Integrating with floor-function:$\int_a^bf'(x)\lfloor x\rfloor dx $ 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? 
 A: 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 $i$.
A: I summarize the answers of others and provide more details that should clarify and drastically simplify the final result:
1) The integral is independent of whether the intervals from $a$ to $b$ considered are open or closed (including $[a,b]$, $(a,b]$, $[a,b)$, $(a,b)$) as long as the function that shall be integrated is integrable, i.e. the other comments on open vs. closed interval are correct.
2) combining the two answers provided [by martini and Guy Fsone] so far, we obtain a complete and finally simple answer:
[following Guy Fsone:]
$$\int_a^b f'(x)\lfloor x\rfloor dx 
=
\int_{a}^{\lfloor a\rfloor+1 } f'(x)\lfloor a \rfloor dx 
+
\sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2}
\int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } 
f'(x)\left(\lfloor a\rfloor +j+1\right) dx\\
+
\int_{\lfloor b \rfloor}^{b} f'(x)\lfloor b\rfloor dx 
$$
Here, as stated before, the sum appears if it makes sense, i.e. if $\lfloor b\rfloor-\lfloor a\rfloor-2\ge 0$; otherwise, the sum is not present.
Moreover, [following martini] we can further simplify the final result by noticing that the term $\left(\lfloor a\rfloor +j+1\right)$ in the integrand does not depend on $x$. So all those terms can be moved out of the integral such that
$$
\int_a^b f'(x)\lfloor x\rfloor dx 
 = 
\lfloor a \rfloor  \int_{a}^{\lfloor a\rfloor+1 } f'(x)dx 
 + 
\sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2} 
\left(\lfloor a\rfloor +j+1\right) 
\int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x) dx \\
+ 
\lfloor b\rfloor\int_{\lfloor b \rfloor}^{b} f'(x) dx
$$ 
and thus 
$$
... = 
\lfloor a \rfloor\left[  - f(a) + f(\lfloor a\rfloor+1) \right] 
+
\sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2} 
\left(\lfloor a\rfloor +j+1\right) 
\left[-f\left(\lfloor a\rfloor+j+1\right)+f\left(\lfloor a\rfloor+ j+2 \right)\right]\\
+
\lfloor b\rfloor~\left[f(b)-f\left( \lfloor b \rfloor\right) \right]
$$
Now this is similar to a telescope sum; if written out explicitly, we observe that 
$$
...  \qquad = 
- \lfloor a \rfloor f(a)  
+ \lfloor a \rfloor f\left(\lfloor a\rfloor+1\right)\\
+ 
\left(\lfloor a \rfloor + 1 \right) 
\left[-f\left(\lfloor a\rfloor+1\right)+f\left(\lfloor a\rfloor+2 \right)\right]\\
+
\left(\lfloor a \rfloor + 2 \right) 
\left[-f\left(\lfloor a\rfloor+2\right)+f\left(\lfloor a\rfloor+3 \right)\right]\\
+ ... \\
+ \left(\lfloor b \rfloor -1 \right) 
\left[-f\left(\lfloor b\rfloor-1\right)+f\left(\lfloor b\rfloor\right)\right]\\
+
\lfloor b\rfloor~\left[f(b)-f\left( \lfloor b \rfloor\right) \right]\\[5mm]
  = \quad
- \lfloor a \rfloor f(a) 
- f\left(\lfloor a\rfloor+1\right)
- f\left(\lfloor a\rfloor+2 \right)
...
- f\left(\lfloor b\rfloor\right)
+ \lfloor b\rfloor~f(b)
$$
where the notation above is slightly sloppy as some terms may not appear if $\lfloor b\rfloor - \lfloor a\rfloor$ is not sufficiently large.
In summary, we obtain
$$\boxed{\int_a^b f'(x)\lfloor x\rfloor dx 
=
\lfloor b\rfloor~f(b) - \lfloor a \rfloor f(a) - \sum_{j=\lfloor a+1\rfloor}^{\lfloor b\rfloor} f(j)}
$$
A: $$\int_a^b f'(x)\lfloor x\rfloor dx =
 \int_{a}^{\lfloor a\rfloor+1 } f'(x)\lfloor a \rfloor dx +
\sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2}\int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x)\left(\lfloor a\rfloor +j+1\right) dx+
\int_{\lfloor b \rfloor}^{b} f'(x)\lfloor b\rfloor dx $$
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
