Representations of the Dirac δ function

vs 0.2, 25/1/08 © Niels Walet

There are many ways to derive the Dirac δ function. A few are sketched below.

1 Gaussian

One of the most useful ones is the limit of a very narrow (and very high!) Gaussian. We first need to figure out how high a Gaussian must be to have unit area. We calculate tjhe area under a curve to be able to divide by it:

Integrate[(E^(-((x^2)/(a^2)))),{x,-∞,∞},Assumptions->a>0]

Now integrate with a general function (Taylor expanded to 4th order)

tst=Integrate[(E^(-((x^2)/(a^2))))Normal[Series[f[x],{x,0,5}]],{x,-∞,∞},Assumptions->a>0]/Integrate[(E^(-((x^2)/(a^2)))),{x,-∞,∞},Assumptions->a>0]

tst/.a->0

Thus as a->0 this goes to f(0), and we have proven that

δ(x)=(lim_(a->0))(1/(a(Sqrt[π])))(e^(-(x^2)/(a^2)))

2 Fourier transform of 1

The definition as a Fourier Transform, δ(x)=(1/(2π))(lim_(L->∞))(∫_(-L)^L)(e^(i k x))dk can be calculated explicitely

(1/(2π))(∫_(-L)^L)(E^(I k x))\[DifferentialD]k

Plot[{((sin( x))/(π x)),((sin( 10x))/(π x))},{x,-π,π},PlotRange->{-1,4}]

The limit isn't so well behaved: we can't Taylor expand f(x) around zero and integrate (why not?)

tst=Integrate[Normal[Series[f[x],{x,0,5}]]((sin(L x))/(π x)),{x,-∞,∞}]

3 High and narrow "tower"

Finally we take a function that is constant for -a<x<a, and zero elsewehere:

g(x_,a_)=(1/(2a))If[Abs[x]<a,1,0]

Plot[{g[x,1/2],g[x,1/10]},{x,-1,1},PlotRange->{0,5}]

Once again, we Taylor expand a general f(x) and integrate:

tst=Integrate[Normal[Series[f[x],{x,0,5}]],{x,-a,a}]/(2a)

And we see how strongly higher derivatives are suppressed as a->0.

Simplify[tst]/.a->0