Integral of $[\frac{1}{x}]^{-1}$ on $(0,1]$ I found in a book this exercise:

Prove that $\int_0^1 \frac{1}{[{\frac{1}{x}}]}dx=+\infty$

In my proof i find that this  integral is finite.
Here is my proof:
By Beppo-Levi:
$\int_0^1 \frac{1}{[{\frac{1}{x}}]}dx=\sum_{n=1}^{\infty}\int_{[\frac{1}{n+1},\frac{1}{n})}\frac{1}{[{\frac{1}{x}}]}dx$
$\frac{1}{n+1} \leq x <\frac{1}{n} \Longrightarrow n<\frac{1}{x} \leq n+1$
So $[\frac{1}{x}]=n \Longrightarrow \frac{1}{[{\frac{1}{x}}]}=\frac{1}{n}$
Thus the integral is equal to $$\sum_{n=1}^{\infty}\frac{1}{n}(\frac{1}{n}-\frac{1}{n+1})<+\infty$$
Am i missing something?

You're right, such integral is finite:
$$\int_{0}^{1}\frac{dx}{\left\lfloor\frac{1}{x}\right\rfloor}=\sum_{m\geq 1}\int_{1/(m+1)}^{1/m}\frac{dx}{\left\lfloor\frac{1}{x}\right\rfloor}=\sum_{m\geq 1}\frac{\frac{1}{m}-\frac{1}{m+1}}{m}=\zeta(2)-1. $$
