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Difficult complex integral

  1. Jun 4, 2015 #1

    ShayanJ

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    How can I solve the integral below?
    ## \int_{-\infty}^{\infty} \sqrt{k^2+m^2} e^{izk} dk ##
    I thought about contour integration but, as you can see, it doesn't satisfy Jordan's lemma. Also no substitution comes to my mind!
     
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  3. Jun 4, 2015 #2

    jasonRF

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    I'm thinking that this integral might exist as a distribution (generalized function), but will not be a nice analytic function.

    jason
     
  4. Jun 4, 2015 #3

    ShayanJ

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    I don't quite understand what you mean. As far as I know, a distribution is a mapping from the space of function to real numbers. But this is a definite integral with no "slot" to put a function in!
     
  5. Jun 4, 2015 #4

    phion

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    You should post this in the really difficult integrals thread too.
     
  6. Jun 4, 2015 #5

    micromass

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  7. Jun 4, 2015 #6

    ShayanJ

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  8. Jun 4, 2015 #7

    jasonRF

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    Are you using equation (5) in the link? If so, according to abramowitz and Stegun (http://people.math.sfu.ca/~cbm/aands/page_376.htm) that is only applicable when ## \nu > -1/2 ## so it doesn't hold in your case. The DLMF (http://dlmf.nist.gov/10.32) also quotes the same restriction, although they may have copied it from Abramowitz and Stegun. For ## \nu = -1 ## I'm not sure integral converges.

    In the special case of ## m=0##, the integral is the Fourier transform of ## | k| ##. Of course this integral probably doesn't converge in the normal sense either, but if you allow yourself to use generalized functions (things like the Dirac delta function) then you find that it is proportional to ## |z|^{-2} ##, as long as you interpret ## |z|^{-2} ## correctly.

    jason
     
  9. Jun 4, 2015 #8

    ShayanJ

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    I checked both ## \int_{-\infty}^{\infty}\sqrt{k^2+m^2} \sin k \ dk ## and ## \int_{-\infty}^{\infty}\sqrt{k^2+m^2} \cos k \ dk ## in wolframalpha.com. Both of them diverge. So the integral is ## \infty\pm i \infty ##.
    Its strange that ## \int_{-\infty}^{\infty}\sqrt{k^2+m^2} \sin k \ dk ## diverges because the integrand is odd and the interval is symmetric.

    I still don't understand. How am I supposed to use the delta? How should I interpret ## |z|^{-2} ##?

    Anyway, this seems strange. The integral is the kernel for another integral and was appeared when I was doing an exercise in a physics text. How can it be ## \infty\pm i \infty ##?! I guess the only explanation is that I did the exercise wrong!!! But I don't know how.
     
  10. Jun 5, 2015 #9

    jasonRF

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    The reason,
    ##\int_{-\infty}^{\infty}\sqrt{k^2+m^2} \sin k \ dk ##
    diverges is because it generally means
    ## \lim_{R \rightarrow \infty} \lim_{S \rightarrow \infty}\int_{-R}^{S}\sqrt{k^2+m^2} \sin k \ dk ##
    where ##R## and ##S## are independent.

    If this is the answer to a physics problem then you may have made a mistake - what is the original problem?

    By the way, the integral you are doing is a Fourier transform. One way to define it would be:
    [tex]
    \mathcal{F}\left[f(k)\right] \equiv F(z) = \int_{-\infty}^{\infty} dk f(k) e^{i k z}
    [/tex]
    Note that with this definition,
    [tex]
    -i \frac{d F(z)}{dz} = \int_{-\infty}^{\infty} dk \, k \, f(k) e^{i k z} = \mathcal{F}\left[k f(k)\right]
    [/tex]
    With these definitions (and a trick!) we can do the ##m=0## case, for which ##f(k) = |k| = k \, sign(k)##, where
    ##sign(k)=1## if ##k>0## and ##sign(k)=-1## if ##k<0##. The trick goes like this:
    [tex]
    \mathcal{F}\left[sign(k)\right] = \lim_{\epsilon\rightarrow 0^+} \mathcal{F}\left[sign(k)e^{-\epsilon |k|}\right]= \lim_{\epsilon\rightarrow 0^+} \int_{-\infty}^{\infty} dk \, e^{i k z} \, e^{-\epsilon |k|}\, sign(k) = i\frac{2}{z}
    [/tex]
    When I first saw this trick as an electrical engineer taking a signals and systems course it was not at all obvious why this was valid. It turns out that it can be justified in the theory of distributions, which I learned from the math department the following year (but it is still not obvious!).
    Anyway, with this result your the ##m=0## integral becomes
    [tex]
    \mathcal{F}\left[|k|\right]=\mathcal{F}\left[k\, sign(k)\right]=-i\frac{d}{dz} \left( i\frac{2}{z}\right) = - \frac{2}{z^2}.
    [/tex]
    Note we have been fast and loose with the rigor, which is usually perfectly fine for an engineer. If that is all you are interested in then don't read the rest of my post please!

    However, just like the dirac delta function is defined by how it operates under an integral, the results above also have that same flavor. Specifically, the ##2i/z## above actually means that when you multiply by a nice function ##g(z)## and integrate, you intepret the result as,
    [tex]
    \lim_{\epsilon\rightarrow 0^+} \left[ \int_{-\infty}^{-\epsilon} dz \frac{2i g(z)}{z} + \int_{\epsilon}^{\infty} dz \frac{2i g(z)}{z}\right] \equiv {PV} \int_{-\infty}^{\infty} dz \frac{2i g(z)}{z}
    [/tex]
    Where the above defines the Cauchy principal value integral. For the reason, sometimes you will see,
    ## \mathcal{F}\left[sign(k)\right] = 2i {PV} \frac{1}{z} ##
    to remind you of how you interpret it. The final result ##-2/z^2 ## would be interpreted,
    [tex]
    -2 \lim_{\epsilon\rightarrow 0^+} \left[\int_{-\infty}^{-\epsilon} dz \frac{g(z)}{z^2} + \int_{\epsilon}^{\infty} dz \frac{g(z)}{z^2} - \frac{2 g(0)}{\epsilon}\right]
    [/tex]
    Sometimes this is referred to as a pseudofunction, and written,
    ## \mathcal{F}\left[|k|\right]=-2 PF \frac{1}{z^2} ##.

    jason
     
  11. Jun 5, 2015 #10

    ShayanJ

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    I've posted the original problem here. It turns out that my calculations were correct and in fact the result should be divergent because of the divergence of the vacuum energy!

    Anyway, the information in your post were interesting. I definitely need more study on this. Thanks.
     
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