TITLE: Show that $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$
QUESTION [21 upvotes]: Show that the improper integral $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$, for $\alpha\in(0,1)$.

This is an integral of Riemann. My work:

The set of discontinuities of the integral is

$$D=\left\{\frac1k:k\in\Bbb N\right\}\cup\left\{\frac{\alpha}{k}:k\in\Bbb N\right\}$$

And when we have that $x>\alpha$ the integral can be simplified to

$$\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\int_0^\alpha \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx-\alpha\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx$$
I dont know how to continue from here, it is not clear how to handle the partition $D$ to simplify the integral. What I did here is just see what is the value of $\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx$ to see if I get some clue.
If there is no weird mistake somewhere:
$$\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx=\int_\alpha^{\frac1{\left\lfloor 1/\alpha\right\rfloor}}\frac{\mathbf 1_{\Bbb N}(1/\alpha)\mathrm dx}{\lfloor 1/\alpha\rfloor}+\sum_{k=1}^{\lfloor1/\alpha\rfloor}\int_{\frac1{k+1}}^{\frac1k}\frac{\mathrm dx}{k}=\\=\mathbf 1_{\Bbb N}(1/\alpha)\frac{1-\alpha\lfloor 1/\alpha\rfloor}{\lfloor 1/\alpha\rfloor^2}+\sum_{k=1}^{\lfloor1/\alpha\rfloor}\frac1{k^2(k+1)}$$
what is not useful at all. So I get stuck with this problem, can you help me to show this identity (not going deeper than a Riemann integral background)? Thank you in advance.

REPLY [9 votes]: I would follow a different route: 
First let us change variables $x\rightarrow1/y$ so the problem is equally stated as 

$$
I(a)=\int_1^{\infty}dy\frac{1}{y^2}\left(\left\lfloor a y\right\rfloor- a\left\lfloor y\right\rfloor\right)
$$

there is a well known Fourier expansion of the floor function which leads to 
$$
I(a)=\frac{1}{\pi}\int_1^{\infty}dy\frac{1}{y^2}\left(\sum_{k\geq1}\frac{\sin(2\pi k a y)-a\sin(2\pi k y)}{k}+\pi\frac{1-a}{2}\right)
$$
by setting $ay=q$ int the first part of this integral we night simplify
$$
I(a)=-\frac{a}{\pi}\int_{1/a}^1dq\frac{1}{q^2}\sum_{k\geq 1} \frac{\sin(2 \pi k q)}{k}+\frac{1-a}{2}
$$
Since the sum is sinply $\Im\sum_{k\geq 1} \frac{e^{2 i \pi k q}}{k}=\frac{\pi}{2}-\pi q$ we are left with trivial integrations
$$
I(a)=-a\int_{1/a}^1dq\left(\frac{1}{2q^2}-\frac{1}{q}\right)+\frac{1-a}{2}
$$
or

$$
I(a)=a\log(a)
$$


A (simple) proof of the aforementioned Fourier expansion might be found here