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I ## \int ~ dx dy dz ~ f(x,y,z)~ \delta (x+y+z-1)##

  1. Jan 26, 2017 #1
    Hi all ,

    I see this integral too much in QFT books when making loop calculations :

    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) = \int_{0}^{1} dz \int_{0}^{1-z} dy ##

    Can any one explain how did we get this ? I mean it's apparent that ##\int_{0}^{1}~ dx \delta (x+y+z-1) ## have been evaluated and equals one, but when applying ##\delta## when integrating ##dy## why it equals ## 1-z## , not ##1-z-x ## ?

    Also suppose now I have the following function:

    ## \int_{0}^{1}~ dx~ dy~ dz~ \log ~\frac{m_1(x+y) + z m_2}{m_1(x+y)+ z(m_2+ m_3)} ##

    how can this evaluated ? or even a simpler function:

    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~ (x+y +z) ##
     
  2. jcsd
  3. Jan 26, 2017 #2

    PeroK

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    If ##y+z > 1## then ##\int_0^1 \delta(x+y+z-1)dx = 0##

    And if ##y+z < 1## then ##\int_0^1 \delta(x+y+z-1)dx = 1##
     
    Last edited: Jan 26, 2017
  4. Jan 26, 2017 #3

    CAF123

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    The delta function constrains x + y + z = 1, and x,y,z in [0,1] so we have an equation for a plane in the positive quadrant intersecting x,y,z axes at (1,0,0), (0,1,0) and (0,0,1) respectively. Draw it out and you'll see we must have 0 < y+z < 1, i.e y < 1-z.
     
  5. Jan 26, 2017 #4
    I think you mean ## \int^1_0 ~ dx ~ δ(x+y+z−1)=1 ##

    What about if I have :

    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~ (x+y +z) ##
     
  6. Jan 26, 2017 #5

    PeroK

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    Yes, I forgot the ##dx##. Corrected now.

    The second one you need to post in the homework section and give it your best attempt.
     
  7. Jan 26, 2017 #6
    Indeed It's not a problem I have, I just confused, now I know that
    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) = \int_{0}^{1} dz \int_{0}^{1-z} dy
    ## because ##\int_{0}^{1}~ dx \delta (x+y+z-1) = 1, ## at ## y+z < 1##, but what if I integrate a function of x, i.e .,
    ## \int_{0}^{1}~ dx \delta (x+y+z-1) ~ x ##? , it seems naive i know !
     
  8. Jan 26, 2017 #7

    PeroK

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    Why don't you try yourself? The delta function simply picks out the function value.
     
  9. Jan 26, 2017 #8
    The problem statement, all variables and given/known data

    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##

    The attempt at a solution

    ## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##
    ## = \int_0^1 ~dz ~\int_0^{1-z}~ dy~ \frac{1}{m_1(y)+m_2 z } , ##

    where I considered ## \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~f(x) = \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~\frac{1}{x m_1} = 1 ##, I don't if this is right ? actually this is my problem .

    Then the integration run normally over z and y .
     
    Last edited: Jan 26, 2017
  10. Jan 26, 2017 #9

    Charles Link

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    The problem requires some clarification in the limits of the integrals. Is it a triple integral with each limit from 0 to 1? I don't have a solution to it yet, but it is interesting. ## \\ ## editing: It looks like you just let ## z=1-(x+y) ## from the dz and delta function integral with the requirement that ## 0<(x+y)<1 ## because ## 0<z<1 ##, (assuming the limits of the z integral are 0 and 1). Outside of the ## 0<(x+y)<1 ## region, the delta function will be zero. The limits of x and y integrals are unclear from the OP. Once you do the dz integration, you would then perform a dx and dy integration.
     
    Last edited: Jan 26, 2017
  11. Jan 26, 2017 #10

    PeroK

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    You have:

    ##\int_0^1 \delta(x+y+z-1)f(x, y, z) dx = f(1-y-z, y, z)## (if ##0 < 1-y-z < 1##)

    And the integral is ##0## otherwise.
     
  12. Jan 26, 2017 #11

    Drakkith

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    Note: The previous 3 posts have been merged into this thread from another.
     
  13. Jan 27, 2017 #12

    ChrisVer

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    fact 1 you are taking the case where the following is true:
    [itex]\int \int \int dx dy dz G(x,y,z) = \int dz \Big( \int dy \big( \int dx G(x,y,z) \big) \Big) [/itex]

    I guess you were tired when you came up with this question? because it looks like you don't follow basic delta "integration" rules... let me clarify the rule that you missed or got confused with:
    [itex]\int_a^b f(x) \delta(x-x_0) = \begin{cases} 0 & \text{if } x_0 \not \in [a,b] \\ f(x_0) & \text{else} \end{cases}[/itex]
    Now the thing is that your [itex]x_0[/itex] is not a constant but a function of [itex]y,z,~~x_0(y,z) \equiv 1-y-z[/itex].
    Then again as given in the integration rule of the delta function, that [itex]x_0[/itex] can't be just anything, it has to be constrained to [itex][0,1][/itex] (integration of x variable), otherwise the integral results to 0. In formula:
    [itex] 0 \le 1 - y -z \le 1 ~~\Rightarrow ~~0 \le 1 - (y+z) \le 1[/itex]
    or
    [itex] 0 \le y+z \le 1[/itex]
    Now for [itex]y \in [0,1][/itex] this can give lower/upper limits for [itex]z[/itex] given by it being 0 to [itex]1-y[/itex] for any value of y...
     
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