# Simple dirac delta question.

1. Jul 24, 2005

### sniffer

just curiosity, if you integrate dirac-delta from exactly zero to infinity, will you get one or a-half?

since it is symmetrical about zero, i think it is a half. is it?

i mean:
$$\int_{0}^{\infty} \delta(x) dx=\frac{1}{2}$$

or is it 1?

thanks.

Last edited: Jul 24, 2005
2. Jul 24, 2005

### lurflurf

Yes it is one half. If one want to get everything one can take one-sided limits.
$$\int_{-\infty}^{\infty} \delta(x) dx=1$$
$$\int_{0-}^{0+} \delta(x) dx=1$$
$$\int_{0-}^{\infty} \delta(x) dx=1$$
$$\int_{0}^{0} \delta(x) dx=0$$
$$\int_{0+}^{\infty} \delta(x) dx=0$$
ans so on.
Where as is usual
$$\int_{0-}^{\infty} \delta(x) dx=\lim_{h\rightarrow 0-}\int_{h}^{\infty} \delta(x) dx=$$
and similar.

3. Jul 24, 2005

### Gokul43201

Staff Emeritus
How did you get 1/2 from this ?

I thought I'd just figured out that the integral was not defined, because I arrived at a contradiction upon assuming it existed. I may have made a mistake somewhere ....I think I'll go to bed.

4. Jul 24, 2005

### Stingray

No, you're right. It isn't well-defined.

Writing $$\int_{a}^{b} \delta(x) dx$$ is technically not meaningful in distribution theory either, but for a and b nonzero, a very slight modification makes it work. Since it's easier not to write out the modified version, nobody does. The fix doesn't work when one of the limits is zero.

5. Jul 24, 2005

### lurflurf

One can use the Riemann-Stieltjes integral, limits, or Dirac sequences, or distribution theory to justify the Dirac Delta function. One can then deal with bothersome technical points as they arise. The Dirac Delta function is often well used in an informal way. Thus we can take
$$\int_{0}^{\infty} \delta(x) dx=\frac{1}{2}$$
as being a sensible result as it is consistant with several things we like to think of as being true.
We want the Dirac Delta to be symetric thus
$$\frac{1}{2}= \int_{0}^{\infty} \delta(x) dx=\int_{-\infty}^{0} \delta(x) dx$$
also if
$$f(y)= \int_{y}^{\infty} \delta(x) dx$$
f(y)=1 y<0
f(y)=0 y>0
thus f(0)=1/2 makes the following true for all y
$$f(y)=\lim_{h\rightarrow 0}\frac{f(y+h)+f(y-h)}{2}$$

6. Jul 24, 2005

### Stingray

If you just want to talk about defining something "nice," then you should have that
$$\int_{0}^{\infty} \delta(x) dx = \int_{0^{-}}^{\infty} \delta(x) dx=\int_{0^{+}}^{\infty} \delta(x) dx$$

I don't think there are any standard definitions of the delta function which do not lead to contradictions if you stand by your claim.

7. Jul 24, 2005

### lurflurf

That would be nice, but it negates the purpose of introducing the delta function to begin with. It comes down to being a useful convention. At times it is good to have integrals involving the Dirac Delta function with 0 as a limit of integration defined, and the above is the usual and most useful definition. It is not unlike the situation with 0^0. It is nice to write
for |x|<1
$$\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$$
Thus invoking 0^0=1 as a useful convention that is not completly consistant
We could refuse to define things when we are unable to be completly consistant, but often a mostly consistant definition is useful.
Pehaps now someone would like to play assail common definitions of the Dirac Delta for sport.
definition Heavyside function
H(x)=0 x<0
H(x)=1/2 x=0
H(x)=1 x>0
we will assume that the Dirac Delta only appears inside integrals.
a & b will stand for any real number, or plus or minus infinity; and when a & b are not finite, as usual it should be viewed as shorthand for a formulation involving limits.
This is so that these definitions will speak to the issue at hand, in the distribution definition in particular it would be more correct and natural to use integrals over all real number exclusively.
For f a suitable function
definition 1: Riemann-Stieltjes integral
$$\int_{a}^{b}f(x)\delta(x) dx=\int_{a}^{b}f(x)dH(x)$$
where the integral on the right hand side is a Riemann-Stieltjes integral
definition 2: Limits
$$\int_{a}^{b}f(x)\delta(x) dx=\lim_{s\rightarrow\infty}\int_{a}^{b}f(x)\delta_s(x)dx$$
where $$\delta_s(x)$$ is a suitable dirac sequence
definition 3: Distributions
$$\int_{a}^{b}f(x)\delta(x) dx=f(0)(H(b)-H(a))$$
The difficulties lie in what we take suitable to mean, though there are meanings that are common, obvious, and useful. Difficulties usually mount when we try to weaken our restrictions.

