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Tricky DeltaFunction integration

  1. Sep 21, 2014 #1
    There is a question.
    Here is the integral:
    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


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    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:
    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


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    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.

    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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