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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).
  2. jcsd
  3. Feb 11, 2009 #2


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    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?

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