A function $\displaystyle f$ defined on [0,1] by $\displaystyle f(x) = 2rx, where \frac{1}{1+r} < x < \frac{1}{r} , (r=1,2,3,......)$ then show that $\displaystyle f \in \Re[0,1]$ and its integral is $\displaystyle \frac{\pi^2}{6}$ where $\displaystyle f \in \Re[0,1]$ denotes that f is Riemann integrable on [0,1].

$\displaystyle {\bf Sol:}$

Here is how I went about solving it firstly the function is discontinuous $\displaystyle \forall x \in S=\{x: x=\frac{1}{r} where (r=1,2,3,......)\}$ and the set $\displaystyle S$ has one limit point namely $\displaystyle \{0\}$ and $\displaystyle f$ is bounded and monotonic in [0,1] hence $\displaystyle f \in \Re[0,1]$

Now the trouble is when I evaluate $\displaystyle \int^1_0 f(x)\,dx$ = $\displaystyle \displaystyle\sum_{r=1}^{\infty}\int^\frac{1}{r}_\ frac{1}{1+r}2rx\,dx$ = $\displaystyle \displaystyle\sum_{r=1}^{\infty}\frac{2r+1}{r(r+1) ^2}$ but I cannot prove it equals $\displaystyle \frac{\pi^2}{6}$ any help is much appreciated.

Regards,

Kalyan.