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Can this integral be computed?

  1. Sep 15, 2012 #1
    Is it possible to compute the precise value of this integral? If not, can we at least show whether it converges or diverges? It seems that the integrand is undefined on both limits of integration.

    [tex] \int^{∞}_{-∞}\frac{1}{x^{2}}dx [/tex]

    Also, does the function being even help us at all?

    BiP
     
  2. jcsd
  3. Sep 15, 2012 #2
    Do you know how to handle improper integrals?
    We have to split it up into two integrals (can you see why? What happens at x=0?), giving us...
    [tex]\int_{a}^{b} \frac{1}{x^2} + \int_{c}^{d} \frac{1}{x^2}[/tex]
    Now integrate and take a, b, c, and d to the appropriate limits and see what happens.
     
  4. Sep 15, 2012 #3

    jedishrfu

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    the general solution is: integral(x^-2) = -1 / x

    seems it would be undefined at zero and hence not be evaluable.
     
  5. Sep 15, 2012 #4

    dextercioby

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    The function being integrated is not even continuos on the domain of integration...
     
  6. Sep 15, 2012 #5
    So is it computable or not? Can't we consider it an improper integral of type II?

    BiP
     
  7. Sep 15, 2012 #6

    chiro

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    If you doing a Riemann integration, you need this to be Riemann-Integrable let alone continuous (which is only one condition of being Riemann-Integrable).
     
  8. Sep 15, 2012 #7
    According to http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx
    the integrals can be computed even if there is a discontinuity in the integrand provided that each of the partitioned integrals converges.

    BiP
     
  9. Sep 15, 2012 #8
  10. Sep 15, 2012 #9
    Can you explain why a function has to be continuous on the domain of integraion?
     
  11. Sep 15, 2012 #10

    chiro

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    I don't know if dextercioby was referring to the more general spaces like L^2 and measures/integrals on those spaces as opposed to the much less general spaces.
     
  12. Sep 16, 2012 #11
    This is a boundless integral. What did you think of when you read that? What do you think of when you're dealing with infinity in calculus??

    Anyway, here's what your first step should look like to see if this improper integral diverges or converges.

    [tex]{\int_{-\infty}^{+\infty}\frac{1}{x^{2}} dx= \lim_{a\to+\infty} \lim_{b\to-\infty}\left. \frac{-1}{x}\right|_{b}^{a}}[/tex].

    I think you can solve it from there.

    OOPSSS I messed up, I forgot to think about x=0 (was too focus on making my latex look cool :/). Anyway, the questions I asked should help develop some more intuition on improper integrals.
     
    Last edited: Sep 16, 2012
  13. Sep 16, 2012 #12

    jgens

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    A function does not have to be continuous on its domain to be Riemann integrable.
     
  14. Sep 16, 2012 #13

    chiro

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    http://en.wikipedia.org/wiki/Riemann_integral#Integrability

    Unless you are talking some really exotic functions, the almost everywhere criteria will not come into play and the normal idea of continuity will be still required.
     
  15. Sep 16, 2012 #14

    jgens

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    The point is that continuity is quite a bit stronger than Riemann integrability. That the integral in question fails to converge is not due to a lack of continuity (since the integrand is continuous almost everywhere), but rather due to its unboundedness.
     
  16. Sep 16, 2012 #15

    chiro

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    You pointed out something about continuity and I replied with that statement. This reply was a general reply about Riemann Integrability, not a reply specific to the statement of the OP.
     
  17. Sep 16, 2012 #16

    HallsofIvy

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    No, it doesn't. The not at all "exotic" function f(x)= 1 for [itex]0\le x< 1[/itex], f(x) [itex]1\le x\le 2[/itex] is integrable from 0 to 2 even though it is not continuous at x= 0. The "almost everywhere" certainly does come into play- the function is discontinuous at a single point which has measure 0 and so is continuous "almost everywhere".
     
  18. Sep 16, 2012 #17

    chiro

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    This is a really trivial example (having the end-points not included or having a discontinuity at an endpoint is something we should all know how to handle). Same reason why Integral over all these regions [a,b], (a,b], [a,b), (a,b) are equal if the measure is the standard infinitesimal one and the function has the expected properties.

    This has more to do with the consideration and clarification of how different intervals can be considered to be equal so that you can focus on the mapping that has all the right properties. It's like the whole 0.9999 (repeated forever) = 1 scenario, and we just use the trick that since the intervals are the same, we can use this to remove these "discontinuities" since it doesn't change the value of the final integral.

    The really tough examples include discontinuities everywhere in any interval or having functions like f(x) = 1 if x irrational, 0 otherwise.
     
  19. Sep 16, 2012 #18
    I think Ivy meant to show a trivial function which was not continuous at x=1 (which is what I was inferring earlier), but wrote it incorrectly.

    Either way, continuity is not a necessity for integrability.

    Is it just me, or does Ivy very often not check her posts before pressing the "Send" button"? I've seen a lot of itex problems recently.

    If you're reading this, Ivy, please accept this as constructive! You are obviously very mathematically intelligent.
     
  20. Sep 17, 2012 #19

    HallsofIvy

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    I have no idea what you are talking about. The whole point was your assertion earlier that integrable functions have to be continuous and then, later, that except for "exotic" functions, integrable functions have to be continous. This example shows that, unless you are going to call piecewise defined functions "exotic", neither of those is true. A function is Riemann integrable if its set of all discontinuities has measure 0 and the discontinuities are no worse than "jump" discontinuities. Any function with a finite (or even countable) number of jump discontinuities is integrable.
     
  21. Sep 17, 2012 #20
    Interesting. A rather simple question and two pages of theoretically dubious digressions about mathematical theory which don't address the question, except for that one useful answer by romsofia:

    Yeah, just to expand on this (taking care of x=0) you just try to calculate [tex]\int_{-\infty}^{0}\frac{1}{x^{2}} dx+\int_{0}^{+\infty}\frac{1}{x^{2}} dx=\left. \frac{-1}{x}\right|_{-\infty}^{0}+\left. \frac{-1}{x}\right|_{0}^{\infty}[/tex].

    Regarding the statement in the OP that "it seems that the integrand is undefined on both limits of integration", that's wrong. The integrand is undefined at [tex]x=0[/tex] but the limits for x=±∞ are equal to zero, so they don't pose a problem. The problem is x=0 where the integrand is undefined. So you can't just integrate over any interval that contains 0 in the naive way that romsofia suggested (although he/she noticed his/her mistake; I'm just trying to clarify).
    "Also, does the function being even help us at all?"
    That just shows that [tex]\int_{-\infty}^{0}\frac{1}{x^{2}} dx=\int_{0}^{+\infty}\frac{1}{x^{2}} dx[/tex] (symmetry).

    I'm always annoyed by smartass people adding more confusion to questions than there originally was. (If you're not a smartass I'm not addressing you. No offense meant.) I hope I'm not making the same mistake.

    About Riemann integrability: Yeah, as others have noticed, chiro was talking ********. A continuous function is Riemann integrable. Continuity is a sufficient condition for Riemann integrability. [tex]\frac{1}{x^2}[/tex] is continuous on [tex]\mathbb{R}-{\{0\}}[/tex], so it's Riemann integrable on every compact interval which does not cross x=0. So you can integrate over such intervals in the "naive" way, (just substracting the indefinite integral at the boundary points). Blah!
     
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