Evaluating $\lim\limits_{R\to +∞}\iint_{x^2+y^2\leq R^2}\left(\frac{1+2x^2}{1+x^4+6x^2y^2+y^4}-\frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy$ 
Evaluate $\lim\limits_{R\to\infty} J(R)$, where
$$J(R) = \iint_{x^2 + y^2\leq R^2} \left(\frac{1+2x^2}{1+x^4+6x^2y^2 + y^4} - \frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy.$$

First note that interchanging $x$ and $y$ does not change the value of the integral, so
$$J(R) = \iint_{x^2 + y^2 \leq R^2}\left(\frac{1+2y^2}{1+x^4+6x^2y^2 + y^4} - \frac{1+x^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy$$
and hence average the two equal expressions for $J(R)$ gives that
$$J(R) = \iint_{x^2 + y^2\leq R^2} (f(x,y) - g(x,y))\,\mathrm dx\mathrm dy,$$
where
$$f(x,y) = \frac{1+x^2 + y^2}{1+x^4+6x^2y^2+y^4},\quad g(x,y) = \frac{1+(x^2+y^2)/2}{2+x^4+y^4}.$$
Note that $f(x,y) = 2g(x+y,x-y)$.

Now, why is it true that
\begin{gather*}\iint_{R^2 \leq x^2 + y^2\leq 2R^2}g(x,y)\,\mathrm dx\mathrm dy\ \text{?}\tag{1}\end{gather*}

I tried using substitutions, but I couldn't get how to prove (1). But assuming (1) does hold, we have by converting to polar coordinates that
\begin{align*}
J(R) &= \int_0^{2\pi} \int_R^{R\sqrt{2}} \frac{1+r^2/2}{2+r^4(\cos\theta)} r\,\mathrm dr\mathrm d\theta\\
&= \int_0^{2\pi}\int_R^{R\sqrt{2}} \frac{1+r^2/2}{2+r^4(1-(\sin^2(2\theta)/2))}r\,\mathrm dr\mathrm d\theta
\end{align*}
Now use the substitution $r\mapsto r/R$ to obtain the equivalent integral
$$\int_0^{2\pi}\int_1^{\sqrt{2}} \frac{1+(Rr)^2/2}{2+(Rr)^4(1-\sin^2(2\theta)/2)}(R^2r)\,\mathrm dr\mathrm d\theta.$$

Why is it true that since integral is uniformly bounded for $R \gg 0$, we can take the limit over $R$ through the integrals to obtain that
$$\lim_{R\to\infty} J(R) = \int_0^{2\pi}\int_1^{\sqrt{2}} \frac{1}{r(2-\sin^2(2\theta))}\,\mathrm dr\mathrm d\theta = \frac{1}2\ln (2)\int_0^{2\pi} \frac{2}{3+\cos(4\theta)}\,\mathrm d\theta\ \text{?}$$

Note that by symmetry,
$$\int_0^{2\pi} \frac{2}{3+\cos(4\theta)}\,\mathrm d\theta = 2\int_0^\pi \frac{2}{3+\cos\theta}\,\mathrm d\theta$$
Now using the half-angle substitution $t = \tan(\theta/2)$, we have
\begin{gather*}
2\int_0^{\pi} \frac{2}{3+\cos\theta}\,\mathrm d\theta = 2\int_{0}^{\infty} \frac{4}{3(1+t^2)+(1-t^2)}\,\mathrm dt\\
= 2\int_0^\infty \frac{2}{2+t^2}\,\mathrm dt = \sqrt{2}\arctan(t/\sqrt{2})\biggr|_0^\infty = \sqrt{2}\pi.
\end{gather*}
Hence we have $\lim\limits_{R\to\infty} J(R) = \dfrac{\sqrt{2}}2\ln(2)\pi$.
 A: $\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\vector#1#2{\begin{pmatrix}#1\\#2\end{pmatrix}}\def\abs#1{\left|#1\right|}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$To prove (1), it suffices to prove\begin{gather*}
\iint\limits_{x^2 + y^2 \leqslant R^2} f(x, y) \,\d x\d y = \iint\limits_{x^2 + y^2 \leqslant 2R^2} g(x, y) \,\d x\d y. \tag{1$'$}
\end{gather*}
Define $φ: \R^2 → \R^2$ by $φ\vector{x}{y} = \vector{x + y}{x - y}$, then $f = 2g \circ φ$. Since $φ^{-1}\vector{u}{v} = \dfrac{1}{2} \vector{u + v}{u - v}$, then\begin{align*}
&\mathrel{\phantom=} φ\paren{ \brace{\vector{x}{y} \in \R^2 \,\middle|\, x^2 + y^2 \leqslant R^2 } }\\
&= \brace{ \vector{u}{v} \in \R^2 \,\middle|\, \paren{ \frac{1}{2}(u + v) }^2 + \paren{ \frac{1}{2}(u - v) }^2 \leqslant R^2 }\\
&= \paren{ \brace{\vector{u}{v} \in \R^2 \,\middle|\, u^2 + v^2 \leqslant 2R^2 } },
\end{align*}
and$$
Jφ(u, v) = \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.
$$
Therefore,\begin{gather*}
\iint\limits_{x^2 + y^2 \leqslant R^2} f(x, y) \,\d x\d y = \iint\limits_{u^2 + v^2 \leqslant 2R^2} f(φ^{-1}(u, v)) |\det Jφ^{-1}(u, v)| \,\d u\d v\\
= \iint\limits_{u^2 + v^2 \leqslant 2R^2} \frac{2g(u, v)}{|\det Jφ(u, v)|} \,\d u\d v = \iint\limits_{u^2 + v^2 \leqslant 2R^2} g(u, v) \,\d u\d v = \iint\limits_{x^2 + y^2 \leqslant 2R^2} g(x, y) \,\d x\d y.
\end{gather*}

Now in order to prove that\begin{gather*}
\small \lim_{R → +∞} \iint\limits_{\substack{1 \leqslant r \leqslant \sqrt{2} \\ 0 \leqslant \leqslant 2π}} \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2θ)) + 4} \,\d r\d θ = \iint\limits_{\substack{1 \leqslant r \leqslant \sqrt{2} \\ 0 \leqslant \leqslant 2π}} \lim_{R → +∞} \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2θ)) + 4} \,\d r\d θ, \tag{2}
\end{gather*}
note that for $1 \leqslant r \leqslant \sqrt{2} \leqslant R$,$$
\abs{ \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2θ)) + 4} } < \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2θ))} \leqslant \frac{R^2 r · 2R^2 r^2}{R^4 r^4} = \frac{2}{r} \leqslant 2,
$$
thus (2) is implied by the dominated convergence theorem.
