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Homework Help: Calculate the Dirac delta function integral

  1. Jan 3, 2018 #21

    PeroK

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    They can be anything, as long as they cover the two zeroes of the quadratic. ##0.9 - 1.1## and ##1.9 - 2.1## would have done just as well.

    Can you see the problem trying to do the quadratic substitution across the whole range? And why you should split the integral first?
     
  2. Jan 3, 2018 #22
    ##-\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy## why did you put a negative sign here?
     
  3. Jan 3, 2018 #23
    yes i think the problem in this case arises because of the two spikes of the function so there is a need to split the integral so that we can solve it separately for each one.
     
  4. Jan 3, 2018 #24

    PeroK

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    The integral with respect to ##y## came out the "wrong" way. The lower limit was greater than the upper limit. In general, you need to look out for this when integrating:

    ##\int_a^b = - \int_b^a##

    So, for example:

    ##\int_1^{-1}\delta(x)dx = -1##

    I could have avoided this by having ##y = -(x^2 -3x +2)## in that integral. Then the limits would have been the right way round (with respect to ##y##) and the negative sign would have come from ##dy##.
     
  5. Jan 3, 2018 #25
    But if we consider ##y = -(x^2 -3x +2)##then our dirac delta function will look like δ(-y) is that okay?
     
  6. Jan 3, 2018 #26

    PeroK

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    @PRASHANT KUMAR even at this level graphs help. I have a nice graph of ##x^2-3x +2##, so I can see clearly what the function is doing about the zeros and what happens when I substitute ##y## It is easy to miss that the function has a negative gradient around ##x=1## so the natural integral substitution comes out the wrong way.
     
  7. Jan 3, 2018 #27

    PeroK

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    Good point. Best to stick with the way I did it first time!
     
  8. Jan 3, 2018 #28
    so where does the solution lie?
     
  9. Jan 3, 2018 #29
    yes it has negative gradient but i could not link these two : natural integral and negative gradient
     
  10. Jan 3, 2018 #30
    ##\int_1^{-1}\delta(x)dx = -1## i think this should be 1
     
  11. Jan 3, 2018 #31

    PeroK

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    Okay. Let's look at

    ##I = \int f(x) \delta((x-a)(x-b))dx## where ##a < b##

    Your problem is a special case of this, but it may be easier to look at the general case to see what is going on.

    If we draw a graph of this quadratic, we get two roots at ##x = a## and ##x = b##. So, we can see that:

    ##I = \int_{a-\epsilon}^{a+\epsilon} f(x) \delta((x-a)(x-b))dx + \int_{b-\epsilon}^{b+\epsilon} f(x) \delta((x-a)(x-b))dx##

    Where we have focused on the two zeroes and taken a small interval around each, using a small ##\epsilon##

    Now, if we do the substitution ##y = (x-a)(x-b)## and ##dy = (2x - a -b) dx## and

    ##I = \int_{y_1}^{y_2} \frac{f(x)}{2x-a-b} \delta(y)dy + \int_{y_3}^{y_4} \frac{f(x)}{2x-a-b} \delta(y)dy##

    Where ##y_1, y_2## etc. are the y-values for ##a-\epsilon, a+\epsilon## etc.

    Now, from the graph, we have to note that ##y_1 > 0## and ##y_2 < 0##. So, we need to get these limits in the correct order. Also, as we have this in the Dirac function for ##y##, we are simply evaluating at y=0, which equates to ##x =a## and ##x=b## respectively. So:

    ##I = -\frac{f(x)}{2x-a-b}|_{x=a} + \frac{f(x)}{2x-a-b}|_{x=b} = \frac{f(a) +f(b)}{b-a}##
     
  12. Jan 3, 2018 #32

    PeroK

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    No.

    ##\int_1^{-1}\delta(x)dx = -\int_{-1}^{1}\delta(x)dx = -1##

    As I said, this is generally true for any integral on any interval. If you swap the limits, you change the sign of the integral.
     
  13. Jan 3, 2018 #33
    yes that is true if we change the sign then the limits are interchanged but how can it imply that the lower limit should always be less than the upper limit
     
  14. Jan 3, 2018 #34

    PeroK

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    It doesn't, but if you use a substitution to solve an integral sometimes the limits come out the wrong way. You might get something like:

    ##\int_{\pi}^{0} \sin(x) dx = -\cos(x)|_{\pi}^0 = -1 -1 = -2##

    And, we know that ##\int_0^{\pi} \sin(x) = 2##

    This is a basic property of the integral we are talking about here.
     
  15. Jan 3, 2018 #35
    while solving the problem if we do such type of substitution then it is necessary to put a negative sign otherwise it will be wrong?
     
  16. Jan 3, 2018 #36

    PeroK

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    No. But, you were going to use:

    (X) ##\int_{-1}^{1}\delta(x)dx = 1##

    And, you were going to assume that the order of the limits doesn't matter. So, you were going to assume that:

    ##\int_1^{-1}\delta(x)dx = 1##

    Which would be wrong. So, you have to get the limits in the right order before you apply equation (X).

    Take a look at the first property of the definite integral here (half way down the page).

    http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx
     
  17. Jan 3, 2018 #37

    Mark44

    Staff: Mentor

    Thread closed, as the OP is now banned.
     
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