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Verifying the integral of a dirac delta function

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    I'll post it as an image since the notation will be tricky to type out. It's problem 4.

    http://img29.imageshack.us/img29/1228/307hw3.jpg [Broken]

    2. Relevant equations

    Not sure this really applies here


    3. The attempt at a solution

    This is for a physics course but as you can see it's pretty much math. If you think it'd do better in a math section please feel free to move it.

    Without really doing much math I considered the integral from -infinity to where the function is not zero, the integral in the region where it's not zero, and the integral from where it starts to be zero again to infinity. Adding them together should give me the total integral. The zero area multiplied by f(x) should still be zero (I think, right?). Therefore the only integral of concern is where the dirac function is not zero.

    Now what I think it's saying is that this integral should be the same as just evaluating f(x) at zero? If that's true I'm not really sure how to do it.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 22, 2011 #2

    CompuChip

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    That's correct.
    So
    [tex]\int_{-\infty}^{\infty} \mathrm dx \, \delta_n(x) f(x) = \int_{-1/(2n)}^{1/(2n)} \mathrm dx \, \delta_n(x) f(x) [/tex]
    and the rest drops out.

    Yep. You will encounter the Dirac "function" in a lot of places when you go and do physics. It doesn't really mean anything outside of an integral sign (although people use it like it does) and has the property that
    [tex]\int_{-\infty}^\infty \mathrm dx \, \delta(x - a) f(x) = f(a)[/tex]
    It is usually represented as an "infinite spike" at x = a, most of the time it represents some particle or source which is localised (i.e. point-like).

    I'm not sure what level of math you are at, but you could try Riemann sums or Taylor expansions, if those things mean anything to you.
     
    Last edited: Sep 22, 2011
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