Calculate double integral with signum-function.

Question

Let

$$f(x, y) := \begin{cases} sgn(xy) \over x^2 + y^2, & \text{$(x, y) \in \Bbb R^2 \setminus ${0}$$} \\ 0, & \text{otherwise} \end{cases}$$

be a function with

$${\rm sgn}(xy) := \begin{cases} 1, & \text{$xy > 0$} \\ -1, & \text{$xy < 0$} \\ 0, & \text{$xy = 0$} \end{cases}$$

Calculate

$$\int_{\Bbb R} \int_{\Bbb R} f(x, y) d\lambda(x) d\lambda(y)$$ and

$$\int_{\Bbb R} \int_{\Bbb R} f(x, y) d\lambda(y) d\lambda(x).$$

Edit:

I deleted my former approach and would like to try a new one:

We are allowed to assume that

$$f = f_+ + f_-.$$

Hence, we are allowed to write the inner integral as

$$\int_{\Bbb R} f(x, y)_+ d\lambda(x) + \int_{\Bbb R} f(x, y)_- d\lambda(x) = \int_{\Bbb R} {1 \over x^2 + y^2} d\lambda(x) + \int_{\Bbb R} {-1 \over x^2 + y^2} d\lambda(x).$$

Using the linearity of the integral, we receive:

$$\int_{\Bbb R} {1 \over x^2 + y^2} d\lambda(x) - \int_{\Bbb R} {1 \over x^2 + y^2} d\lambda(x).$$

Both integrals are identical, hence their difference is $0$, and so is the outer integral then.

Is that a valid answer?

Answer 1

Hint

See, that for $(x,y)\in \mathbb{R}^2$ we have $f(x,y)=-f(-x,y)$

Hint 2 $$\int_\mathbb{R} f(x,y) dx =\int_\mathbb{R_-} f(x,y) dx +\int_\mathbb{R_+} f(x,y) dx = \\ =\int_\mathbb{R_+} f(-x,y) dx +\int_\mathbb{R_+} f(x,y) dx$$

Hint 3 $$\text{sgn}(\cos(\varphi)\sin(\varphi)) = \text{sgn}(\sin(2\varphi)) = \begin{cases}1&, \varphi \in (0,\frac{\pi}{2} + k\pi)&, k\in \mathbb{Z}\\ -1&, \varphi \in (\frac{\pi}{2} + k\pi, (k+1)\pi)&, k\in \mathbb{Z}\\ 0&, x= \frac{k\pi}{2} &, k\in \mathbb{Z}\end{cases}$$

Hint 4 $$\int_0^{\pi} \frac{ \text{sgn}(\cos \varphi \sin \varphi)}{r} d\varphi = \int_{\pi}^{2\pi} \frac{ \text{sgn}(\cos \varphi \sin \varphi)}{r} d\varphi = 0$$

Answer 2

1. OP's measurable function $f:\mathbb{R}^2\to \mathbb{R}$ is not Lebesgue integrable $$ \int_{\mathbb{R}^2}\! d\lambda(x,y)~|f(x,y)|~\stackrel{r=\sqrt{x^2+y^2}}{=}~2\pi \int_{\mathbb{R}_+} \! \frac{\mathrm{d}r}{r}~\stackrel{\text{hint}}{=}~\infty. $$ cf. Tonelli's theorem. Therefore we cannot use Fubini's theorem. In particular, OP's successive integrations in polar coordinates (of the function $f$, as opposed to the function $|f|$) are unjustified, i.e. not relevant for OP's mentioned exercise.

2. In rectangular coordinates for fixed $y$, we calculate: $$\forall y\in\mathbb{R}\backslash\{0\}:~~ \int_{\mathbb{R}}\! d\lambda(x)~|f(x,y)|~=~\frac{\pi}{|y|}~<~\infty,$$ so the function $x\mapsto f(x,y)=-f(-x,y)$ is a Lebesgue integrable odd function. Hence its integral vanishes $$\forall y\in\mathbb{R}\backslash\{0\}:~~ \int_{\mathbb{R}}\! d\lambda(x)~ f(x,y) ~=~0.$$

3. Therefore the function $$ y~~\mapsto~~ \int_{\mathbb{R}}\! d\lambda(x)~ f(x,y) ~=~0 ~~\text{ a.e.} $$ vanishes almost everywhere, so that the sought-for double integral becomes $$ \int_{\mathbb{R}}\! d\lambda(y)\int_{\mathbb{R}}\! d\lambda(x)~ f(x,y) ~=~0. $$

4. There is a similar conclusion for the function $y\mapsto f(x,y)$ for fixed $x$ because of the symmetry $f(x,y)=f(y,x)$.