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Improper integral of odd integrand

  1. Feb 28, 2012 #1

    Wondering whether I am right that [itex]\int_{-1}^{1} \frac{dx}{x} = 0[/itex], and therefore it's convergent, because my teacher insists on splitting into two divergent improper integrals then he says it's divergent.

    Thanks in advnace,,
  2. jcsd
  3. Feb 28, 2012 #2


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    No, amazingly, you are wrong and your teacher is right!

    Since that integrand does not exist at x= 0, you have to use the definition:
    [tex]\int_{-1}^1 \frac{dx}{x}= \lim_{\alpha\to 0^-}\int_{-1}^\alpha \frac{dx}{x}+ \lim_{\beta\to 0^+}\int_\beta^1\frac{dx}{x}[/tex]
    and those limits do not exist.
    You cannot just evaluate the anti-derivative at 1, -1, and then subtract- that's ignoring the whole problem of what happens at 1.
    Last edited by a moderator: Feb 29, 2012
  4. Feb 28, 2012 #3
    Review the fundamental theorem. It requires that the integrand be continuous on your interval.
  5. Feb 28, 2012 #4
  6. Feb 29, 2012 #5
    OK.. But in the case of 1/x, in what way is it explained that the area at the right of the y-axis is equal to the negative of the left area, and regardless of the fact that they're infinities. Aren't these two infinities equal to each other.
  7. Feb 29, 2012 #6


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    You can talk about two "infinities" (I would prefer to say "infinite sets") having the same cardinality but it make no sense to say two "infinities" are equal.
  8. Feb 29, 2012 #7
    You're right, the infinities may not be comparable, but I just want to be convinced why it's wrong to think this way: we know that the area at the right of the y-axis is infinite, if we suppose I take any portion of that infinite area, I can take an exactly equal area at the left of the y-axis that cancels it, and therefore the integral converges to zero.
  9. Feb 29, 2012 #8
    Here is why it doesn't work: we don't take limits in improper integrals simultaneously, we do them one at a time. If we don't, then we're taking a Cauchy principal value, which I mentioned above, and which is *different* from a Riemann integral.

    There are two directions to go to infinity towards, and we have to do them one at a time.
    + If we go to the right first, we get +∞. Then, as we go left, it does not go to zero, it will simply stay at positive infinity.
    + If we do them in reverse order, we get -∞, and again it will remain at -∞.

    Now for the integral to be defined, it's value has to be independent of arbitrary choices like the order in which we take limits. Since this one depends on that order, it is not defined.
  10. Feb 29, 2012 #9
    Correction, I was thinking of

    [itex]\int_{-\infty}^{+\infty} \frac{dx}{x}[/itex].

    In fact, in my case there are four limits to consider.

    In your case, the idea is the same except that the limits are towards zero, not towards infinity. You'll find the same result in either case, however.
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