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Improper Integral

  1. Nov 4, 2014 #1
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    State whether the integral converges or diverges and if it converges state the value it converges to.

    Integral from 0 to 2 of 1/(1-x)dx



    I broke it up into 2 integrals (0 to 1) and (1 to 2) set up the limit for both using variables instead of 1 and I evaluated the integral to equal 0 so I figured it converges to 0 but in my book it says it diverges...it also says it diverges on wolfram alpha so I'm sure that the book is correct, therefore I must be missing something.

    Additionally, I took a look at the integral from -1 to 1 of 1/x (which seems very similar to the first problem I stated), yet wolfram alpha says this one is 0.
    Why does the second converge to 0 and the first one diverge???

    I just learned about improper integrals today so I figure it is likely I am unaware of some detail of problems like these so I would like some help. Thanks.
     
  2. jcsd
  3. Nov 4, 2014 #2

    Matterwave

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    I think wolfram should tell you the integral of -1 to 1 of 1/x diverges as well. In fact that integral is formally divergent, but there is a special case in which you can make it converge, called the Cauchy principle value integral.

    Describe in more detail how you set up "the limit for both using variables instead of 1".
     
  4. Nov 4, 2014 #3
    I'll try to explain how I attempted to solve it:

    I saw that there would be an infinite discontinuity at 1 so I first broke the integral up into lim b-> 1- of the integral from o to b of 1/(1-x)dx + lim a->1+ of the integral from a to 2 of 1/(1-x)dx I determined the anti derivative to be -ln(1-t) and then plugged in the bounds of as I would for a normal integral problem. After I had done that I took the limit and what I had at that point was lim b->1- of -ln(1-b) + lim a->1+ of ln(1-a). I figured that since it looked like these expressions would reach -infinity and infinity respectively at the same rate then they would cancel out to be 0. This is what I did to solve the problem.

    Additionally, the graph looks like it should be 0 I think...It looks like they cancel out perfectly, so I am very confused.
     
  5. Nov 4, 2014 #4

    PeroK

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    What you've done is essentially calculate the Cauchy Principle Value (CPV). This is, in fact, different from an improper integral. For an improper integral, all limits must exist (and be finite) otherwise the integral is not defined. For the CPV, you can combine limits in the way you did.

    One good example is:

    ##lim_{a \rightarrow \infty} \int_{-a}^{+a}sin(x)dx = 0##

    This is the CPV.

    But, the improper integral would be defined as:

    ##\int_{-\infty}^{+\infty}sin(x)dx = lim_{a \rightarrow -\infty} \int_{a}^{0}sin(x)dx + lim_{a \rightarrow \infty} \int_{0}^{a}sin(x)dx##

    As neither of these limits exists (and is finite), the improper integral is not defined.
     
  6. Nov 4, 2014 #5

    pasmith

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    You can't cancel infinities like that.

    You have to take the limits separately; if either limit diverges then the integral diverges.
     
  7. Nov 4, 2014 #6
    What you are describing (that you are attempting to do) is the CPV of the integral and, unfortunately, Cauchy beat you to it.
    If you look closely at the definition of Riemann integrals in terms of limits of Riemann sums over ever finer partitions, you should see that the Riemann integral does not exist as defined.

    This speaks directly to your example:
    http://en.wikipedia.org/wiki/Cauchy_principal_value
     
  8. Nov 4, 2014 #7
    Thank you for the replies this is much more clear to me now
     
  9. Nov 4, 2014 #8

    Matterwave

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    I'd like to add that the CPV integral requires you to take the limit on both sides at the same rate. So instead of 2 limits with a and b separately approaching 1, you need to have limit as a approaches 1 and as (2-a) approaches 1. Otherwise, the two "infinities" will not cancel at every step and you end up with a divergence. The Cauchy Principle value says you take the limits in such a way that the two pieces cancel at every step and arrive at an answer of 0. But like others have said, a CPV integral is not the same as a regular integral.
     
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