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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?

  • 1
    What does that symbol $\;d\lambda(x)\;$ mean?? – 2017-01-30
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    You integrate with respect to $\lambda(x)$. – 2017-01-30
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    @B Thanks, I already knew that: **what is** $\;\lambda\;$ , though?? – 2017-01-30
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    It's the Lebesgue Measure. – 2017-01-30
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    I think that the point of the exercise is that the *iterated* integrals are both zero while the function isn't integrable. Cf. https://en.wikipedia.org/wiki/Fubini%27s_theorem. – 2017-01-31
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    Maybe. But where are my calculations wrong then? – 2017-01-31
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    Can anyone explain to me why I get downvoted? – 2017-02-01
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    The hint appears useless or worse. – 2017-02-03
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    Note that $\int_0^{2\pi} \operatorname{sgn} a\,\mathrm d\phi=0$. So with your transformations you actually get $0\cdot\infty$ which is an indeterminate form. Note that if you integrate over any finite ball, you'll get $0$. – 2017-02-04
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    Thanks for your comment! Do you possibly have another approach then the two that were already given in the other answers? – 2017-02-04

2 Answers 2

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  1. OP's measurable function $f:\mathbb{R}^2\to \mathbb{R}$ is not Lebesque 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 Lebesque 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 $f(x,y)=f(y,x)$.

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    For showing that the function is not Lebesgue-integrable, I would still prefer switching to polar coordinates. It is way easier than thinking about a clever substitution. By switching to polar coordinates and applying Tonelli, you receive the integral $\int_{\Bbb R} \int_{\Bbb R} {1 \over r^2} r dr d\phi$, thus, you can apply the hint directly. – 2017-02-03
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    Besides that, I have never encountered the term "odd function" or how it is related to an Lebesgue-integral. – 2017-02-03
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    I updated the answer. – 2017-02-03
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    I understand what odd functions are, but I believe that there is a simpler way to do this since it is an old exam question. Since we never talked about integrating odd functions, I don't think that this is what our professor wants us to achieve. Do you know any other solution? And wouldn't it still work with polar coordinates somehow? – 2017-02-03
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    Or applying $f = f_+ + f_-$? – 2017-02-03
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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 $$\int_0^{\pi} \frac{ \text{sgn}(\cos \varphi \sin \varphi)}{r} d\varphi = \int_{\pi}^{2\pi} \frac{ \text{sgn}(\cos \varphi \sin \varphi)}{r} d\varphi = 0$$

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    Could you please elaborate what you are refering to? Are my calculations wrong in general? – 2017-01-30
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    I don't see what you are up to with your hints. Can't you just tell me where my calculations went wrong first? – 2017-01-30