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Is this an improper integral?

  1. Oct 1, 2011 #1
    Hello everyone! My first time on this forum :).

    Could anybody please help me with the following? I need to know whether this: ∫abs(2x-1)dx is an improper integral and if so, why?

    I use abs( as a notation for absolute value.

    Thanks for the help!
     
  2. jcsd
  3. Oct 1, 2011 #2
    Well, what are the limits of integration?
     
  4. Oct 1, 2011 #3
    Sorry for forgetting to mention that. The lower limit is 0 and the upper limit is 3.
     
  5. Oct 1, 2011 #4
    No, it is not an improper integral. If you look at the function on the graph, you can see that it's area is finite.
     
  6. Oct 1, 2011 #5
    One minor correction:

    It is not improper, but not because the "area is finite." It's rather because the function is integrable. An improper integral occurs when the function is not bounded on the interval we're consider (it goes to infinity somewhere) or the interval itself is not bounded.

    The very reason improper integrals are useful and interesting is that, even though the region isn't bounded, they still sometimes do have a finite area!
     
  7. Oct 2, 2011 #6
  8. Oct 2, 2011 #7
    Thanks for the answers. Great explanation! Thanks to Austintrose as well for the very interesting link.
     
  9. Oct 2, 2011 #8
    The ABS function means that you will always have a positive result and the limits mean that x is always positive as well.

    An improper integral of 0 would be present in something like y = x (45 degree line through 0,0) when your limits were -1 to +1 and you considered the positive area (0 to +1) was equal to the negative area (-1 to 0) apart from the sign.
     
  10. Oct 2, 2011 #9
    Why would that be an improper integral? Wouldn't it just be 0...?
     
  11. Oct 2, 2011 #10

    micromass

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    That's not an improper integral...
    An improper integral occurs when you're not working on a compact set.
     
  12. Oct 2, 2011 #11
    H.J. Keisler, in his 1976 textbook 'Elementary Calculus', describes an improper integral in the way that I have outlined. The integral is improper because on one side of 0,0 the area is measured above the line (above the curve) while on the other side the area is measured below the line (below the curve).
     
  13. Oct 2, 2011 #12
    I've never heard that definition before, but look again: the bounds of integration are 0 and 3, so the integral doesn't cross 0 anyways.
     
  14. Oct 2, 2011 #13
    I said that in post 8 above.
     
  15. Oct 2, 2011 #14

    Mute

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    Would that book be the same as the one posted on this website? If so, the defintion of "Improper Integral", as given in chapter 6, is

    "Suppose f is continuous on the half-open interval (a,b]. The improper integral from a to b is defined by the limit

    [tex]\int_a^b dx~f(x) = \lim_{u \rightarrow a^+} \int_u^b dx~f(x).[/tex]

    If the limit exists the improper integral is said to converge. Otherwise the integral is said to diverge."

    Perhaps the author revised his definition of "Improper Integral" since the first (1976) edition of his book so as to conform with the current meaning.
     
  16. Oct 3, 2011 #15
    I'd say so Mute, the original text said that the area under the curve on the negative side was undefined, as per other posts in this thread, and as a result you should not consider the answer as 0 but leave it undefined. The current text is different.

    That is very interesting because Black and Scholes won the nobel prize for economics around that time.
     
  17. Oct 4, 2011 #16
    Actually Black and Scholes presented their first paper in 1973 and received the Nobel economics prize in 1997.

    http://en.wikipedia.org/wiki/Black–Scholes

    Watch out if you try to analyse the calculus. Selling before buying is allowed in financial maths but I'm not quite sure if it is in pure calculus.

     
  18. Oct 4, 2011 #17
    I found my reference to the 1976 version.


    The main difference between the second edition and the first was that the definition of an integral as the area beneath the curve was redefined to mean the area between the curve and the x axis.
     
    Last edited: Oct 4, 2011
  19. Oct 4, 2011 #18

    Mute

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    Are you sure the book isn't referring to the following kind of situation:

    [tex]\int_{-\infty}^\infty dx~x = ?[/tex]

    The integral is improper because the limits are both infinite. If the limits of integral were finite then the integral is certainly zero. I would think the author is explaining that although it seems like even the improper integral should be zero, it is best to leave it undefined.
     
  20. Oct 6, 2011 #19
    Thanks for your input Mute, it has certainly cleared up things in my mind.

    The original had a half page plot with the two areas shaded and I cannot remember what the limits were but I didn't get the impression that they were infinity. I remember now that both shaded areas were between the curve and the x axis but as the negative area was half of a bisected rectangle the area is the same eitherway.

    In the second version HJ Keisler says,
    The new version just says that there are finite limit improper integrals that converge and other integrals with infinite limits that diverge. Your example was a divergent integral with infinite limits.

    This quote and figure was in relation to a cyclic function with a changing variable and a sub part with infinite limits similar to the divergent integral you described. So divergent indefinite integrals can equal 0 if their infinite limits are equal and they are part of a higher level cycle but convergent integrals are only improper and equal to 0 when they have finite limits.

    The interesting bit is that the cyclic field constructs that have limits from - infinity to + infinity and represent one complete cycle of a higher level process count, that can only equal 1, have the same properties as defined limit improper integrals.

    I'll remember that this type of cyclic construct can be used to represent a time based continuum that changes a linear limit from - infinity to + infinity into 2 cyclic limits from 0 to very large numbers and the reverse. If I can find a multiple of Pi as a constant artifact within the calculations or results of the construct I can also expect that a cyclic construct is being used to mask infinity where a linear construct would lead to a paradox and an entirely different answer.
     
    Last edited: Oct 6, 2011
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