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Difficult integral

  1. Sep 26, 2006 #1
    Let [itex]\alpha(t)[/itex] be monotically increasing on [0,1]. Prove that
    [itex]\lim_{n \rightarrow \infty} \int_0^1 t^n d\alpha(t)=\alpha(1)-\alpha(1-)[/itex] where [itex]\alpha(1-)=\lim_{t \rightarrow 1^{-}} \alpha(t)[/itex].

    Here's what I have so far. I know that [itex]\alpha(t) [/itex] is monotonically increasing, so it has at most countably many points of discontinuity. So it is continuous almost everywhere which implies that it is Riemann integrable. That means that [itex]\int_0^1 t^n d\alpha(t)=\int_0^1 t^n \alpha ' (t) dt[/itex] where the second integral is just a plain Riemann integral.

    Then integrating by parts with [itex]u=t^n[/itex] and [itex]dv=\alpha ' (t)dt[/itex], I get that [itex]\int_0^1 t^n d\alpha(t)=\int_0^1 t^n \alpha ' (t) dt= \alpha(1) - \int_0^1 \alpha(t) n t^{n-1} dt[/itex]. This is where I'm stuck. I can't get that [itex]\lim_{n \rightarrow \infty} \int_0^1 \alpha(t) n t^{n-1} dt = \alpha(1-) [/itex] In fact, it looks like it should blow up to me.

    Any help would be appreciated.
    Last edited: Sep 26, 2006
  2. jcsd
  3. Sep 26, 2006 #2


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    You can't assume [itex]d\alpha(t)=\alpha'(t)dt[/itex]. For example, if [itex]\alpha(t)[/itex] is a http://en.wikipedia.org/wiki/Devil%27s_staircase" [Broken] function, its derivative is zero where it's defined (almost everywhere). However it is true that:

    [tex]\int_a^b f(x) dg(x)=f(b)g(b)-f(a)g(a)-\int_a^b g(x) df(x)[/tex]

    This gives the same result, but you should be more careful. You can write:

    [tex]\int_0^1 \alpha(t) n t^{n-1} dt = \int_0^a \alpha(t) n t^{n-1} dt + \int_a^1 \alpha(t) n t^{n-1} dt [/tex]

    where 0<a<1. Since addition is continuous (and assuming everything converges), you can take the limits of each integral on the RHS seperately. The limit of the first integral is zero by uniform conintuity. So try to bound the second one, taking a limit as a->1 at the end.
    Last edited by a moderator: May 2, 2017
  4. Sep 26, 2006 #3
    I see. The theorem I was trying to use requires that [itex]\alpha '[/itex] is Riemann integrable. What are the conditions for the first equation you gave? Either I don't remember it, or it's not in baby Rudin (and I'm guessing on the former).

    Also, and I'm sorry for asking, how does uniform continuity give the first one 0? For the second one, I should try to bound it above and below and apply the squeeze theorem?

    Analysis has always been my weakest area, so thanks for trying to help me.
  5. Sep 26, 2006 #4


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    According to wikipedia that identity is true whenever either of the integrals exist, but I'll try to find a more reliable source.

    Next, I meant uniform convergence, sorry. This can be proved using the fact that alpha is bounded and a<1. Finally, yes, the squeeze theorem should work for the other one.
  6. Sep 26, 2006 #5


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    It's in Foundations of Mathematical Analysis, Johnsonbaugh and Pfaffenberger theorem 53.3 (it's in google book search if you don't have it). It requires f and g to be bounded, and f to be integrable with respect to g (g integrable w.r.t. f follows).

    Baby rudin has a couple integration by parts, but I don't think one as general as this.
  7. Sep 27, 2006 #6
    Now that makes sense. Thanks.

  8. Sep 27, 2006 #7


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    Yea, that sounds right. Thanks.
    Last edited: Sep 27, 2006
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