Double integral

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In mathematical analysis, there is an important distinction between a double integral and an iterated integral. To one who has had an advanced calculus course but not a measure-theoretic real analysis course, the difference may seem subtle.

1 Definitions
2 Counter example
3 Explanation via Lebesgue theory
4 In the positive sense
5 See also
6 External links

[edit] Definitions

On the one hand, a double integral

$\iint_{[a,b]\times[c,d]} f(x,y)\,d(x,y)$

is defined via a 2-dimensional measure in the plane, rather than by integrating twice (see Lebesgue integral).

On the other hand, if we define

$g(y)=\int_a^b f(x,y)\,dx$

then

$I=\int_c^d g(y)\,dy=\int_c^d\int_a^b f(x,y)\,dx\,dy$

is an iterated integral, so called because one integrates, and then integrates again.

[edit] Counter example

Does it matter whether one integrates first with respect to x and then with respect to y or vice-versa?

Perhaps surprisingly, in some cases yes, as an example shows:

$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy.$

Obviously the sign gets reversed if the order of iterated integration gets reversed, i.e., if "dy dx" replaces "dx dy". But the value of the integral is not zero, and so the values of the two iterated integrals differ from each other. For the details of the evaluation of this integral, see a counterexample related to Fubini's theorem.

[edit] Explanation via Lebesgue theory

To give the analytic explanation: the double integral exists only if

$\iint_{[a,b]\times[c,d]} \left|f(x,y)\right|\,d(x,y)<\infty,$

and in that case, the double integral coincides in value with either of the two iterated integrals. Thus, whenever the two iterated integrals differ in value from each other, the double integral of the absolute value of the function must be infinite. See Fubini's theorem.

[edit] In the positive sense

One can give a further explanation, however from the other direction, based on the special role of functions f(x)g(y).

These, in which the roles of the two variables are uncoupled, present no problem in this context; and neither do their linear combinations. Quite generally, given compact spaces X and Y, we can use the Stone-Weierstrass theorem to show that such functions give a subalgebra of C(X×Y) that is dense in the uniform norm: or in other words any continuous function on X×Y can be uniformly approximated by sums of functions f(x)g(y).