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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

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

Show that

$$\int_{\Bbb R} \int_{\Bbb R} f(x, y) dx dy = 0.$$

My attempt:

First, we assume that $y > 0$. This means that $sgn(xy) = 1$ for $x \in (0, \infty)$ and $sgn(xy) = -1$ for $x \in (-\infty, 0)$. We are then allowed to write the inner integral like this:

$$\int_0^\infty {1 \over x^2 + y^2} dx + \int_{-\infty}^0 {-1 \over x^2 + y^2} dx = \int_0^\infty {1 \over x^2 + y^2} dx - \int_{-\infty}^0 {1 \over x^2 + y^2} dx.$$

Both integrals are identical, and it can be shown that they are finite. Hence, the value of the inner integral is $0$ and so is the value of the double integral.

Now we need to assume that $y < 0$. This means that $sgn(xy) = -1$ for $x \in (0, \infty)$ and $sgn(xy) = 1$ for $x \in (-\infty, 0)$. Now we get:

$$\int_{-\infty}^0 {1 \over x^2 + y^2} dx + \int_0^\infty {-1 \over x^2 + y^2} dx = \int_{-\infty}^0 {1 \over x^2 + y^2} dx - \int_0^\infty {1 \over x^2 + y^2} dx.$$

Once again, both integrals are identical and finite, hence, the inner integral equals $0$, and so does the double integral.

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    Seems the attempt written is sound. So what's the issue here? – 2017-02-17
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    I'm just wondering whether it's correct or not. :) – 2017-02-17
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    I think this is correct. Put this up as an answer. "Self-learner" :) – 2017-02-17
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    I will wait a little bit, maybe someone else finds a mistake. :) But thank you very much so far! – 2017-02-17
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    My solution is wrong. $f = f_+ - f_-$, not $f = f_+ + f_-$ – 2017-02-18
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    ... but one doesn't need it, I can simply split up the sgn-function in the intervals. – 2017-02-18
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    I added the tag [[proof-verification]] – 2017-03-07

1 Answers 1

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This looks correct, well done! This is a pretty general argument, as you might notice that if we replace $sgn(xy)/(x^2+y^2)$ with any odd function of $x$ (or of $y$) the argument works just as well.