654 Diracδ-function
Let us employ the normalized Gaussian sequence in eqn. (A.5) to prove eqns. (A.2)
and (A.3). Eqn. (A.2) follows immediately from the fact that eqn. (A.5) is properly
normalized so that
∫∞
−∞dx δ(x) = lim
σ→ 0∫∞
−∞dx δσ(x)= lim
σ→ 01
√
2 πσ^2∫∞
−∞dxe−x(^2) / 2 σ 2
= 1. (A.8)
To prove eqn. (A.3), we need to perform the integral
∫∞
−∞dx δ(x)f(x) = lim
σ→ 0∫∞
−∞dx δσ(x)f(x)= lim
σ→ 01
√
2 πσ^2∫∞
−∞dxe−x(^2) / 2 σ 2
f(x). (A.9)
This integral can be carried out by introducing a Taylor series expansion forf(x)
aboutx= 0 into eqn. (A.9):
1
√
2 πσ^2
∫∞
−∞dxe−x(^2) / 2 σ 2
f(x) =
1
√
2 πσ^2∫∞
−∞dxe−x(^2) / 2 σ 2 ∑∞
n=0
f(n)(0)
n!
xn
=
1
√
2 πσ^2∑∞
n=0f(n)(0)
n!∫∞
−∞dx xne−x(^2) / 2 σ 2
,(A.10)
wheref(n)(0) is thenth derivative off(x) evaluated atx= 0. Ifnis odd, the integral
vanishes because the product of an odd and an even function yieldsequal areas of
opposite sign forx >0 andx <0. Forneven, the integrals are just the moments of
the Gaussian distribution. These moments have the general form
1
√
2 πσ^2
∫∞
−∞dx xne−x(^2) / 2 σ 2
= (n−1)!!σn, (A.11)
where (n−1)!! = 1· 3 · 5 ···(n−1) and−1!!≡1. Since we need the limitσ→0, the
only term that does not vanish in this limit is then= 0 term. Thus, we find
lim
σ→ 0
∫∞
−∞dx δσ(x)f(x) =f(0). (A.12)In a similar manner, it is also possible to prove the more general property
∫∞
−∞dx δ(x−a)f(x) =f(a), (A.13)which is left as an exercise for the reader.