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Integration over delta distribution

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    a) ∫δ( x² cos x) dx with x from -∞ to +∞
    b) Integral over C, with C = unit circle
    C δ(α-α0) a ds
    a
    =(1,1)

    2. Relevant equations
    a) δ(g(x)) =∑δ(x-xi)/|g'(xi)|
    b)?
    3. The attempt at a solution
    a) g(x) = x² cosx
    g'(x) = 2x cos x - x² sin x
    xi = 0 and (2n+1)π/2
    g'(0)=0 then what is 1/|g'(0)|??
     
  2. jcsd
  3. Jan 27, 2016 #2

    andrewkirk

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    For (a), why are you calculating ##g'(x)##? The integral is simply equal to 1 if ##x^2\cos x## takes the value zero for any ##x\in(-\infty,\infty)## and 0 otherwise.

    The same applies to (b). Assuming you are using ##\alpha## to represent the angle in a polar representation of the number plane (##\theta## would be more conventional, by the way), is there any value of ##\alpha## in ##[0,2\pi)## that is equal to ##\alpha_0##?
     
  4. Jan 28, 2016 #3
    Thx, but I don't get it!! Is there a way to show this? Or at least an explanation? I know the relation ∫δ(x-a) dx= θ(x-a) but I've never seen

    ∫δ(g(x)-a) dx =
    • 0 g(x) ≠a
    • 1 g(x)=a
     
  5. Jan 28, 2016 #4

    Ray Vickson

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    There is a serious issue with (a), but first, let's simplify it. Around the positive zero ##x_n = (2n+1)\pi/2## of ##f(x) = x^2 \cos(x)## we have ##f(x) \approx f'(x_n) (x-x_n)##, so integration of ##\delta(f(x))## near there gives just ##1/f'(x_n)##, as you said. However, since f(x) is even, the corresponding term for ##x = -x_n## has ##f'(-x_n) = -f'(x_n)##, so the contributions from ##x = x_n## and ##x = -x_n## cancel.

    So, all that matters is the value of ##\int_{-a}^a \delta(f(x)) \, dx## for small ##a > 0##, and this is where we run into trouble. For small ##|x|##, ##f(x)## looks like just plain ##x^2##, but the point is that its graph does not cross the x-axis; it grazes the x-axis from above, so one is essentially looking at ##2\int_0^a \delta(f(x)) \, dx##. This integrates only half of the ##\delta## function, so can either be regarded as non-existent, or can be looked at through a limiting procedure that gives ##\delta(\cdot)## in the limit. If you regard ##\delta(u)## as a limit of even functions ##f_n(u)##, then your "half-##\delta##" integral would be one half of the usual. However, one can equally well regard ##\delta(u)## as a limit of unsymmetrical (not even) functions ##f_n(u)##, in which case the "half-##\delta##" integral can be a fraction other than 1/2---basically, you can make it equal anything from 0 to 1 if you want to.

    However, let's not overlook the fact that f(0) = 0 in your case.
     
  6. Jan 28, 2016 #5
    ok, thx, and how do I take into account that f(0)=0?

    And b)? I have still no clue
     
  7. Jan 28, 2016 #6

    mfb

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    This is not true, and I don't see how you got that.


    Another issue with (a) are the zeros of the cosine. We get a sum over 1/n which does not converge. You can find limits where the contributions cancel in a nice way, but that is an arbitrary limit procedure, and different ways would lead to different results

    You can convert it to a one-dimensional integral and solve that.
    (b) doesn't have the issues (a) has.
     
  8. Jan 28, 2016 #7

    Ray Vickson

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    In (a) we get a convergent series. At ##x = x_n = (2n+1)\pi/2## we get the contribution ##1/f'(x_n)##, where ##f(x) = x^2 \cos(x)##. Since ##f'(x_n) = -x_n^2 \sin(x_n) + 2 x_n \cos(x_n) = (-1)^n (\pi /2)^2 (2n+1)^2## we get the convergent series ##\sum c(-1)^n/(2n+1)^2##. Then, as I said in #4, the terms from positive ##x_n## are cancelled by those at ##-x_n##, so the whole thing comes out as 0 (except, perhaps, for the part from ##x = 0##).
     
  9. Jan 28, 2016 #8

    mfb

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    Ah right, the sine term contributes, not the cosine. Sorry, my mistake. That just leaves the weird x=0 thing.
     
  10. Jan 28, 2016 #9

    Ray Vickson

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    For (a): you tell me.
     
  11. Jan 28, 2016 #10

    andrewkirk

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    Perhaps I have misunderstood your question. When you write:
    do you mean the definite integral
    $$\int_{-\infty}^\infty\delta(x^2\cos x)\,dx$$
    or are you looking for an indefinite integral?
     
  12. Feb 2, 2016 #11
    Andrewkirk: Yes, right, that's the one.
    Ray vickson, I get the same series but still have this x=0 problem.
    I still don't know what is the correct answer to this cuestion a)

    b) I tried to convert it in an one-dimensional integral with polar coord.:
    [tex]\int_C \delta (\vec{\alpha_0}-\vec{\alpha}) (1,1) d\vec{s} = \int_0^1 \int_0^{2 \pi}\delta (\vec{\alpha_0}-\vec{\alpha}) d\alpha dr = \int_0^{2 \pi}\delta (\vec{\alpha_0}-\vec{\alpha}) d\alpha [/tex]... and again without knowing if this is correct and how to proceed.
     
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