What is the integral of the floor function and any power of the floor function?

Question

What is the integral of the floor of x? Furthermore, what is the integral of a power of the floor of x? I saw in one of the answers here that:

$$ \int \;\lfloor x \rfloor dx = x\lfloor x \rfloor - \frac12 \lfloor x \rfloor(\lfloor x \rfloor + 1)\; $$

But the person did not provide a proof. I do not quite understand how they arrived at that answer. I tried Googling it, but I can't seem to find a proof. Can anyone provide one please? Additionally, can you provide how you would take the integral of a power of the floor of x, such as:

$$ \int \;\lfloor x \rfloor^n dx $$

where n is a natural number. I'd appreciate it.

Answer 1

If $F(b)$ is the antiderivative of $\lfloor b\rfloor$, then

$$F(b)-F(0)=\int_0^b\lfloor x\rfloor dx$$ $$=\sum_{n=0}^{\lfloor b\rfloor-1}\int_n^{n+1}\lfloor x\rfloor dx+\int_{\lfloor b\rfloor}^b\lfloor x\rfloor dx$$ $$=\sum_{n=0}^{\lfloor b\rfloor-1}n+\int_{\lfloor b\rfloor}^b\lfloor b\rfloor dx$$ $$=\frac{\lfloor b\rfloor(\lfloor b\rfloor-1)}2+\lfloor b\rfloor(b-\lfloor b\rfloor)$$ $$=\frac{\lfloor b\rfloor(\lfloor b\rfloor-1)+2\lfloor b\rfloor(b-\lfloor b\rfloor)}2$$ $$=\frac{\lfloor b\rfloor(2b-\lfloor b\rfloor-1)}2$$ $$=\lfloor b\rfloor b-\frac{\lfloor b\rfloor(\lfloor b\rfloor+1)}2$$

If we define $F(0)=0$, then we have $F(b)=\lfloor b\rfloor b-\frac{\lfloor b\rfloor(\lfloor b\rfloor+1)}2$.

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Similarly, if $F(b)$ is the antiderivative of $\lfloor b\rfloor^n$, then:

$$F(b)-F(0)=\int_0^b\lfloor x\rfloor^n dx$$ $$=\sum_{i=0}^{\lfloor b\rfloor-1}\int_i^{i+1}\lfloor x\rfloor^n dx+\int_{\lfloor b\rfloor}^b\lfloor x\rfloor^n dx$$ $$=\sum_{i=0}^{\lfloor b\rfloor-1}i^n+\int_{\lfloor b\rfloor}^b\lfloor b\rfloor^n dx$$ $$=\sum_{i=0}^{\lfloor b\rfloor-1}i^n+(b-\lfloor b\rfloor)\lfloor b\rfloor^n$$

Using robjob's Hockey-Stick Identity: $$\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}$$ $$\sum_{i=0}^{\lfloor b\rfloor-1}i^n=\sum_{j=0}^n\binom{\lfloor b\rfloor}{j+1}\stirtwo{n}{j}j!$$

Where $\stirtwo{m}{j}$ are Stirling Numbers of the Second Kind.

Hence,

$$F(b)-F(0)=\sum_{j=0}^n\left[\binom{\lfloor b\rfloor}{j+1}\stirtwo{n}{j}j!\right]+(b-\lfloor b\rfloor)\lfloor b\rfloor^n$$

If $F(0)=0$, then

$$F(b)=\sum_{j=0}^n\left[\binom{\lfloor b\rfloor}{j+1}\stirtwo{n}{j}j!\right]+(b-\lfloor b\rfloor)\lfloor b\rfloor^n$$

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Alternately, applying the more well-known but less compact Faulhaber's formula:

$$\sum_{i=0}^{\lfloor b\rfloor-1}i^n= \frac{(\lfloor b\rfloor-1)^{n+1}}{n+1}+\frac12(\lfloor b\rfloor-1)^n+\sum_{k=2}^n \frac{B_{k}}{k!}\frac{n!}{(n-k+1)!}(\lfloor b\rfloor-1)^{n-k+1}$$

Where $B_k$ is the $k$th Bernoulli number.

Hence,

$$F(b)-F(0)=\frac{(\lfloor b\rfloor-1)^{n+1}}{n+1}+\frac12(\lfloor b\rfloor-1)^n+\sum_{k=2}^n\left[\frac{B_{k}}{k!}\frac{n!}{(n-k+1)!}(\lfloor b\rfloor-1)^{n-k+1}\right]+(b-\lfloor b\rfloor)\lfloor b\rfloor^n$$

If we let $F(0)=0$, then

$$F(b)=\frac{(\lfloor b\rfloor-1)^{n+1}}{n+1}+\frac12(\lfloor b\rfloor-1)^n+\sum_{k=2}^n\left[\frac{B_{k}}{k!}\frac{n!}{(n-k+1)!}(\lfloor b\rfloor-1)^{n-k+1}\right]+(b-\lfloor b\rfloor)\lfloor b\rfloor^n$$

Answer 2

When you are integrating the floor function, you are essentially adding rectangles

$$\int_0^x \lfloor t\rfloor dt = \int_0^10dt + \int_1^21dt + \int_2^32dt + ... + \int_{{\lfloor x \rfloor}-1}^{\lfloor x \rfloor}(\lfloor x \rfloor -1)dt + \int_{\lfloor x \rfloor}^{x} \lfloor x \rfloor dt$$

$$ = 0 + 1 + 2 + ... + \lfloor x \rfloor-1 + \lfloor x \rfloor(x-\lfloor x \rfloor) $$