Results 1 to 5 of 5

Math Help - Integration

  1. #1
    Junior Member
    Joined
    Oct 2013
    From
    Australia
    Posts
    45

    Integration



    Can someone help me with this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,935
    Thanks
    784

    Re: Integration

    \begin{align*}\log(e^{2t}+5) & = \log \left[ 5 \left( \dfrac{ e^{2t} }{5} + 1 \right) \right] \\ & = \log(5) + \log\left( \dfrac{ e^{2t} }{5} + 1 \right)\end{align*}

    So, \int_2^x \log(e^{2t}+5)dt = (x-2)\log(5) + \int_2^x \left( \sum_{n\ge 1} (-1)^{n+1}\dfrac{e^{2nt}}{n5^n} \right)dt

    Note: I replaced \log\left(\dfrac{e^{2t}}{5}+1\right) with its power series. The radius of convergence of the power series is \left| \dfrac{e^{2t} }{5} \right| < 1, but over the complex numbers, it can be analytically extended uniquely to the original function. Similarly, once you integrate, the radius of convergence of the series you obtain will also only be finite, but can also be analytically extended (uniquely) over the complex numbers to an expression containing the dilogarithm function (the polylogarithm function where n=2).
    Last edited by SlipEternal; November 15th 2013 at 07:24 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2013
    From
    Australia
    Posts
    45

    Re: Integration

    Could you break down how you arrived at the 3rd line of reasoning from the second?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,935
    Thanks
    784

    Re: Integration

    Quote Originally Posted by Darrylcwc View Post
    Could you break down how you arrived at the 3rd line of reasoning from the second?
    The paragraph directly below the third line explains in a lot of detail precisely how I got from the second line to the third line. Which part is confusing you?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2013
    From
    Australia
    Posts
    45

    Re: Integration

    I think I got it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum