Integrating the floor of a function The integral that I'm trying to simplify is this: (both $x$ and $c$ are natural numbers, if that helps)
$$
\mathrm{F}\left(x,c\right) \equiv
\int_{0}^{c}\left\lbrace\vphantom{\LARGE A}%
\left\lfloor 10^{-\lfloor t \rfloor} x \right\rfloor -
10\left\lfloor 10^{-\lfloor t \rfloor - 1}x\right\rfloor \right\rbrace^{2}\,
\mathrm{d}t
$$
I know this is fairly ugly, please let me know if I need to modify it for the question. Thank you.
Edit: This integral came from the following sum:
$$
\sum_{n = 0}^{c}\left\lbrace\vphantom{\LARGE A}%
\left\lfloor 10^{-n} x \right\rfloor -
10\left\lfloor 10^{-n - 1}x \right\rfloor \right\rbrace^{2}
$$
Is there a way to simplify either that does not lead to a different sum ?.
 A: As $\lfloor t\rfloor$ remains constant between two integers, you can rewrite the integral as a simple sum
$$\int_0^c \left(\left\lfloor 10^{-\lfloor t \rfloor} x \right\rfloor -10\left\lfloor 10^{-\lfloor t \rfloor-1}x \right\rfloor \right)^2 dt=\sum_{t=0}^{c-1} \left(\left\lfloor 10^{-\lfloor t \rfloor} x \right\rfloor -10\left\lfloor 10^{-\lfloor t \rfloor-1}x \right\rfloor \right)^2.$$
Now you can find the relation of the summand to the decimal representation of $x$.
A: The integrand is
$$
f(x,t) = 
\left\lfloor 10^{-\lfloor t \rfloor} x \right\rfloor -10\left\lfloor 10^{-\lfloor t \rfloor-1}x \right\rfloor
$$
so for $x \in \mathbb{N}$, $x = (d_{m-1}\dotsb d_0)_{10}$ this gives
$$
f(x,t)\vert_{t\in[0,1)} = 
\left\lfloor x \right\rfloor -10\left\lfloor 10^{-1}x \right\rfloor
= d_0
$$
which is the last digit of $x$.
Further
$$
f(x,t)\vert_{t\in[1,2)} = 
\left\lfloor 10^{-1} x \right\rfloor -10\left\lfloor 10^{-2}x \right\rfloor = d_1
$$
to
$$
f(x,t)\vert_{t\in[m-1,m)} = 
\left\lfloor 10^{-(m-1)} x \right\rfloor -10\left\lfloor 10^{-(m-1)-1}x \right\rfloor = d_{m-1}
$$
and
$$
f(x,t)\vert_{t\in[m,m+1)} = 
\left\lfloor 10^{-m} x \right\rfloor -10\left\lfloor 10^{-m-1}x \right\rfloor = 0
$$
This gives
$$
F(x,c) = \int\limits_0^c f(x,t)^2 \,dt
= \sum_{i=0}^{c-1} \left( f(x,t)\vert_{t\in[i,i+1)} \right)^2
= 
\begin{cases}
\sum\limits_{i=0}^{c-1} d_i^2 & ; c \le m \\
\sum\limits_{i=0}^{m-1} d_i^2 & ; c > m
\end{cases}
$$
