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Integral involving Dirac Delta generalized function

  1. Feb 12, 2015 #1
    • Member warned about posting with no effort
    1. The problem statement, all variables and given/known data
    Evaluate the integrals in the attached image

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2015 #2

    RUber

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    The dirac delta function has the great property that ##\int_{-\infty}^{\infty} \delta(x) dx = 1 ##.
    Couple that with what you know about when it is not equal to zero, and these integrals are quick and easy.
    In the future, please put a little more meat into you post, so we can better understand what you already know and what you need help fixing.
     
  4. Feb 12, 2015 #3

    thanks . I already know this property and many others (including the one in this new attached picture) . If I apply that rule (image) , i would get 1 in the first integral while many others told me 0 . I have problem with the first 2 integrals only .
     

    Attached Files:

  5. Feb 12, 2015 #4

    RUber

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    Using the rule you attached, you should get 1 in the first integral. What about the second?
     
  6. Feb 12, 2015 #5
    The second one would be 0 since -0.5 is not in the interval of integration .
    I found a solution to the first integral which totally confused me ( in the attached image) claiming that the answer should be 0 for 1st integral .
     

    Attached Files:

  7. Feb 12, 2015 #6

    Matterwave

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    Where did you get that image? That is not how the Dirac Delta is defined...
     
  8. Feb 12, 2015 #7
    It is a Chegg solution . And it confused me .
    I believe that the first integral should be 1 , i just need more proof
     
  9. Feb 12, 2015 #8

    Dick

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    I think the proof is the rule you quoted in post 3. The Chegg solution appears to be for a unit step function, not for a delta function. It's wrong.
     
  10. Feb 12, 2015 #9

    Matterwave

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    Using the fact that $$\int_a^b f(x)\delta(x-c)dx = f(c),~\text{iff}~~a<c<b$$ we can get immediately $$\int_0^5 \cos(2\pi t)\delta(t-2)dt=\left.\cos(2\pi t)\right|_{t=2}=\cos(4\pi)=1$$

    The Chegg solution is clearly wrong. The function used there is not a dirac delta function.
     
  11. Feb 13, 2015 #10

    stevendaryl

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    There is a related function, called the "Heaviside step function" or just the "step function":

    [itex]x < 0 \Rightarrow \Theta(x) = 0[/itex]

    [itex]x = 0 \Rightarrow \Theta(x) = 1/2[/itex]

    [itex]x > 0 \Rightarrow \Theta(x) = 1[/itex]

    The definition in that image seems correct for the step function, but not the delta function. The two are related by:

    [itex]\Theta(x) = \int_{-\infty}^x \delta(s) ds[/itex]
     
  12. Feb 13, 2015 #11

    RUber

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    I agree with Dick and Matterwave. They were defining the Heavyside function which is related to the dirac delta by:
    ##H(x) = \int_{-\infty}^x \delta(t) dt## or ##\frac{d}{dx} H(x) = \delta(x) ##

    Edit: Got scooped by stevendaryl...Exactly.
     
  13. Feb 13, 2015 #12
    Thank you guys for your contribution :)
    I'll be posting a new topic regarding convolution integral :)
     
  14. Feb 13, 2015 #13

    Mark44

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    @Legend101, if you start another thread with no effort shown, it will be deleted.
     
  15. Feb 13, 2015 #14
    No worries . I showed my efforts in attached file
     
  16. Feb 13, 2015 #15

    Mark44

    Staff: Mentor

    You need to show what you have tried in the first post of the thread. And it would be better to include the work directly in the post, rather than an image of the work.
     
  17. Feb 13, 2015 #16
    The work involves many integrals and mathematical notations . It would be hard to show all the work directly written on the post especially if someone is using a smartphone .
     
  18. Feb 13, 2015 #17

    Mark44

    Staff: Mentor

    See my post in that thread.
     
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