2 variable delta function integration

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SUMMARY

The discussion focuses on the integration of a two-variable delta function, specifically the integral \(\int^{A}_{-A}\int^{Bx}_{-Bx}c\delta(x\cos\varphi+y\sin\varphi-d)dydx\). Participants highlight the challenge of integrating delta functions with variable limits and constants A, B, c, and d. A key insight shared is the property \(\delta(af(x)) = \frac{1}{|a|}\delta(f(x))\), which allows for factoring out the cosine term before performing the x integration. Caution is advised regarding the limits of integration and the behavior of the delta function within those limits.

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Homework Statement



[tex]\int^{A}_{-A}[/tex][tex]\int^{Bx}_{-Bx}c\delta(xcos\varphi+ysin\varphi-d)dydx[/tex]
where A, B, c, d are constant

Homework Equations





The Attempt at a Solution


I have tried a few different ways to integrate this, but am completely confused with what happens to this kind of delta function when you integrate it. I know integrating a delta function usually gives you 1 but I don't think this can work in this case. The answer has A,B,c and d in it, so the limits must be used somewhere. This is one step in a much longer problem, but it is frustrating to get close to the end and get stuck because I can't find anything anywhere about delta functions of 2 variables. Some help would really be appreciated.
 
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Did you or anyone else figure out how to deal with this? I have the same problem.
 
One could, for example, use the following property of a delta function:

[tex]\delta(af(x)) = \frac{1}{|a|}\delta(f(x))[/tex]

to factor out the cosine in the argument of the delta function and then perform the x integration. The x integration is then easy, but there's a trick - you don't know for sure if the delta function argument is zero inside the limits of x integration, so you'll have to think carefully about that.
 

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