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Integral of derivative

  1. Jun 11, 2013 #1
    Assume that a function [itex]f:[a,b]\to\mathbb{R}[/itex] is differentiable at all points in [itex][a,b][/itex] (we accept left and right sided derivatives at the end points). Will

    [tex]
    \int\limits_{[a,b]}f'(x)dm_1(x) = f(b)-f(a)\quad\quad\quad\quad (1)
    [/tex]

    hold, where the integral is the Lebesgue integral?

    Now, becareful with this thing. I know it looks simple, but I was unable to find an answer after going through my pedagogical material. The question contains different assumptions than the most commonly known theorems.
     
    Last edited: Jun 11, 2013
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  3. Jun 11, 2013 #2

    micromass

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    Billingsley gives the following counterexample:

    Define ##f(x) = x^2 \sin(1/x^2)## for ##0< x\leq 1/2##. Define ##f(x) = 0## for ##x\leq 0## and ##x>1##. Define ##f## on ##1/2<x<1## in such a way that ##f## is continuously differentiable on ##(0,+\infty)##. Then ##f## is everywhere differentiable. But ##f^\prime## is not even integrable, which makes the equality impossible for ##a=0##.
     
    Last edited: Jun 11, 2013
  4. Jun 11, 2013 #3
    I had never seen [itex]\frac{1}{x^2}[/itex] being put inside a trigonometric function. Only [itex]\frac{1}{x}[/itex]... :redface:
     
  5. Jun 11, 2013 #4

    micromass

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    The issue here is clearly that ##f^\prime## doesn't need to be Lebesgue integral. This has to be seen as a shortcoming of the Lebesgue integral, and not of our function ##f^\prime##. Luckily, there is a more general, and better behaved notion of integral of the real line. This is the Henstock-Kurzweil integral. This generalizes Lebesgue integrals and extended Riemann integrals. We have a really nice fundamental theorem in that case:

    If ##f## is a differentiable function on ##[a,b]##, then ##f^\prime## is Henstock-Kurzweil integrable and

    [tex]\int_a^b f^\prime = f(b) - f(a)[/tex]

    See Bartle "A modern theory of integration". There is even an extension of the theorem where we allow countable many points where ##f^\prime## does not exist!
     
  6. Jun 11, 2013 #5
    I follow you up to your last sentence, at which point I'm lost. Why is ##f^\prime## not integrable?
     
  7. Jun 11, 2013 #6
    Suppose [itex]f:\mathbb{R}\to\mathbb{R}[/itex] is defined by

    [tex]
    f(x) = \left\{\begin{array}{ll}
    0,\quad &x=0\\
    x^2\sin\Big(\frac{1}{x^2}\Big),\quad &x\neq 0\\
    \end{array}\right.
    [/tex]

    Now the function is differentiable everywhere, and the derivative is

    [tex]
    f'(x) = \left\{\begin{array}{ll}
    0,\quad &x=0\\
    2x\sin\Big(\frac{1}{x^2}\Big)-\frac{2}{x}\cos\Big(\frac{1}{x^2}\Big),\quad &x\neq 0\\
    \end{array}\right.
    [/tex]

    The derivative is not integrable over intervals containing the origo, and also is not integrable overs sets [itex][a,0][/itex] or [itex][0,b][/itex] with any [itex]a<0<b[/itex].

    However, the limits

    [tex]
    \lim_{\delta\to 0^-}\int\limits_{a}^{\delta} f'(x)dx = f(0)-f(a)
    [/tex]

    and

    [tex]
    \lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f'(x)dx = f(b)-f(0)
    [/tex]

    still hold. Suppose we define the integral over [itex][a,b][/itex] (with [itex]a<0<b[/itex]) by

    [tex]
    \int\limits_{a}^{b}f'(x)dx \;:=\; \lim_{\delta\to 0^-}\int\limits_{a}^{\delta}f'(x)dx \;+\; \lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f'(x)dx
    [/tex]

    Now the formula

    [tex]
    \int\limits_{a}^{b}f'(x)dx = f(b)-f(a)
    [/tex]

    holds. No ordinary theorem would tell that it holds, but it holds anyway.

    --------------

    I was just going to complain here that I have a feeling that there is some result missing from my pedagogial materials. But now I see that while writing this message, micromass has already written something about more general integrals.... anyway, I'll submit this example still...
     
  8. Jun 11, 2013 #7

    micromass

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    jostpuur, what you wrote is exactly the idea and motivation of the Henstock-Kurzweil integral!
     
