Results 1 to 5 of 5

Math Help - Riemann Integrability of f(x)

  1. #1
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    143
    Thanks
    14

    Riemann Integrability of f(x)

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

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

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

    Regards,
    Kalyan.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    A hint: use partial fractions

    \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 \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    143
    Thanks
    14
    Yeah thats true I tried that but I did not know that \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 \frac{\pi^2}{6} I can prove that its convergent but what about the limit of convergence of the series.

    Regards,
    Kalyan.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by kalyanram View Post
    Yeah thats true I tried that but I did not know that \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 \frac{\pi^2}{6} I can prove that its convergent but what about the limit of convergence of the series.
    Notice that \displaystyle\sum\limits_{r = 1}^\infty  {\left( {\frac{1}{r} - \frac{1}{{r + 1}}} \right)}  = 1.

    Further notice that \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}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    Quote Originally Posted by kalyanram View Post
    Yeah thats true I tried that but I did not know that \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 \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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Riemann Integrability
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 11th 2010, 08:31 PM
  2. Riemann Integrability
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 23rd 2010, 11:12 AM
  3. Replies: 1
    Last Post: January 16th 2010, 03:31 AM
  4. Riemann integrability
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: December 7th 2009, 12:12 PM
  5. Riemann Integrability
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 15th 2008, 04:22 PM

/mathhelpforum @mathhelpforum