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Delta Function Integrals

  1. Jan 1, 2016 #1
    Hello,

    I feel like I am fudging these integrals a bit and would like some concrete guidance about what's going on.

    1. The problem statement, all variables and given/known data

    Evaluate ##I = \int_{-1}^{1} dx \delta'(x)e^3{x} ##

    2. Relevant equations


    3. The attempt at a solution

    I use integration by parts as follows,
    ##u = e^{3x}##
    ##du = 3e^{3x} dx##

    ##dv = \delta'(x)##
    ##v = \delta(x)##

    ##I = uv|^{1}_{-1} - \int^{1}_{-1} v du##

    Evaluating ##uv## gives zero because the delta function is zero at both limits.

    ##I = - \int^{1}_{-1} dx \delta(x) 3e^{3x}##

    Clearly, the only non zero value of this function is at ##x = 0##, making me think the answer to my problem is ##I = -3##, but this is the step that I think i'm 'fudging'. I feel like i'm disregarding the integration sign...

    Is this justified?

    Thanks for any guidance you can give.
     
  2. jcsd
  3. Jan 1, 2016 #2

    blue_leaf77

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    That's justified because ##\delta(x)## is centered in ##x=0##, which is contained by the integration limits.
     
  4. Jan 1, 2016 #3

    PeroK

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    That looks correct. It sometimes helps to think of a finite spiked function instead of delta. Let's say it's 0 outside ##[-a, a]## for some small a, and has a value of ##1/2a## in that interval. The integral is then approx: ##2a \frac{1}{2a} e^0 = e^0##
     
  5. Jan 1, 2016 #4
    I don't quite follow.

    Where do you evaluate the limits?

    I see that the spiked function is ##\frac{1}{2a}## inside this interval, and since (I assume) it's centered on ##0##, I see where ##e^0## comes from, but not the ##2a##.
     
  6. Jan 1, 2016 #5

    PeroK

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    2a is the width of the interval. An integral is the area under a curve!
     
  7. Jan 1, 2016 #6
    Ah!

    I'm with you now. So an integral of a function multiplied by the dirac delta function is like evaluating the function itself at the point where the delta function is non-zero?
     
  8. Jan 1, 2016 #7

    PeroK

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    That's exactly what it is. You can think of the delta function as the limit of a sequence of rectangular functions with area 1. That gives some justification for things.
     
  9. Jan 1, 2016 #8
    Cool - Thanks for your help!
     
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