  9. Jun 11, 2013 #8

    micromass

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    So we care about the integral

    [tex]\int_0^a \frac{\sin(1/x^2)}{x} dx[/tex]

    By substitution ## t = 1/x^2##, we transform this into an integral of the form

    [tex]\int_b^{+\infty} \frac{\sin(t)}{t}dt[/tex]

    This integral is well known not to be Lebesgue integrable. A proof can be found in several textbooks. The proof consists of

    [tex]\int_{b}^{+\infty} \frac{|\sin(t)|}{t}dt = \sum_{n=c}^{+\infty} \int_{(n-1)\pi}^{n\pi}\frac{|\sin(t)|}{t}dt \geq \sum_{n=c}^{+\infty} \frac{1}{n\pi}\int_0^\pi |\sin(t)|dt[/tex]

    But this last series diverges. Again, this is a shortcoming of the Lebesgue integrable.
     
  10. Jun 11, 2013 #9
    I have a vague memory, that Baire's theorem can be used to prove something about how discontinuous derivatives can be, but I cannot find the theorem anywhere anymore. Do you know anything about it?
     
  11. Jun 11, 2013 #10
    Sorry. I thought you were saying that ##f^\prime## was not integrable at all. For some reason, I forgot we were talking specifically about Lebesgue integrals. Sorry. :redface:
     
  12. Jun 11, 2013 #11

    micromass

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  13. Jun 11, 2013 #12

    micromass

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    Bartle deals with this example in his book "A modern theory of integration". The difference between a Riemann integral, is that there you had to make the partition "uniformly small". Now you also have access to a gauge, which means you can make some parts smaller than other parts.

    Here is what Barle writes in his book:

    Let ##\{r_k~|~k\in \mathbb{N}\}## be an enumeration of the rational numbers in ##[0,1]## and let ##\varepsilon >0##. We define the gauge

    [tex]\delta_\varepsilon = \left\{\begin{array}{l}\varepsilon/2^{k+1}~\text{if}~t= r_k\\ 1~\text{if}~t~\text{is irrational}\end{array}\right.[/tex]
    Now let ##\{(I_i,t_i)\}_{i=1}^n## be a ##\delta_\varepsilon##-fine partition of ##[0,1]##. IF the tag ##t_i\in I_i## is rational, then ##f(t_i)=1##, but the length ##l(I_i)=x_i - x_{i-1}## is small. More precisely, if ##t_i = r_k##, then ##I_i \subseteq [r_k -\delta_\varepsilon(r_k), r_k + \delta_\varepsilon (r_k)]## so that ##l(I_i)\leq \varepsilon 2^k##. Further, if ##r_k## is the tag for two consecutive subintervals, the sum of the lengths of these two nonoverlapping subintervals is ##\leq \varepsilon /2^k##. We conclude that the rational ##r_k##can contribute at most ##\varepsilon /2^k## to the Riemann sum. Since only rational tags make a nonzero contribution to the Riemann sum, we have
    [tex]|\sum_P f| \leq \sum \frac{\varepsilon}{2^k} \leq \varepsilon[/tex]
    Since ##\varepsilon## is arbitrary, the function is integrable and ##\int_0^1 f = 0##.
     
  14. Jun 14, 2013 #13
    This is still unclear: Assume that a function [itex]f:[a,b]\to\mathbb{R}[/itex] is differentiable at all points in [itex][a,b][/itex]. Also assume that

    [tex]
    \int\limits_{[a,b]}|f'(x)|dm_1(x) < \infty
    [/tex]

    holds. Is it now possible that

    [tex]
    f(x) = f(a) + \int\limits_{[a,x]}f'(u)dm_1(u)
    [/tex]

    would not hold for some [itex]x[/itex]?

    I know that if I define a new function

    [tex]
    g(x) = f(a) + \int\limits_{[a,x]}f'(u)dm_1(u)
    [/tex]

    then [itex]g[/itex] will be absolutely continuous, and it will have a derivative (at least) almost everywhere. Also [itex]g'=f'[/itex] will hold almost everywhere, according to some theorems. But I couldn't find theorems that would imply [itex]g=f[/itex].
     
  15. Jun 14, 2013 #14
    I don't understand. Maybe I'm just unfamiliar with some of your notation and think it means something else, but can't we just evaluate it? You could say it is a special case of the fundamental theorem of calculus, but...

    $$g(x)=f(a)+\int\limits_{[a,x]}f'(u) \, du = f(a)+f(x)-f(a)=f(x).$$
     
  16. Jun 14, 2013 #15
    You need to use some theorem when you "evaluate" the integral like that, and the theorem will involve some assumptions.

