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Calculating an integral.

  1. Jan 19, 2008 #1
    I probably need to calculate this integral with complex residues calculus, but i'm having difficulty with it:
    where 0<a<1 is real number.
    I mean trying to find a closed contour here seems to be a pain in the neck, I mean how do i calculate the residue of 1/(x^a), assuming my contour runs with a circle around it (not from inside the curve but outside it, i hope it's clear.

    thanks in advance.
  2. jcsd
  3. Jan 19, 2008 #2


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    if a where an integer I would know how to do it, not by residues.

    If you want to solve this by residues, I have no idea.

    It have to bee solved analytically?

    And the pole in x= 0 is of the order a, which is a real number, not an integer. No ideas.
  4. Jan 19, 2008 #3
    yes, analytically.
  5. Jan 19, 2008 #4
    You need to put a cut from the singularity at the origin to infinity. Have it go along the real axis but skirting round the pole at x=4. Now look for a closed contour that does not cross the cut. When you go round the singularity at the origin in a small circle you pick up a factor....
  6. Jan 19, 2008 #5


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    you will need a so-called "keyhole contour" (i guess that's what the previous poster is suggesting). I've got a feeling that it may also work with a half-disk "indented contour" with the help of fractional residue theorem (but not tried that before).

    let me give you a hint by giving you a standard result:
    [tex]\int_0^{\infty}\frac{1}{z^a (z+1)} dz = \frac{\pi}{\sin(\pi a)},\quad 0<a<1[/tex]
  7. Jan 19, 2008 #6


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    After thought: you may need to check the convergence of the integral in [tex]\left[0,\infty\right)[/tex] since my example involves "+1" while yours is "-4". the pole location is different and i've go the feeling that it may have problems. if you have have
    [tex]\left[0,-\infty\right)[/tex] instead, it would be just as my example.

    EDIT: now I understand you difficulty. you can't have a branch cut from 0 to +ve infinity.
    Last edited: Jan 19, 2008
  8. Jan 19, 2008 #7
    Try the following contour, and the integral

    [tex]I=\oint_C\frac{d\,z}{z^\alpha(z-4)}=\oint_C f(z)\,d\,z[/tex]

    Since there are no ploles inside [tex]C[/tex], we have [tex]I=0[/tex].

    It is easy to prove that
    [tex]\lim_{\epsilon \rightarrow 0}\int_{C_3}f(z)\,d\,z=-4^{-\alpha}\,i\,\pi[/tex]

    Now write down the rest of the integrals, and equate real and imaginary parts. You should arrive to


    Be careful for the AB integral. There [tex]z=x\,e^{i\,\pi}[/tex]

    EDIT Change sign in [tex]C_3[/tex] integral.

    Attached Files:

    Last edited: Jan 20, 2008
  9. Jan 20, 2008 #8
    shouldn't the integral of C_3 when epsilon approaches zero equal minus what youv'e wrriten, we learned it should be this way.

    besides, this thanks, i did it before i asked, i just wasn't sure what is a_-1 of 1/z^a, but from writing i guess it should be zero, correct? (when a is between 0 and 1).
  10. Jan 20, 2008 #9
    how do i calculate the integral of AB?
    it goes from -R to -epsilon
  11. Jan 20, 2008 #10
    You are right for the integral of [tex]C_3[/tex]. I changed it in my previous post. Calculations are not my best thing! :smile:

    Now let's denote [tex]I[/tex] your integral. The integral on AB is

    [tex]\int_{-R}^{-\epsilon}\frac{e^{-i\,\pi\,\alpha}}{x^\alpha(x+4)}\,d\,x \xrightarrow[\substack{\epsilon \to 0\\ R \to \infty}]{} e^{-i\,\pi\,\alpha}\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}=e^{-i\,\pi\,\alpha}\,J [/tex]

    Thus we have

    [tex]\oint_C f(z)\,d\,z=0\Rightarrow I+e^{-i\,\pi\,\alpha}\,J-i\,4^{-\alpha}\,\pi=0 \quad (I,J) \in \mathbb{R}[/tex]

    In this equation equate real and imaginary parts, in order to arrive to the wanted result.
    Last edited: Jan 21, 2008
  12. Jan 21, 2008 #11


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    I am puzzled as to how you get this result... either there is a typo here somewhere or you have done something wrong.

    but regardless of that, how did you manage to pick out a piece that corresponds to integration from 0 to infinity from your "half-disk indented contour"?

    I've got a feeling that you have either done the following
    [tex]\int_{0}^{\infty} \frac{dx}{x^a (x+4)} = \frac{4^{-a}\pi}{\sin(a\pi)}, \quad 0<\text{Re}(a)<1[/tex]
    [tex]\int_{-\infty}^{0} \frac{dx}{x^a (x-4)}=-\frac{(-4)^{-a}\pi}{\sin(a\pi)}, \quad 0<\text{Re}(a)<1[/tex]

  13. Jan 21, 2008 #12
    Do we agree that in contour at hand

    [tex]\oint_C f(z)d\,z=0 \, ?[/tex]



    At the limit [tex]\epsilon\to 0,R \to \infty[/tex]
    • The 1st and the 3rd integral vanishes.
    • The 4th and the 6th integral combine to
    • The 5th integral gives
    • The 2nd integral
      [tex]e^{-i\,\pi\,\alpha}\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}=e^{-i\,\pi\,\alpha}\,J [/tex]

    Now apply what is said in my previous post and you will calulate both

    [tex]I=\int_{0}^{\infty}\frac{dx}{x^a(x-4)},\, J=\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}[/tex]

    This is a general trick, applied when you have a pole on the real line and the integral "passes" through it, i.e. Cauchy principal value.

    Does this clears up things? :smile:
  14. Jan 21, 2008 #13


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    I think I know why I had a concern that you didn't have. just a matter of how we interpreted the question:
    you took it as "calculating the given integral, if it is not absolutely convergent then just find the PV of it"
    I took it as "calculating the given integral as it is, if it is not absolutely convergent then...say it is not doable..."

    ok, fair enough, i guess the PV would be useful anyway. But I did thought that since the singularity at +4 is non-removable, you probably can't evaluate the original integral in the same sense as you would evaluate, say

    [tex]\int_{0}^{\infty} \frac{1}{x^a(x+1)}dx = \frac{\pi}{\sin(\pi a)}[/tex]

    which can be done (by using a keyhole contour which picks up the residue at -1) without resorting to just finding the PV.

    Perhaps you see where I was coming from before, when I commented divergence etc. And since in the original post there is no mention of PV of the integral... I thought you are not meant to do that.. but in hindsight it is probably what the intention was.
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