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Integrating through singularities

  1. Nov 17, 2008 #1
    I am a little bit confused about dealing with integrals around singularities because my professor seems to treat some situations more rigorously than others.

    We talked this integral and said

    [tex] \int_{-\infty}^{\infty}\frac{1}{x^3} dx = undefined [/tex]

    This seems a little bit unintuitive to me because clearly we have an odd function, so no matter what happens as you approach 0, you should get equal cancellations from the left and right sides... but since you cannot say EXACTLY what happens at x = 0, I can see calling this indeterminate.

    But then when we came across this integral

    [tex] \int_{-\infty}^{\infty}\frac{sin(x)}{x} dx [/tex]

    and integrated right through the removable singularity without any problems. It seems hypocritical to deal with this one by just appealing to intuition and saying that the part near x=0 doesn't "blow up" so it has a negligible contribution when we don't in fact know what it does AT x=0.

    Is there a good reason why in this case we can just integrate straight through the singularity even though we have an undefined value at x = 0?
  2. jcsd
  3. Nov 17, 2008 #2


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    I don't know if this answer is satisfactory to you, but

    [tex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/tex]
    [tex]\lim_{x \to 0} \frac{1}{x^3} = \infty[/tex]

    So I don't agree with your argument that
    You even said it yourself: it is a removable singularity, while in the first integral it is not. The word removable already indicates that it's not a real "hard" singularity.

    In fact, your idea of integrating 1/x^3 while leaving out some small neighbourhood of the origin and then shrinking that to zero is very much possible, it is called principle value. In this case, you would find
    [tex]\mathcal P \int_{-\infty}^{\infty}\frac{1}{x^3} dx = \lim_{\epsilon \downarrow 0} \left[
    \int_{-\infty}^{-\epsilon}\frac{1}{x^3} dx + \int_{\epsilon}^{\infty}\frac{1}{x^3} dx
    \right] = \lim_{\epsilon \downarrow 0} 0 = 0.[/tex]
    Last edited: Nov 17, 2008
  4. Nov 17, 2008 #3


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    To add to what CompuChip said about the principal value integral: note that the resulting value you get depends on how you take the limit of the integral bounds. Because the limit was taken symmetrically in the example he gave the result was zero. However, I could just as well do

    [tex]\lim_{\epsilon \rightarrow 0} \left[ \int_{-\infty}^{-2\epsilon}\frac{1}{x^3} dx + \int_{\epsilon}^{\infty}\frac{1}{x^3} dx\right] = \lim_{\epsilon \rightarrow 0} -\frac{1}{2}\left[\frac{1}{4\epsilon^2} - \frac{1}{\epsilon^2}\right] \neq 0[/tex].

    This is why the integral of 1/x^3 is typically left as undefined: in general you DON'T get equal contributions to the integral from both sides! Taking the limit in different ways can give different results! The principal value integral does turn out to be useful in physics, but as it doesn't give the only possible result of an integral with a singularity, such integrals are typically regarded as undefined.
  5. Nov 17, 2008 #4
    I have done principle parts, but even though sin(x)/x converges to 1 in the limit that x approaches 0, if we wish to integrate directly through the point x=0 how can we ignore the fact that it is undefined at 0. It is a removable discontinuity, but it is still a discontinuity.

    Isn't integrating through the removable discontinuity as if it weren't there, essentially taking a principle part?
  6. Nov 17, 2008 #5

    Ben Niehoff

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    No, because the two limits

    [tex]\lim_{a \rightarrow 0^+} \int_{a}^{\infty} \frac{\sin x}{x} \; dx[/tex]


    [tex]\lim_{b \rightarrow 0^-} \int_{-\infty}^{b} \frac{\sin x}{x} \; dx[/tex]

    converge independently.
  7. Nov 18, 2008 #6
    Instead of thinking of the integral as a procedure (going along x in a certain direction and adding up rectangles), it may be useful to think of the integral as the (signed) measure of the 2D region between 0 and the curve.
  8. Nov 18, 2008 #7
    These replies are all helpful. Thanks for the feedback.
  9. Nov 18, 2008 #8
    Log(|x|) also "blows up" at x=0, but the integral [tex]\int_{-1}^1 Log{|x|}dx[/tex] is well defined. Anything less singular than 1/x is integrable.
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