Prove that $f^y$ and $f_x$ are Lebesgue-integrable

Question

Let $f:\Bbb R^2\to \Bbb R$ given by: $$f(x,y) = \begin{cases} \frac{x^2-y^2}{(x^2+y^2)^2} & \text{if $(x,y)\in(0,1)\times(0,1)$} \\ 0 & \text{if $(x,y)\not\in(0,1)\times(0,1)$} \\ \end{cases}$$ Prove that $f^y:=f(\cdot,y)$ and $f_x:=f(x,\cdot)$ are Lebesgue-integrable $\forall x,y\in\Bbb R$, and that $$\int_{\Bbb R}\left(\int_{\Bbb R}f(x,y)dx\right)dy\not=\int_{\Bbb R}\left(\int_{\Bbb R}f(x,y)dy\right)dx$$

I want to prove first that $f$ is continuous, so I can use something I've already proved $^\star$, however I'm having a lot of problems when $(x,y)\to(0,0)$, since in that case $f(x,y)\to(\frac{-1}{y^2},\frac{1}{x^2})$. Besides, the integral of $f^y$ and $f_x$ near $0$ is not even finite, do I use a different thing to prove this?

$\star$
Suppose $f:(a,b)\to \Bbb R$ continuous, and $|f|$ $Riemann-integrable$, i.e., $$\lim_{s\to a^+}\operatorname{\mathfrak R-\int_s^c}f \;\;\;\;\text{ and }\;\;\;\; \lim_{t\to b^-}\operatorname{\mathfrak R-\int_c^t}f $$ exist and are finite, then $$\operatorname{\mathfrak R-\int_a^b}f=\lim_{s\to a^+}\operatorname{\mathfrak R-\int_s^c}f+\lim_{t\to b^-}\operatorname{\mathfrak R-\int_c^t}f $$ Then, $f$ is Lebesgue-integrable in $(a,b)$, and $$\int_{(a,b)}=\operatorname{\mathfrak R-\int_a^b}f$$

Answer 1

Hint:

it's easy to see that:

$ \bullet$ if $0<y<1$ then $f^y(x)=0$ if $x \not \in(0,1)$ else $f^y(x)=\frac{x^2-y^2}{(x^2+y^2)^2}$

That means : $$\int_{\Bbb R} f^y(x) dx=\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}dx$$

since $x \mapsto \frac{x^2-y^2}{(x^2+y^2)^2}$ is continuous in $(0,1]$ and $\displaystyle \lim_{x \to 0^+} \frac{x^2-y^2}{(x^2+y^2)^2}=-\frac 1{y^2}$ we can say that $x\mapsto \frac{x^2-y^2}{(x^2+y^2)^2}$ is Lebesgues-integrable on $(0,1]$ so $f^y$ is is Lebesgues-integrable on $\Bbb R$.

$ \bullet$ The case $y \not \in (0,1)$ is easy since, in this case we have: $f^y(x)=0$ for all $x \in \Bbb R$

Answer 2

You have to integrate $$ \int^{1}_{0}\frac{x^{2}-c}{(x^{2}+c)^{2}}dx, c=y^{2} $$ is Lesbegue integrable for any $c$. But the inside function is continuous and well defined for any $x$, so it must be bounded on $[0,1]$. Then you get it is in $L^{1}$ for free. This will not change if $c\rightarrow 0$ as long as it is not $0$. And for $c=0$ you know the original function is $0$. So $f^{y}$ is well defined and in $L^{1}$. The same now works for $f^{x}$.