Re: DiracDelta Integrals

• To: mathgroup at smc.vnet.net
• Subject: [mg9356] Re: DiracDelta Integrals
• From: "Stephen P Luttrell" <luttrell at signal.dra.hmg.gb>
• Date: Sat, 1 Nov 1997 03:33:42 -0500
• Organization: Defence Evaluation and Research Agency
• Sender: owner-wri-mathgroup at wolfram.com

```> just out of curiosity, i tried
>
>    Integrate[ Exp[I k x], {x, -Infinity, Infinity}]
>
> and Mathematica said "the integral is 0 if Im[k]==0, otherwise, i give
> up".   so, thinking i was being slick, i loaded the
> Calculus`DiracDelta` package.
>
>...DELETIA...

In Mathematica 3 the following input:

<<Calculus`FourierTransform`
FourierTransform[1, t, w]

Gives the following output:

2 \[Pi] DiracDelta[w]

If you want to obtain this result using Integrate, then here is a rather
indirect
way of doing it.

The following inputs:

b =Integrate[Exp[I w t-a Abs[t]],{t,-Infinity,Infinity},
Assumptions->{Im[w]==0,Re[a]>0}] b/.{a->0}
Integrate[b,{w,-w0,+w0}]//Limit[#,a->0]&//PowerExpand

produces the following outputs:

\!\(\(2\ a\)\/\(a\^2 + w\^2\)\)
0
2 \[Pi]

Here I use the parameter "a" as a regulariser, which ensures that the
integral converges
as Abs[t]->Infinity. The integral has poles at w==+I a and w==-I a,
which pinch the integration contour
as a->0. For w^2>0 the output from second line of the input shows that
the integral is 0, whereas the
output from third line of the input shows that the integral (from w==-w0
to w==+w0) of the integral
is 2 \[Pi].

That demonstrates that the integral is 2 \[Pi] DiracDelta[w], as
required.

----------------------------------------------------------------------------
---------------------------------------- Stephen P Luttrell
luttrell at signal.dra.hmg.gb