Order of Indefinite Double Integrals

Kushwoho44

Hi,

Rather simple question here, just want to confirm:

When we are dealing with indefinite double integrals, it's true to say

∫∫ f(x,y) dx dy = ∫∫ f(x,y) dy dx

i.e, order of integration doesn't matter right?

Vargo

Lets try it with a function f(x,y)=1. Its indefinite integral with respect to x is x+c. The indefinite integral of that with respect to y is xy+cy+d.

If you reverse the order of the variables, you get xy+cx+d, a different answer.

The term "indefinite double integral" is not really used in mathematics. Probably for this reason. You can find the antiderivative with respect to a given variable, but there is no "double antiderivative" as this calculation shows. On the other hand, you can calculate definite integrals of multivariable functions. And you can reverse the order of integration when you calculate those (Fubini's Theorem).

Michael Redei

You can't always reverse that order, e.g.
$$
\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\mathrm dx\,\mathrm dy = -\frac\pi4 \quad\mbox{ but } \quad \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\mathrm dy\,\mathrm dx = +\frac\pi4
$$

Only when the two definite integrals
$$
\int_{y_1}^{y_2}\int_{x_1}^{x_2}\Big|f(x,y)\Big|\, \mathrm dx\,\mathrm dy <+\infty \quad\mbox{ and } \quad \int_{x_1}^{x_2}\int_{y_1}^{y_2}\Big|f(x,y)\Big|\, \mathrm dy\,\mathrm dx <+\infty
$$
exist, then they are equal, and also equal to
$$
\int_{y_1}^{y_2}\int_{x_1}^{x_2}f(x,y)\,\mathrm dx\,\mathrm dy = \int_{x_1}^{x_2}\int_{y_1}^{y_2}f(x,y)\,\mathrm dy\,\mathrm dx
$$

Kushwoho44

Interesting, thanks for both replies guys.