Integration of funtion multiplied by shifted heavyside step

  • #1

Main Question or Discussion Point

As I understand it, the following is true:

[tex]
\int_{0}^{\infty}{u(t - \lambda) d\lambda} =
\int_{0}^{t}{d\lambda}
[/tex]


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

[tex]
\int_{\lambda}^{\infty}{d\lambda}
[/tex]

since

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

I obviously don't understand this correctly. What am I not doing right?
 

Answers and Replies

  • #2
I guess there is not a function multiplied by the unit step function here. But if there were, like:
[tex]

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

[/tex]

then it would be equal to

[tex]
\int_{0}^{t}{f( \lambda) d\lambda}
[/tex]

Right?
 
  • #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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