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I have the following integrals $\int_0^1 \int_1^\infty e^{-xy} - 2 e^{-2xy} \lambda(dx) \lambda(dy)$ and $\int_1^\infty \int_0^1 e^{-xy} - 2 e^{-2xy} \lambda(dy) \lambda(dx)$

and I shall prove, that both are not equal. What I did so far, is to calculate the inner Integral as a Rieman integral (But I am not sure why i should be allowed to do in case of the integral from 1 to infinity).

Then I look at the difference of the two with gives me $\int _0^1 \frac{e^{-y}}{y} -\frac{e^{-2y}}{y} \lambda \left( dy\right) + \int _1^\infty \frac{e^{-x}}{x} - \frac{e^{-2x}}{x} \lambda \left( dx\right)$ then I say that I can rename the variable $y$ and then have two integrals over the same functions on disjunkt intervals which therefore can be combined to $\int _{\left( 0,\infty \right) } \frac{e^{-x}}{x} -\frac{e^{-2x}}{x} \lambda \left( dy\right)$ Now I say, that the the function is positiv and $>0$ one a arbitary compact intervall therefore the integrals is for sure $>0$ and there not equals $0$, so the integrals are not equal.

However all this is quite a mess in my opinion. Maybe anyone has an idea how to show the integrals are not equal in a more elegant way?

And the next question is why Fubini is not appliable here. I first though that the function is not integrable, but it looks like it is. So I am somehow lost here too.

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Write your integrand as $ \left(1-2 e^{-xy} \right) e^{-xy}. $ So, what goes wrong with Fubini? Easy! The integrand changes sign! Try to compute $ \int_0^1 \left| e^{-xy}-2e^{-2xy} \right|\, dy, $ which splits as $ \int_0^{\frac{\log 2}{x}} \left( 2e^{-xy}-1 \right) e^{-xy}\, dy + \int_{\frac{\log 2}{x}}^1 \left( 1-2e^{-xy} \right) e^{-xy}\, dy. $ Unluckily, the first integral behaves like $\frac{1}{x}$, which is not integrable as $x \to +\infty$. Isn't this nice?

Edit: I am not making fun of anybody. I myself spent half an hour to understand why the integrals are different. I just mean that we tend to underrate the issue of positivity, when applying Fubini's theorem.

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    Thanks! I probably would have spend hours finding this...2012-06-27