Dirac delta function, change of variable confusion

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Discussion Overview

The discussion revolves around the properties and implications of the Dirac delta function, particularly in the context of change of variables in integrals. Participants explore the mathematical definition and interpretation of the Dirac delta function as a generalized function or distribution, and the implications of these definitions on the validity of u-substitution in integrals involving the delta function.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the definition of the Dirac delta function and questions the validity of the result when applying a change of variable, specifically regarding the integral involving δ(cx).
  • Another participant argues that the concept of a "graph" for δ(x) is misleading, as δ(x) is not a function but a generalized function or distribution, which requires a different interpretation.
  • A further reply challenges the justification for using u-substitution with the Dirac delta function, questioning whether such substitutions are valid when dealing with distributions rather than traditional functions.
  • In response, a participant asserts that the definition of δ(cx) allows for the application of u-substitution, stating that it is defined such that the integral involving δ(cx) yields a specific result.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the Dirac delta function and the validity of applying u-substitution in integrals involving it. There is no consensus on the justification for these mathematical operations.

Contextual Notes

The discussion highlights the complexities and nuances involved in the mathematical treatment of the Dirac delta function, particularly regarding its classification as a generalized function and the implications for integration techniques.

bob900
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The Dirac delta "function" is often given as :

δ(x) = ∞ | x = 0
δ(x) = 0 | x \neq 0

and ∫δ(x)f(x)dx = f(0).

What about δ(cx)? By u=cx substitution into above integral is, ∫δ(cx)f(x)dx = ∫δ(u)f(u/c)du = 1/c f(0).

But intuitively, the graph of δ(cx) is the same as the graph of δ(x)! At x=0, cx=0 so δ(cx)=∞, and everywhere else cx\neq0 so δ(cx)=0.

So how can it be that ∫δ(x)f(x)dx = 1/c ∫δ(cx)f(x)dx ?
 
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The difficulty is talking about the "graph" of \delta(x) at all- which is a consequence of talking about \delta(x) as a function. There is, in fact, NO function of x that satisfies the conditions you give. It is a "generalized function" (also called "distribution"). A "generalized" function has to be interpreted as an operator on functions rather than a function itself.
 
HallsofIvy said:
The difficulty is talking about the "graph" of \delta(x) at all- which is a consequence of talking about \delta(x) as a function. There is, in fact, NO function of x that satisfies the conditions you give. It is a "generalized function" (also called "distribution"). A "generalized" function has to be interpreted as an operator on functions rather than a function itself.

Right, but then what justification do we have to do u-substitution to get the value of ∫δ(cx)f(x)dx, if u-substitution only applies to integrals of only functions, and not to integrals of functions multiplied with 'distributions'?
 
bob900 said:
Right, but then what justification do we have to do u-substitution to get the value of ∫δ(cx)f(x)dx, if u-substitution only applies to integrals of only functions, and not to integrals of functions multiplied with 'distributions'?

Because that is the definition of \delta(cx) it is defined exactly as the distribution such that

\int \delta(cx)f(x)=\frac{1}{c}f(0)
 

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