# Property of the Dirac Delta Function

1. Apr 8, 2009

### jaejoon89

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

How do you show that int[delta(t)]dt from negative infinity to infinity is 1?

2. Relevant equations

Dirac delta function defined as infinity if t = 0, 0 otherwise

3. The attempt at a solution

My teacher said that it has to do with m->infinity for the following function
f_m(t) = m*exp(-(mt)^2)/sqrt(pi)

I don't understand - what is this function exactly and how does the delta function property follow?

2. Apr 8, 2009

### Hurkyl

Staff Emeritus
The bad news: delta(t) isn't a function, and $\int_{-\infty}^{+\infty} \quad \, dx$ isn't the integration operation you learned about in your elementary calculus class.

The good news: they're close enough for many purposes.

One of the methods to work with these "distributions" is to name them by means of a limit. (But this limit isn't calculated in the set of functions, so don't try to apply what you know of limits of functions) delta(t) can be named as the "limit" of the sequence of functions you mentioned.

Once you've named a distribution (names aren't unique), $f(x) = \lim_{n \rightarrow +\infty} f_n(x)$ it's "integral" $\int_{-\infty}^{+\infty} f(x) \, dx$ is defined to be:

$$\lim_{n \rightarrow +\infty} \int_{-\infty}^{+\infty}f_n(x) \, dx$$

This last expression has finally been written in terms of ordinary limits and ordinary integration of ordinary functions, and so it can be computed by ordinary means.

(For the record I haven't given all details about what's going on)

3. Apr 8, 2009

### dx

That is not the definition of the delta function. Did your teacher actually say that? For most purposes, the defining property of the delta function can be taken as

$$\int_{-\infty}^{\infty} \delta(x-x') f(x') dx' = f(x).$$ ​

Your problem can be solved by choosing a suitable f in the integral.

(You posted this in "Advanced Physics", so I'm assuming you encountered this in a physics course. If you encountered it in a math course, the definition I gave is probably not precise enough. As Hurkyl said, the rigorous way to think about delta functions is in terms of what are called "distributions")

Last edited: Apr 8, 2009
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