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

'delta' integral?

  1. Aug 14, 2010 #1
    I'm not even sure if that's the right name, but my question is when you have a [tex]\delta[/tex] under the integral.

    For example,
    [tex]\int\limits_{-\infty}^{\infty} ln(x+3) \delta (x+2) \, dx[/tex]

    Without the [tex]\delta[/tex] the integral is easy enough (I think) using a u-substitution (u=x+3) then it is [tex](x+3) \ln (x+3) - (x+3) +C [/tex] but I don't know between the limits..
  2. jcsd
  3. Aug 14, 2010 #2
    Actually, without the delta-function, the integral is not "simple", because it doesn't have meaning! (the logarithm is not defined for x < -3). Ignoring this, the integral would be even more simple: the delta-function has the property:


    so your integral is log(-2 + 3) = 0.
  4. Aug 17, 2010 #3
    So in general you just take the function f of the sum of the things in brackets, where the second must be in the form [tex](x-x_0)[/tex] and ignore the x. But that brings two questions:

    1. If you have a factor for x which are different for the f and the delta, like say nx or [tex] x^m [/tex] ?

    2. It looks like the integral limits don't affect anything except as a definition of the above property, but what happens if you have finite limits? For example

    \int_{a}^{b}f(x)\delta(x-x_0)= ?
    Last edited: Aug 17, 2010
  5. Aug 17, 2010 #4
    1) I don't understand this question.

    2) if x_0 belongs to the interval [a,b], then the integral is f(x_0), otherwise it's zero.
  6. Aug 17, 2010 #5


    User Avatar
    Science Advisor

    Do you mean something like [itex]\int x^n \delta(x- a)dx[/itex]. That's no different. x- a= 0 when x= a so the integeral is [itex]a^n[/itex]. If you have, say [itex]\int f(x)\delta(g(x))dx[/itex], where f(x) and g(x) can be any reasonable functions, then its a bit more complicated. The integral will be [itex]\sum f(x_i)[/itex] where the sum over all [itex]x_i[/itex] within the interval of integration such that [itex]g(x_i)= 0[/itex].

    [itex]\int_a^b f(x)\delta(x-x_0)dx= f(x_0)[/itex] if [itex]a\le x_0\le b[/itex], 0 otherwise.
  7. Aug 17, 2010 #6
    Sorry I didn't give an example for 1, but what I meant is if you have a general form [tex]\int_{a}^{b} f(x) \delta (mx^n + c)
    how do you relate your [tex]x[/tex] and [tex]x - x_0[/tex]?

    Will the integral be non-zero if [tex]-c \in \[ a,b \] [/tex] and the result just be [tex] f(mx^n) [/tex]?

    So that's basically asking if its also linear; if
    [tex]\delta (x+2) = \delta x \plus \delta 2[/tex] and
    [tex]\delta 3x = 3 \delta x [/tex]
    Last edited: Aug 17, 2010
  8. Aug 17, 2010 #7


    User Avatar
    Homework Helper

    You missed writing a factor of [itex]1/|g'(x_i)|[/itex] in the sum. It should be

    [tex]\int_a^b dx f(x)\delta(g(x)) = \sum_i \frac{f(x_i)}{|g'(x_i)|}[/tex]
    where the x_i are the values of x in the interval [itex]a \leq x \leq b[/itex] such that [itex]g(x_i) = 0[/itex], so long as g'(x_i) is non-zero.

    For the OP, this answers your question: g(x) = mx^n + c, so the x_i are [itex]x_i = (-c/m)^{1/n}[/itex] when they are real. g'(x) = mnx^(n-1), so for all real x_i in the inteval, you get

    [tex]\sum_i \frac{f(x_i)}{|mn x_i^{n-1}|}[/tex]
  9. Aug 17, 2010 #8
    Looks to me the OP is missing the whole concept of the delta function. It's a limiting process of a set of functions which tend to a particular form as some parameter is varied. First study the Wikipedia reference:


    then re-interpret your integral from that perspective. In fact I think it would be a good exercise to numerically compute the integral for various cases, say let a go from 1 to 1/20 for appropriate choice of the sequence of functions which tend to delta and then show how the integral tends to f(x_0): I'll leave interpretation of the following Mathematica code to the interested reader where I used log(x+5) so it's different from zero.

    Code (Text):

    mydelta[a_, x_] := Exp[-(x^2/a^2)]/

    mytable = Table[
       {a, Re[NIntegrate[Log[x + 5]*
           mydelta[a, x + 2], {x, -100, -5,
           100}, Method ->
       {a, 1/20, 1, 1/20}]

    myval = Log[3];

    Show[{ListPlot[mytable, Joined -> True],
       Graphics[{Red, Line[{{0, myval},
           {1, myval}}]}]}]
    Last edited: Aug 17, 2010
  10. Aug 17, 2010 #9
    Thanks that helps a lot

    My only other question is a solution for higher order delta function integrals
    [tex]\int f(x) {\delta}^3 d^3 x [/tex]
  11. Aug 18, 2010 #10


    User Avatar
    Science Advisor

    And that is simply f(0, 0, 0).
  12. Aug 18, 2010 #11


    User Avatar
    Homework Helper

    What do you mean by "[itex]\delta^3[/itex]"? You need to write the arguments. Typically one uses the notation [itex]\delta^3(\mathbf{x}-\mathbf{x}_0)[/itex] to mean the 3d dirac delta function [itex]\delta(x-x_0)\delta(y-y_0)\delta(z-z_0)[/itex]. This is what HallsofIvy took your question to mean. However, it seems to me like you're asking about the actual cube of a delta function: [itex][\delta(x-x_0)]^3[/itex]. The answer to this interpretation is that the delta function is not an actual function. It is a "distribution" or "generalized function", and multiplying distributions is typically not well defined. So, the answer to your question is that your integral has no meaning. If you use the property of the delta function on your integral you find

    [tex]\int dx f(x)[\delta(x-x_0)]^3 = f(x_0)[\delta(0)]^2[/tex]
    which is meaningless unless regulated somehow (e.g., using a limiting process like jackmell described, in which case the integral would likely diverge). The delta function really only exists properly inside of integrals.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook