6Fourier transforms
IB Methods
6.5 Fourier transformation of distributions
We have
F[δ(x)] =
Z
∞
−∞
e
ikx
δ(x) dx = 1.
Hence we have
F
−1
[1] =
1
2π
Z
∞
−∞
e
ikx
dk = δ(x).
Of course, it is extremely hard to make sense of this integral. It quite obviously
diverges as a normal integral, but we can just have faith and believe this makes
sense as long as we are talking about distributions.
Similarly, from our rules of translations, we get
F[δ(x − a)] = e
−ika
and
F[e
−i`x
] = 2πδ(k − `),
Hence we get
F[cos(`x)] = F
1
2
(e
i`x
+ e
−i`x
)
=
1
2
F[e
i`x
]+
1
2
F[e
−i`x
] = π[δ(k−`)+δ(k+`)].
We see that highly localized functions in
x
-space have very spread-out behaviour
in k-space, and vice versa.