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How to do this integral ?

  1. Jun 8, 2014 #1
    hi,

    I have difficulty in figuring out the following integral:

    [itex] I(l,m;z) = \int^1_0 dx~\frac{x^l}{z - x^m} [/itex],

    where [itex] l [/itex] and [itex] m [/itex] are integers, while [itex] z = \omega + i0_+[/itex] is a complex number that is infinitely close to the real axis. What is interesting to me is when [itex] \omega [/itex] is close to zero, so that the integrand bears a singularity in the domain.

    Could somebody help me out ?

    Thanks a lot !

    hiyok
     
    Last edited by a moderator: Jun 8, 2014
  2. jcsd
  3. Jun 8, 2014 #2

    Simon Bridge

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    You mean:$$z=\lim_{k\rightarrow 0^+}\omega + ik$$... but since z is not a function of x, what is the problem?

    I'm more concerned with what ##x^\prime## represents.

    The point x=0 is not part of the integral if that's what you were worried about. The integration is from 0<x<1. z does not take part in the integral anyway. If you are interested in what happens to the result for ω=0 put it in and see.

    What have you tried so far?
    How does this integral come up in the first place?
     
  4. Jun 8, 2014 #3

    D H

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    It's not x', it's xl (x raised to the power of l).
     
  5. Jun 8, 2014 #4
    Thanks for response.

    1. Yes, I mean [itex] \lim_{k\rightarrow 0}(\omega+ik)[/itex].

    2. As pointed out by D H, it is [itex]x^l[/itex], x raised to the power of l.

    3. Initially, I tried to do it by this formula, [itex]\frac{1}{\omega+i0_+-x^m} = \mathcal{P}\left(\frac{1}{\omega-x^m}\right)-i\pi \delta(\omega - x^m)[/itex], with [itex]\delta(x)[/itex] denoting the Dirac function and [itex]\mathcal{P}[/itex] indicating the principal value. The imaginary part can thus be easily worked out. But I do not know how to handle the real part (i.e., the principal value part), which is supposed to contain a singularity at [itex]x^m = \omega[/itex].

    4. I have also tried to make a change of variable, [itex]x^m \rightarrow y[/itex]. In terms of [itex]y[/itex], the integral becomes something like [itex]\int dy ~ \frac{y^{\nu}}{z-y}[/itex], with [itex]\nu = l/m<1[itex] (assumption). However, the same problem exists.

    Then, how to do the principal value?
     
    Last edited: Jun 8, 2014
  6. Jun 8, 2014 #5

    Simon Bridge

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    Have you tried rationalizing the denominator or converting it to a complex exponential?
    have you tried u=z-x^m ... of course this makes u a complex number...
     
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