How can I calculate the following integral?

Question

$$ \int_{a}^{b} \left \lfloor \frac{1}{u} \right \rfloor u du $$

I can't really tell what I should do in this case. I have calculated integrals of floor functions before, but this one is really not working...

Answer 1

The integrand $f:\mathbb{R} \to\mathbb{R}$ defined as

$$\tag{1} f(x)~:=~ \left\{\begin{array}{lcl} x \left\lfloor \frac{1}{x} \right\rfloor &{\rm for}& x\neq 0 ,\cr 1 &{\rm for}& x= 0,\end{array} \right. $$

satisfies the double inequality

$$\tag{2} 1-x^+ ~\leq~ f(x)~ \leq~ 1+ x^-,$$

and hence $f\in{\cal L}^1_{\rm loc}(\mathbb{R})$ is locally integrable. The integral

$$\tag{3} I(a,b)~:=~ \int_a^b \!dx~ f(x)$$

is therefore a continuous function of the two endpoints $a$ and $b$. We assume that $a\leq b$. We claim that the integral (3) is

$$\tag{4} I(a,b) ~=~ F(b) - F(a) + \sum_{n\in \mathbb{Z}\backslash \{0\}}^{a\leq \frac{1}{n}<b}\frac{1}{2n^2} ,$$

where we have defined a function $F:\mathbb{R} \to\mathbb{R}$ by

$$\tag{5} F(x)~:=~ \left\{\begin{array}{lcl} \frac{x^2}{2} \left\lfloor \frac{1}{x} \right\rfloor &{\rm for}& x\neq 0 ,\cr 0 &{\rm for}& x= 0.\end{array} \right. $$

Sketched proof: One may check that eq. (4) is differentiable in $a$ and $b$; that $I(a,a)=0$; and that

$$ \tag{6} \frac{\partial I(a,b)}{\partial b}~=~ f(b) .$$