# Homework Help: 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.