Riemann integration of an odd function 
$f(x) = {\rm sgn}({\rm sin}(\frac{\pi}{x}))$ if $x \neq 0$ and $f(0) = 0$

where ${\rm sgn}(x) = 1$ if $x > 0$, ${\rm sgn}(x) = −1$ if $x < 0$ and ${\rm sgn}(0) = 0$.
Show $f$ is Riemann integrable on $[0,1]$
Is there a way to do this without actually solving the upper and lower sums? If not, then how do you evaluate the upper and lower sums?
 A: You can prove in general that if you have a bounded function on $[a,b]$ which is integrable on $[a+\varepsilon,b]$ for all $\varepsilon >0$ then it is integrable on $[a,b]$. 
To prove this claim just consider a partition of $[a,b]$ where the first interval is $[a,a+\varepsilon]$. The area on that interval is bounded by $\varepsilon L$ where $L$ is the bound of the function. On the rest you know that the upper and lower sums are adjacent. 
A: $ f(x)=-1$ on $[\frac{1}{2n},\frac{1}{2n-1}]$
and  $ f(x)=1$ on $[\frac{1}{2n-1},\frac{1}{2n-2}]$
So $$\int_0^1 f(x)\ dx = \sum_{n=1}^\infty \frac{1}{2n}-\frac{1}{2n-1} +\sum_{n=2}^\infty \frac{1}{2n-2} - \frac{1}{2n-1}$$
 $$=\sum_{n=1}^\infty \frac{-1}{(2n-1)2n} +\sum_{n=2}^\infty \frac{1}{(2n-1)(2n-2)}=\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} $$
$ |\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} |<\infty$. So integrable
Detail :
   Given $\epsilon >0$ there exists $N$ s.t. $$ \sum_{n \geq
   N}^\infty
   \frac{(-1)^n}{n(n+1)} <\epsilon $$
Let $\delta = \frac{\epsilon}{N}$ and consider partition $P,\ P_N$ s.t.
   $$ \| P\| < \delta,\ P_N \ :\ 0 < \frac{1}{N} < \frac{1}{N-1}< \cdots < 1 $$
Then let $P_0=P\cup P_N$ In this partition, riemann sum $I$
satisfies $$ |I- \sum_{n=1}^\infty
   \frac{(-1)^n}{n(n+1)} | < N\delta + \epsilon $$
Surely if $Q$ is finer than $P_0$,
$$ |I(Q)- \sum_{n=1}^\infty
   \frac{(-1)^n}{n(n+1)} | < N\delta + \epsilon $$
