Show that $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$

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 $\alpha \log \alpha$.

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

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

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