What is the result of this double integral?

Click For Summary

Discussion Overview

The discussion revolves around the evaluation of a double integral expressed as the product of two separate integrals, one of which converges to zero and the other diverges to infinity. Participants explore the implications of this situation and the appropriate methods for addressing the resulting indeterminate form.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the expression is not a true double integral but rather a product of two integrals, questioning how to resolve the 0 times infinity form.
  • Another participant mentions L'Hospital's rule as a potential method for resolving limits but acknowledges that it may not apply directly to this case due to the lack of a single parameter.
  • There is a discussion about the definition of improper integrals, with one participant asserting that the expression represents the Cauchy Principal Value, while another insists on a different limit definition for improper integrals.

Areas of Agreement / Disagreement

Participants express disagreement regarding the classification of the integral and the appropriate methods for evaluation. There is no consensus on how to interpret or resolve the situation presented.

Contextual Notes

Participants highlight the complexity of the problem, noting the absence of a single parameter that leads to the 0 times infinity form and the differing definitions of improper integrals that may apply.

arroy_0205
Messages
127
Reaction score
0
Consider the double integral
<br /> \int_{-\infty}^{\infty}dx f(x) \, \int_{-\infty}^{\infty}dy g(y) <br />
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?
 
Physics news on Phys.org
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?
 
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.
 
Last edited:
arroy_0205 said:
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?
 
D H said:
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
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K