Page 2 of 2 FirstFirst 12
Results 16 to 20 of 20

Math Help - Proof that sum^infty of 1/n^2 = ...

  1. #16
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    No, I didn't, I thought you referenced it because it contained a proof--I didn't realize it had six. Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    This is a quality thread! I now know 3 more proofs of this!

    To simplependulum: did you think of that yourself?! If so, how long did that take you and what led you to it? Kudos!
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by chiph588@ View Post
    This is a quality thread! I now know 3 more proofs of this!

    To simplependulum: did you think of that yourself?! If so, how long did that take you and what led you to it? Kudos!
    At first i was thinking how to solve  \int_1^{\infty} \frac{\ln(x)}{x^2 - 1 }~dx without using the fact  \zeta(2) = \frac{\pi^2}{6} , in other words , I shall give another way to prove this fact starting at this point .

    However , i failed , all my attempts eventually led to other troublesome integrals so I gave up , until I was asked to solve another integral  \int_0^1 \frac{ \tan^{-1}(x)}{\sqrt{1-x^2 }}~dx which brought some ideas to me in solving the integral I was struggling two weeks before .
    Last edited by simplependulum; August 8th 2010 at 10:33 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Junior Member
    Joined
    Dec 2008
    Posts
    62
    Quote Originally Posted by simplependulum View Post
      \int_1^{\infty} \frac{\ln(x)}{ x^2 - 1  }~dx = \frac{3}{4} \zeta(2)

     \vdots

     = \frac{\pi^2}{8}
    Very creative! I'm impressed!

    Two questions though:

    First, how do we know  \displaystyle \int_1^\infty \frac{\ln(x)}{x^2-1}dx = \frac34 \zeta(2)?

    Also for  \displaystyle \int_0^1 \left[ \tan^{-1}\left(t\sqrt{1-y^2}\right) \right]_0^{\infty} \frac{dy}{\sqrt{1-y^2} } , we have  \displaystyle y\in[0,1] .

    So  \displaystyle \lim_{t\to\infty} \tan^{-1}\left(t\sqrt{1-y^2}\right) = \begin{cases} \frac\pi2 ,\text{ if } y\neq1\\0,\text{ if } y=1 \end{cases} .

    Therefore, what do we do when  \displaystyle y=1?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Here is my derivation of the first fact, although we need this definition of the zeta function

    [LaTeX ERROR: Convert failed]

    So, in the initial integral, let u = \ln(x). Then dx = x du = e^{u} du. u(\infty) = \infty and u(1) = 0. This last bit gives you the new limits of integration.

    [LaTeX ERROR: Convert failed]

    Where on the RHS, I've let u = x so that i don't have to use u from now on. Now lets focus on the integrand.

    \displaystyle \frac{xe^x}{e^{2x}-1} = \frac{x(e^x+1-1)}{e^{2x}-1} = \frac{x}{e^{x}-1} - \frac{x}{e^{2x}-1}.

    The integral becomes

    \displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx<br />
    provided both converge. Now on the second piece, let  u = 2x, dx = du/2 and x = u/2 to give that the piece is

    \displaystyle \frac{1}{4}\int_{0}^{\infty}\frac{u}{e^u-1}\,du

    Combining both pieces now we get

    \displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx = \frac{3}{4}\int_{0}^{\infty}\frac{x}{e^x-1}\,dx = \frac{3}{4}\Gamma(2)\zeta(2) = \frac{3}{4}\zeta(2)
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. \int_0^{\infty}\frac{\sin x}{x}\ dx
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 21st 2011, 01:16 AM
  2. Replies: 2
    Last Post: March 1st 2011, 04:04 AM
  3. [SOLVED] P(s,t)=\sum_{0}^{\infty}\sum_{0}^{\infty}a_{m,n}s^ mt^n
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: January 11th 2011, 01:46 PM
  4. Integrals \int^{\infty}_0 \ln(x)/\cosh^2(x) dx
    Posted in the Calculus Forum
    Replies: 0
    Last Post: June 4th 2008, 09:28 PM
  5. Behavior as x,y -> infty
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 5th 2008, 08:36 PM

/mathhelpforum @mathhelpforum