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!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor

    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]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook