How can $\int_0^1 \lfloor ax^2+bx+c \rfloor \,\Bbb dx$ be generalized?

Question

Let $$I = \int_0^1 \lfloor -8x^2+6x-1 \rfloor \,\Bbb dx$$ (where $\lfloor \cdot \rfloor$ represents the floor function/greatest integer function) which on solving gives the value $$I = \frac{\sqrt{17}-13}{8} = -0.876.$$

Solving this integral took me a lot of time (manually graphing the quadratic and breaking it at integral values) and I was wondering if there can be a generalized result possible for, let $a$, $b$, $c$ be integers, $$J = \int_0^1 \lfloor ax^2+bx+c \rfloor \,\Bbb dx$$ If evaluating $J$ is not conceptually possible/correct, can there be generalized solution for [1] or [2]?

[1] $\displaystyle\int \lfloor ax^2+bx+c \rfloor \,\Bbb dx$

[2] $\displaystyle\int_n^m \lfloor ax^2+bx+c \rfloor \,\Bbb dx$ ($m$ and $n$ are integers)

Answer 1

The function $ax^2+bx+c$ can be converted into either $x^2+r$ if $a>0$ or $-x^2+r$ if $a<0$. We therefore calculate the integrals $\int_0^x \lfloor u^2+r\rfloor du$ and $\int_0^x \lfloor -u^2+r\rfloor du$ when $x\ge 0$.

Since $r=\lfloor r\rfloor +\{r\}$, without loss of generality, we assume $0\le r<1$. For calculating $\int_0^x \lfloor u^2+r\rfloor du$, we have $$ \int_0^x \lfloor u^2+r\rfloor du=\begin{cases} 0&,\quad 0\le x\le \sqrt{1-r}\\ x-\sqrt{1-r}&,\quad \sqrt{1-r}\le x\le \sqrt{2-r}\\ \sqrt{2-r}-\sqrt{1-r}+2(x-\sqrt{2-r})&,\quad \sqrt{2-r}\le x\le \sqrt{3-r}\\ \vdots\\ i(x-\sqrt{i-r})+\sum_{j=1}^{i-1}j(\sqrt{j+1-r}-\sqrt{j-r})&,\quad \sqrt{i-r}\le x\le \sqrt{i+1-r}\\ \vdots \end{cases}. $$ Therefore, for any $r\in \Bbb R$ and $x\ge 0$ we have $$ \int_0^x \lfloor u^2+r\rfloor du{ =\int_0^x \lfloor u^2+\lfloor r\rfloor +\{r\}\rfloor du \\=\int_0^x \lfloor u^2 +\{r\}\rfloor +\lfloor r\rfloor du \\=\int_0^x \lfloor u^2 +\{r\}\rfloor du +\lfloor r\rfloor x \\=\lfloor r\rfloor x+x\lfloor x^2+\{r\}\rfloor-\sum_{j=1}^{\lfloor x^2+\{r\}\rfloor} \sqrt{j-\{r\}} \\=x\lfloor x^2+r\rfloor-\sum_{j=1}^{\lfloor x^2+\{r\}\rfloor} \sqrt{j-\{r\}}. } $$ Similarly, for any $r\in \Bbb R$ and $x\ge 0$ we have $$ \int_0^x \lfloor -u^2+r\rfloor du=-x\lfloor x^2-r+1\rfloor +\sum_{j=0}^{\lfloor x^2-\{r\}\rfloor}\sqrt{j+\{r\}}. $$ By using $\int_0^x f(u)du=\text{sgn}(x)\int_0^{|x|} f(u)du$ for even $f(u)$, we can write $${ \int_0^x \lfloor u^2+r\rfloor du=x\lfloor x^2+r\rfloor-\text{sgn}(x)\sum_{j=1}^{\lfloor x^2+\{r\}\rfloor} \sqrt{j-\{r\}}, \\ \int_0^x \lfloor -u^2+r\rfloor du=-x\lfloor x^2-r+1\rfloor +\text{sgn}(x)\sum_{j=0}^{\lfloor x^2-\{r\}\rfloor}\sqrt{j+\{r\}}. } $$

Answer 2

I do not know if this helps, but this was too long for a comment.

