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Example of f integrable but |f| not integrable

  1. Nov 20, 2007 #1
    Define F(x) = integral (from x to 1) f

    take limF(x) as x goes to 0.

    I would like to give an example of a function on (0,1] so that the above limit exists but if we replace f with |f| the limit does not exist.

    So I came up with f= [(2^n)/(n)](-1)^(n+1) for x in (1/(2^n),1/[2^(n-1)]] (which geometrically) at these endpoints gives us the alternating harmonic series, that is the limit above with this function is the limit of the alternating harmonic series.

    Is there a simpler example?
     
  2. jcsd
  3. Nov 20, 2007 #2
    f(x)=sin(1/x)/x.
     
  4. Nov 20, 2007 #3

    Gib Z

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    [tex]\int^{\infty}_0 \frac{ \sin x}{x} dx = \frac{\pi}{2}[/tex]

    [tex]\int^{\infty}_0 |\frac{ \sin x}{x}| dx = \infty[/tex]
     
  5. Nov 21, 2007 #4
    i'm looking for examples on the domain [0,1].
     
  6. Nov 21, 2007 #5

    morphism

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    This one works on [0,1] (it's not Lebesgue integrable there IIRC).
     
  7. Nov 21, 2007 #6
    If f is a bounded function on [0,1] which is integrable then |f| is integrable on [0,1]. It is a simple theorem to prove.
     
  8. Nov 21, 2007 #7

    morphism

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    While this is true, it's not really what the OP asked for. On the other hand, it does tell us that if we are going to find such a function f, it's not going to be (properly) Riemann integrable on [0,1].
     
  9. Nov 22, 2007 #8
    I'm aware of this, but this isn't my question. I'm asking for a function so that f is integrable for every c>0 on [c,1] and with the following properties:

    (1) the limit of the integrals as c goes to zero exists.
    (2) the same limit as (1) replaced with f with |f| does NOT exist.

    Although I did come up with one, I'm asking for more, possibly simpler, ones.

    The theorem you mentioned tells us that any such function should NOT be bounded on [0,1].
     
  10. Nov 23, 2007 #9
    My solution, perhaps not the best.

    This is problem 7 in Rudin, Chapter 6. I used a harmonic series:
    [tex]f(x) = (n+1)(-1)^{n+1},\ x \in (\frac{1}{n+1},\frac{1}{n}],\ n = 1,2,...[/tex]. Then [tex]\int_{\frac{1}{n+1}}^{\frac{1}{n}} f(x) dx = \frac{(-1)^{n+1}}{n}[/tex]. The integral [tex]\int_{c}^{1} f(x) dx[/tex] converges to [tex]\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}[/tex] as [tex]c \rightarrow 0[/tex]. But it's not absolutely convergent.
     
  11. Nov 23, 2007 #10
    About sin(1/x)/x

    [tex]\int_{0}^{1}\frac{sin(\frac{1}{x})}{x}dx[/tex] does not appear to me to converge. Set [tex]\frac{sin(\frac{1}{x})}{x} = x\frac{sin(\frac{1}{x})}{x^2}[/tex], and integrate by parts, with [tex]G(x)=x, \ f(x) = -sin(u(x))u'(x) = \frac{sin(\frac{1}{x})}{x^2}[/tex], then [tex]\int_{e}^{1}x\frac{sin(\frac{1}{x})}{x^2}dx = 1 \cdot cos(\frac{1}{1}) - e \cdot cos(\frac{1}{e}) - \int_{e}^{1}cos(\frac{1}{x})dx [/tex] does not converge as [tex]e \rightarrow 0+[/tex]. Using the same integration by parts, you can show [tex]\int_{0}^{1}\frac{sin(\frac{1}{x})}{\sqrt{x}}dx[/tex] does converge, but it also converges absolutley, since [tex]\int_{0}^{1}\frac{1}{\sqrt{x}}dx[/tex] also converges.
     
    Last edited: Nov 23, 2007
  12. Nov 23, 2007 #11
    OK I think I'm wrong there... [tex]\int_{e}^{1}cos(\frac{1}{x})dx[/tex] must indeed converge as [tex]e \rightarrow 0[/tex], if the identity [tex]\int_{e}^{1}\frac{sin(\frac{1}{x})}{x}dx = \int_{1}^{\frac{1}{e}}\frac{sin(u)}{u}du [/tex] holds. Then I suppose that's an easier to construct example of an integral that is convergent but not absolutely convergent.
     
    Last edited: Nov 23, 2007
  13. Nov 23, 2007 #12
    I think I've done enough for now...

    What the hell was I thinking?!? Of course the integral of cos(1/x) converges on [0,1] because it's a bounded function that's only discontinuous at one point!

    However, the problem with sin(1/x)/x is that doesn't seem like a trivial function to deal with. The limit is in the residue integration section of complex analysis books. But nonetheless I went ahead and calculated that it does not converge absolutely. It was laborous.

    The approach I began above was mistaken but in retrospect yields that [tex]\int_{0}^{1}\frac{sin(\frac{1}{x})}{x}dx = cos(1) - \int_{0}^{1}cos(\frac{1}{x})dx [/tex]. In particular, it converges.

    Using the integration of parts for even n, [tex]\int_{\frac{1}{n\pi}}^{\frac{1}{(n-1)\pi}}|\frac{sin(\frac{1}{x})}{x}|dx \geq \frac{1}{n\pi} + \frac{1}{(n-1)\pi} - \frac{1}{n^2\pi^2}}[/tex], hence [tex]\sum_{n}\int_{\frac{1}{n\pi}}^{\frac{1}{(n-1)\pi}}|\frac{sin(\frac{1}{x})}{x}|dx \geq \sum_{even \ n}\frac{1}{n\pi} + \frac{1}{(n-1)\pi} - \frac{1}{n^2\pi^2}} = +\infty[/tex]
     
    Last edited: Nov 23, 2007
  14. Nov 23, 2007 #13
    rudinreader: your solution is similar to mine in that it converges to the alternating harmonic series although i think I like yours better! check out my post on uniform convergence it is in regards to problem 4 chapter 7.
     
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