# Why $\int_{0}^{\infty} \lfloor x \rfloor x dx= \sum_{0}^{\infty}n\int_{n}^{n+1}xdx$

I am trying to learn how to work with the floor function, and I would really appreciate if someone can explain me the reason behind:

$$\int_{0}^{\infty} \lfloor x \rfloor x dx= \sum_{0}^{\infty}n\int_{n}^{n+1}xdx$$

It sort of makes sense in my head, but I am confused - how can I simply change the integral in that way? since when (kidding) separating a product of two functions is even legal?

I guess it has something to do with the fact that $$\lfloor x \rfloor$$ is constant for every $$n\leq x \leq n+1$$ and n is natural, but I can't justify that with real rigorous claims.

Thanks.

• Both sides are equal to $\infty$. Are you familiar with basic concepts of measure theory? Feb 12, 2019 at 8:44
• @KaviRamaMurthy not so much, but actually the above was an example, I am mostly interested in the moving from the left integral to the right one, I don't mind if they converge or not Feb 12, 2019 at 8:46

For $$x \in [n,n+1)$$ we have $$\lfloor x \rfloor x=nx$$. Hence $$n\int_{n}^{n+1}xdx=\int_{n}^{n+1}\lfloor x \rfloor x dx$$.
As mentioned in the comment: $$\int_{0}^{\infty} \lfloor x \rfloor x dx= \infty =\sum_{0}^{\infty}n\int_{n}^{n+1}xdx$$
For each $$n\in \mathbb{N}$$, define a function $$f_n(x)= \begin{cases}nx & \mbox{ if }x\in [n,n+1), \\ 0& \mbox{ else.}\end{cases}.$$
Show that $$\lfloor x \rfloor x=\sum_{n\in \mathbb{N}}f_n(x)$$ for all $$x\geq 0$$. The result follows.
P.S. There are minor issues with defining an infinite sum of functions, however in this particular case everything is fine as $$f_n(x)$$ is zero except on the interval $$[n,n+1)$$ and $$f_n(x)=f_m(x)\neq 0$$ implies that $$n=m$$.