Integral depending on coordinate differences

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Homework Help Overview

The discussion revolves around a double integral of a function that depends on the difference between two variables, specifically the expression \(\int_a^b \int_a^b f(x-y)\, dxdy\). Participants are exploring potential simplifications, particularly the possibility of reducing the expression to a one-dimensional integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using a change of variables to simplify the integral, proposing a transformation involving \(u = x - y\) and \(v = x + y\). Others question the effectiveness of this approach and raise concerns about the proper application of the Jacobian in the context of double integration.

Discussion Status

The discussion is ongoing, with participants providing guidance on the change of variables technique while also expressing uncertainty about its application. There is recognition of the need to correctly apply the Jacobian factor and adjust the limits of integration, indicating a productive exploration of the topic.

Contextual Notes

Participants are navigating the complexities of double integrals and the implications of variable transformations, highlighting a lack of consensus on the best approach to simplify the integral.

Irid
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Homework Statement


Consider a function which depends only on a difference between two variables, and integrate it with respect to both:
[tex] \int_a^b \int_a^b f(x-y)\, dxdy[/tex]
Is there any way to simplify this expression, like reducing it into a 1-D integral?
 
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Irid said:

Homework Statement


Consider a function which depends only on a difference between two variables, and integrate it with respect to both:
[tex] \int_a^b \int_a^b f(x-y)\, dxdy[/tex]
Is there any way to simplify this expression, like reducing it into a 1-D integral?

Use a change of variables like u=x-y, v=x+y. That will reduce it to a single integration over u after you do the dv integration.
 
This gives me
[tex]dxdy = (dv^2-du^2)/4[/tex]
and I don't see how this makes the integral any easier
 
Irid said:
This gives me
[tex]dxdy = (dv^2-du^2)/4[/tex]
and I don't see how this makes the integral any easier

That's not how you do change of variables in double integration. dxdy is equal to dudv times a Jacobian factor, remember? http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Then you have to change the limits.
 
Oh, thanks Dick, I wasn't aware of this...
 

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