Editing Dirac delta function (section)

==Properties==

===Scaling and symmetry===
The delta function satisfies the following scaling property for a non-zero scalar {{mvar|α}}:{{sfn|Dirac|1930|loc=§22 The ''δ'' function}}{{sfn|Strichartz|1994|loc=Problem 2.6.2}}

<math display="block">\int_{-\infty}^\infty \delta(\alpha x)\,dx
=\int_{-\infty}^\infty \delta(u)\,\frac{du}{|\alpha|}
=\frac{1}{|\alpha|}</math>

and so
{{NumBlk2|:|<math>\delta(\alpha x) = \frac{\delta(x)}{|\alpha|}.</math>|4}}

Scaling property proof:
<math display="block">\int\limits_{-\infty}^{\infty} dx\ g(x) \delta (ax) = \frac{1}{a}\int\limits_{-\infty}^{\infty} dx'\ g\left(\frac{x'}{a}\right) \delta (x') = \frac{1}{   a  }g(0).
</math>
where a change of variable {{math|1=''x&prime;'' = ''ax''}} is used. If {{mvar|a}} is negative, i.e., {{math|1=''a'' = &minus;{{!}}''a''{{!}}}}, then
<math display="block">\int\limits_{-\infty}^{\infty} dx\ g(x) \delta (ax) = \frac{1}{-\left \vert a \right \vert}\int\limits_{\infty}^{-\infty} dx'\ g\left(\frac{x'}{a}\right) \delta (x') = \frac{1}{\left \vert a \right \vert}\int\limits_{-\infty}^{\infty} dx'\ g\left(\frac{x'}{a}\right) \delta (x') = \frac{1}{\left \vert a \right \vert}g(0).
</math>
Thus, {{nowrap|<math>\delta (ax) = \frac{1}{\left \vert a \right \vert} \delta(x)</math>.}}

In particular, the delta function is an [[even function|even]] distribution (symmetry), in the sense that

<math display="block">\delta(-x) = \delta(x)</math>

which is [[homogeneous function|homogeneous]] of degree {{math|&minus;1}}.

===Algebraic properties===
The [[distribution (mathematics)|distributional product]] of {{mvar|δ}} with {{mvar|x}} is equal to zero:

<math display="block">x\,\delta(x) = 0.</math>

More generally, <math>(x-a)^n\delta(x-a) =0</math> for all positive integers <math>n</math>.

Conversely, if {{math|1=''xf''(''x'') = ''xg''(''x'')}}, where {{mvar|f}} and {{mvar|g}} are distributions, then

<math display="block">f(x) = g(x) +c \delta(x)</math>

for some constant {{mvar|c}}.{{sfn|Vladimirov|1971|loc=Chapter 2, Example 3(d)}}

===Translation===
The integral of any function multiplied by the time-delayed Dirac delta <math> \delta_T(t) {=} \delta(t{-}T)</math> is

<math display="block">\int_{-\infty}^\infty f(t) \,\delta(t-T)\,dt = f(T).</math>

This is sometimes referred to as the ''sifting property''<ref>{{MathWorld|urlname=SiftingProperty|title=Sifting Property}}</ref> or the ''sampling property''.<ref>{{Cite book|last=Karris|first=Steven T.|url={{google books |plainurl=y |id=f0RdM1zv_dkC}}| title=Signals and Systems with MATLAB Applications|date=2003|publisher=Orchard Publications|isbn=978-0-9709511-6-8|language=en| page=[{{google books |plainurl=y |id=f0RdM1zv_dkC&pg=SA1-PA15 }} 15]}}</ref> The delta function is said to "sift out" the value of ''f(t)'' at ''t'' = ''T''.<ref>{{Cite book|last=Roden|first=Martin S.|url={{google books |plainurl=y |id=YEKeBQAAQBAJ}}|title=Introduction to Communication Theory|date=2014-05-17|publisher=Elsevier|isbn=978-1-4831-4556-3|language=en|page=[{{google books |plainurl=y |id=YEKeBQAAQBAJ|page=40}}]}}</ref>

