# Integration of funtion multiplied by shifted heavyside step

1. Sep 26, 2010

### SpaceDomain

As I understand it, the following is true:

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

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

$$\int_{\lambda}^{\infty}{d\lambda}$$

since

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

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

2. Sep 26, 2010

### SpaceDomain

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

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

Right?

3. Sep 26, 2010

### SpaceDomain

Oh, I think I am seeing my mistake in logic here.

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

I should be thinking of this as a graph of $$\lambda$$ vs. $$u(\lambda)$$.

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.