Floor function as derivative Does there exist a function f such that $f'(x)=\lfloor x \rfloor$ for all $x$ in $\mathbb{R}$?
I was thinking no, since if you draw it out and look at the integral, it should be an imaginary number, right?
But how do I explain it? 
 A: The integral certainly won't be an imaginary number, but you'll have an issue at the integers because the floor function has a jump discontinuity there, which will manifest itself as a "kink" in the integral. The function below will have $f^\prime=\lfloor x\rfloor$ at all noninteger points: 
$$
f(x)=nx,\; x\in [n,n+1)\quad n\in\Bbb{Z}
$$
However, this function is not differentiable if $x=k$ where $k\in\Bbb{Z}$.  As suggested by the comment, it can be shown that in fact no such function can exist that has $f^\prime(x)=\lfloor x\rfloor$ for all $x$.
A: Well, no if you wish for $f$ to be differentiable (as pointed out by David Mitra, Darboux's theorem can be used to formalize this). However, you could consider weak derivatives. Then $f$ could be a picewise linear function, the step of which is $\lfloor x \rfloor$, e.g. (there are many because of the integration constant):
$\hspace{40pt}$
Be aware that this is not regular antiderivative, on the other hand it's quite a common occurence in practical applications. The shape is in fact almost the same as the example of @icurays1, still it might be just more intuitive to make it continous. I hope I didn't confused you ;-)