It follows that the effect of [[Convolution|convolving]] a function {{math|''f''(''t'')}} with the time-delayed Dirac delta  is to time-delay {{math|''f''(''t'')}} by the same amount:<ref>{{Cite book|last1=Rottwitt|first1=Karsten|url={{google books |plainurl=y |id=G1jSBQAAQBAJ}}|title=Nonlinear Optics: Principles and Applications|last2=Tidemand-Lichtenberg|first2=Peter| date=2014-12-11| publisher=CRC Press|isbn=978-1-4665-6583-8|language=en|page=[{{google books |plainurl=y |id=G1jSBQAAQBAJ|page=276}}] 276}}</ref>

<math display="block">\begin{align}
(f * \delta_T)(t) \ &\stackrel{\mathrm{def}}{=}\  \int_{-\infty}^\infty f(\tau)\, \delta(t-T-\tau) \, d\tau \\
&= \int_{-\infty}^\infty f(\tau)  \,\delta(\tau-(t-T)) \,d\tau \qquad \text{since}~ \delta(-x) = \delta(x) ~~ \text{by (4)}\\
&= f(t-T).
\end{align}</math>

The sifting property holds under the precise condition that {{mvar|f}} be a [[Distribution (mathematics)#Tempered distributions and Fourier transform|tempered distribution]] (see the discussion of the Fourier transform [[#Fourier transform|below]]). As a special case, for instance, we have the identity (understood in the distribution sense)

<math display="block">\int_{-\infty}^\infty \delta (\xi-x) \delta(x-\eta) \,dx = \delta(\eta-\xi).</math>

===Composition with a function===
More generally, the delta distribution may be [[distribution (mathematics)#Composition with a smooth function|composed]] with a smooth function {{math|''g''(''x'')}} in such a way that the familiar change of variables formula holds (where <math>u=g(x)</math>), that

<math display="block">\int_{\R} \delta\bigl(g(x)\bigr) f\bigl(g(x)\bigr) \left|g'(x)\right| dx = \int_{g(\R)} \delta(u)\,f(u)\,du</math>

provided that {{mvar|g}} is a [[continuously differentiable]] function with {{math|''g&prime;''}} nowhere zero.{{sfn|Gelfand|Shilov|1966–1968|loc=Vol. 1, §II.2.5}} That is, there is a unique way to assign meaning to the distribution <math>\delta\circ g</math> so that this identity holds for all compactly supported test functions {{mvar|f}}.  Therefore, the domain must be broken up to exclude the {{math|1=''g&prime;'' = 0}} point.  This distribution satisfies {{math|1=''δ''(''g''(''x'')) = 0}} if {{mvar|g}} is nowhere zero, and otherwise if {{mvar|g}} has a real [[root of a function|root]] at {{math|''x''<sub>0</sub>}}, then

<math display="block">\delta(g(x)) = \frac{\delta(x-x_0)}{|g'(x_0)|}.</math>

It is natural therefore to {{em|define}} the composition {{math|''δ''(''g''(''x''))}} for continuously differentiable functions {{mvar|g}} by

<math display="block">\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}</math>

where the sum extends over all roots of {{mvar|''g''(''x'')}}, which are assumed to be [[simple root|simple]].  Thus, for example

<math display="block">\delta\left(x^2-\alpha^2\right) = \frac{1}{2|\alpha|} \Big[\delta\left(x+\alpha\right)+\delta\left(x-\alpha\right)\Big].</math>

In the integral form, the generalized scaling property may be written as

<math display="block"> \int_{-\infty}^\infty f(x) \, \delta(g(x)) \, dx = \sum_{i}\frac{f(x_i)}{|g'(x_i)|}. </math>

===Indefinite integral===
For a constant <math>a \isin \mathbb{R}</math> and a "well-behaved" arbitrary real-valued function {{math|''y''(''x'')}},
<math display="block">\displaystyle{\int}y(x)\delta(x-a)dx = y(a)H(x-a) + c,</math>
where {{math|''H''(''x'')}} is the [[Heaviside step function]] and {{math|''c''}} is an integration constant.

===Properties in ''n'' dimensions===
The delta distribution in an {{mvar|n}}-dimensional space satisfies the following scaling property instead,
<math display="block">\delta(\alpha\boldsymbol{x}) = |\alpha|^{-n}\delta(\boldsymbol{x}) ~,</math>
so that {{mvar|δ}} is a [[homogeneous function|homogeneous]] distribution of degree {{math|&minus;''n''}}.

