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Beginner Improper Integral

  1. Sep 17, 2011 #1
    [tex]\int_{0}^{2} \frac{1}{1-x^{1/3}} dx[/tex]
    I then would break up into:
    [tex]\int_{0}^{1} \frac{1}{1-x^{1/3}} dx + \int_{1}^{2} \frac{1}{1-x^{1/3}} dx[/tex]

    I'm lost on where to go from here. The integral looks so simple but I'm not sure if I should make a u sub or a trig sub. Could someone give me a hint on how to solve it?
  2. jcsd
  3. Sep 17, 2011 #2
    You don't need to break it up. Use the u-sub u=x1/3.
  4. Sep 17, 2011 #3
    Why wouldn't you split it up if x is unbounded at 1?
  5. Sep 17, 2011 #4


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    Here's a small hint for dealing with such situations: keep in mind that the "1" in the numerator is also "x0"...
  6. Sep 17, 2011 #5
    It cancels out unless the limit of the integral doesn't exist there.
  7. Sep 17, 2011 #6
    Thanks dynamic, but I still have no idea where to start. and what cancels out Harris?
  8. Sep 17, 2011 #7


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    If u = 1 - x1/3 , what is du? You've taken care of the original denominator then, but picked up something new down there. What would go in the numerator to get rid of it? You should have an integral in u that you can deal with. (This is called a "rationalizing substitution" by some authors.)

    Don't forget to transform your limits in the definite integral if you finish the integral in u without transforming back to x.
  9. Sep 17, 2011 #8
    Alright, one more question. Is there any shortcut to finding out of a function diverges or converges or do you have to wait until you get to the end of solving the equation.
  10. Sep 17, 2011 #9
    ac f'(x) dx
    = ∫ab f'(x) dx + ∫bc f'(x) dx
    = [f(b)-f(a)] + [f(c)-f(b)]
    = f(c)-f(a)

    Usually, you split integrals up when lim f(x) as x→b doesn't exist, but lim f(x) as x→b+ and lim f(x) as x→b- do exist and they're not the same. Then the integral becomes:

    ac f'(x) dx
    = ∫ab- f'(x) dx + ∫b+c f'(x) dx
    = [f(b-)-f(a)] + [f(c)-f(b+)] (Keep in mind that f(b-) and f(b+) are limits.)
    In this case, you should wait for the end. Since the indefinite integral can be found, it's much easier to wait for the end.
    Last edited: Sep 17, 2011
  11. Sep 17, 2011 #10


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    In the case of an integral like this, there is something you can look at. As x approaches 1 "from the left", the integrand runs off to "positive infinity"; however, as x approaches 1 "from the right", the integrand goes to "negative infinity". Most authors I've run into reject the total as a convergent sum, even if the individual integrals converge (but they don't for this integral); many physicists might say otherwise...

    There are things like the "p-test" for convergence/divergence, which can be adapted to Type II improper integrals like the one you're working on. For this particular integrand, we have to be somewhat careful, since it doesn't have quite the right form; the treatment is probably a bit beyond what you'll see in your course. In any case, for now, you'll be expected to show the full-out integration in homework or exams.

    (Let's say I wasn't surprised to find that this integral diverges...)
  12. Sep 17, 2011 #11
    I'm going to disagree here. This might not sound very mathematical or rigorous, but let's consider the function f'(x)=1/x. From a geometric standpoint,

    -11 1/x dx = 0

    because the function is symmetrical and it just so happens that

    -b-a 1/x dx
    = -∫ab 1/x dx

    The function in the original post might not be symmetrical, but I think the same logic applies.
    Last edited: Sep 17, 2011
  13. Sep 17, 2011 #12


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    I'll say, go argue with the analysts. Many physicists will say the "opposing infinities" should cancel, but books I've dealt with would reject, for instance, even [itex] \int_{-a}^{a} \frac{dx}{x^{1/3}} [/itex] as being convergent...
    Last edited: Sep 17, 2011
  14. Sep 17, 2011 #13


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    There's really no argument. Evaluating the integral outside of a symmetrical interval around the singularity and letting the interval go to zero gives you the Cauchy principal value of the integral. It's useful for some problems like in distribution theory. But the integral in the mathematical sense of Riemann or Lebesgue (where you have to be able to sum in any possible way, not just symmetrically) still doesn't exist.
  15. Sep 18, 2011 #14
    If I find [tex]ln(0)[/tex] when evaluating the limit would that mean the integral is diverging.
  16. Sep 18, 2011 #15


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    Yes, it does.
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