Confusion with Delta Dirac Function's First Property: Why Does Infinity Equal 1?

[danong]

I had just reviewed back the properties of Delta Dirac Function, however I'm having a little confusing about the first property as stated :

\delta\left(x-a)\right = 0 if x \neq a,

\delta\left(x-a)\right = \infty if x = a;Here is my problem :
when integrate over the entire region (ranging from negative infinity to positive),
the total area is summed to be 1,

but from the property above, if x is allocated at x = a, it gives infinity value,
which means the area should be infinity as well, but why again it sticks to 1?

This is where i was confused about the property stated.
Thanks in advance.

 

[nicksauce]

Well that's not really the definition of the Delta function, although it helps to visualize it. A more correct definition would be

<br /> \int_{a}^{b}dxf(x)\delta(x-c) = f(c)\textnormal{ if }c\in(a,b)\textnormal{,}=0\textnormal{ otherwise}

 

[HallsofIvy]

That's because those properties are a very rough way of defining the delta function. (Dirac was, after all, a physicist, not a mathematician.)

The delta "function" is not actually a function- it makes no sense to talk about a function having a value of "infinity" for a function and it makes no sense to talk about the integral being 0 over any set that does not contain 0 and 1 over any set that does contain 0.
The delta function is actually a "distribution" or "generalized function". There are several ways of defining that the most widely used one is Schwartz definition as a "functional"- that is, a "distribution" is a function of functions, not numbers. A distribution is a functional that assigns a number to every function. Regular functions can be thought of as a subset of distributions: the function f(x) is the functional, over set C, that assigns to every function g(x) the number \int_S f(x)g(x)dx where S is whatever set we are basing our distribution on.

In particular, the delta function is the functional, over the real line, that, to any function f(x), assigns the number f(0). The idea that the delta function is "0 for any x\ne 0" while \delta(0) is infinity, in such a way that \int_{-\infty}^\infty \delta(x)dx= 1 comes from trying to fit that into the integral I mentioned in the previous paragraph. If f(x) is a function defined on any set containing 0, the \delta(f)= f(0) and we try to force that to be \int_a^b \delta(x)f(x)dx= f(0).

Blast! Nicksauce beat me to it again!

 

[danong]

wahahaha amazing explanation, thanks for the detailed explanation afterall. appreciated ;)

 

[lstellyl]

just to add:

nicksauce and hallsofivy are indeed correct, and the dirac delta function can often times not be treated the same way as a function (but instead as a distribution). It only makes physical sense when integrated over, and is often used in physics to simplify problems that have the nature of an "impulse" because it simplifies the math.

It is even used very profitably for forces and such that are NOT impulsive in nature in the case of Green's Functions and such.

However, to get back to your original question, if you want an easy way to sort of visualize it, you can think of the delta function as a square function defined as:

<br /> f(x)=0 \:for \: \(|x|&gt;\frac{1}{2a} \)<br />
<br /> f(x)=a \: for \: \(|x|&lt;\frac{1}{2a} \)<br />

where

<br /> \lim_{a \rightarrow \infty} f(x) = \delta(x)<br />


and you can see that the area under this function is a*(1/a)=1. this definition has the disadvantage of not being continuous, however.

The point being, the integral over the delta function can be one because, while the value of the function at x=0 is infinity, it is infinity for an infinitesimal period, and "the infinities cancel out" so to speak.

A similar definition is often used with the limit of a gaussian wave packet, which has the advantage of having continuous derivatives

 

[HallsofIvy]

Or, if you prefer continuous functions, the delta function can be written as the limit of
f_n(x)= 0 if |x|> |1/n|
f_n(x)= x+ 1/n if -1/n\le x&lt; 0
f_n(x)= -x+ 1/n if 0\le x\le 1/n[/itex]<br /> as n goes to infinity.

 

[lurflurf]

If we just put the Riemann–Stieltjes integral in first year calculus we could avoid all this silliness.

 

[danong]

ohh... impressive. thanks ;D

 

[arildno]

I had just reviewed back the properties of Delta Dirac Function, however I'm having a little confusing about the first property as stated :

\delta\left(x-a)\right = 0 if x \neq a,

\delta\left(x-a)\right = \infty if x = a;


Here is my problem :
when integrate over the entire region (ranging from negative infinity to positive),
the total area is summed to be 1,

but from the property above, if x is allocated at x = a, it gives infinity value,
which means the area should be infinity as well, but why again it sticks to 1?

This is where i was confused about the property stated.



Thanks in advance.

That's because those properties are a very rough way of defining the delta function. (Dirac was, after all, a physicist, not a mathematician".

The delta "function" is not actually a function- it makes no sense to talk about a function having a value of "infinity" for a function and it makes no sense to talk about the integral being 0 over any set that does not contain 0 and 1 over any set that does contain 0.
The delta function is actually a "distribution" or "generalized function". There are several ways of defining that the most widely used one is Schwartz definition as a "functional"- that is, a "distribution" is a function of functions, not numbers. A distribution is a functional that assigns a number to every function. Regular functions can be thought of as a subset of distributions: the function f(x) is the functional, over set C, that assigns to every function g(x) the number \int_S f(x)g(x)dx where S is whatever set we are basing our distribution on.

In particular, the delta function is the functional, over the real line, that, to any function f(x), assigns the number f(0). The idea that the delta function is "0 for any x\ne 0" while \delta(0) is infinity, in such a way that \int_{-\infty}^\infty \delta(x)dx= 1 comes from trying to fit that into the integral I mentioned in the previous paragraph. If f(x) is a function defined on any set containing 0, the \delta(f)= f(0) and we try to force that to be \int_a^b \delta(x)f(x)dx= f(0).

Blast! Nicksauce beat me to it again!

To follow up HallsofIvy's post; what does it mean to say that "we try to force" the Dirac functional into behaving as a factor of the integrand?

In the following thread, I went into this in more depth:
https://www.physicsforums.com/showthread.php?t=73447

 

[danong]

Thanks arildno for the full documentation, appreciated :)

 

[danong]

sorry but i have another question when i spot another equation stating about the integral property of delta dirac function, i just not quite understand to the notation the equation used.

\int f(T)\delta (t - T) dT = f(t)

what does it says when the integral is taking over the region of dT instead of dt?
i understand \int f(t)\delta (t - T) dt = f(T) which means the contribution of the function f(t) with the distribution of delta is equal to the contribution of f(t=T).
but i don't seem to understand the first one which takes over the derivation of dT instead of dt.


Thanks in advance.

 

[mikeph]

The symbols are changed but there is no mathematical difference.

The delta function is even, so delta(t - T) = delta (T - t).

 

[danong]

thanks for reply,
but what does dT means then?

if t is my variable of time, then T is specified at a certain position where peak comes, then by taking dT seems to have different meaning.

again and by taking f(T) * delta (t - T) seems to be different by taking f(t) * delta(t - T);
i just couldn't figure out how f(T) * delta(t - T) could be distributed over the region.

 

[mikeph]

T is not a single position, t is. T is the domain: read it like delta(T - t). In the top line, the delta function is a continuous function of T with parameter t, that selects the value of f(t) during the integration process.

dT means exactly what dt did in the other line.

 

[danong]

ohh i see ... thanks for the information :)