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Thread: Integral of Dirac delta distribution over finite region

  1. #1
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    Integral of Dirac delta distribution over finite region

    Hi there. I was wondering about what happens if one integrates the Dirac delta in a finite region.

    It is well known that:
    \int_{-\infty}^{\infty} \delta (x)dx=1

    and:
    \int_{-\infty}^{\infty} f(x) \delta (x)dx=f(0)

    Now, what happens if the integration is defined in a finite region, for example:

    \int_{-5}^{5} \delta (x)dx=?

    \int_{-5}^{5} f(x) \delta (x)dx=?

    Or even:
    \int_{0}^{x_f} f(x) \delta (x)dx=?

    Does this has a meaning at all?
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  2. #2
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    Re: Integral of Dirac delta distribution over finite region

    $\forall \epsilon > 0,~\displaystyle \int_{-\epsilon}^\epsilon~\delta(\tau)~d\tau = 1$

    $\forall \epsilon > 0,~\displaystyle \int_{-\epsilon}^0~\delta(\tau)~d\tau = \int_0^\epsilon~\delta(\tau)~d\tau = \dfrac 1 2$

    $\forall \epsilon > 0, ~\displaystyle \int_{-\epsilon}^\epsilon~f(\tau) \delta(\tau)~d\tau = \int_0^\epsilon~f(\tau) \delta(\tau)~d\tau = \int_{-\epsilon}^0~f(\tau) \delta(\tau)~d\tau=f(0)$
    Last edited by romsek; Jul 6th 2017 at 11:05 AM.
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  3. #3
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by romsek View Post
    \forall \epsilon > 0,~\displaystyle \int_{-\epsilon}^0~\delta(\tau)~d\tau =  \int_0^\epsilon~\delta(\tau)~d\tau = \dfrac 1 2

    \forall \epsilon > 0, ~\displaystyle \int_{-\epsilon}^\epsilon~f(\tau) \delta(\tau)~d\tau =  \int_0^\epsilon~f(\tau) \delta(\tau)~d\tau =  \int_{-\epsilon}^0~f(\tau) \delta(\tau)~d\tau=f(0)
    With f(x)=1, you have a contradiction between the two lines here.

    \int_{-a}^a f(x)\delta(x)\,\mathrm d x = f(0) is a trivial result when you consider the value of \delta(x) on the intervals (-\infty,-a) and (a,\infty).
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  4. #4
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by Archie View Post
    With f(x)=1, you have a contradiction between the two lines here.

    \int_{-a}^a f(x)\delta(x)\,\mathrm d x = f(0) is a trivial result when you consider the value of \delta(x) on the intervals (-\infty,-a) and (a,\infty).
    yeah... my mistake

    $\displaystyle \int_{-\epsilon}^0 f(\tau)\delta(\tau)~d\tau = \int_0^\epsilon f(\tau)\delta(\tau)~d\tau = \dfrac 1 2 f(0)$
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  5. #5
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    Re: Integral of Dirac delta distribution over finite region

    Are there different interpretations of this integral? Wolframalpha seems to think:

    $\displaystyle \int_0^\epsilon \delta(\tau)d\tau = \int_{-\epsilon}^0 \delta(\tau)d\tau = 1$

    Wolfram|Alpha: Computational Knowledge Engine

    Wolfram|Alpha: Computational Knowledge Engine

    Is that a mistake? I'm sure Wolfram can fix it if it is.
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  6. #6
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by romsek View Post
    yeah... my mistake

    $\displaystyle \int_{-\epsilon}^0 f(\tau)\delta(\tau)~d\tau = \int_0^\epsilon f(\tau)\delta(\tau)~d\tau = \dfrac 1 2 f(0)$
    Now that I think about it this result was key to showing that the Fourier Series converges to the average of the left and right hand limits at discontinuities.
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  7. #7
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by SlipEternal View Post
    Are there different interpretations of this integral? Wolframalpha seems to think:

    $\displaystyle \int_0^\epsilon \delta(\tau)d\tau = \int_{-\epsilon}^0 \delta(\tau)d\tau = 1$

    Wolfram|Alpha: Computational Knowledge Engine

    Wolfram|Alpha: Computational Knowledge Engine

    Is that a mistake? I'm sure Wolfram can fix it if it is.
    every representation of the delta function I've ever seen has shown it as symmetric with respect to the origin.

    It's considered the limit of a symmetric distribution where the standard deviation tends to zero.

    If you integrate one side or the other you get 1/2

    You can stuff your snide little attitude asshole.
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  8. #8
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by SlipEternal View Post
    Are there different interpretations of this integral? Wolframalpha seems to think:

    $\displaystyle \int_0^\epsilon \delta(\tau)d\tau = \int_{-\epsilon}^0 \delta(\tau)d\tau = 1$

    Wolfram|Alpha: Computational Knowledge Engine

    Wolfram|Alpha: Computational Knowledge Engine

    Is that a mistake? I'm sure Wolfram can fix it if it is.
    When I entered the link it says that the integral from 0 to 1 gives \theta(0), where theta is the heaviside step function. The heaviside step function is 1/2 at zero: https://en.wikipedia.org/wiki/Heaviside_step_function

    It actually depends on the definition of the Heaviside function, but I'm pretty sure that the definition wolfram handles is the one with 1/2 at zero.
    Last edited by Ulysses; Jul 6th 2017 at 03:38 PM.
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  9. #9
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by Ulysses View Post
    When I entered the link it says that the integral from 0 to 1 gives \theta(0), where theta is the heaviside step function. The heaviside step function is 1/2 at zero: https://en.wikipedia.org/wiki/Heaviside_step_function

    It actually depends on the definition of the Heaviside function, but I'm pretty sure that the definition wolfram handles is the one with 1/2 at zero.
    Nope. I tried heavysidetheta(0) and it gave one. I think it is an error at wolframalpha.
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  10. #10
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    Re: Integral of Dirac delta distribution over finite region

    Quote Originally Posted by romsek View Post

    You can stuff your snide little attitude asshole.
    I was actually asking in earnest, Mr. Paranoid.
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  11. #11
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    Re: Integral of Dirac delta distribution over finite region

    Relax guys , everything is fine. Thank you both.
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  12. #12
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    Re: Integral of Dirac delta distribution over finite region

    I contacted Wolframalpha and asked that they include both answers as a result for Heavisidetheta[0].
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