How can I prove that the function $f(x) = \cos (\pi/x)$ when $0 < x \leq 1$, and $f(x)= 0$ when $x=0$ is integrable on $[0,1]$?
I know I can use that the funtion is continous in the interval and using the Riemann's criterion for integrability but I don't know exactly how to start it or if it makes sense.
On
According to Calculus by Michael Spivak (1994), a function is integrable if the lower sum $L$ and the $U$ upper sum converges to same value for any partition $P_n$ when $n \rightarrow\infty$
It means, Let $P_n$ a partition on the interval $[a,b]$, where $n$ is the number of parts of the partition. A function $f$ is integrable, if
$$\lim_{\substack{n\to\infty\\\Vert P_n \Vert \to 0}} U(f,P_n)=\lim_{\substack{n\to\infty\\\Vert P_n \Vert \to 0}} L(f,P_n)$$
Where $\Vert P_n \Vert$ is the norm of the partition.
Here's one thing you could do: fix $\varepsilon > 0$, and partition the domain into $[0, \varepsilon / 4]$ and $[\varepsilon / 4, 1]$. On the latter interval, the function is continuous, and hence integrable.
Therefore, we may find a partition $\mathcal{P}$ of $[\varepsilon/4, 1]$ such that both the upper and lower sums, $U(f, \mathcal{P})$ and $L(f, \mathcal{P})$ respectively, on this partition are within $\varepsilon / 2$ of each other.
Extend $\mathcal{P}$ to a new partition $\mathcal{P}'$ on $[0, 1]$, by taking $\mathcal{P}$ and including $[0, \varepsilon / 4]$. Note that the function is bounded by $1$ and $-1$, and in fact, attains these extreme values infinitely often on $[0, \varepsilon / 4]$. Therefore, the upper sum $U(f, \mathcal{P}')$ will just be $U(f, \mathcal{P}) + \varepsilon/4$. Similarly, $L(f, \mathcal{P}') = L(f, \mathcal{P}) - \varepsilon/4$.
Hence, for any $\varepsilon > 0$, we have constructed a partition $\mathcal{P}'$ on $[0, 1]$ such that $$U(f, \mathcal{P}') - L(f, \mathcal{P}') = U(f, \mathcal{P}) - L(f, \mathcal{P}) + \varepsilon / 2 < \varepsilon.$$ Thus, the function must be integrable.