Integrating with floor-function:$\int_a^bf'(x)\lfloor x\rfloor dx $

Question

I have an expression I want to evaluate

$$\int_a^bf'(x)\lfloor x\rfloor dx $$

where $f'(x)$ is continuous over the interval. I am looking for a solution without using integration by parts.

My attempt to find the value of this integral is to divide it into parts:

$$\int_a^{\lfloor a+1\rfloor}f'(x)\lfloor x\rfloor dx+\int_{\lfloor a+1\rfloor}^{\lfloor a+2\rfloor}f'(x)\lfloor x\rfloor dx +... +\int_{\lfloor b\rfloor}^{b}f'(x)\lfloor x\rfloor dx.$$

With the reasoning being that $\lfloor x\rfloor$ should be constant under these intervals. But I can't really convince myself of why this should be the case. The intervals $[a,\lfloor a+1\rfloor),[\lfloor a+1\rfloor,\lfloor a+2\rfloor),...$ do the trick but when we split up the intergrals as above my concern is that the intervals we get are actually the closed ones $[a,\lfloor a+1\rfloor],[\lfloor a+1\rfloor,\lfloor a+2\rfloor],...$ over which the floor function is not constant. So what gives?

Answer 1

The point is: Is does not make any difference if you consider the half-open intervals or the closed ones, because the integral is stable against changing the value of the function at one point. That means even if you integrate over a closed intervall, you have $$ \def\f#1{\left\lfloor#1\right\rfloor}\int_{\f{a+i}}^{\f{a+i+1}}\f{x}f'(x)\,dx = \f{a+i} \Bigl(f\bigl(\f{a+i+1}\bigr) - f\bigl(\f{a+i}\bigr)\Bigr) $$ for each $i$.

Answer 2

$$\int_a^b f'(x)\lfloor x\rfloor dx = \int_{a}^{\lfloor a\rfloor+1 } f'(x)\lfloor a \rfloor dx + \sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2}\int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x)\left(\lfloor a\rfloor +j+1\right) dx+ \int_{\lfloor b \rfloor}^{b} f'(x)\lfloor b\rfloor dx $$ Where the terms under summation hold only when it makes sens. Namely $\lfloor b\rfloor-\lfloor a\rfloor-2\ge 0$. The remaining computations are left to the OP