# The primitive of a discontinuous function?

I'm thinking about the primitive of an arbitrary function. As we all know, every continuous function has a primitive according to the Newton-Leibniz formula, $\int_{x_0}^{x} f(x) dx$ is the primitive of function $f(x)$. However, if $f(x)$ is discontinuous, say has a point of discontinuity of the first kind, the the integral has both left and right derivative, but they don't coincide. So the question is whether we are able to find a primitive for those kind of functions. Or more generally, what kind of function has a primitive?

-
What you call "primitive" here is usually known as an "antiderivative" in English. – Henning Makholm Jan 2 '12 at 2:18

To complete Henning answer, two small things:

Darboux Theorem States that if $f$ admits primitives, then $f$ satisfies the Intermediate Value Theorem.

This Theorem is a powerful tool to prove that certain functions cannot have primitives.

Also,

Let

$$F(x) = \cases{x^2 \cos\frac{1}{x} & \text{for } x\ne 0 \\ 0&\text{for }x=0} \,.$$

Then, its derivative is

$$f(x) = \cases{2xcos\frac{1}{x}-\sin\frac{1}{x} &\text{for } x\ne 0 \\0&\text{for }x=0}$$

But the function

$$g(x) = \cases{2xcos\frac{1}{x} &\text{for } x\ne 0 \\0&\text{for }x=0}$$ is continuous, thus admits primitive.

Hence, their difference

$$g-f= \cases{\sin \frac{1}{x} &\text{for } x\ne 0 \\0&\text{for }x=0}$$ admits a primitive.

I find this example interesting since for all $-1<a <1$, the function $$\cases{\sin \frac{1}{x} &\text{for } x\ne 0 \\a&\text{for }x=0}$$

verifies the Intermediate Value Theorem, but it only admits primitives for $a=0$ (it cannot admit primitives for two different values of $a$, since otherwise their difference would also admit primitives, thus satisfy the IVT)..

The functions which admit primitives are between continuous functions and functions which satisfy the IVT....

-
Good points both. – Henning Makholm Jan 2 '12 at 3:11

The Fundamental Theorem of Calculus tells us that if $f$ has an antiderivative, then the antiderivative equals $b\mapsto \int_a^b f(x)dx$ plus some constant. So in the case of an isolated jump discontinuity, your argument shows that there cannot be any antiderivative.

More complex discontinuities can allow antiderivatives, however, but to the best of my knowledge there is no nice characterization of the functions that happen to have one.

Some strange-looking specimens can be constructed by taking standard examples of a differentiable function with discontinuous derivative, such as $$F(x) = \cases{x^2 \sin\frac{1}{x^2} & \text{for } x\ne 0 \\ 0&\text{for }x=0}$$ and differentiating to get $$f(x) = \cases{2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2} &\text{for } x\ne 0 \\0&\text{for }x=0}$$ which exists everywhere but is wildly discontinuous at 0. It seems to be quite difficult to point to a property of this $f$ that would tell us that it has an antiderivative.

-
The validity of the first sentence may depend on context. Derivatives are not always Riemann integrable, e.g., see mathoverflow.net/questions/6711/integrability-of-derivatives, so one may need to go to Lebesgue or Henstock–Kurzweil. – Jonas Meyer Jan 2 '12 at 5:27