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

Question

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.

Answer 1

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

Answer 2

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 $n=m$.