what we had before. The coefficient of $\frac{1}{R^{\frac{a}{a+1}}}$ is $$\begin{aligned}\frac{\partial}{\partial a} \frac{\partial}{\partial b} \frac{\alpha+1}{a \left((\alpha+1)(b-1)-\alpha a\right)} \bigg|_{(a,b)=(\alpha,\alpha)}&= \frac{\partial}{\partial a} \frac{-(\alpha+1)^2}{a \left((\alpha+1)(b-1)-\alpha a\right)^2}\bigg|_{(a,b)=(\alpha,\alpha)}\\ &= \frac{(\alpha+1)^2}{a^2 \left((\alpha+1)(b-1)-\alpha a\right)^2}\bigg|_{(a,b)=(\alpha,\alpha)} + \frac{2 (\alpha+1)^2\cdot (-\alpha)}{a \left((\alpha+1)(b-1)-\alpha a\right)^3}\bigg|_{(a,b)=(\alpha,\alpha)} \\ &= \frac{(\alpha+1)^2}{\alpha^2} + 2 (\alpha+1)^2 = 2\alpha^2 + 4 \alpha+3 + \frac{2\alpha+1}{\alpha^2}, \end{aligned}$$ which is exactly what we had before.
So all is well. But, as you can see, this is not really shorter or much easier than what we had before. (It seems to be a bit longer and a bit easier.)
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
$$x-y = \left(\frac{u}{v}\right)^{\frac{1}{2}}(v-1)$$