# 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 ?.

• If I am right, this computes the sum of the $c$ rightmost digits of $x$, squared.
– user65203
May 31 '16 at 20:38
• Have you tried out a few test values of $x$ and $c$ to see what's happening? May 31 '16 at 20:38
• I think you'll find this answer very helpful. math.stackexchange.com/questions/360323/… May 31 '16 at 20:47
• Yves Daoust, that is precisely what this does. C would be the number of non-fractional digits of the number Jun 1 '16 at 19:03

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 $x$.

• This is really the critical fact you need to realize that isn't stressed in basic calculus-namely a "discrete" integral is just a sum over the integer values of the domain.It really SHOULD be and it should be repeated in basic analysis because it's THAT important. This fact comes up regularly in generating functions,probability theory computations and other settings. May 31 '16 at 21:24

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 $$F(x,c) = \int\limits_0^c f(x,t)^2 \,dt = \sum_{i=0}^{c-1} \left( f(x,t)\vert_{t\in[i,i+1)} \right)^2 = \begin{cases} \sum\limits_{i=0}^{c-1} d_i^2 & ; c \le m \\ \sum\limits_{i=0}^{m-1} d_i^2 & ; c > m \end{cases}$$

• the condition $x \in \mathbb{N}$ can be relaxed, and consider the digits of the decimal representation of $x$ before the point. May 31 '16 at 22:42
• Yes, I think it would work for non-negative real $x$.
– mvw
May 31 '16 at 23:02