Under any [[reflection (mathematics)|reflection]] or [[rotation (mathematics)|rotation]] {{mvar|ρ}}, the delta function is invariant,
<math display="block">\delta(\rho \boldsymbol{x}) = \delta(\boldsymbol{x})~.</math>

As in the one-variable case, it is possible to define the composition of {{mvar|δ}} with a [[Lipschitz function|bi-Lipschitz function]]<ref>Further refinement is possible, namely to [[submersion (mathematics)|submersions]], although these require a more involved change of variables formula.</ref> {{math|''g'': '''R'''<sup>''n''</sup> → '''R'''<sup>''n''</sup>}} uniquely so that the following holds
<math display="block">\int_{\R^n} \delta(g(\boldsymbol{x}))\, f(g(\boldsymbol{x}))\left|\det g'(\boldsymbol{x})\right| d\boldsymbol{x} = \int_{g(\R^n)} \delta(\boldsymbol{u}) f(\boldsymbol{u})\,d\boldsymbol{u}</math>
for all compactly supported functions {{mvar|f}}.

Using the [[coarea formula]] from [[geometric measure theory]], one can also define the composition of the delta function with a [[submersion (mathematics)|submersion]] from one Euclidean space to another one of different dimension; the result is a type of [[current (mathematics)|current]]. In the special case of a continuously differentiable function {{math|''g'' : '''R'''<sup>''n''</sup> → '''R'''}} such that the [[gradient]] of {{mvar|g}} is nowhere zero, the following identity holds{{sfn|Hörmander|1983|loc=§6.1}}
<math display="block">\int_{\R^n} f(\boldsymbol{x}) \, \delta(g(\boldsymbol{x})) \,d\boldsymbol{x} = \int_{g^{-1}(0)}\frac{f(\boldsymbol{x})}{|\boldsymbol{\nabla}g|}\,d\sigma(\boldsymbol{x}) </math>
where the integral on the right is over {{math|''g''<sup>−1</sup>(0)}}, the {{math|(''n'' − 1)}}-dimensional surface defined by {{math|1=''g''('''x''') = 0}} with respect to the [[Minkowski content]] measure.  This is known as a ''simple layer'' integral.

More generally, if {{mvar|S}} is a smooth hypersurface of {{math|'''R'''<sup>''n''</sup>}}, then we can associate to {{mvar|S}} the distribution that integrates any compactly supported smooth function {{mvar|g}} over {{mvar|S}}:
<math display="block">\delta_S[g] = \int_S g(\boldsymbol{s})\,d\sigma(\boldsymbol{s})</math>

where {{mvar|σ}} is the hypersurface measure associated to {{mvar|S}}.  This generalization is associated with the [[potential theory]] of [[simple layer potential]]s on {{mvar|S}}.  If {{mvar|D}} is a [[domain (mathematical analysis)|domain]] in {{math|'''R'''<sup>''n''</sup>}} with smooth boundary {{mvar|S}}, then {{math|''δ''<sub>''S''</sub>}} is equal to the [[normal derivative]] of the [[indicator function]] of {{mvar|D}} in the distribution sense,

<math display="block">-\int_{\R^n}g(\boldsymbol{x})\,\frac{\partial 1_D(\boldsymbol{x})}{\partial n}\,d\boldsymbol{x}=\int_S\,g(\boldsymbol{s})\, d\sigma(\boldsymbol{s}),</math>

where {{mvar|n}} is the outward normal.{{sfn|Lange|2012|loc=pp.29–30}}{{sfn|Gelfand|Shilov|1966–1968|p=212}} For a proof, see e.g. the article on the [[surface delta function]].

In three dimensions, the delta function is represented in spherical coordinates by:

<math display="block">\delta(\boldsymbol{r}-\boldsymbol{r}_0) = 
\begin{cases} 
    \displaystyle\frac{1}{r^2\sin\theta}\delta(r-r_0) \delta(\theta-\theta_0)\delta(\phi-\phi_0)& x_0,y_0,z_0 \ne 0 \\
    \displaystyle\frac{1}{2\pi r^2\sin\theta}\delta(r-r_0) \delta(\theta-\theta_0)& x_0=y_0=0,\ z_0 \ne 0 \\ 
    \displaystyle\frac{1}{4\pi r^2}\delta(r-r_0) & x_0=y_0=z_0 = 0  
\end{cases}</math>