# Is this a justified expression for the integral of the floor function?

Mathematica seems to agree with me in general with saying that $\displaystyle\int \lfloor x \rfloor dx = \frac{\lfloor x\rfloor (\lfloor x\rfloor-1)}{2}+\lfloor x\rfloor \{ x \}+C = \frac{\lfloor x\rfloor(2 x-\lfloor x\rfloor-1)}{2}+C$; that is, letting $I(x) = \frac{\lfloor x\rfloor(2 x-\lfloor x\rfloor-1)}{2}$ and checking whether $\displaystyle\int_a^b \lfloor x \rfloor dx = I(b)-I(a)$ returned true for all counts, decimals included.

What exactly is preventing this from being true in general? Is the reason there is no indefinite integral due to the discontinuity in the floor function, even though Mathematica defines the definite integral along any interval?

-
What makes you say there is no indefinite integral? –  Gerry Myerson Apr 17 '11 at 23:56

The floor function is a piecewise constant function with a finite number of discontinuities, so it is (Riemann) integrable on any interval $[0,x]$. Indefinite integrals $\int f(x)dx$ can be regarded as equivalent classes represented by $\displaystyle \int_0^x f(t)dt$. So the floor function does have an indefinite integral, which can be computed by adding area of a few rectangles: \begin{eqnarray} \int \lfloor x \rfloor dx &=& \int_0^x \lfloor t \rfloor dt +C\\ &=& 1+2+\cdots+(\lfloor x \rfloor-1)+ \{x\}\lfloor x\rfloor+C \\ &=&\frac{\lfloor x\rfloor (\lfloor x\rfloor-1)}{2}+\lfloor x\rfloor \{ x \}+C. \end{eqnarray}
To add: sometimes one has to use things like Assuming[] in Mathematica if you're sure you only need results over the reals. –  Guess who it is. Apr 18 '11 at 0:09