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Homework Help: Calculate the Dirac delta function integral

  1. Jan 3, 2018 #1
    Last edited by a moderator: Jan 3, 2018
  2. jcsd
  3. Jan 3, 2018 #2

    PeroK

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  4. Jan 3, 2018 #3
  5. Jan 3, 2018 #4

    PeroK

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  6. Jan 3, 2018 #5
    yes it has
     
  7. Jan 3, 2018 #6

    PeroK

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    Okay, so does that not give you a clue as to how to evaluate the integral? Hint: trying splitting the integral.

    Note: you'll also have to think about a substitution to get the ##\delta## function into a simpler format.
     
  8. Jan 3, 2018 #7
  9. Jan 3, 2018 #8
    I did not get this Note: you'll also have to think about a substitution to get the ##\delta## function into a simpler format.[/QUOTE]
     
  10. Jan 3, 2018 #9

    PeroK

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    [/QUOTE]

    Think about how to calculate, for example:

    ##\int_{-\infty}^{+\infty} f(x)\delta(x^2)dx##
     
  11. Jan 3, 2018 #10
    it will be f(0)
     
  12. Jan 3, 2018 #11

    PeroK

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    Try the substitution ##y = x^2##.

    In fact, a simpler example to try first is:

    ##\int_{-\infty}^{+\infty} f(x)\delta(2x)dx##
     
  13. Jan 3, 2018 #12
    okay i got it
    when y =2x then dy = 2 dx so the integral will be 1/2 f(0)
     
  14. Jan 3, 2018 #13
    is the discussion over?
     
  15. Jan 3, 2018 #14

    PeroK

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    I was at the dentist!

    So, you have to do something with that quadratic in the delta function.
     
  16. Jan 3, 2018 #15
    what is wrong if i do it like this

    ##\int_∞^1 (x^2+1) δ(x^2-3x+2) \, dx##+##\int_1^2(x^2+1) δ(x^2-3x+2) \, dx##+##\int_2^\infty(x^2+1) δ(x^2-3x+2) \, dx## (in the first integral the limits will be from - ∞ to 1)
    = 0+2+5=7
    since in the first integral the dirac delta function is zero and in the second and third it will be one
     
  17. Jan 3, 2018 #16

    PeroK

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    You have the same problem that ##\delta(x^2 - 3x + 2) \ne \delta(x)##. That first function is not the Dirac Delta function, it's a composition function and doesn't have exactly the same properties as the Dirac Delta function.
     
  18. Jan 3, 2018 #17
    so now if i consider ##x^2-3x+2 = y## then it will give me (2x-3)dx=dy and if i substitute dx=##\frac{dy}{2x-3}##
    x will become ##\frac{3\pm\sqrt{1+y}}{2}## and ##x^2##= ##\frac{(10+y)\pm6\sqrt{1+y}}{4}##


    then my integral will look like
    ##\int_{-\infty}^{+\infty}\frac{(\frac{(10+y)\pm6\sqrt{1+y}}{4}+1)δ(y)}{\pm\sqrt{1+y}}\,dy##
    here if we take + signs into account then the answer comes out to be 5 and if we consider - signs then the answer is -2 which is not appropriate because the integral of dirac delta function should be a unit step function and which is one if y is greater than or equal to zero and everywhere else it is zero so the positive 5 should be considered as correct answer . i think so
    is this correct?
     
  19. Jan 3, 2018 #18
    ##δ(x^2-3x+2)≠δ(x)## because δ(x) has only one spike at zero while ##δ(x^2-3x+2)## has two spikes at 1 and 2 .
    But i have a doubt what is the need of equating these two?
     
  20. Jan 3, 2018 #19

    PeroK

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    This was better. You need to split the integral before you look at substitutions. I would do the first integral from ##0## to ##3/2## and the second from ##3/2## to ##3##, say. That seems a lot neater.

    Then you can use the ##y## substitution. Although, there is also a trick to leave the x-function as it is. Since you are not going to integrate it, it won't change. This is what I mean:

    With ##y = x^2 - 3x + 2 = (x - 3/2)^2 - 1/4## and ##dy = (2x - 3)dx##

    ##\int_0^{3/2}(x^2 + 1) \delta(x^2 - 3x + 2)dx = \int_2^{-1/4} \frac{x^2 +1}{2x-3} \delta(y) dy = -\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy = -\frac{x^2 +1}{2x-3}|_{y=0} = -\frac{x^2 +1}{2x-3}|_{x=1} = 2##

    Notice that because ##2x-3 = -1## you get same answer,. This is a coincidence. If you change the quadratic it won't be the same.
     
  21. Jan 3, 2018 #20
    how did you chose the limits of integration from 0 to 3/2 and 3/2 to3
     
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