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The Dirac Delta Function

  1. Jan 24, 2006 #1
    I've recently come across this function in one of my science classes and am wondering were this identity comes from:
    [tex]\displaystyle{\int{\delta(t-\tau)f(\tau)d\tau}=f(t)} [/tex]
    Where [tex]\delta(t)[/tex] is the dirac delta function and f(t) is any (continuous?) function.
    Last edited: Jan 24, 2006
  2. jcsd
  3. Jan 24, 2006 #2


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    You only know how to deal with dirac deltas of the form [itex]\int \delta(x) g(x) \, dx[/itex], right? Well, there is an obvious substitution to try...
  4. Jan 24, 2006 #3
    Are you familiar with the properties of the Dirac delta function? This one is called the sampling property.
    Its comes from the fact that for any continuous function [itex]f(t)[/itex], the multiplication with the Dirac delta function yields
    [itex]f(t) \delta (t-t_0) = f(t_0) \delta(t-t_0)[/itex]

    Also the area under the curve of the Dirac Delta function is 1.

    [itex]\int_a^b f(t) \delta (t-t_0) dt = f(t_0) \int_a^b \delta (t-t_0) dt= f(t_0)[/itex]

    for [itex]a < t_0 < b[/itex]
  5. Jan 24, 2006 #4
    Actually, I don't know how to work with dirac deltas of any form...

    My professor just did some hand-waving about taking the limit of a sequence of functions with one peak and defining this to be the dirac delta function. Then he quoted the two properties we needed and proceeded. (This was one of them)

    Things I am not sure about:
    1). What is meant by the "limit of a seqence of functions"?
    2). How do we know this limit exists?
    3). How do we know it is a function... is it a function?
  6. Jan 24, 2006 #5
    Delta function is represented in several ways. For example

    [tex]\delta(x) = \lim_{n \rightarrow \infty} \frac {\sin nx} {\pi x}[/tex]
    [tex]\delta(x) = \lim_{n \rightarrow \infty} \frac n {\sqrt\pi} \exp(-(nx)^2)[/tex]

    It looks like [tex]\delta(0)=\infty[/tex], but actually you can integrate this function [tex]-\infty \rightarrow \infty[/tex]
    Last edited: Jan 24, 2006
  7. Jan 24, 2006 #6
    I know that the delta function can be represented in that way, my question is perhaps "what is special about those functions that causes their limit to be a function?"

    Certainly it is not true in general that the limit of a sequence of functions would be a function. (consider the limit as n-->oo of f(x)=nx. This would be a "function" with a vertical line for a graph, i.e. not a function at all).
    Last edited: Jan 25, 2006
  8. Jan 25, 2006 #7


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  9. Jan 25, 2006 #8

    matt grime

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    It isn't a function, at least in the sense of a map from R to R that you're used to. It is properly called a distribution, not a function. Think of it as an operator which when put inside an integral does something nice.

    The problem is the odd way applied maths abuses the word function, not in your understanding.

    It is bizzarely often called the derivative of the Heaviside step function which at least is a function though not a differentiable one (0 for x<=0, 1 for x>0)
  10. Jan 25, 2006 #9

    George Jones

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    There is a natural injection from the space of functions that can be integrated against test functions [**] into the space of distributions. When considered as distributions, the derivative of the Heaviside step function is the Dirac distribution.

    [**] This is the inspiration for the fictional, but extremely useful, integral notation for the Dirac distribution.

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