I have to prove that the function : $f(x)=0$, if $x=0$ and $f(x)=\frac {1}{[\frac {1}{x}]}$, if $x\in (0,1]$ is integrable and to find the integral $I=\int_0^1 f(x)dx$.
I define $f_n(x)= 0, x\in [0,\frac {1}{n+1})$ and $f_n(x)=\frac {1}{k}, x\in (\frac {1}{k+1},\frac {1}{k}], k=1,2, ..., n$.
Now there are some problems : (a) Does the sequences converge to f uniformly?
(b) Trying the theorem, I find that I is equal to $\frac {\pi ^2}{6}-1$, but is that correct ? Thank you in advance !
Yes the sequence converges uniformly. Indeed, let $\epsilon>0$. Let $N \in \mathbb{N}$ be such that $\frac{1}{N+1} < \epsilon$.
Also, notice that by definition, $f_n(x)=f(x)$ on $\big{(}\frac{1}{n+1},1]$. Also, for $x \in [0,n+1]$, $0 \leq f(x) \leq \frac{1}{n+1}$.
Therefor, for $n \geq N$ and for $x \in [0,1]$, we have that $|f_n(x)-f(x)|\leq \frac{1}{n+1}<\epsilon$. We deduce uniform convergence.
For your evaluation of the integral, that is correct according to my calculations, but you should add how you came to that value in your post!
Hope this helps!