Last edited: Jul 24, 2005
8. Jul 24, 2005

### sniffer

probably this problem will never arise in physics.

i never see it myself in any physics textbooks.

9. Jul 24, 2005

### George Jones

Staff Emeritus
The Dirac delta function is really a distribution $\delta$ for which $\delta\left[ f\right] = f(0)$ for all nice functions $f$. This is the definition of the Dirac delta function. From where does the usual notation come? Any function $g$ integrable over $\mathbb{R}$ naturally defines a distribution $G$:

$$\begin{equation*} G\left[ f\right] =\int_{-\infty}^{\infty}g\left( x\right) f\left( x\right) dx \end{equation*}$$

for all nice functions $f$. This motivates the very convenient pretense that $\delta$ is a funtion such that

$$\begin{equation*} f\left( 0\right) =\int_{-\infty}^{\infty}\delta\left( x\right) f\left( x\right) dx. \end{equation*}$$

If $g$ is any function with

$$\begin{equation*} \int_{-\infty}^{\infty}g\left( x\right) dx=1, \end{equation*}$$

then the family of functions $g_{\varepsilon}\left( x\right) :=g\left( x/\varepsilon\right) /\varepsilon$ defines a family of distributions $G_{\varepsilon}$, with $G_{\varepsilon}\rightarrow\delta$ as $\varepsilon \rightarrow0$ (weak convergence). This is often written as

$$\begin{equation*} \delta\left( x\right) =\lim_{\varepsilon\rightarrow0}g_{\varepsilon}\left( x\right) \end{equation*}$$

or

$$\begin{equation*} f\left( 0\right) =\lim_{\varepsilon\rightarrow0}\int_{-\infty}^{\infty }g_{\varepsilon}\left( x\right) f\left( x\right) dx. \end{equation*}$$

Definr the functionals $G_{a}$ by

$$\begin{equation*} G_{a}\left[ f\right] =\lim_{\varepsilon\rightarrow0}\int _{a}^{\infty}g_{\varepsilon}\left( x\right) f\left( x\right) dx. \end{equation*}$$

My guess is that

$$\begin{equation*} G_{a}\left[ f\right] =\left\{ \begin{array} f f\left( 0\right) \text{; }a<0\\ \frac{1}{2}f\left( 0\right) \text{; }a=0\\ 0\text{; }a > 0 \end{array} \right. \end{equation*}$$

This is just a guess!

Define a function $h$ by $h\left( a\right) =G_{a}\left[ 1\right]$. Then,

$$\begin{equation*} h \left( a \right) =\left\{ \begin{array} 1 1\text{; }a<0\\ \frac{1}{2}\text{; }a=0\\ 0\text{; }a>0 \end{array} \right. \end{equation*}$$

From these definitions, it seems that

$$\begin{equation*} \int_{0}^{\infty}\delta\left( x\right) dx=\frac{1}{2}. \end{equation*}$$

Other definitions might give different results. Note also that

$$\begin{equation*} \int_{0}^{\infty}\delta\left( x\right) dx=\frac{1}{2}\left( \int _{0-}^{\infty}\delta\left( x\right) dx+\int_{0+}^{\infty}\delta\left( x\right) dx\right), \end{equation*}$$

which is a definition often used in analysis for piecewise continuous (as $h$ is) functions.

Regards,
George

10. Jul 24, 2005

### Gokul43201

Staff Emeritus
Moving this to Calc. so the mathematicians can chime in.

11. Jul 24, 2005

### Stingray

Ok, I'll be more explicit this time. As George mentioned, it is usual to define the delta function in distribution theory as a functional $$\delta[f(x)]=f(0)$$ for all (test) functions f(x) which have compact support, and are $$C^{\infty}$$. But it is standard to rewrite this as (the integral shouldn't be taken too literally)
$$\delta[f(x)]= \int_{-\infty}^{\infty} \delta(x) f(x) dx$$

It is now reasonable to define
$$\int_{a}^{b} \delta(x) f(x) dx =: \delta[ \left( H(x-a)-H(x-b) \right) f(x) ]$$
with H(x) the Heaviside step function. But this isn't strictly meaningful. The function inside the argument here is no longer continuous (let alone $$C^{\infty}$$). In general, though, it's a bad thing to multiply two distributions (H is itself a distribution in standard definitions).

Anyway, it is still possible to approximate H(x) by a sequence of test functions which come arbitrarily close to it. And multiplying two test functions together produces a third, so everything's ok. Now what happens if (say) a=0? We can approximate
$$H(x) \rightarrow H_{\epsilon} := \exp\left(-\left(\epsilon/x\right)^2 \right)$$
for x>0, and zero otherwise. Then the integral in question would be identically zero.

