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:

  1. The set of discontinuities of the integral is

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

  1. 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.


2 Answers 2


The improper integral can be evaluated as the limit of an integral over $[1/n,1]$ as $n \to \infty.$

Making the change of variables $u = 1/x,$ we get

$$\alpha\int_{1/n}^1 \left\lfloor\frac{1}{x}\right\rfloor dx = \alpha\int_{1}^{n } \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1}\int_{k}^{k+1 } \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1}\int_{k}^{k+1 } \frac{k}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1} k\left(\frac{1}{k} - \frac{1}{k+1} \right) \\ = \alpha \sum_{k=2}^{n} \frac{1}{k},\tag{1}$$

and, making the change of variables $u = \alpha/x,$ we get

$$\int_{1/n}^1 \left\lfloor\frac{\alpha}{x}\right\rfloor dx= \alpha \int_{\alpha}^{n \alpha} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \int_{\alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du - \alpha \int_{n \alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \int_{1}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du - \alpha \int_{n \alpha}^{n } \frac{\left\lfloor u\right\rfloor}{u^2} \, du\tag{2}$$

Subtracting (1) from (2) we eliminate the divergent harmonic sum and obtain

$$\int_{1/n}^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor - \alpha \left\lfloor\frac{1}{x}\right\rfloor \right) dx = - \alpha \int_{n \alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = - \alpha \int_{1}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du + \alpha \int_{1}^{\lfloor n \alpha \rfloor} \frac{\left\lfloor u\right\rfloor}{u^2} \, du + \alpha \int_{\lfloor n \alpha \rfloor}^{n \alpha} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = -\alpha \sum_{k=2}^{n} \frac{1}{k} + \alpha \sum_{k=2}^{\lfloor n \alpha \rfloor}\frac{1}{k} + \alpha \frac{n \alpha - \lfloor n \alpha \rfloor}{n \alpha} \\ = -\alpha \left(\sum_{k=1}^{n} \frac{1}{k} - \log n \right) + \alpha \left( \sum_{k=1}^{\lfloor n \alpha \rfloor}\frac{1}{k} - \log\lfloor n \alpha \rfloor \right) - \alpha \log n + \alpha \log \lfloor n \alpha \rfloor + \alpha \frac{n \alpha - \lfloor n \alpha \rfloor}{n \alpha}$$

From here, it should be relatively straightforward to take the limit as $ n \to \infty$ to obtain $\alpha \log \alpha$.

  • $\begingroup$ @Masacroso: Everything is integrable on $[1/n,1]$ -- so we can split up, change variables and add it back together as we wish. Later you are going to take the limit as $n \to \infty$. $\endgroup$
    – RRL
    Jul 4, 2016 at 4:10
  • $\begingroup$ I see, they are partial sums/finite integrals, after you go the limit, right? $\endgroup$
    – Masacroso
    Jul 4, 2016 at 4:11
  • 1
    $\begingroup$ I reduced the problem to those three final integrals which you should be able to do. Split the first two into sums. The third one is easy since $\lfloor u \rfloor = \lfloor n \alpha \rfloor$ on that interval. $\endgroup$
    – RRL
    Jul 4, 2016 at 4:16
  • $\begingroup$ Another question: the change of variable is defined over continuous functions, so I must understand that if we divide the integral on the intervals where is continuous, making the change of variable on each one is the same that make the change of variable directly over the discontinuous function, at least for this problem. $\endgroup$
    – Masacroso
    Jul 4, 2016 at 4:22
  • $\begingroup$ @Masacroso: I'm not sure I understand the last question. I added to the answer and, hopefully, this is now clear. In general, evaluation of such integrals involving floors or fractional parts usually proceeds in this way. Partition the interval to eliminate the appearance of the floor function and then, hopefully, reduce the problem to a sum that you recognize. $\endgroup$
    – RRL
    Jul 4, 2016 at 5:47

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

$$ I(a)=-a\int_{1/a}^1dq\left(\frac{1}{2q^2}-\frac{1}{q}\right)+\frac{1-a}{2} $$


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

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


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