Results 1 to 3 of 3

Math Help - a couple more definite integrals

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    a couple more definite integrals

     \displaystyle \int_{0}^{1} \ln(1+x) \ln(1-x) \ dx

    If the above integral is too easy, also try to evaluate  \displaystyle \int_{0}^{1} \ln(1+x) \ln (x) \ln(1-x) \ dx (though I personally wouldn't recommend it ).
    Last edited by Random Variable; March 29th 2011 at 08:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    With the substitution t=1-x the integral becomes...

    \displaystyle I= \int_{0}^{1} \ln t\ \ln (2-t)\ dt (1)

    Now is...

    \displaystyle  \ln (2-t)= \ln 2 - \sum_{n=1}^{\infty} \frac{t^{n}}{n\ 2^{n}} (2)

    ... and remembering the general formula...

    \displaystyle  \int_{0}^{1} t^{n}\ \ln t\ dt = -\frac{1}{(n+1)^{2}} (3)

    ... You obtain...

    \displaystyle  I= -\ln 2 + \sum_{n=1}^{\infty} \frac{1}{n\ (n+1)^{2}\ 2^{n}} (4)

    To extablish if the sum of the series in (4) can be expressed as function of known constants it is necessary some more work...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    If you make a slightly different substitution, you'll end up having to evaluate a much simpler series.


     \displaystyle \int^{1}_{0} \ln (1+x) \ln(1-x) \ dx = \frac{1}{2} \int^{1}_{-1} \ln (1+x) \ln(1-x) \ dx


    Now let  \displaystyle 2t = 1+x


     \displaystyle = \int^{1}_{0} \ln (2t) \ln(2-2t) \ dt = \int^{1}_{0} \big(\ln 2 + \ln t\big) \big(\ln 2 + \ln(1-t) \big) \ dt

     \displaystyle = \ln^{2}(2) \int ^{1}_{0} \ dt + \ln 2 \int^{1}_{0} \ln(1-t) \ dt + \ln 2 \int^{1}_{0} \ln t \ dt + \int^{1}_{0} \ln t \ln(1-t) \ dt

    The first three integrals are straightforward, and the last one is equivalent to evaluating \displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^2} which is much easier to evaluate than  \displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^{2}2^{n}} .


    The final answer is  \displaystyle \ln^{2}(2) - 2 \ln 2 + 2 - \frac{\pi^{2}}{6}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 06:21 PM
  2. Replies: 4
    Last Post: April 13th 2011, 03:08 AM
  3. Replies: 14
    Last Post: April 6th 2011, 11:22 AM
  4. Need help with couple of integrals!
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 1st 2010, 11:32 AM
  5. a couple integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 18th 2008, 01:20 PM

Search Tags


/mathhelpforum @mathhelpforum