# The Dirac delta function question

1. Nov 30, 2003

### MathematicalPhysicist

in the attatch file there is the dd function.
what i want to know is: when x doesnt equal 0 the function equals 0 and the inegral is the integral of the number 0 which is any constant therefore i think the integral should be equal 0.
can someone show me how this integral equals 1?

for your convinience here is the website that the gif was taken from:
http://www.engr.unl.edu/~glibrary/home/DefineG/node6.html

another thing that i dont understand is how can you restrict an integral with b and a when it equals a constant for example int(b-a):0dx. the anti derevative of 0 is some C now how can you put into a number numbers that propety is only for variables.

Last edited: Jan 4, 2006
2. Nov 30, 2003

### Hurkyl

Staff Emeritus
The dirac delta function isn't really a function! Basically, the dirac delta function is really a gadget that modifies how the integral works rather than being something you integrate.

This gadget can be approximated by functions, though. For example, we can define the class of functions:

$$\delta_n(x) = \frac{n}{\sqrt{\pi}} e^{-(nx)^2}$$

And then we have, under reasonable circumstances:

$$\int_a^b f(x) \delta(x) \, dx = \lim_{n \rightarrow \infty} \int_a^b f(x) \delta_n(x) \, dx$$

If you plot a few of the functions $\delta_n(x)$, you'll notice that for $x \neq 0$ these functions converge to $0$, and for $x = 0$ these functions diverge to $+\infty$. Also, the area under each of these curves is $1$. This is why you think intuitively of the dirac delta "function" as being infinite at $x = 0$ and $0$ everywhere else in such a way that its integral is $1$ iff the region contains $0$.

Last edited: Nov 30, 2003
3. Nov 30, 2003

### HallsofIvy

Staff Emeritus
To back up Hurkyl's answer, the delta "function" is defined as the "function" such that the integral of &delta;(x) over any set not containing 0 is 0 and the integral of &delta;(x) over any set containing 0 is 1.

Of course, there is no such function. It is, more correctly, a "distribution" or "generalized function". One can also define it as the "operator" on functions that, to any function f(x), assigns the value f(0). This is true because the integral of f(x)&delta;(x) over all real number is f(0).

If you don't like exponentials, instead of the approximation Hurkyl gave, you can use "dn(x)= 0 for x< -1/n,
n for -1/n<= x<= 1/n
0 for x> 1/n.
dn has the property that the integral from -1/n to 1/n is 1. The "limit" as n goes to infintiy is &delta;(x). I put limit in quotes because, of course, that doesn't actually converge but you can do things like find the Fourier transform of the delta function by finding the Fourier transform of each dn and then take the limit of that.

Hurkyl's sequence consists of differentiable functions, mine, only continuous functions. They would both be referred to as "delta sequences".

4. Nov 30, 2003

### Hurkyl

Staff Emeritus
Minor correction: HallsofIvy's sequence should be either

$$d_n(x) = \left\{ \begin{array}{ll} 0, & x < -\frac{1}{2n} \\ n, & -\frac{1}{2n} \leq x \leq \frac{1}{2n} \\ 0, & \frac{1}{2n} < x \end{array} \right$$

or

$$d_n(x) = \left\{ \begin{array}{ll} 0, & x < -\frac{1}{n} \\ n (1 - |x|), & -\frac{1}{n} \leq x \leq \frac{1}{n} \\ 0, & \frac{1}{n} < x \end{array} \right$$

(I'm not sure which one he intended)

Last edited: Nov 30, 2003
5. Dec 1, 2003

### HallsofIvy

Staff Emeritus
The first was what I intended and you are right- I forgot the "1/2" needed since the rectangle extends a distance 1/n on both sides (and it's not continuous). Thanks.

6. Dec 6, 2003

### BigRedDot

Well, yes it is. It is not a real-function of a real variable, but it is certainly a continuous linear scalar function on some appropriate space of test functions (usually either the Schwarz space of smooth functions of rapid decrease, or the space D of smooth functions with compact support). It is defined by $$\delta(f) \equiv f(0)$$. Although it cannot be represented as integration against some locally integrable function:
$$T_g(f) = \int g(x)f(x)$$
it is the limit of such function(als) in the topology of distributions, hence often the suggestive if slightly misleading notation:
$$\delta(f)\equiv f(0)= \int \delta(x)f(x)$$

Last edited: Dec 6, 2003