# Antiderivative of the greatest integer function

One of my homework problem wants me to prove that the greatest integer function $f(x)= \lfloor x \rfloor$ does not have an antiderivative. While thinking, I got to this expression,

$$\int_0 ^x \lfloor t \rfloor \, dt = \frac{1}{2} \lfloor x \rfloor(2x- \lfloor x \rfloor -1)$$

Is there an antiderivative for the greatest integer function?

The first part of the fundamental theorem of calculus implies the existence of an antiderivative for continuous functions. But what about discontinuous functions? Is there an antiderivative for any discontinuous functions?

And thank you for any hints or ideas for my homework problem.

• hint: a function f with a discontinuity of the first kind does not have an antiderivative. – Noam Dolovich May 28 '16 at 7:03
• What is 'discontinuity of the first kind'? – zxcvber May 28 '16 at 7:18
• when at a point x0 the one sided limits L+ and L- are finite and different – Noam Dolovich May 28 '16 at 7:41
• Strange... I see a related topic that tells how to integrate this function... math.stackexchange.com/q/408953/17976 – Mike May 28 '16 at 7:47
• You define an antiderivative in terms of a limit, not an integral. Show that it is not possible for this limit to hold because of the jump discontinuities. – Someguy May 28 '16 at 8:49

The function $f(x) = \lfloor x \rfloor$ can be integrated, but it cannot have an antiderivative on the entire real line if we require that an antiderivative must be differentiable everywhere.

Suppose, towards a contradiction, that $F(x)$ were an antiderivative for $f(x)$, differentiable everywhere. By Darboux's theorem, the derivative $F'(x) = f(x)$ satisfies the intermediate value property, so since $f(0) = 0$ and $f(1) = 1$ there has to be some $a \in (0,1)$ such that $f(a) = \frac12$. So $\lfloor a \rfloor = \frac12$, but this is impossible.