# Antiderivative question

1. Jul 8, 2011

### pc2-brazil

1. The problem statement, all variables and given/known data

Show that the function U(x) defined by
$$U(x)=\begin{cases} 0 & \text{ if } x<0 \\ 1 & \text{ if } x\geq0 \end{cases}$$
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.

2. Jul 8, 2011

### Hariraumurthy

U(x) is discontinuous at x=0 and therefore by the Fundemental Theorem of Integral Calculus, U(x) is not integrable.

3. Jul 8, 2011

### micromass

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

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

To pc2-brazil. Your proof seems to be correct.

4. Aug 1, 2011

### pc2-brazil

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.