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arroy_0205
Jun23-08, 01:57 AM
Consider the double integral

\int_{-\infty}^{\infty}dx f(x) \, \int_{-\infty}^{\infty}dy g(y)

The first one gives 0 the second one gives infinity (diverges). Then how to express the result of the integral? Is it 0 or infinity or neither (indeterminate)? Any other comments about the integration?

D H
Jun23-08, 06:59 AM
What you wrote isn't so much a double integral as it is a product of two integrals. What rule from basic calculus can be used to resolve things that tend to 0\cdot\infty as some parameter tends to zero or infinity?

arroy_0205
Jun23-08, 10:15 AM
It was really wrong to call that a double integral.

L'Hospital rule come to my mind as the answer to your question but that is applicable in calculating limit problems. This is case different. I do not know of any method applicable here.

Also there is no "single parameter" in the problem which gives rise to 0\cdot\infty form.

D H
Jun23-08, 10:28 AM
L'Hospital rule come to my mind as the answer to your question but that is applicable in calculating limit problems. This is case different. I do not know of any method applicable here.
The improper definite integral
\int_{-\infty}^{\infty}f(x)\,dx
is shorthand for
\lim_{L\to\infty}\int_{-L}^{L}f(x)\,dx
So, how is this case any different?

HallsofIvy
Jun23-08, 10:37 AM
The improper definite integral
\int_{-\infty}^{\infty}f(x)\,dx
is shorthand for
\lim_{L\to\infty}\int_{-L}^{L}f(x)\,dx
So, how is this case any different?
NO! that is the "Cauchy Principal Value". The correct definition is
\lim_{A\to\infty}\lim_{B\to\infty}\int_{B}^{A}f(x) \,dx