Talk:Dirac delta function

From the article:

If you integrate the delta function between ANY limits a and b, then the integral is:

0 if a,b > 0 or a,b < 0
1 if a < 0 < b
0.5 if a = 0 or b = 0

Really? I'm not sure about the last of these lines, the one with value 0.5. Surely this contradicts the "compact support" bit?

Looks right to me -- I remember the delta func being defined as the limit of a sequence of functions, each getting pointier. A pointy function of full integral 1, centred on 0 clearly has a half-integral of 1/2. with Lebesgue integration (which I think is the only thing you can use for the delat function, you can't use Riemann), there's something about the limit of the integrals is the integral of the limits (uniform convergences is probably a requirement too) -- Tarquin 20:44 Nov 22, 2002 (UTC)

The difference is that in Lebesgue integration you really integrate over a set. For Lebesgue measure it doesn't matter whether that set is an open interval or its closure, but for the Dirac measure it does. Thus when you integrate over [0, b] you get 1 and when you intergrate over (0, b) you get zero. There is no way to get 1/2. Just as there is no set of which 0 is half an element. -MarSch 17:57, 5 May 2005 (UTC)

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The sequence of functions that go into the delta function does not necessarily have to be centered at zero. I conjecture the sufficient condition is simply that the center approaches 0 as the function approaches an infinite spike. For example, consider the rectangular function

<math>f_{\epsilon}(x)=\left\{\begin{matrix} 1/\epsilon, & \mbox{if } (c - 1/2)\epsilon < x < (c + 1/2)\epsilon \\ 0, & \mbox{otherwise} \end{matrix}\right. <math>

This function is centered at cε. For any real c,

<math>\lim_{\epsilon \to 0}f_{\epsilon}(x) = \delta(x)<math>

However, the value of the half-integral depends on c.

Cyan 05:58 4 Jul 2003 (UTC)

I don't think I really understand this discussion. It is morally equivalent to trying to convolve the Dirac delta-function with the Heaviside function, no? Which is like trying to specify the value of the Heaviside function at 0. Saying it is 0.5, i.e. halfway up, is sort of the right answer in Fourier theory - but I doubt it is the right way to say it.

Charles Matthews 07:48 4 Jul 2003 (UTC)

I think what you are trying to say is: if you evaluate the Fourier series of a discontinuous periodic function at a discontinuity, it converges to the average of the original function's limiting values at the discontinuity. But all that means is that the original function and the Fourier series representation can disagree at discontinuities.

The delta function can be defined in various ways, e.g. as a measure in measure theory, or as a linear functional, or as an integral satisfying certain properties under a limiting operation. I don't know squat about measure theory or functional analysis, so I go with the third definition:

if

<math> \lim_{\epsilon \to 0}\int_{-\infty}^{\infty} f_{\epsilon}(x)g(x)dx = g(0)<math>

we say that <math>f_{\epsilon}(x)<math> is a delta sequence, and for shorthand, we abuse proper notation by writing

<math> \lim_{\epsilon \to 0}f_{\epsilon}(x) = \delta(x)<math>

(Lists of delta sequences may be found at [1] (http://mathworld.wolfram.com/DeltaFunction.html) and [2] (http://mathworld.wolfram.com/DeltaSequence.html).)

Here's the issue: is the following statement true for all delta sequences?

<math> \lim_{\epsilon \to 0}\int_{0}^{\infty} f_{\epsilon}(x)g(x)dx = 0.5 \cdot g(0)<math>

Now, some delta sequences are symmetric about the y-axis, and would yield a half-integral of 0.5*g(0). But other delta sequences, like the one I defined above, are not necessarily symmetric about the y-axis. The half-integral is really indeterminate, because the definition of a delta sequence doesn't constrain it to any particular value.

Cyan 05:18 7 Jul 2003 (UTC)

We Japanese think that

<math>

   \int^{0}_{-\epsilon} \delta(x) dx = \int^{+\epsilon}_{0} \delta(x) dx = 1/2 

<math> for all <math>\epsilon > 0 <math>

and every image <math> H(x) <math> of Heaviside step function <math> H : \mathbb{R} \ni x \longrightarrow H(x) \in H(\mathbb{R}) \subset \mathbb{R} <math> is

<math>

   H(x) = \int^{x}_{-\infin} \delta(\xi) d\xi = \frac{1+{\rm sgn} x}{2} = \left\{\begin{matrix} 0 & \left( x < 0 \right) \\ 1/2 & \left( x = 0 \right) \\ 1 & \left( x > 0 \right) \end{matrix}\right. 

<math> .

User:Koiki Sumi 00:00, 15 Sep 2003 (UTC) & 00:30, 18 Sep 2003 (UTC)

I believe that my off-center rectangular function (see above) is a counter-example to the idea that

<math> \int^{0}_{-\epsilon} \delta(x) dx = \int^{+\epsilon}_{0} \delta(x) dx = 1/2 <math> for all <math>\epsilon > 0 <math>. -- Cyan 04:47, 15 Sep 2003 (UTC)

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The indefinite integral of a distribution is another distribution, not a pointwise valued function. So, the whole question about <math>\int_0^a \delta(x) dx<math> just isn't well-defined. Phys 17:44, 6 Nov 2003 (UTC)

Japanese

Cannot understand the sense of that paragraph... Pfortuny 11:29, 13 Sep 2004 (UTC)

the statement that <math>\delta(\mathbb{R}) \subset \mathbb{R}<math> is probably false. for example, <math>\delta(0)\not\in\mathbb{R}<math>. I am going to remove that part.

Anyway, in short, this japanese version (i have never heard this version referred to as "japanese", can someone attest that usage?) simply says that the delta function is the derivative of the Heaviside step function. In other words, let

<math>\theta(x)=\begin{cases}

0& x<0\\ 1& x\leq0 \end{cases}<math>

(this is what is known as the Heaviside step function. It can also be written in terms of the signum function, as is done in the article). Then you can simple define the delta function to be

<math>

\delta(x)=\frac{d\theta}{dx} <math>

Your homework: figure out how what the article says is the same thing as what I said. -Lethe | Talk

sequences

Well, I just want to add this admittedly pedantic piece of comment: From the point of view of mathematical exactness and completeness it should be emphasized that the parameter a shares the property of all epsilon-small quantities in maths: It is positive! Otherwise, many given formulae are incorrect in the sense that they in fact define sign(a)*delta(x).

The support of the delta function

Ok, maybe the support of the delta function is zero, but consider this:

<math>\lim_{a\rightarrow 0}\frac{\textrm{sinc}(x/a)}{\pi a}=\delta(x)<math>

where sinc(x)=sin(x)/x is the sinc function. This is indeed a delta function, but the reason it yields zero when integrated over any interval not containing zero is NOT because it goes to zero, but because its period of oscillation goes to zero, while its amplitude stays between ±1/x. This means that it goes negative sometimes, and is therefore not a probability density function. Only if you specify that the Dirac delta function be further constrained to be a probability density function can it be thought of as approaching zero except at x=0. That is why I question the support of δ(x) being zero, and the inclusion of the probability infobox in the article. PAR 01:50, 24 Mar 2005 (UTC)