# Homework Help: Dirac delta

1. Sep 25, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
How would one show that dirac delta is the limit of the normal distribution?
http://en.wikipedia.org/wiki/Dirac_delta
using the definition $$\delta(k) = 1/(2\pi)\int_{-\infty}^{\infty}e^{ikx}dx$$

2. Relevant equations

3. The attempt at a solution

Last edited: Sep 26, 2007
2. Sep 25, 2007

### Hurkyl

Staff Emeritus
Equality for distributions is defined pointwise. You just have to prove you get the same value if you convolve either one with a test function. I.E. for any test function f, you have to prove

$$\int_{-\infty}^{+\infty} \delta(k) f(k) \, dk = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{ikx} f(k) \, dk \, dx$$

3. Sep 26, 2007

### Avodyne

One way (not so rigorous mathematically) to define the delta function is that it is a function that satisfies $$\delta(x)=0$$ if $$x\ne 0$$, and $$\textstyle \int_{-\infty}^{+\infty}dx\;\delta(x)=1.$$ So you need to show (1) that the limit of the normal distribution has these properties, and (2) that $$\textstyle{1\over2\pi}\int_{-\infty}^{+\infty}dk\;e^{ikx}$$ has these properties. Part (1) is easy. Amusingly, the easiest way to do part (2) is to define it by inserting a convergence factor of $$\exp(-\epsilon^2 k^2/2)$$ into the integrand, which turns it into a normal distribution that becomes a delta function in the limit $$\epsilon\to 0.$$

4. Sep 26, 2007

### Hurkyl

Staff Emeritus
Oh, hah, I misread the problem. I thought the equation the OP posted was the equation he wanted to prove.

The idea is the same, though. For a distribution F(_) and a family of distributions G(_, y), to prove

$$F(x) = \lim_{y \rightarrow 0} G(x, y)$$

you have to show

$$\int F(x) f(x) \, dx = \lim_{y \rightarrow 0} \int G(x, y) f(x) \, dx$$

5. Sep 26, 2007

### ehrenfest

Can I choose any f(x)?

6. Sep 26, 2007

### Avodyne

It has to be infinitely differentiable, or something like that (not up on my rigorous defs, sorry), but otherwise yes.