Let $$f_a(x,y)=\frac{(x+y)\log(y/x)}{(xy)^{a}(1+x^2+y^2)}=\frac{1}{x^a}\frac{(x+y)\log(y/x)}{y^a(1+x^2+y^2)} $$ defined on $$O=\{(x,y):x>0,y>0\}.$$
I need to find all $a\in \mathbb{R}$ such that $f_a\in L^1(O)$ and evaluate the integral for such $a.$
My attempt
I have $f(x,y)\geq 0$ for all $(x,y)$ such that $y>x$ and $f(x,y)\leq 0$ for $(x,y)$ such that $y<x$ as in this case we have $\log(y/x)< 0.$ I can divide the domain of integration as the disjoint union $$O=O^+ \cup O^- =\{(x,y):0<x<y\}\cup\{(x,y):0<y<x\}.$$ Then I have $$\int_O |f_a|=\int_{O^+}f_a+\int_{O^-}|f_a|.$$
I start by evaluatuating $$\int_{O^+}f_a=I=\int_0^y \frac{1}{x^a} \ dx \int_x^{+\infty}\frac{(x+y)\log(y/x)}{y^a(1+x^2+y^2)} \ dy $$
But I am already not sure if this is even a correct approach, because the result of the integral would now depend on $x$ or $y$ depending on the order of integration...
What am I doing wrong?
If I manage to prove that the integral, as a function of, say, the variable $y$, is less or equal than a function $f(y)$ which is bounded on $O$ maybe I can conclude?
Let's go to polar coordinates: I'll write $c$ for $\cos \theta$ and $s$ for $\sin \theta.$ After some factoring the integral becomes
$$\int_0^{\pi/2}\int_0^\infty\frac{r(c+s)\ln (s/c)}{c^as^ar^{2a}(1+r^2)}\,r\,dr\,d\theta.$$
That splits nicely into
$$ = \int_0^{\pi/2}\frac{(c+s)\ln (s/c)}{c^as^a}\,d\theta\cdot \int_0^\infty\frac{r^2}{r^{2a}(1+r^2)}\,dr.$$
For the $r$ integral, let's look at $0<r\le 1$ first. Since $1/2<1/(1+r^2)<1,$ we can ignore it. We're left with
$$\int_0^1r^{2-2a}\,dr.$$
That converges iff $2-2a >-1,$ or $a<3/2.$
For the interval $(1,\infty),$ we can ignore $r^2/(1+r^2)$ and so the integral here converges iff $2a>1$ or $a>1/2.$
Conclusion: The full $r$ integral converges iff $1/2<a<3/2.$
See if you can make progress on the $\theta$ integral. A rubbing-the-chin estimate says to me it converges iff $a<1.$ If that's true then the convergence range of $a$ for the full integral is $1>a>1/2.$