But we can also use a sequence
$$H_{\epsilon} := \exp\left(-\left(\epsilon/(x+\epsilon) \right)^2 \right)$$
for $$x>-\epsilon$$, and zero otherwise. In this case, the integral is always 1/e. It's obvious that this game can be played to get any number you want.

All of this can also be reversed. Consider H(x) to be a distribution in the usual sense, and use sequences to approximate the delta function. Again, you can get out arbitrary answers by using creative sequences.

12. Jul 24, 2005

### lurflurf

Indeed it is and it implies
$$\int_{0}^{\infty}\delta(x) dx=\frac{1}{2}$$
Which is why it is usual to place restrictions on the sequences. In particular one usually requires that Dirac sequences be symmetrical. Any of the three approches are useful in that they allow continous and point sources to be treated in a unified manner. Often such formulations are later reduced to integrals and sums. The convention that the Dirac Delta should be symmetric
is the most sensible and widely applicable. One can refuse to define the Dirac Delta in such situations or adopt nonstandard conventions, but doing so is not generally helpful.
It can be bad if the distributions have singularities in common. There are singularities and there are singularities. Simple jump discontinuities are not big problems. In fact for nice functions
$$\int_{-\infty}^{\infty}f(x)H(x) dx=\int_{0}^{\infty}f(x) dx$$
If we give this property to the Dirac Delta we have another reason
As we would like
$$\int_{-\infty}^{\infty}\delta(x)H(x) dx=H(0)=1/2$$
Physically is we consider torques on a beam in equilibium we will need in certain situation the Dirac delta to generate no net torque. We can focus on the integral over all real number, but if we do define the half line integral we want the torques to balance. It is important to focus on the concept we are trying to express with an equation rather than being overly technical. Often for instance one will write sin(x)/x when one means
$$\lim_{y\rightarrow x}\frac{\sin(y)}{y}$$
This is why the Laplace transform is defined as
$$\int_{o^-}^{\infty}f(t)e^{-s t} dt$$
when one intends to take laplace transforms of the Dirac delta.

Last edited: Jul 24, 2005
13. Jul 24, 2005

### Stingray

Eh, ok. Your definition seems overly contrived. Why should the sequences be symmetric? I'd rather go with the usual definitions (which are those that I used). Even if you do require that, it should be possible to use sequences to approximate the step function. And that does not give you a unique answer.

What do you mean by this? I don't understand.

14. Jul 24, 2005

### lurflurf

The sequences should be symmetric for physical reasons and to reduce some complications that arise. What is your "usual" definition? I have used the three common definitions, with the usual convention that Dirac sequences are symmetric, as is the Dirac delta, and H(0)=1/2. I think your position is one of these 3. Which one is it? Or what is it if not one of these three?
1) The object in question should never be defined.
Alright but then we can't use it.
2) The object should be defined, but with some other value than 1/2 thus violating symmetry.
O.k., but symmetry is nice.
3) The object should be defined with different values at different times
Alright, but confusing.

Re: beams
Say we push down on a beam with force exp(-x^2)/sqrt(pi) and up with force diracdelta(x). We expect that forces and torques balance. And they do. This implies that either the dirac delta is symmetric, or has some other feature to explain this. We chose symmety. The same type of thing suggest that the Dirac delta should be symmetric in other situations. If we have a stationary string that we pluck with a dirac delta the strings movement should be symmetric. If we cause waves in a still body of water with a dirac delta they should be symmetric. If we heat a body with a dirac delta the tempeture should be symetric.

What types of dirac sequences are you thinking of that are symmetric, but for which
$$\lim_{s\rightarrow\infty}\int_0^{\infty}\delta_s(x) dx$$
is not 1/2?
I will state the conditions Lang gives as best as I recall them. He notes that some of them can be relaxed, if one deals with the complications that result.
1)Normalization
$$\int_{-\infty}^{\infty}\delta_s(x) dx=1$$
for all s under consideration
2)Positivity
$$\delta_s(x)>0$$
for all x
3)Symmetry
$$\delta_s(-x)=\delta_s(x)$$
4)Concentration
for any a,h>0
$$1-\int_{-a}^{a} \delta_s(x) dx<h$$
for any s suffciently large

15. Jul 25, 2005

### Stingray

My "usual" definition is the one in terms of functionals that I outlined above (i.e. distribution theory). I think it is the one preferred by mathematicians, and does not give a unique answer for the integral we've been discussing.

I also don't usually consider H(0)=1/2, but I can accept that definition for certain uses. If you instead want to define H(x) as a distribution, then the same problem appears. But unlike the delta function, it can simply be treated as a discontinuous function.

