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Tricky DeltaFunction integration

  1. Sep 21, 2014 #1
    Hey!
    There is a question.
    Here is the integral:
    13613fa34e60.jpg
    What i'm trying to do is starting with the third integral over zi and with the help of integral definition of DeltaFunction i want to calculate it. As you can see f(yi) has no influence on the integral. Am i right here?
     
  2. jcsd
  3. Sep 21, 2014 #2

    ShayanJ

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    Gold Member

    No, you're wrong

    First, in the step [itex] \frac{1}{2\pi}\int_0^{f(y_i)} dz_i \int_{-\infty}^{\infty} dk \exp{(i(z^{'}-z_i)k)}=\frac{1}{2\pi}\int_0^{f(y_i)} dz_i \exp{(i(z^{'}-z_i)k)} \int_{-\infty}^{\infty} dk [/itex], you can't take the exponential out of the integral w.r.t. k because it contains a k.

    Second, you don't need such calculations. The integral [itex] \int_0^{f(y_i)} dz_i \delta(z^{'}-z_i) [/itex] is zero if [itex] -z^{'} \epsilon\!\!/\ (0,f(y_i))[/itex] and is one otherwise.
     
  4. Sep 22, 2014 #3
    I don't think you are right. If you've got something like that:
    6c1dc3d304da.jpg
    you can calculate the integral over x thinking that y is a constant. Then you come back to integral over y. Maybe i misunderstood something?

    I know, but i'd like to write it down analytically. That's why i use integral definition of DeltaFunction.
     
  5. Sep 22, 2014 #4

    ShayanJ

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    Gold Member

    Oh yeah...you're right. I thought you were going to integrate w.r.t. k.

    Well, then you certainly should know that there is something wrong with your calculations. Because [itex] f(y_i) [/itex] does affect the value of the integral. But I don't see any problem!
    I should say such calculations can be very tricky but since we know that your result is wrong, you should either find what's wrong with your calculations or abandon them.

    EDIT:
    One more piece of information.
    Dirac delta is an even distribution meaning [itex] \delta(-x)=\delta(x) [/itex], so you can write [itex] \int_0^{f(y_i)} dz_i \delta(z^{'}-z_i)=\int_0^{f(y_i)} dz_i \delta(z_i-z^{'}) [/itex]. The first is non-zero only when [itex] -z^{'}\epsilon [0,f(y_i)] \Rightarrow z^{'}\epsilon [-f(y_i),0] [/itex] and the second one is non-zero only when [itex] z^{'}\epsilon[0,f(y_i)] [/itex], So the integral is non-zero only when [itex] z^{'}=0 [/itex]. That's what you should assume from the first place, before doing any other calculation.
     
    Last edited: Sep 22, 2014
  6. Sep 22, 2014 #5
    if the integral is non-zero only when z′=0 z^{'}=0 , it is a little bit wierd . and f(yi) has no influence on the integral as it was supposed, but i doubt it.
     
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