1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trying to integrate a summation of a unit step function.

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Define I(x)= I( x - x_n ) =

    { 0 , when x < x_n
    { 1, when x >= x_n.

    Let f be the monotone function on [0,1] defined by

    [tex] f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} I ( x - x_n) [/tex]

    where [tex] x_n = \frac {n}{n+1} , n \in \mathbb{N} [/tex].

    Find [tex] \int_0^1 f(x) dx [/tex].

    Leave your answer in the form of an infinite series.

    2. Relevant equations



    3. The attempt at a solution

    I know the theorem, if f is monotone on [0,1], then f is Riemann integrable on [0,1]. I am also familiar with what the graph of this function looks like.

    My calculation of what is under the function is,

    [tex] \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1}{n^2 +3n +2} [/tex].


    I need some advice for completing the problem (assuming that sum is correct).
    Thanks
     
  2. jcsd
  3. Feb 11, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you mean this to be f(x), shouldn't it be a function of x? If you mean it to be the integral, then aren't you done?


     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trying to integrate a summation of a unit step function.
Loading...