Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac delta function, change of variable confusion

  1. Aug 2, 2012 #1
    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 ?
     
  2. jcsd
  3. Aug 2, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. Aug 2, 2012 #3
    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'?
     
  5. Aug 2, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dirac delta function, change of variable confusion
  1. Dirac Delta function (Replies: 44)

  2. Dirac Delta function (Replies: 1)

Loading...