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Given the function

$f: \Bbb R^2 \rightarrow \Bbb R$,

$f(x, y) :=$ $1 \over y^2$, $0 < x < y < 1$,

$f(x, y) :=$ $- 1 \over x^2$, $0 < y < x < 1$,

$f(x, y) := 0$, otherwise,

I would like to calculate

$\int_0^1 \int_0^1 f(x, y) d\lambda(x) d\lambda(y)$.

I know that I can apply the theorem of Fubini in this case. Futhermore, since $[0, 1]$ is compact, we can treat the Lebesgue-integral as an Riemann-integral. I guess we need to do a case analysis, so let's assume that $x < y$ first. We receive:

$\int_0^1 \int_0^1 f(x, y) d\lambda(x) d\lambda(y)$ $= \int_0^1$ $(\int_0^1$ $1 \over y^2$ $dx) \ dy$ $= \int_0^1$ $1 \over y^2$ $dy =$ [$-1 \over y$]$_0^1$.

The problem is that I am not allowed to divide by zero, so I guess I have to write the integral differently from the very beginning. But how do I need to do that?

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Julian
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  • "I know that I can apply the theorem of Fubini in this case" Can you? Why not rather compute $$\int_0^1 f(x, y) d\lambda(x)$$ for every $y$ in $(0,1)$, and see what happens? – Did Jan 06 '17 at 12:17
  • Which of $x$ and $y$ is the smaller one changes at some point of the inner integral. Split it, $$\int_0^1 f(x,y),d\lambda(x) = \int_0^y f(x,y),d\lambda(x) + \int_y^1 f(x,y),d\lambda(x).$$ – Daniel Fischer Jan 06 '17 at 12:35
  • How would I compute them now though? – Julian Jan 06 '17 at 12:38
  • You know, you will have to do something yourself at one point or another... Already my hint was quite specific, but now that @DanielFischer made it even more explicit, nothing is preventing you to make some progress. – Did Jan 09 '17 at 11:04

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