Dirac delta function, change of variable confusion

  • Thread starter bob900
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  • #1
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The Dirac delta "function" is often given as :

δ(x) = ∞ | x [itex]=[/itex] 0
δ(x) = 0 | x [itex]\neq[/itex] 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[itex]\neq[/itex]0 so δ(cx)=0.

So how can it be that ∫δ(x)f(x)dx = 1/c ∫δ(cx)f(x)dx ?
 

Answers and Replies

  • #2
HallsofIvy
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The difficulty is talking about the "graph" of [itex]\delta(x)[/itex] at all- which is a consequence of talking about [itex]\delta(x)[/itex] 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.
 
  • #3
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The difficulty is talking about the "graph" of [itex]\delta(x)[/itex] at all- which is a consequence of talking about [itex]\delta(x)[/itex] 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'?
 
  • #4
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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 [itex]\delta(cx)[/itex] it is defined exactly as the distribution such that

[tex]\int \delta(cx)f(x)=\frac{1}{c}f(0)[/tex]
 

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