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Antiderivative question

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the function U(x) defined by
    [tex]U(x)=\begin{cases}
    0 & \text{ if } x<0 \\
    1 & \text{ if } x\geq0
    \end{cases}[/tex]
    has not an antiderivative in (-∞, +∞).
    (Suggestion: Assume U has an antiderivative F in (-∞, +∞) and obtain a contradiction, showing that this contradiction follows from the mean value theorem, according to which there is a number K such that F(x) = x + K if x > 0, and F(x) = K if x < 0).

    2. The attempt at a solution

    I guess that, if there is an antiderivative for this function, it would have to be some constant C1 for x < 0, and x + C2 for x ≥ 0.
    If there is a function F(x) whose derivative is U(x) in (-∞, +∞), then F(x) must be differentiable at all points. So, at any particular value of x, the one-sided derivative coming from the left must be equal to the one-sided derivative coming from the right. But, at x = 0, the one-sided derivatives of F(x) are, from the definition of U(x), 0 (coming from the left) and 1 (coming from the right). The derivative of F(x) doesn't exist at x = 0, so U(x) can't be the derivative of F(x).
    I'm not sure whether this is correct, but I didn't use the suggestion. I don't know exactly what this suggestion is asking me to do.

    Thank you in advance.
     
  2. jcsd
  3. Jul 8, 2011 #2
    U(x) is discontinuous at x=0 and therefore by the Fundemental Theorem of Integral Calculus, U(x) is not integrable.
     
  4. Jul 8, 2011 #3

    micromass

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    Staff Emeritus
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    Firstly, the fundamental theorem of calculus has nothing to do with this.

    Secondly, your claim is wrong. The function [itex]2x\sin(1/x)-\cos(1/x)[/itex] (and which attains 0 in 0) is discontinuous in 0. But it is integrable, and it has an antiderivative: [itex]x^2\sin(1/x)[/itex].

    To pc2-brazil. Your proof seems to be correct. :smile:
     
  5. Aug 1, 2011 #4
    Thank you for confirming it.
    I hadn't noticed that a new answer had appeared. I was just about to send another question here arguing that U(x) is, indeed, integrable.
     
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