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Limits and integrating over a singularity

  1. Oct 4, 2012 #1
    Suppose one needs to evaluate a definite integral over a singularity, like: [itex]-\int_{-1}^3 \frac{1}{x^2} dx[/itex]

    The textbook way to do so is to split the integral into two parts around the singularity and take the limit, like so:

    [tex]\lim_{b\rightarrow 0} -\int_{-1}^b \frac{1}{x^2} dx[/tex]

    and:

    [tex]\lim_{c\rightarrow 0} -\int_{c}^3 \frac{1}{x^2} dx[/tex]

    and then add the results of the two expressions above. In this case, the integral is not convergent, and one obtains an infinite result from those evaluations.

    So far, so good. But here's a question:

    What's wrong with just using the same limit for both integrals? Like this:

    [tex]-\int_{-1}^3 \frac{1}{x^2} dx = \lim_{c\rightarrow 0} \Bigr( -\int_{-1}^c \frac{1}{x^2} dx -\int_{c}^3 \frac{1}{x^2} dx ~\Bigr)[/tex]

    [tex]= \lim_{c\rightarrow 0}\Bigr(\frac{1}{x}\Bigr|_{-1}^c + \frac{1}{x}\Bigr|_c^3~\Bigr) [/tex]

    The 1/c terms cancel each other out, and a finite result remains. In fact, the divergent terms would always cancel out when trying to integrate over a singularity in this way. Clearly, this is a wrong answer, so an invalid step has been performed above. But what part was wrong, and why? Perhaps it is an error to apply the same limit process to both sides of the split integral at the same time. If so, why can't you do that?
     
  2. jcsd
  3. Oct 4, 2012 #2

    Mark44

    Staff: Mentor

    This should be
    $$ \lim_{b\rightarrow 0^-} -\int_{-1}^b \frac{1}{x^2} dx$$

    Similar for the one just below, except that the limit is as c → 0 from above.
    The two limits are different, and are both one-sided, which you don't show above. In the first integral, the limit is as x → 0 from the left, while in the second integral, the limit is as x → 0 from the right.
     
    Last edited: Oct 4, 2012
  4. Oct 5, 2012 #3
    Thanks for the reply, Mark!

    There's something I still don't get, though: shouldn't the limiting behavior of the function be exactly symmetrical on the left and right sides of the singularity (for some functions, like this one)? And as such, wouldn't those 1/c and 1/b terms still cancel each other out?
     
  5. Oct 5, 2012 #4
    I think your arithemetic is wrong:

    [tex]-\int_{-1}^{2} \frac{1}{x^2}dx=-\lim_{\epsilon\to 0} \left(\frac{1}{x}\biggr |_{-1}^{-\epsilon}+\frac{1}{x}\biggr |_{\epsilon}^{2}\right)= \lim_{\epsilon\to 0}\left(-\frac{1}{\epsilon}+1+1/2-\frac{1}{\epsilon}\right)\to \infty[/tex]

    Now use the same approach with

    [tex]\int_{-1}^{2} \frac{1}{x} dx[/tex]
     
    Last edited: Oct 5, 2012
  6. Oct 5, 2012 #5
    Ahh, perhaps that's it - I see you set up your inner integration bounds as -e and +e, whereas I just used +c in both cases. So, your range of integration never actually includes the singularity, while mine implicitly did. That sounds like a good reason why I'd get the wrong answer.

    Hopefully someone will correct me if my interpretation is wrong, but I think I get it now.
     
  7. Oct 5, 2012 #6
    That's called a Cauchy Principal-valued integral and I left off the PV just to be annoying. Really, I should have written it as:

    [tex]P.V. \int_{-1}^{2} \frac{1}{x^2} dx[/tex]
     
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