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Validity of integral involving delta function

  1. Aug 23, 2014 #1
    Hi,

    Is the following integral well defined? If it is, then what does it evaluate to?

    [tex] \int_{-1}^{1} \delta(x) \Theta(x) \mathrm{d}x [/tex]

    where [itex]\delta(x)[/itex] is the dirac delta function, and [itex]\Theta(x)[/itex] is the the Heaviside step function.

    What about if I choose two functions [itex]f_k[/itex] and [itex]g_k[/itex], which are such that [itex] f_k \rightarrow \delta(x) [/itex] and [itex] g_k \rightarrow \Theta(x) [/itex]? Will this integral converge? I understand that if i choose [itex]f_k[/itex] and [itex]g_k[/itex] such that [itex]f_k = g'_k[/itex] then this integral will evaluate to [itex]1/2[/itex], but what if i choose other independent representation of these functions?
     
    Last edited: Aug 23, 2014
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  3. Aug 23, 2014 #2

    Simon Bridge

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  4. Aug 23, 2014 #3
    It isn't really translation. the Theta function makes a reasonable difference. And i'm evaluatig both funcitons without any delay. Could you elaborate how it is a case of translation?

    The integral is coming about in my attempts to solve a differential equation of a particular form (I had asked a question about this earlier: https://www.physicsforums.com/showthread.php?t=760961). In trying to evaluate a some boundary conditions of this equation i'm ending up evaluating integrals of this form.
     
  5. Aug 23, 2014 #4

    Simon Bridge

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    $$\int_{-1}^1 \Theta(x)\delta(x)\; dx = \int_{-\infty}^\infty \Theta(x)\delta(x)\; dx - \int_{-\infty}^1\Theta(x)\delta(x)\; dx - \int_1^\infty \Theta(x)\delta(x)\; dx$$
     
  6. Aug 23, 2014 #5
    Um.. Could you elaborate on how this helps?
     
  7. Aug 23, 2014 #6

    pwsnafu

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    IIRC, we have ##\delta \cdot H = \frac{1}{2}H##, so
    ##\int_{-1}^{1} \delta(x)\, H(x) \, \phi(x)\, dx = \int_{-1}^{1}\frac{1}{2}H(x) \, \phi(x) = \frac{1}{2}\int_{0}^{1}\phi(x)\, dx##
    for any test function ##\phi \in \mathcal{D}(-1,1)##. Set ##\phi(x)=1## and we get 1/2.
     
  8. Aug 23, 2014 #7
    And how do you prove this statement? I would be more inclined to believe it to be ##\frac{1}{2}\delta##, instead. But i can't prove that either.
     
  9. Aug 23, 2014 #8

    pwsnafu

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    Doh! You're right it would be ##\frac{1}{2}\delta## wouldn't it?
    supp##(\delta\cdot H)## = supp##(\delta)\cdot##supp(H) ##=\{0\}##, so it can't be Heaviside.

    When I go to uni on Monday I'll look it up for you.
     
  10. Aug 23, 2014 #9

    Simon Bridge

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  11. Aug 23, 2014 #10

    HallsofIvy

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    The [itex]\delta[/itex] "function" is NOT a true function. It is a "distribution" or "generalized function". Those are defined as operators on functions. Specifically, if f is any function defined on a closed and bounded measurable set that includes "0" then [tex]\delta(f)= f(0)[/tex]. Given a "regular function, g(x), we can define a corresponding generalized function using and integral- the generalized function G is given by [itex]G(f)= \int_{-\infty}^{\infy}f(t)f(t)dt[/itex]. But most generalized functions, including the "[itex]\delta[/itex] function" cannot be written that way. The Polish mathematician, Mikuzinski, showed how to "construct" the generalized functions from regular functions in much the way the real numbers can be "constructed" from the rational numbers using equivalence classes of sequences.
     
  12. Aug 25, 2014 #11

    pwsnafu

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    Okay it is indeed ##\frac{1}{2}\delta##. A proof sketch follows:

    Let ##\rho_n(x)## be a delta sequence. Then define
    ##H(x) = \int_{-\infty}^x \rho_n(t) \, dt##.
    We see that ##\lim_{n\to\infty}H^2_n = H## and hence
    ##\lim_{n\to\infty} 2 H_n(x) \rho_n(x) = \delta(x)##.
     
  13. Sep 4, 2014 #12
    The problem is ill-defined. Considering such well-defined “deformations” of the expression under integral as [itex]\delta(x-x_0) \Theta(x),\ x\in{\mathbb R}\setminus\{0\} [/itex], we’ll obtain 0 for negative x0 and [itex]\delta(x-x_0)[/itex] for positive x0.
     
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