1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [/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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...