collection of delta function forms
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Phil's brief working note collecting delta function sequences, dated 4.12.11, with only the first example visible. It integrates the Gaussian exp(-x^2/(4t)) over a symmetric interval, gets an error function, and takes t to 0 to identify the limit as a multiple of delta(x). A corollary multiplies by the step function theta(x) and concludes delta(x)theta(x) = delta(x)/2. The extraction has lost the normalization factors and integral limits.
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Collection of Delta Function Sequences PhL 4.12.11
1. ft(x) = exp(-x2/(4t))/ () limt→0 ft(x) = 2 δ(x)
We start with
ft(x) = exp(-x2/(4t))/ ()
!Syntax Error, I ft(x)dx = !Syntax Error, I exp(-x2/(4t))/ ()dx = (1/ ())!Syntax Error, I exp(-x2/(4t)) dx
= (2/ ())!Syntax Error, I exp(-x2/(4t)) dx since integrand is even in x
= (2/ ())erf(a/(2) // Maple
= 2 erf(a/(2)
We have thus shown that
!Syntax Error, I ft(x)dx = 2 erf(a/(2)
Now we take the limit t→0 on both sides
limt→0 {!Syntax Error, I ft(x)dx } = 2 erf(0) = 2
Since this result is independent of a, it must be true no matter how small a is. Therefore we conclude this symbolic function fact
limt→0 ft(x) = 2 δ(x)
2. Corollary. Now suppose we consider instead gt(x) = ft(x)θ(x) :
gt(x) = θ(x) exp(-x2/(4t))/ ()
Then if we carry out the steps above, we will find that
!Syntax Error, I gt(x)dx = 1 erf(a/(2)
limt→0 gt(x) = δ(x)
Really we are just claiming here that
δ(x)θ(x) = δ(x)θ(0) = δ(x)/2