Dirac Delta Integration Problem

[md5fungi]

Homework Statement

$$\int_{-\infty}^t (cos \tau)\delta(\tau) d\tau$$

Evaluate the integral. I'm supposed to evaluate this for all t I believe, so I'm concerned with t<0, t=0, t>0.

Homework Equations

$$\int_{-\infty}^{\infty} f(x)\delta(x) dx = f(0)$$

The Attempt at a Solution

I understand the second equation will give me some of my answer. When t is infinity, cos 0 is to be evaluated, and I get 1. The problem is, I don't really understand how the Dirac Delta function works with integrals. I have no idea what would happen if I let t = -5, or t = 0, with merely the one equation I am given. The textbook I am using doesn't seem to give much information besides the above equation, and how the Delta function and Unit Step function relate.

Can anyone help me understand this better?

The answer is given to me, as well: 1 for t > 0, 0 for t < 0, not defined for t = 0.

 

[md5fungi]

Figured it out. :)

The integral of the Delta function is the Unit Step function, and then cos 0 evaluates to 1... so the answer is basically the Unit Step.

 

[diazona]

I find it convenient to remember this: the Dirac delta function is just a spike at x=0, and zero everywhere else. So the integral of anything times a delta function is obviously zero if the range of integration doesn't include the spike. That is,
$$\int_a^b f(x) \delta(x) = 0$$
if a and b are either both positive, or both negative.

 

[Hurkyl]

Be aware that there are several conflicting definitions for what it might mean to take an integral involving the Dirac delta whose domain isn't everywhere.

For example, one could adopt the convention that $\int_S f(x) \, \delta(x) \, dx$ is f(0) if $0 \in S$ and 0 otherwise. However, the $\int_0^a f(x) \delta(x) \, dx$ is now ambiguous -- is the interval of integration supposed to be $(0, a)$? Or is it $[0, a]$? Or something else? What if a were negative?

Normally, individual points are irrelevant, but when you start working with things like the Dirac delta that are not functions, you have to deal with the oddities this new type of object demands.

One specific oddity is the following: what is $\int_{-1}^0 f(x) \delta(x) \, dx$? What is $\int_0^1 f(x) \delta(x) \, dx$? What is $\int_{-1}^1 f(x) \delta(x) \, dx$? Have you worked that out (using whatever definition you're using)?

Now check if the following equation is valid:
$$\int_{-1}^0 f(x) \delta(x) \, dx + \int_0^1 f(x) \delta(x) \, dx = \int_{-1}^1 f(x) \delta(x) \, dx$$


For that reason (and others), another convention that people use is that a half-interval only gets half of delta: for any positive t:
$$\int_{-t}^0 f(x) \delta(x) \, dx = \frac{1}{2} f(0)$$
and
$$\int_0^t f(x) \delta(x) \, dx = \frac{1}{2} f(0)$$


Other times, people add an extra annotation, whether the limit stops "before" 0 or "after" 0:
$$\int_{-1}^{0^-} f(x) \delta(x) \, dx = 0$$
$$\int_{-1}^{0^+} f(x) \delta(x) \, dx = f(0)$$
and in this convention, the following is simply illegal
$$\int_{-1}^{0} f(x) \delta(x) \, dx$$
because you didn't annotate 0 with a direction.