I don't see any physical reasons to require the Dirac sequences to be symmetric. The whole point of using Dirac deltas in physical problems is that something is happening on so short a time (or spatial) scale that its precise structure is irrelevant. Certainly, the convolution of the sequences with 'nice' functions (and all physically relevant functions are nice) does not depend on the symmetry of the sequence being used.

Anyway, I pulled out my book on distribution theory (Friedlander), and it gives a class of Dirac sequences as an example. They are not symmetric.

None of that has anything to do with the symmetry of the sequences used. The delta function itself is symmetric even though sequences converging to it may not be.

None. They all have that property. I was saying to let the delta function be in its exact form (or approximated some symmetric function), but define the step function as a limit of continuous functions. The integral we're talking about can then be given arbitrary values depending on the sequence chosen. If we're still making definitions with physics in mind, this would not be acceptable.

16. Jul 25, 2005

### reilly

A particularly nice sequence for defining the delta function is one of gaussians with decreasing variances -- the normal curve gets narrower as the sequence moves along.
Gaussians, centered at zero, are even and entire functions, so under virtually all circumstances of interest to physicists, "LIM" gaussian -> delta; and so
integral (0, infinity) of delta * f = (1/2) integral (-infinity, infinity) of delta*f.

Or, one could as easily use a proof by contradiction.

Regards,
Reilly Atkinson

Note: Lighthill's Fourier Analysis and Generalized Functions is a great basic, practical, text on distributions. he does the Gaussian thing.

17. Jul 27, 2005

### George Jones

Staff Emeritus
This post gives the differences and similarities of the approaches used by Stingray and me. Any answer to the question posed in the original post probably does not have many applications, but I find the material interesting.

For a function $g$ that is integrable over the interval $\left[ 0,\infty\right)$,

$$\int_{0}^{\infty}g\left( x\right) dx=\int_{-\infty}^{\infty}H\left( x\right) g\left( x\right) dx,$$

$$\int_{0}^{\infty}\delta\left( x\right) f\left( x\right) dx=\int_{-\infty }^{\infty}\delta\left( x\right) H\left( x\right) f\left( x\right) dx.$$

But, as Stingray noted, when $f$ is a test function, $H\left( x\right)f\left( x\right)$ is not a valid test function, and therefore cannot be used as an argument of the $\delta$ functional. Let $g_{e}$ be a family of functions that converge weakly to $\delta$. Then, because of the jump discontinuity in the integrand,

$$\lim_{\varepsilon\rightarrow0}\int_{-\infty}^{\infty}g_{\varepsilon}\left( x\right) H\left( x\right) f\left( x\right) dx$$

can converge to $H\left( 0\right) f\left( 0\right)$, or to anything else, depending on which representation $g_{\varepsilon}\left( x\right)$ for $\delta$ is used.

One interpretation of the problem posed in the original post is to find a distribution that gives meaning to the symbolic

$$\int_{0}^{\infty}\delta\left( x\right) f\left( x\right) dx.$$

Stingray used the above for motivation for the following definiition. Let $H_{\varepsilon}$ be a family of $C^{\infty}$ functions that converge to $H$ as $\varepsilon\rightarrow0$. Then $D_{\varepsilon}=H_{\varepsilon}\delta$ is a family of distributions that converge weakly to a distribution $D$. Define

$$\int_{0}^{\infty}\delta\left( x\right) f\left( x\right) dx:=D\left( f\right) =\lim_{\varepsilon_{1}\rightarrow0}\delta\left( H_{\varepsilon_{1} }f\right) =\lim_{\varepsilon_{1}\rightarrow0}\lim_{\varepsilon_{2} \rightarrow0}\int_{-\infty}^{\infty}g_{\varepsilon_{2}}\left( x\right) H_{\varepsilon_{1}}\left( x\right) f\left( x\right) dx.$$

This definition gives different answers depending on the representation $H_{\varepsilon}$ of $H$ used. I gave the following definition

$$\int_{0}^{\infty}\delta\left( x\right) f\left( x\right) dx:=\lim _{\varepsilon_{2}\rightarrow0}\int_{0}^{\infty}g_{\varepsilon_{2}}\left( x\right) f\left( x\right) dx=\frac{1}{2}f\left( 0\right) .$$

The difference between the two definitions is clearly an extra limiting process.

As Stingray has noted, members of the family $g_{\varepsilon}$ do not have to be even functions.

Regards,
George

18. Jul 27, 2005

### Stingray

Why should the second equality hold unless $$g_{\epsilon_{2}}$$ is even?

Edit: I guess it doesn't have to be even. It only needs to satisfy
$$\int_{0}^{\infty} g_{\epsilon}(x) dx = \int_{-\infty}^{0} g_{\epsilon}(x) dx ,$$
but that's still not completely general.

Last edited: Jul 27, 2005