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- Thread starter SW VandeCarr
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tiny-tim

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Hi SW VandeCarr!

The Dirac delta function, despite its name,

it's a

see http://en.wikipedia.org/wiki/Distribution_(mathematics)

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Thanks tiny-tim. I thought the modern definition of a function is any mapping from one mathematical space to another. I guess that's what you mean by a generalized function. Nevertheless the integral of the Dirac delta function is given as unity in number of sources and I suppose it wouldn't work if this weren't true. It's possible to integrate, for example, the Gaussian distribution (whose integral is also unity). Is there no analytic way to integrate this "function."? If not, where does this result come from?Hi SW VandeCarr!

The Dirac delta function, despite its name,isn'ta function …

it's adistribution(a generalised function), used only in combination with other functions (usually to help in difficult integrations) …

see http://en.wikipedia.org/wiki/Distribution_(mathematics)

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arildno

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You might read my tutorial on this:

https://www.physicsforums.com/showthread.php?t=73447

https://www.physicsforums.com/showthread.php?t=73447

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Let me put more precisely what tiny-tim said. The Dirac Delta is not R->R. Meaning, it is not a map from the real numbers into the real numbers. It IS however, a plain old function in the set-theoretic sense.I thought the modern definition of a function is any mapping from one mathematical space to another. I guess that's what you mean by a generalized function.

There are a few formalisms of the Dirac Delta. The branch which studies it is called Distribution Theory or something like that. A "generalized function" or a distribution is actually a function (R->R) ->R, which maps real functions to real numbers. So the Dirac Delta would be defined as [tex]D(f) = f(0)[/tex]. It takes an R->R function as it's input, f, and returns f evaluated at 0. We can define all sorts of distributions this way, often using integrals. Another common one would be [tex]G(f) = \int_{-\infty}^\infty f(x) e^{-x^2} dx[/tex], which takes a function f and returns the infinite integral of f times the gaussian function.

The nice thing about Distribution Theory is it gives a very clean way of handling Fourier Transforms without sweeping infinities "under the rug."

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Ben Niehoff

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It's not a result. It is theNevertheless the integral of the Dirac delta function is given as unity in number of sources and I suppose it wouldn't work if this weren't true. It's possible to integrate, for example, the Gaussian distribution (whose integral is also unity). Is there no analytic way to integrate this "function."? If not, where does this result come from?

[tex]\int_{-\infty}^{\infty} \delta(x) f(x) \; dx = f(0)[/tex]

for

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Distributions are basically "integral factors" of other functions. It works kinda like this:

You choose a suitable space of ("test") functions - for example, the space of smooth real functions R^n -> R with compact support or smooth complex functions R^n -> C which decrease fast enough near infinity (any derivative of a test function times any polynomial converges to zero). The latter are called "tempered distributions". Then you define distributions as*continuous linear functionals* on that space of test functions. ("Continuous" in the sense that, roughly, the derivatives of f_n [times any polynomial] uniformly approach the derivatives of f [times the polynomial].)

Clearly, the functional [itex]\delta_a[/itex] that assigns to a test function f its value f(a) at some point a is continuous and linear, so is the functional generated by any locally integrable function f that grows slowly near infinity (i.e. at most polynomially, along with its derivatives):

[itex]\varphi \rightarrow \int_{R^n} f(x) \varphi(x) \textrm{d}V[/itex]

So we can look at "well-behaved" functions as both regular functions and distributions. One can then define for example the derivative and the Fourier/Laplace transform of a distribution so that the derivative/transform of a distribution generated by f is the distribution generated by the derivative/transform of f. In particular, <d',f> := - <d,f'> for any distribution d and test function f (<d,f> meaning d applied to f). You can try proving for yourself that the Dirac delta distribution is the derivative of the Heaviside step function.

ETA: note that the integral in

[itex]\int_R \delta(x) \varphi(x) dx[/itex]

is just a notation for [itex]\delta(\varphi)[/itex], i.e. the application of the distribution delta to the test function phi,*not* an actual Lebesgue integral of some function.

You choose a suitable space of ("test") functions - for example, the space of smooth real functions R^n -> R with compact support or smooth complex functions R^n -> C which decrease fast enough near infinity (any derivative of a test function times any polynomial converges to zero). The latter are called "tempered distributions". Then you define distributions as

Clearly, the functional [itex]\delta_a[/itex] that assigns to a test function f its value f(a) at some point a is continuous and linear, so is the functional generated by any locally integrable function f that grows slowly near infinity (i.e. at most polynomially, along with its derivatives):

[itex]\varphi \rightarrow \int_{R^n} f(x) \varphi(x) \textrm{d}V[/itex]

So we can look at "well-behaved" functions as both regular functions and distributions. One can then define for example the derivative and the Fourier/Laplace transform of a distribution so that the derivative/transform of a distribution generated by f is the distribution generated by the derivative/transform of f. In particular, <d',f> := - <d,f'> for any distribution d and test function f (<d,f> meaning d applied to f). You can try proving for yourself that the Dirac delta distribution is the derivative of the Heaviside step function.

ETA: note that the integral in

[itex]\int_R \delta(x) \varphi(x) dx[/itex]

is just a notation for [itex]\delta(\varphi)[/itex], i.e. the application of the distribution delta to the test function phi,

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Thanks all. I will need to do more research based on your responses. If I can prove that the Dirac delta distribution is a derivative of the Heaviside step function, will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity, or is Ben Neihoff correct when he states ( if I understand correctly) that this integral of the Dirac delta is simply defined this way and not derived analytically?

