Results 1 to 5 of 5

Math Help - Tough Integral

  1. #1
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2

    Tough Integral

    For m\in\mathbb{N}, find the following integral:

    I_{m} =  \int_{0}^{1}\left(\frac{\ln{x}}{1-x}\right)^{m}\;{dx}.

    Source: a cool guy called Drexel28 left it as an exercise for an unsuspecting student in mathlinks.
    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

    Re: Tough Integral

    Quote Originally Posted by TheCoffeeMachine View Post
    For m\in\mathbb{N}, find the following integral:

    I_{m} = \int_{0}^{1}\left(\frac{\ln{x}}{1-x}\right)^{m}\;{dx}.

    Source: a cool guy called Drexel28 left it as an exercise for an unsuspecting student in mathlinks.
    Several years ago I 'discovered' that is...

    \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}} (1)

    Now if we consider the 'binomial expansion'...

     (1-x)^{-m} = 1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!}\ x^{n} (2)

    ... combining (1) and (2) we obtain...

    I_{m}= \int_{0}^{1} (\frac{\ln x}{1-x})^{m}\ dx = (-1)^{m}\ m!\ \{1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!\ (n+1)^{m+1}} \} (3)

    At this point some more work on the series (3) is useful!...

    Kind regards

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

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Tough Integral

    Quote Originally Posted by chisigma View Post
    Several years ago I 'discovered' that is...

    \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}} (1)

    Now if we consider the 'binomial expansion'...

     (1-x)^{-m} = 1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!}\ x^{n} (2)

    ... combining (1) and (2) we obtain...

    I_{m}= \int_{0}^{1} (\frac{\ln x}{1-x})^{m}\ dx = (-1)^{m}\ m!\ \{1 + \sum_{n=1}^{\infty} \frac{m\ (m+1)\ ...\ (m+n-1)}{n!\ (n+1)^{m+1}} \} (3)

    At this point some more work on the series (3) is useful!...
    For some values of m...

    I_{1} =  - \{1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}}\} = - \zeta(2) (1)

    I_{2} =  2 \{1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}}\} = 2\ \zeta(2) (2)

    I_{3} =  - 3! \{1 + \sum_{n=1}^{\infty} \frac{n+2}{2\ (n+1)^{3}}\} = - 3! \{1 + \sum_{n=1}^{\infty} \frac{1}{2\ (n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{2\ (n+1)^{3}} \} =

    = -3! - 3\ \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}} - 3\ \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} = -3\ \{\zeta(2) + \zeta(3)\} (3)

    Does anyone want to proceed?...

    Kind regards

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

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2

    Re: Tough Integral

    Quote Originally Posted by chisigma View Post
    Several years ago I 'discovered' that is...

    \int_{0}^{1} x^{n}\ \ln ^{m} x\ dx = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}} (1)
    Nice. I've seen you use this before. Did you prove by setting up a recurrence relation?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12

    Re: Tough Integral

    Gamma function,

    \int_{0}^{1}{{{x}^{n}}{{\ln }^{m}}x\,dx}={{(-1)}^{m}}\int_{0}^{\infty }{{{t}^{m}}{{e}^{-(n+1)t}}\,dt}=\frac{{{(-1)}^{m}}}{{{(n+1)}^{m+1}}}\int_{0}^{\infty }{{{t}^{m}}{{e}^{-t}}\,dt}=\frac{{{(-1)}^{m}}m!}{{{(n+1)}^{m+1}}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tough Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 3rd 2010, 12:22 AM
  2. Tough Integral.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 19th 2009, 10:23 PM
  3. tough integral
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 11th 2009, 11:27 PM
  4. tough integral?.
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 31st 2008, 11:36 PM
  5. tough integral?.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 19th 2008, 05:17 PM

Search Tags


/mathhelpforum @mathhelpforum