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Suppose $f(x,y)$ is a function such that the double integral $$\int_{X\times Y} |f(x,y)|\,\mathrm dA$$ is infinite.

Then $\displaystyle \int_X\int_Y f(x,y)\,\mathrm dy\mathrm dx$ and $\displaystyle \int_Y\int_X f(x,y)\,\mathrm dx\mathrm dy$ may disagree. Wikipedia demonstrates this fact here.

But I don't understand how the double integral is computed.. $$\int_{X\times Y} |f(x,y)|\,\mathrm dA$$ In the article they just seem to do the iterated integral where they integrate in $x$ first. what am I getting wrong? How does anyone compute a double integral without doing iteration???

Git Gud
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michael_carbon
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2 Answers2

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There is no need to compute the double integral explicitly. The fact that the two iterated integrals are different guarantees (via Fubini's Theorem) that the double integral is infinite.

Martin Argerami
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  • But if I wanted to compute it directly, how can I do so using the product measure? – michael_carbon Dec 26 '13 at 21:12
  • There is no obvious way. One-dimensional integrals (and so iterated integrals) can be calculated by this miracle called the "Fundamental Theorem of Calculus", that lets you use antiderivatives to calculate integrals. There is no such thing in dimensions 2 and up. Of course you can always try and take a look at the Riemman sums and maybe get lucky that for some functions the limit can be inferred from there, but that will be the exception more than the rule. – Martin Argerami Dec 26 '13 at 21:32
  • Hmm. If there is no obvious way, then doesn't that mean that we can never really check the premise of Fubini theorem..? I guess one requires a bound like $|f(x,y)| \leq \text{const}$ to make it simple. – michael_carbon Dec 26 '13 at 21:37
  • Bounding an integral is most of the time way easier than calculating the exact value. For example, any continuous function on a compact subset of $\mathbb R^n$ will have finite integral. – Martin Argerami Dec 26 '13 at 22:08
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To get rid of the absolute value signs in $$\frac{x^2-y^2}{(x^2+y^2)^2}$$ we need to rule out where in $[0,1]\times [0,1]$ this is positive and negative. But $x^2-y^2=(x-y)(x+y)$. The term $x+y$ is always nonnegative, while $x-y$ is positive or negative according as $x>y$ or $x<y$. This is why in the article, they split the integral along the two triangles the line $x=y$ divides $[0,1]\times [0,1]$ into, and leave the appropriate sign in the integrand.

Pedro
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  • But, in the equation in the subsection "Proof", on the second line, haven't they integrated over x? But they cannot do this since they are only allowed to integrate with respect to the product measure as Fubini does not apply. – michael_carbon Dec 26 '13 at 21:15
  • @michael_carbon Sure, you can compute the iterated integrals. What you cannot assert is they equal the double integral. See Martin's answer. – Pedro Dec 27 '13 at 03:04