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?
 A: You're right, such integral is finite:
$$\int_{0}^{1}\frac{dx}{\left\lfloor\frac{1}{x}\right\rfloor}=\sum_{m\geq 1}\int_{1/(m+1)}^{1/m}\frac{dx}{\left\lfloor\frac{1}{x}\right\rfloor}=\sum_{m\geq 1}\frac{\frac{1}{m}-\frac{1}{m+1}}{m}=\zeta(2)-1. $$
A: Maybe there's a typo in the orginal excercise, so that the r.h.s. should read $\lt \infty$.
But why not calculate the integral $i = \int_0^1 \frac{1}{\left[\frac{1}{x}\right]}\,dx$ explicitly?
Here $\left[x\right]$ is the integer part of x.
Subsituting $x\to \frac{1}{t}$, $dx\to - \frac{dt}{t^2}$ the integral becomes
$$i = \int_{1}^\infty \frac{1}{t^2} \frac{1}{\left[t\right]}\,dt$$
Splitting the integration region into parts from $k$ to $k+1$ we get the sum
$$i = \sum_{k=1}^\infty \int_{k}^{k+1} \frac{1}{t^2} \frac{1}{\left[t\right]} \,dt $$
Since $\left[t\right]=k$ for $k\le t \lt{k+1}$ the sum becomes
$$i = \sum_{k=1}^\infty \frac{1}{k} \int_{k}^{k+1} \frac{1}{t^2} \,dt = \sum_{k=1}^\infty \frac{1}{k} \left(\frac{1}{k}-\frac{1}{k+1}\right) =\sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{k(k+1)}= \zeta(2) - 1 \simeq 0.644934$$
