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Changing the order of integration and summation

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

    I want to justify that [itex] \int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x) \sum_{k=0}^{\infty} x^{n} \ dx = \sum_{k=0}^{\infty} \int_{0}^{1} f(x) x^{n} \ dx [/itex]

    2. Relevant equations

    3. The attempt at a solution

    I always thought changing the order of summation and integration could be justified by uniform convergence. But the geometric series [itex] \sum_{k=0}^{\infty} x^{n} [/itex] does not converge uniformly for [itex] x= 1 [/itex]. In fact, it doesn't converge at all for [itex] x=1[/itex]. Is that an issue?
  2. jcsd
  3. Mar 6, 2012 #2
    That is an issue. But there are certain other theorems which assure you that you can switch integral and limit (a sum is a special kind of limit of course).

    See this page: http://planetmath.org/encyclopedia/CriterionForInterchangingSummationAndIntegration.html [Broken]

    You shouldn't really care about "measurable functions". If your functions are continuous, then everything holds.

    There are other criterions of course.
    For example, if all [itex]f_n[/itex] are positive, then [itex]\sum_n \int f_n(x)dx=\int \sum_n f_n(x)dx[/itex] always holds, for example. So what you wrote above is certainly allowed if your f is positive.
    Last edited by a moderator: May 5, 2017
  4. Mar 6, 2012 #3
    Say that [itex] f(x)= \ln x [/itex].
  5. Mar 6, 2012 #4
    Then the question becomes whether

    [tex]\int_0^1\sum_{n=1}^{+\infty}|ln(x)x^n| dx<+\infty[/tex]

    This is equivalent to showing that

    [tex]\int_0^1 \frac{|ln(x)|}{1-x}dx<+\infty[/tex]

    Perhaps you could split the integrals and try a comparison test?
  6. Mar 6, 2012 #5
    Could you say the following?

    [itex] \int_{0}^{1} |\ln x \ x^{n} | \ dx = - \int_{0}^{1} \ln x \ x^{n} = \frac{1}{(n+1)^{2}} [/itex]

    and [itex] \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}} = \zeta(2) = \frac{\pi}{6} < \infty[/itex]
  7. Mar 6, 2012 #6
    Ah, yes. That is good as well.

    Now, you've actually proven that [itex]\sum\int |ln(x)x^n|dx<+\infty[/itex]. But that's ok.

    So good, here it is alright to skip sum and integral.

    (and by the way, [itex]\zeta(2)=\frac{\pi^2}{6}[/itex]. You forgot the square)
  8. Mar 6, 2012 #7
    So you're saying that I didn't actually have to evaluate it directly. I just needed to show that it converges.

    Do you have a link for that second criterion? Is it also something from measure theory?
  9. Mar 6, 2012 #8
    Yes, it's measure theory. I don't have a direct link. But I can tell you it either follows from the measure-theoretic Fubini theorem, or from the theorem of monotone/dominated convergence.

    The monotone and dominated convergence theorems can be found here:


    But it's all measure theory and Lebesgue integration.
  10. Mar 6, 2012 #9
    Thanks. I'm familiar with both convergence theorems, although my knowledge of Lebesgue integration is limited.

    I have one more question.

    If [itex] \sum_{n=0}^{\infty} x^{n} [/itex] does not converge for [itex] x = 1 [/itex], why is it necessarily true that [itex] \int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x) \sum_{k=0}^{\infty} x^{n} \ dx [/itex]?
  11. Mar 6, 2012 #10
    Note that the function [itex]\frac{f(x)}{1-x}[/itex] is also not defined in 0. If you're dealing with Lebesgue integration then this doesn't hurt. In Lebesgue integration you only have to have functions which are defined almost everywhere. That is, if a function is not defined in a point, then this causes no harm.

    If you're dealing with Riemann integration, then the situation is a bit more serious. In that case, you're dealing with an improper integral.
  12. Mar 6, 2012 #11
    Is something like this what you had in mind?

    [itex]\int_0^1 \frac{|ln(x)|}{1-x}dx = \int_0^\frac{1}{2} \frac{|ln(x)|}{1-x}dx + \int_\frac{1}{2}^{1} \frac{|\ln (x)|}{1-x} \ dx [/itex]

    [itex]\int_0^\frac{1}{2} \frac{|ln(x)|}{1-x}dx < \int_{0}^{\frac{1}{2}} \frac{1}{(1-x)\sqrt{x}} \ dx < \infty [/itex] since the integral behaves like [itex]\frac{1}{\sqrt{x}} [/itex] near zero.

    And the second integral converges since [itex] 1 [/itex] is a removable singularity.
    Last edited: Mar 6, 2012
  13. Mar 6, 2012 #12
    Looks good.
  14. Mar 7, 2012 #13
    I'm a bit confused. You said that we need to show that [itex] \int_0^1\sum_{n=1}^{+\infty}|ln(x)x^n| dx<+\infty[/itex]. I originally showed that [itex] \sum_{n=1}^{+\infty}\int_0^1|ln(x)x^n| dx<+\infty [/itex] (because that's what it says needs to be shown on the web page to which you linked). But since one implies the other (due to the monotone convergence theorem), does it not matter which one we choose to show?
  15. Mar 7, 2012 #14
    Indeed. If all terms are positive (and you tool the absolute value, so all terms ARE positive), then you can change the integral and the sum around with no problems. So both ways of showing this are good.
  16. Mar 7, 2012 #15
    But even if [itex] \sum \ln x \ x^{n}[/itex] did converge uniformly for [itex]x=1[/itex] we would still have a problem at [itex]x=0[/itex], right?

    And since we are not integrating over a closed interval, we couldn't invoke uniform convergence to justify that [itex] \int_{0}^{\infty} \frac{f(x)}{1+e^{x}} \ dx = \int_{0}^{\infty} \sum_{k=1}^{\infty} f(x) (-1)^{n-1} e^{-nx} \ dx = \sum_{k=1}^{\infty} \int_{0}^{\infty} f(x) (-1)^{n-1} e^{-nx} \ dx [/itex] either, correct?
  17. Mar 8, 2012 #16
    Now I'm not so sure if [itex] [0, \infty)[/itex] is considered to be a closed interval or not.

    But regardless of that, do we also have an issue at [itex] x=0 [/itex] even if [itex] f(x) = 0 [/itex]?
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