For $ x \in \mathbb{R} $, let us denote $ [x] \in \mathbb{Z} $ the integer part of $x$ and define $ f : (0,1) \rightarrow \mathbb{R} $ as follows: $$ f(x) = \begin{cases} 2^{\left[\frac{1}{x}\right]} \,,& x\in \mathbb{Q} \\ 2^{-\left[\frac{1}{x}\right]} \,,& x\not\in \mathbb{Q} \end{cases} $$
show that $f$ is Lebesgue-integrable and calculate its integral.
Some idea? In the beginning, I thought that since $f(x)$ is not bounded on $(0,1)$ it cannot be integrable. But it seems that spite of ot being bounded a function $f$ could be integrable.
Note that the function has the value $2^{-n}$ for almost all $x$ between $\frac 1 {n+1}$ and $\frac 1 n$. [ $\mathbb Q$ has measure $0$]. The integral is just $\sum 2^{-n} (\frac 1 n -\frac 1 {n+1})$. You can compute this by splitting this into two sums.
[Let $g(x)=\sum_{n=1}^{\infty} \frac {x^{n}} n$ where $0 <x <1$. Then $g'(x)=\sum_{n=1}^{\infty} x^{n-1} =\frac 1 {1-x}$. This gives $g(x)=-\log(1-x)$. This gives the value of $\sum 2^{-n} \frac 1 n$ (by taking $x=\frac 1 2$). Now can you find the value of $\sum 2^{-n} \frac 1 {n+1}$?]