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Help me to survive from a big dilemma!(Unit step function)

  1. Oct 9, 2014 #1
    Consider:

    [tex] u(t)=\begin{cases} 1\quad \quad \quad \quad t>0 \\ 0\quad \quad \quad \quad t<0 \end{cases} [/tex]
    Now I want to calculate this:

    [tex] \int _{ 0 }^{ a }{ \frac { u(t)-u(t-a) }{ { t }^{ 2 } } } dt [/tex]
    whereas: a>0
    What is confusing me is this point that should our answer for the integral include the step function again?
     
  2. jcsd
  3. Oct 9, 2014 #2

    SteamKing

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    It's not clear what you mean when you say, "should our answer for the integral include the step function again?"

    What I see is a step function which is active when t > 0 but which is turned off for t > a, so that there is a non-zero integrand in the interval 0 < t < a.
     
  4. Oct 9, 2014 #3

    HallsofIvy

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    In this integral t is always between 0 and a so t is positive while t- a is negative. u(t)= 1 and u(t- a)= 0.

    Your integral is just [itex]\int_0^a \frac{1}{t^2} dt[/itex]
     
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