# Riemann Integrability of f(x)

• Mar 30th 2011, 08:30 PM
kalyanram
Riemann Integrability of f(x)
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.
• Mar 30th 2011, 09:49 PM
roninpro
A hint: use partial fractions

$\displaystyle \displaystyle \frac{2r+1}{r(r+1)^2}=\frac{1}{r}-\frac{1}{1+r}+\frac{1}{(1+r)^2}$

and the fact that

$\displaystyle \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$
• Mar 31st 2011, 09:33 AM
kalyanram
Yeah thats true I tried that but I did not know that $\displaystyle \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

Can you tell me how to prove that the sum of the series is in fact $\displaystyle \frac{\pi^2}{6}$ I can prove that its convergent but what about the limit of convergence of the series.

Regards,
Kalyan.
• Mar 31st 2011, 11:13 AM
Plato
Quote:

Originally Posted by kalyanram
Yeah thats true I tried that but I did not know that $\displaystyle \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$
Can you tell me how to prove that the sum of the series is in fact $\displaystyle \frac{\pi^2}{6}$ I can prove that its convergent but what about the limit of convergence of the series.

Notice that $\displaystyle \displaystyle\sum\limits_{r = 1}^\infty {\left( {\frac{1}{r} - \frac{1}{{r + 1}}} \right)} = 1$.

Further notice that $\displaystyle \displaystyle\sum\limits_{r = 1}^\infty {\frac{1}{{\left( {r + 1} \right)^2 }}} = \sum\limits_{r = 2}^\infty {\frac{1}{{\left( r \right)^2 }} = \frac{{\pi ^2 }}{6} - 1}$
• Apr 4th 2011, 09:28 PM
roninpro
Quote:

Originally Posted by kalyanram
Yeah thats true I tried that but I did not know that $\displaystyle \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

Can you tell me how to prove that the sum of the series is in fact $\displaystyle \frac{\pi^2}{6}$ I can prove that its convergent but what about the limit of convergence of the series.

Regards,
Kalyan.

I know this is a little late, but:

Basel problem - Wikipedia, the free encyclopedia