# How do you prove this?

1. Mar 11, 2005

### Reshma

Let $$\theta(x)$$ be a Step function:

$$\theta(x) = 1$$ if x>0
$$\theta(x) = 0$$ if x=<0

Show that $$\frac{d\theta}{dx}=\delta(x)$$

$$\delta(x)$$ is a Dirac Delta function.

2. Mar 11, 2005

### arildno

Let f(x) be a comparison function so that $$\lim_{x\to\pm\infty}f=0$$
Furthermore, we have:
$$\int_{-\infty}^{\infty}\theta(x)\frac{df}{dx}dx=\int_{0}^{\infty}\theta(x)\frac{df}{dx}dx+\int_{-\infty}^{0}\theta(x)\frac{df}{dx}dx=-f(0)$$
However, by using integration by parts and the infinity conditions on f, we have:
$$\int_{-\infty}^{\infty}\theta(x)\frac{df}{dx}dx=-\int_ {-\infty}^{\infty}\frac{d\theta}{dx}fdx$$
That is, for arbitrary f, we have:
$$f(0)=\int_{-\infty}^{\infty}\frac{d\theta}{dx}fdx$$
which suggests the introduction of the delta function.
(Of course, this "proof" is as unrigourous as it can be..)

Last edited: Mar 11, 2005
3. Mar 11, 2005

### Reshma

Thanks a million!

4. Mar 12, 2005

### Reshma

What do you mean-->is there a more rigourous proof?

5. Mar 12, 2005

### arildno

To focus on one aspect, Dirac's delta function isn't a proper function.
Its very definition is, from a strict, mathematical perspective, meaningless.
In order to make a mathematically sensible&rigourous definition of the Dirac "function", mathematicians have developed the tools of "generalized functions"&"distributions.