Well, using the fact that:

$$\left\lfloor x\right\rfloor=x-\frac{1}{2}+\frac{1}{\pi}\sum_{n\ \geq\ 1}\frac{\sin\left(2\pi nx\right)}{n}\tag1$$

So, we get:

\begin{align} \mathcal{I}\left(\alpha,\beta,\gamma\right)&=\int\limits_0^1\left\lfloor \alpha x^2+\beta x+\gamma\right\rfloor\,\Bbb dx\\ &=\int\limits_0^1\left(\alpha x^2+\beta x+\gamma-\frac{1}{2}+\frac{1}{\pi}\sum_{n\ \geq\ 1}\frac{\sin\left(2\pi n\left(\alpha x^2+\beta x+\gamma\right)\right)}{n}\right)\,\Bbb dx\\ &=\frac{\alpha}{3}+\frac{\beta}{2}+\gamma-\frac{1}{2}+\frac{1}{\pi}\sum_{n\ \geq\ 1}\frac{1}{n}\int\limits_0^1\sin\left(2\pi n\left(\alpha x^2+\beta x+\gamma\right)\right)\,\Bbb dx\tag2 \end{align}

Answer 3

Let $f(x)=ax^2+bx+c$ be a quadratic. The floor function can be written as

$$\lfloor x \rfloor=\sum_{n=1}^\infty \theta(x-n)-\sum_{n=0}^\infty \theta(-x-n)$$ where $\theta(x)=1$ if $x\ge0 $ and $\theta(x)=0$ otherwise.

This transforms the integral into a sum

$$\int_\alpha^\beta \lfloor f(x) \rfloor dx= \sum_{n=1}^\infty \int_\alpha^\beta\theta(f(x)-n) dx-\sum_{n=0}^\infty \int_\alpha^\beta\theta(-f(x)-n) dx$$

Each integral appearing in this sum measures the size of the solution set (in $[\alpha,\beta]$) to $f(x)\ge n$ or $f(x)\le -n$ for $n>0$. For a quadratic, $f(x)=ax^2+bx+c$ let $r_1,r_2$ be the two roots of $f(x)-n=0$. Then the solutions to $f(x)\ge n$ or $f(x)\le -n$ are either $[r_1,r_2]$ or $\mathbb{R}\setminus[r_1,r_2]$ according to the sign of $a$. The roots can of course be written in terms of the coefficients $a,b,c$ using the quadratic formula.

Let's see how this works in your example, $f(x)=-8x^2+6x-1$. In this case, the stationary point is at $(\frac{3}{8},\frac{17}{8})$ so $f(x)\ge n$ has no solutions for $n>0$. Moreover, for $n>2$, $f(x)\le -n$ has no solutions in $[\alpha,\beta]=[0,1]$. Therefore, the whole thing is a sum of just a few terms:

\begin{align*} -8x^2+6x-1\le 0 & \iff x\in [0,1]\setminus(\frac{1}{4},\frac{1}{2})\\ -8x^2+6x-1\le -1 & \iff x\in [0,1]\setminus(0,\frac{3}{4})\\ -8x^2+6x-1\le -2 & \iff x\in [0,1]\setminus(\frac{3-\sqrt{17}}{8},\frac{3+\sqrt{17}}{8})\\ \end{align*} To get the answer, you add the negative length of each solution-set. So, \begin{align*} \int_\alpha^\beta \lfloor -8x^2+6x-1 \rfloor dx&=-\left(1-\left(\frac{1}{2}-\frac{1}{4}\right)\right)-\left(1-\left(\frac{3}{4}-0\right)\right)-\left(1-\left(\frac{3+\sqrt{17}}{8}-\frac{3-\sqrt{17}}{8}\right)\right)\\ &=-\frac{3}{4}-\frac{1}{4}-\frac{4-\sqrt{17}}{4}\\ &=\frac{-8+\sqrt{17}}{4} \end{align*}