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Thread: Dirac delta function

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Dirac delta function

    Could someone explain to me why $\displaystyle \int_{-\infty}^\infty f(x)\delta(x)\, dx = f(0) $.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Could someone explain to me why $\displaystyle \int_{-\infty}^\infty f(x)\delta(x)\, dx = f(0) $.
    Which definition of the Dirac delta function are you using? The one with limits?
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Which definition of the Dirac delta function are you using? The one with limits?
    $\displaystyle \delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases} $

    s.t. $\displaystyle \int_{-\infty}^\infty \delta(x) \, dx = 1 $.

    I think I have a solution, but I'm afraid the way $\displaystyle \delta $ is defined makes my argument invalid.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    $\displaystyle \delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases} $

    s.t. $\displaystyle \int_{-\infty}^\infty \delta(x) \, dx = 1 $.

    I think I have a solution, but I'm afraid the way $\displaystyle \delta $ is defined makes my argument invalid.
    Well, to be honest with you I am no expert on this field. But, I was under the impression that the integral you quoted is an abuse of notation considering the way $\displaystyle \delta$ is defined makes it impossible for $\displaystyle \int_{-\infty}^{\infty}f(x)\delta(x)dx$ to be an actual Riemann or Lebesgue integral.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Well I suppose $\displaystyle \delta(x) = \lim_{a\to 0}\frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2} $.
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