Signals - Integration of Heavyside Step & Dirac Delta Functions

1. Sep 6, 2010

SpaceDomain

1. The problem statement, all variables and given/known data

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

2. Relevant equations

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

$$\int_{-\infty}^{\infty}{f(t)\delta(t-a)dt} = f(a)$$

3. The attempt at a solution

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

I obviously could use the second relevant equation if the u(t) term was not in these integrals.

I am stuck. Could someone point me in the right direction?

Last edited: Sep 6, 2010
2. Sep 6, 2010

SpaceDomain

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

Is that correct?

3. Sep 6, 2010

vela

Staff Emeritus
The upper limit of the integral should be t, not ∞.
Why does the presence of u(t) stop you? What's the definition of the Heaviside step function?

4. Sep 6, 2010

vela

Staff Emeritus
Yes.

5. Sep 6, 2010

SpaceDomain

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

$$= [e^{-t}]_{t=-1} + [e^{-t}]_{t=1}$$

$$[e^{-t}]_{t=-1} = 0$$ because $$t=-1$$ is out of the limits of integration.

$$[e^{-t}]_{t=-1} + [e^{-t}]_{t=1} = \frac{1}{e}$$

Does that look right?

6. Sep 6, 2010

vela

Staff Emeritus
Yes and no. You figured out the answer correctly, but what you wrote isn't correct. For one thing, the first integral isn't equal to $[e^{-t}]_{t=-1}$ since $[e^{-t}]_{t=-1}=e$. Second, $[e^{-t}]_{t=-1}$ is always equal to e; it's never equal to 0. What you should have written is simply that the first integral is equal to 0 because the delta function is zero over the interval of integration.