Results 1 to 6 of 6

Math Help - [SOLVED] Integration of floor functions

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    [SOLVED] Integration of floor functions

    If \left \lfloor x \right \rfloor denotes the greatest integer not exceeding x, then \int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}dx.

    \int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}dx=\sum_{x=0}^{\infty}\left \lfloor x \right \rfloor e^{-x}

    I know this can be manipulated into a telescoping series but I am stuck here.
    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
    Is...

    \int_{0}^{\infty} \lfloor x \rfloor \cdot e^{-x} dx = \int_{0}^{1} 0\cdot e^{-x} dx + \int_{1}^{2} e^{-x} dx + \int_{2}^{3} 2\cdot e^{-x} dx + \dots + \int_{n}^{n+1} n\cdot e^{-x} dx + \dots =

    e^{-1} - e^{-2} + 2\cdot e^{-2} -2\cdot e^{-3} + \dots + n\cdot e^{-n} - n\cdot e^{-(n+1)} + \dots = e^{-1} + e^{-2} + \dots + e^{-n} + \dots =

    = \frac{e^{-1}}{1-e^{-1}} = .581976706869...

    Kind regards

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

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by chisigma View Post
    Is...
    e^{-1} - e^{-2} + 2\cdot e^{-2} -2\cdot e^{-3} + \dots + n\cdot e^{-n} - n\cdot e^{-(n+1)} + \dots = e^{-1} + e^{-2} + \dots + e^{-n} + \dots =

    = \frac{e^{-1}}{1-e^{-1}} = .581976706869...

    Kind regards

    \chi \sigma
    I am confused though on how you went from the series to fraction
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The expression \sum_{n=1}^{\infty} e^{-n} is a 'geometric series' and its sum is given by a well known formula...

    Kind regards

    \chi \sigma
    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
    13
    in short way, it's \int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{je^{-x}\,dx}}, so the computation of last integral and the remaining series gives the result.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    in short way, it's \int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{je^{-x}\,dx}}, so the computation of last integral and the remaining series gives the result.
    I was hoping to see the integral set up as a series minus a series which turns out to be telescoping since I know it can be done that way.
    Last edited by dwsmith; May 18th 2010 at 09:02 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 11
    Last Post: November 15th 2009, 11:22 AM
  2. Floor functions
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: November 12th 2009, 11:15 AM
  3. Help with floor and ceiling functions
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: August 9th 2009, 10:30 PM
  4. floor functions.
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: March 16th 2009, 01:11 PM
  5. with floor functions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 8th 2008, 09:43 AM

Search Tags


/mathhelpforum @mathhelpforum