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A differentiable function whose derivative is not integrable

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose g is a differentiable on [a,b] and f = g', then does there exist a function f which is not integrable?


    2. Relevant equations



    3. The attempt at a solution

    I've tried to look at pathological functions such as irrational, rational piecewise functions. but the g would not be differentiable. Moreover, it seems intuitive that the derivative of a function is integrable. Please help.
     
  2. jcsd
  3. Feb 4, 2009 #2

    Dick

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    How about a function that oscillates REALLY FAST as it approaches an endpoint but does it in such a way as to remain differentiable?
     
  4. Feb 4, 2009 #3

    quasar987

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    Can such a function exist and be differentiable at the endpoint? :eek:
     
  5. Feb 4, 2009 #4

    Dick

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    f(x)=x^2*sin(1/x^3) for x in (0,1], f(0)=0? What's wrong with that?
     
  6. Feb 4, 2009 #5
    Thanks for the advice Dick. I thought of a function g(x) = x^2sin (1/x) at x not equals to 0 and g(x) = 0 when x = 0

    Such a function is differentiable and g'(x) = f(x) = 2x sin (1/x) - cos (1/x) when x not equal 0 and 0 when x = 0

    Yet, can't we say that such a function f is integrable on [-1,1]? since it is continuous at all points except at 0 and it is bounded on [-1,1] since both sin and cos functions are bounded. Thus, can't we choose a partition such that U(f,P)-L(f,P) < epsilon for any epsilon > 0?
     
  7. Feb 4, 2009 #6

    Dick

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    Yes. I think so. If you look at my response to quasar987, that's why I picked one that oscillates even faster and is obviously unbounded and not integrable. I can make it oscillate even faster if you want.
     
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