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Integration of funtion multiplied by shifted heavyside step

  1. Sep 26, 2010 #1
    As I understand it, the following is true:

    \int_{0}^{\infty}{u(t - \lambda) d\lambda} =

    But I do not understand why. It seems to me that the left side above should equal



    u(t - \lambda) =
    \left\{\begin{array}{cc}0,&\mbox{ if }
    t< \lambda \\ 1, & \mbox{ if } t> \lambda \end{array}\right.

    I obviously don't understand this correctly. What am I not doing right?
  2. jcsd
  3. Sep 26, 2010 #2
    I guess there is not a function multiplied by the unit step function here. But if there were, like:

    \int_{0}^{\infty}{f( \lambda ) u(t - \lambda) d\lambda}


    then it would be equal to

    \int_{0}^{t}{f( \lambda) d\lambda}

  4. Sep 26, 2010 #3
    Oh, I think I am seeing my mistake in logic here.

    The unit step [tex] u(t - \lambda) [/tex] is a shift the unit step [tex] u(\lambda) [/tex] and then symmetric about the vertical axis.

    I should be thinking of this as a graph of [tex] \lambda [/tex] vs. [tex] u(\lambda) [/tex].

    I was getting thrown off by the variable t here because I am so used to it being the independent variable. Okay, okay. Never mind.
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