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Apparently there are integrals which you can express as $\int_A \int_B f(x, y) \mathop{}\!\mathrm{d}x \mathop{}\!\mathrm{d}y$ but not $\int_C \int_D f(x, y) \mathop{}\!\mathrm{d}y \mathop{}\!\mathrm{d}x$. When would this be the case? Is it to do with their limits not being invertible functions?

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    $\begingroup$ See Fubini's theorem. Most treatment include (or at least should include) a counterexample when its hypotheses fail. $\endgroup$ – anomaly Oct 31 '17 at 4:28
  • $\begingroup$ See also the list of related questions on this page. $\endgroup$ – Gerry Myerson Oct 31 '17 at 4:30
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There are pathological examples when the integral of the absolute value of $f$ over the whole space diverges. For instance, the example given in this question: if $$ f(x,y) = \begin{cases} e^{y-x}, & x > y \geq 0 \\ -e^{x-y}, & 0 \leq x \leq y \end{cases}$$ then $$ \int_0^\infty \left( \int_0^\infty f(x,y) \, dx \right) dy = 1$$ but $$ \int_0^\infty \left( \int_0^\infty f(x,y) \, dy \right) dx = -1.$$ The theorem used to determine whether you can switch the order of integration is Fubini's Theorem, which gives the condition that $\int_X\lvert f\rvert$ must converge, where $X$ is the whole space (in this case, $\mathbb{R}^2$).

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  • $\begingroup$ I meant something slightly different: when you can express the integral as dy dx in the first place, rather than whether the two integrals are not equal. $\endgroup$ – Bluefire Oct 31 '17 at 4:37
  • $\begingroup$ That is indeed slightly different. I don't know any cases offhand where one order of integration converges to the proper value but not the other, so I don't know that I can help. Fubini's theorem only deals with whether both orders converge, as far as I know. $\endgroup$ – Andrew Tindall Oct 31 '17 at 4:43

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