Is $f(x,y)\in L^1([0,1]\times[0,1])$?

Question

I have to study the Lebesgue integrability of $f(x,y)=\dfrac{x-y}{(x+y)^3}$ on $Q=[0,1]^2\subseteq\mathbb R^2$.
The sign of $f$ is variable in $Q$, so I studied $|f(x,y)|=\dfrac{|x-y|}{(x+y)^3}$ in order to apply Tonelli's theorem (because $|f|$ is positive and measurable in $Q$ and I can evaluate the iterated integrals).
$$\iint_{Q}|f(x,y)|dxdy=\iint_{[0,1]^2}|f(x,y)|dxdy=\int_0^1dy\int_0^1\dfrac{|x-y|}{(x+y)^3}dx$$ with the change of variable $\begin{cases}u=x-y\\v=x+y \end{cases}\implies\begin{cases}x=\dfrac{u+v}{2} \\y=\dfrac{v-u}{2}\end{cases}$ I have $|\det(J)|=\dfrac{1}{4}$, with $\begin{cases}0<u+v<2\\0<v-u<2 \end{cases}$ as new set of integration $\tilde Q$ in $(u,v)$-plane.
So we obtain $$\dfrac{1}{4}\Bigg(\int_{-1}^0|u|du\int_{-u}^{2+u}\dfrac{dv}{v^3}+\int_{0}^1|u|du\int_{u}^{2-u}\dfrac{dv}{v^3}\Bigg)=$$ $$=\dfrac{1}{4}\Bigg(\int_{-1}^0\dfrac{u}{2(2+u)^2}-\dfrac{1}{2u}du+\int_{0}^1\dfrac{-u}{2(2-u)^2}+\dfrac{1}{2u}du\Bigg)$$ and I think that this shows that $f\notin L^1([0,1]^2)$.

Answer 1

I don't see how you apply the Fubini-Tonelli's theorem as I don't manage to write $\tilde Q$ as a rectangle. Anyway here is a way out :

First notice that $|f|$ is symmetric so we can put ourself in the case $x < y$ in order to get ride of the absolute value : $$\int_{[0,1]^2}|f(x,y)|dxdy = 2 \int_{[0,1]^2}1_{x < y}\frac{y-x}{(x+y)^3}dxdy.$$ Now I apply the Fubini-Tonelli's theorem to rewright the last integral as \begin{align} \int_0^1 \int_0^y \frac{y-x}{(x+y)^3}dxdy &= \int_0^1 \int_y^{2y} \frac{2y-x}{x^3} dxdy \\ &= \int_0^1 2y \left[ \frac{x^{-2}}{-2} \right]_y^{2y}-\left[ \frac{x^{-1}}{-1} \right]_y^{2y}dy \\ &= \int_0^1 y\frac{1}{y^2} - y\frac{1}{(2y)^2} + \frac{1}{2y} - \frac{1}{y}dy \\ &= \int_0^1 \frac{1}{4y}dy = + \infty. \end{align} Now $f$ is not integrable over $[0,1]^2$.