    For example, if [itex]f'[/itex] is Riemann-integrable, then we can evaluate like that. If we only assume [itex]f'[/itex] to be Lebesgue-integrable, then it's not obvious.

    I have also found a theorem that says that if [itex]f[/itex] is absolutely continuous, then the result will follow without Riemann-integrability.
     
  17. Jun 14, 2013 #16
    Here's one example that can be seen as a motivation for these questions.

    First let's define [itex]\phi:\mathbb{R}\to\mathbb{R}[/itex]

    [tex]
    \phi(x) = \left\{\begin{array}{ll}
    0,\quad\quad &x=0\\
    |x|^{3/2}\sin\Big(\frac{1}{x}\Big),\quad\quad &x\neq 0\\
    \end{array}\right.
    [/tex]

    Now [itex]\phi[/itex] is differentiable everywhere, and the derivative is

    [tex]
    \phi'(x) = \left\{\begin{array}{ll}
    0,\quad\quad &x=0\\
    \frac{3}{2}\sqrt{|x|}\sin\Big(\frac{1}{|x|}\Big) - \frac{1}{\sqrt{|x|}}\cos\Big(\frac{1}{x}\Big),\quad\quad &x\neq 0\\
    \end{array}\right.
    [/tex]

    Now [itex]\phi'[/itex] is not Riemann-integrable over intervals that contain the origo, but still

    [tex]
    \int\limits_{[-1,1]}|\phi'(x)|dm_1(x) < \infty
    [/tex]

    and

    [tex]
    \int\limits_{[a,b]}\phi'(x)dm_1(x) = \phi(b)-\phi(a)
    [/tex]

    holds for all [itex]a<0<b[/itex]. So we see that sometimes Lebesgue-integral works without limits while Riemann-integral doesn't.

    Next, let's define

    [tex]
    \phi_n(x) = \sum_{j=1}^{2^{n-1}} \phi\Big(x - \frac{2j-1}{2^n}\Big)
    [/tex]

    for all [itex]n=1,2,3,\ldots,[/itex] so that

    [tex]
    \phi_1(x) = \phi\Big(x - \frac{1}{2}\Big)
    [/tex]
    [tex]
    \phi_2(x) = \phi\Big(x - \frac{1}{4}\Big) + \phi\Big(x - \frac{3}{4}\Big)
    [/tex]
    [tex]
    \phi_3(x) = \phi\Big(x - \frac{1}{8}\Big) + \phi\Big(x - \frac{3}{8}\Big) + \phi\Big(x - \frac{5}{8}\Big) + \phi\Big(x - \frac{7}{8}\Big)
    [/tex]

    and so on...

    Then set

    [tex]
    \Phi(x) = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x)
    [/tex]

    What kind of differentiability properties will [itex]\Phi[/itex] have? What happens, when you integrate its derivative?
     
  18. Jun 14, 2013 #17

    micromass

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    We have the following theorem:

    If ##f,g:[a,b]\rightarrow \mathbb{R}## are absolutely continuous and if ##f^\prime = g^\prime## a.e., then ##f-g## is constant on ##[a,b]##.

    This is theorem 20.16 in "Real Analysis" by Carothers. Does that help?
     
  19. Jun 14, 2013 #18
    $$\frac{d}{dx}\Phi(x) = \sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x).$$

    Then,

    $$\int\frac{d}{dx}\Phi(x)\, dx = \int\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x)\, dx = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x) + C$$

    In fact,

    $$\int\limits_{[0,x]}\frac{d}{dx}\Phi(x)\, dx = \int\limits_{[0,x]}\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x)\, dx = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x)$$

    By Stokes' Theorem, we have that ##\displaystyle \int\limits_{[a,b]}\, df = \int\limits_{\partial[a,b]}f = f(b)-f(a)##. I don't see why it has to be Riemann integrable.
     
  20. Jun 14, 2013 #19
    This leads to the following question: If [itex]f:[a,b]\to\mathbb{R}[/itex] is differentiable at all points in [itex][a,b][/itex] and [itex]\int\limits_{[a,b]}|f'(x)|dm_1(x)<\infty[/itex], will [itex]f[/itex] be absolutely continuous?
     
  21. Jun 14, 2013 #20

    micromass

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    The following are equivalent:
    (1) ##f## is absolutely continuous
    (2) ##f^\prime## exists a.e., ##f^\prime \in L^1[a,b]## and ##f(x) - f(a) = \int_a^x f^\prime##.


    So I'd say no. I'll try to look for a counterexample if you want.
     
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