As an aside, the function F(x)=log(x) (for all positive real bases) is infinite at x=0 and arguably the sum for all non-zero positive real test values 'a' of F(x) is zero. I don't know if this has anything to do with the Dirac delta, but I thought of it when trying to think of any common continuous symmetrical distributions that are infinite at x=0.

As an aside, the function F(x)=log(x) (for all positive real bases) is infinite at x=0 and arguably the sum for all non-zero positive real test values 'a' of F(x) is zero. I don't know if this has anything to do with the Dirac delta, but I thought of it when trying to think of any common continuous symmetrical distributions that are infinite at x=0.

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arildno

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You won't be able to do that, since the Heaviside function is discontinuous, and hence not differentiable in the strict meaning of that term.Thanks all. I will need to do more research based on your responses. If I can prove that the Dirac delta distribution is a derivative of the Heaviside step function,

No.will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity, or is Ben Neihoff correct when he states ( if I understand correctly) that this integral of the Dirac delta is simply defined this way and not derived analytically?

As has been said, the Dirac delta "function" is not a "normal" function at all, it is what we call a "generalized function", or distribution, which can be regarded as a proper function on a function space.

The legitimacy of the abuse of notation of the integral representation of the effect of the Dirac delta "function", comes from the provable fact that we may form a sequence of integrals with respect to a class of test functions, so that in the limit of this integral sequence, the end result is to get out f(0), i.e, the "output" of the Dirac delta distribution.

I advise you to read my tutorial on this, or some other work.

I have no idea of what you are trying to say here, it sounds like complete nonsense.As an aside, the function F(x)=log(x) (for all positive real bases) is infinite at x=0 and arguably the sum for all non-zero positive real test values 'a' of F(x) is zero. I don't know if this has anything to do with the Dirac delta, but I thought of it when trying to think of any common continuous symmetrical distributions that are infinite at x=0.

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No,Thanks all. I will need to do more research based on your responses. If I can prove that the Dirac delta distribution is a derivative of the Heaviside step function, will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity, or is Ben Neihoff correct when he states ( if I understand correctly) that this integral of the Dirac delta is simply defined this way and not derived analytically?

[itex]\int_{-\infty}^\infty \delta(x) \textrm{d}x = \delta(1) = 1(0) = 1,[/itex]

where 1 is the function f: f(x) = 1 for all x.

It's ok to think of the Dirac delta as being "infinite x=0", but be aware that this is just a metaphor, distributions don't actually have "values" at different points.As an aside, the function F(x)=log(x) (for all positive real bases) is infinite at x=0 and arguably the sum for all non-zero positive real test values 'a' of F(x) is zero. I don't know if this has anything to do with the Dirac delta, but I thought of it when trying to think of any common continuous symmetrical distributions that are infinite at x=0.

Anyway, "symmetric" distributions "infinite" at x=0 (other than multiple of delta and delta + a symmetric distribution) are for example the even derivatives of delta. An odd distribution (odd distributions are naturally defined by d(f(-x))=-d(f(x)) ) "infinite at x=0" (or rather "zero but behaving weirdly at x=0") is for example:

[itex]\left(\textrm{v.p.} \frac{1}{x}\right)(\varphi) = \lim_{\epsilon\rightarrow 0+} \int_{|x|\geq \epsilon} \frac{\varphi(x)}{x} \textrm{d}x[/itex]

We're talking about distributional derivatives.You won't be able to do that, since the Heaviside function is discontinuous, and hence not differentiable in the strict meaning of that term.

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arildno

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That is, however, quite clearly not the meaning by which the OP imbued the word "derivative", that is why it is correct to answer his question in the negative on that point.

Please try to include the OP next time you utilize the pronoun in "we were talking about.."-sentences.

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The OP was responding to

That is, however, quite clearly not the meaning by which the OP imbued the word "derivative", that is why it is correct to answer his question in the negative on that point.

Please try to include the OP next time you utilize the pronoun in "we were talking about.."-sentences.

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arildno

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Really? I thought the folllowing sentrence should make it clear that the OP is rather uncertain about what a "distribution" is:The OP was responding tomy suggestionthat he test his understanding of the definition of the distributional derivative by proving that the Dirac delta is the derivative of the Heaviside function. That was the context, and I think I know what I meant when I said that.

Just because OP utilizes the word "distribution", doesn't mean he has properly learnt its import.will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity,

This is not to denigrate either you or the OP, but that the OP first and foremost needs to understand how a distribution differs from a normal function.

Speaking then of "distributional derivatives" is way too early.

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True.Just because OP utilizes the word "distribution", doesn't mean he has properly learnt its import.

Possibly. I don't have didactic experience with explaining distribution theory to people.This is not to denigrate either you or the OP, but that the OP first and foremost needs to understand how a distribution differs from a normal function.

Speaking then of "distributional derivatives" is way too early.

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arildno

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Anyhow, just to make something clear:

If OP reads your comments closely, he'll learn a lot.

If OP reads your comments closely, he'll learn a lot.

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It was poorly stated. All I was really trying to say is that the logarithms for the totality of positive real numbers sum to zero and that the logarithm of zero is negative infinity (which Preno addressed) . Thanks to all for your help. I need to work on this.I have no idea of what you are trying to say here, it sounds like complete nonsense.

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