Integral and derivative of $\lfloor x\rfloor$ and $x - \lfloor x\rfloor$

Question

I've always assumed by graphical inspection that

$\int (x - \lfloor x\rfloor)\mathrm dx = \dfrac{(x - \lfloor x\rfloor)^2 + \lfloor x\rfloor}{2}$ (W|A) and

$\int \lfloor x\rfloor\mathrm dx = x\lfloor x\rfloor - \dfrac{\lfloor x\rfloor(\lfloor x\rfloor + 1)}{2}$ (W|A)

Why does Wolfram|Alpha say for each integral, "no result found in terms of standard mathematical functions"?

I also assumed that $\frac{\mathrm d}{\mathrm dx} \lfloor x \rfloor = 0$, yet according to Wolfram|Alpha $\frac{\mathrm d}{\mathrm dx} \lfloor x \rfloor = \mathop {\rm floor}'(x)$—which is not not explained, but the graph looks very strange. What is going on here?

Answer 1

Let me start with the last question. $\lfloor x\rfloor$ is a piece-wise constant. It's is not differentiable at $x \in \mathbb{Z}$, which is why W|A gives Floor'[x] for an answer. Floor'[x] is a short form for Derivative[1][Floor][x], i.e. derivative remains unevaluated in agreement with Mathematica's evaluation principles.

When W|A says integral can not expressed in terms of known function, it means that Integrate could not provide a solution for your integral.

The expressions you wrote are correct for a definite integral, i.e. $$ \int_0^x \left( y - \lfloor y \rfloor \right) \mathrm{d} y = \frac{1}{2} \left( \left( x - \lfloor x \rfloor \right)^2 + \lfloor x \rfloor \right) $$ in doing so you completely fixed an additive piece-wise constant, ortherwise you should write $ \int \left( x - \lfloor x \rfloor \right) \mathrm{d} x = \frac{x^2}{2} - x \lfloor x \rfloor + C$.