Definite integral concerning the greatest integer function

Question

Evaluate the integral $$I=\int_{0}^{x} \lfloor t+1 \rfloor^3 dt$$where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

The answer is given as $$\Bigg[\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\Bigg]^2+(\lfloor x \rfloor+1)^3 \{x\}$$ where $\{x\}$ denotes the fractional part of $x$.

What I have done so far is this:

I used the property $\lfloor x+a \rfloor = a+\lfloor x \rfloor$ where $a$ is a positive integer and expanded the binomial.

$\int_{0}^{x}\lfloor t+1 \rfloor^3 dt = \int_{0}^{x} \lfloor t \rfloor^3 dt + 3\int_{0}^{x} \lfloor x \rfloor^2 dt + 3\int_{0}^{x} \lfloor x \rfloor dt + \int_{0}^{x} dt$

Where

$\int_{0}^{x}\lfloor t \rfloor^3 dt = \int_{1}^{2} 1 dt + \int_{2}^{3} 2^3 dt + \cdots +\int_{\lfloor x \rfloor -1}^{\lfloor x \rfloor}(\lfloor x \rfloor -1)^3+\int_{\lfloor x \rfloor}^{x}\lfloor x\rfloor^3 dt=\Big[\frac{\lfloor x \rfloor(\lfloor x \rfloor -1)}{2}\Big]^2 + \lfloor x \rfloor^3 \{x\} $

$\int_{0}^{x}\lfloor t \rfloor^2 dt = \int_{1}^{2} 1 dt + \int_{2}^{3} 2^2 dt + \cdots +\int_{\lfloor x \rfloor -1}^{\lfloor x \rfloor}(\lfloor x \rfloor -1)^2+\int_{\lfloor x \rfloor}^{x}\lfloor x\rfloor^2 dt=\lfloor x \rfloor^2(\lfloor x \rfloor -1) + \lfloor x \rfloor ^2 \{x\} $

And $\int_0^{x} \lfloor t \rfloor dt = \frac{\lfloor x \rfloor (\lfloor x\rfloor -1)}{2} + \lfloor x \rfloor\{x\}$

Therefore $$I=\Big[\frac{\lfloor x \rfloor(\lfloor x \rfloor -1)}{2}\Big]^2 + \lfloor x \rfloor^3 \{x\} + 3(\lfloor x \rfloor^2(\lfloor x \rfloor -1) + \lfloor x \rfloor ^2 \{x\}) + 3 \Big(\frac{\lfloor x \rfloor (\lfloor x\rfloor -1)}{2} + \lfloor x \rfloor\{x\}\Big)+x$$

But I am having trouble in reducing it further. I might be missing something, I am sorry about that. Please help me, thank you.

Answer 1

According to OP's stated answer we have \begin{align*} \color{blue}{\int_{0}^{x} \lfloor t+1 \rfloor^3\,dt=\left(\frac{\lfloor x \rfloor(\lfloor x \rfloor +1)}{2}\right)^2+(\lfloor x \rfloor+1)^3 \{x\}}\tag{1} \end{align*}

This can be shown as follows \begin{align*} \int_{0}^{x}\lfloor t+1\rfloor ^3\,dt&=\int_{1}^{x+1}\lfloor u\rfloor ^3\,du\tag{2}\\ &=\sum_{j=1}^{\lfloor x\rfloor}j^3+\int_{\lfloor x\rfloor+1}^{x+1}\lfloor u\rfloor ^3\,du\tag{3}\\ &=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor+1\right)\right)^2+\left(\lfloor x\rfloor+1\right)^3\left\{x\right\} \end{align*}

Comment:

• In (2) we substitute $t+1=u, dt=du$ and arrange the limits accordingly.

• In (3) we use the same representation as in OPs post regarding $\int_{0}^{x}\lfloor t \rfloor^3 dt$.

We can also show the result (1) using OPs calculation, with the correction regarding $\int_{0}^{x}\lfloor t \rfloor^2 dt$ \begin{align*} \int_{0}^{x}\lfloor t \rfloor^3 dt &=\sum_{j=1}^{\lfloor x\rfloor-1}j^3\int_{j}^{j+1}\,dt+\lfloor x\rfloor^3\{x\}\\ &=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)\right)^2+\lfloor x\rfloor^3\{x\}\tag{4}\\ \int_{0}^{x}\lfloor t \rfloor^2 dt &=\sum_{j=1}^{\lfloor x\rfloor-1}j^2\int_{j}^{j+1}\,dt+\lfloor x\rfloor^2\{x\}\\ &=\frac{1}{6}\left(\lfloor x\rfloor-1\right)\lfloor x\rfloor\left(2\lfloor x\rfloor-1\right)+\lfloor x\rfloor^2\{x\}\tag{5}\\ \int_{0}^{x}\lfloor t \rfloor dt &=\sum_{j=1}^{\lfloor x\rfloor-1}j\int_{j}^{j+1}\,dt+\lfloor x\rfloor\{x\}\\ &=\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)+\lfloor x\rfloor\{x\}\tag{6} \end{align*}

With (4)-(6) we obtain according to OPs calculation

\begin{align*} \color{blue}{\int_{0}^{x}\lfloor t+1 \rfloor^3 dt} &= \int_{0}^{x} \lfloor t \rfloor^3 dt + 3\int_{0}^{x} \lfloor x \rfloor^2 dt + 3\int_{0}^{x} \lfloor x \rfloor dt + \int_{0}^{x}\,dt\\ &=\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)\right)^2+\lfloor x\rfloor^3\{x\}\\ &\qquad +3\left(\frac{1}{6}\left(\lfloor x\rfloor-1\right)(\lfloor x\rfloor\left(2(\lfloor x\rfloor-1\right)+\lfloor x\rfloor^2\{x\}\right)\\ &\qquad +3\left(\frac{1}{2}\lfloor x\rfloor\left(\lfloor x\rfloor-1\right)+\lfloor x\rfloor\{x\}\right) \\ &\qquad +\lfloor x\rfloor+\{x\}\\ &\,\,\color{blue}{=\frac{1}{4}\lfloor x\rfloor ^4+\frac{1}{2}\lfloor x\rfloor ^3+\frac{1}{4}\lfloor x\rfloor ^2+(\lfloor x \rfloor+1)^3 \{x\}} \end{align*} corresponding to the claim in (1).