Understanding the Dirac Delta Function and its Role in Generalized Functions

[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.

 

[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..)

 

[Reshma]

Thanks a million!

 

[Reshma]

Of course, this "proof" is as unrigourous as it can be..

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

 

[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.