5Transform theory

IB Complex Methods


5.1 Fourier transforms
Definition
(Fourier transform)
.
The Fourier transform of a function
f
(
x
) that
decays sufficiently as |x| → ∞ is defined as
˜
f(k) =
Z
∞
−∞
f(x)e
−ikx
dx,
and the inverse transform is
f(x) =
1
2π
Z
∞
−∞
˜
f(k)e
ikx
dk.
It is common for the terms
e
−ikx
and
e
ikx
to be swapped around in these
definitions. It might even be swapped around by the same author in the same
paper — for some reason, if we have a function in two variables, then it is
traditional to transform one variable with
e
−ikx
and the other with
e
ikx
, just to
confuse people. More rarely, factors of 2π or
√
2π are rearranged.
Traditionally, if
f
is a function of position
x
, then the transform variable is
called
k
; while if
f
is a function of time
t
, then it is called
ω
. You don’t have to
stick to this notation, if you like being confusing.
In fact, a more precise version of the inverse transform is
1
2
(f(x
+
) + f(x
−
)) =
1
2π
PV
Z
∞
−∞
˜
f(k)e
ikx
dk.
The left-hand side indicates that at a discontinuity, the inverse Fourier transform
gives the average value. The right-hand side shows that only the Cauchy principal
value of the integral (denoted PV
R
, P
R
or
−
R
) is required, i.e. the limit
lim
R→∞
Z
R
−R
˜
f(k)e
ikx
dk,
rather than
lim
R→∞
S→−∞
Z
R
S
˜
f(k)e
ikx
dk.
Several functions have PV integrals, but not normal ones. For example,
PV
Z
∞
−∞
x
1 + x
2
dx = 0,
since it is an odd function, but
Z
∞
−∞
x
1 + x
2
dx
diverges at both −∞ and ∞. So the normal proper integral does not exist.
So for the inverse Fourier transform, we only have to care about the Cauchy
principal value. This is convenient because that’s how we are used to compute
contour integrals all the time!
Notation.
The Fourier transform can also be denoted by
˜
f
=
F
(
f
) or
˜
f
(
k
) =
F
(
f
)(
k
). In a slight abuse of notation, we often write
˜
f
(
k
) =
F
(
f
(
x
)), but this
is not correct notation, since
F
takes in a function at a parameter, not a function
evaluated at a particular point.
Note that in the Tripos exam, you are expected to know about all the
properties of the Fourier transform you have learned from IB Methods.
We now calculate some Fourier transforms using the calculus of residues.
Example. Consider f(x) = e
−x
2
/2
. Then
˜
f(k) =
Z
∞
−∞
e
−x
2
/2
e
−ikx
dx
=
Z
∞
−∞
e
−(x+ik)
2
/2
e
−k
2
/2
dx
= e
−k
2
/2
Z
∞+ik
−∞+ik
e
−z
2
/2
dz
We create a rectangular contour that looks like this:
−R
γ
1
R
γ
+
R
γ
0
γ
−
R
ik
The integral we want is the integral along
γ
0
as shown, in the limit as
R → ∞
. We
can show that
R
γ
+
R
→
0 and
R
γ
−
R
→
0. Then we notice there are no singularities
inside the contour. So
Z
γ
0
e
−z
2
/2
dz = −
Z
γ
1
e
−z
2
/2
dz,
in the limit. Since γ
1
is traversed in the reverse direction, we have
˜
f(k) = e
−k
2
/2
Z
∞
−∞
e
−z
2
/2
dz =
√
2πe
−k
2
/2
,
using a standard result from real analysis.
When inverting Fourier transforms, we generally use a semicircular contour
(in the upper half-plane if x > 0, lower if x < 0), and apply Jordan’s lemma.
Example. Consider the real function
f(x) =
(
0 x < 0
e
−ax
x > 0
,
where a > 0 is a real constant. The Fourier transform of f is
˜
f(k) =
Z
∞
−∞
f(x)e
−ikx
dx
=
Z
∞
0
e
−ax−ikx
dx
= −
1
a + ik
[e
−ax−ikx
]
∞
0
=
1
a + ik
.
We shall compute the inverse Fourier transform by evaluating
1
2π
Z
∞
−∞
˜
f(k)e
ikx
dk.
In the complex plane, we let
γ
0
be the contour from
−R
to
R
in the real axis;
γ
R
be the semicircle of radius
R
in the upper half-plane,
γ
0
R
the semicircle in
the lower half-plane. We let γ = γ
0
+ γ
R
and γ
0
= γ
0
+ γ
0
R
.
γ
0
γ
R
γ
0
R
−R R
×
i
We see
˜
f(k) has only one pole, at k = ia, which is a simple pole. So we get
I
γ
˜
f(k)e
ikx
dk = 2πi res
k=ia
e
ikx
i(k − ia)
= 2πe
−ax
,
while
I
γ
0
˜
f(k)e
ikx
dk = 0.
Now if
x >
0, applying Jordan’s lemma (with
λ
=
x
) to
C
R
shows that
R
C
R
˜
f(k)e
ikx
dk → 0 as R → ∞. Hence we get
1
2π
Z
∞
−∞
=
1
2π
lim
R→∞
Z
γ
0
˜
f(k)e
ikx
dk
=
1
2π
lim
R→∞
Z
γ
˜
f(k)e
ikx
dk −
Z
γ
R
˜
f(k)e
ikx
dk
= e
−ax
.
For
x <
0, we have to close in the lower half plane. Since there are no singularities,
we get
1
2π
Z
∞
−∞
˜
f(k)e
ikx
dk = 0.
Combining these results, we obtain
1
2π
Z
∞
−∞
˜
f(k)e
ikx
dk =
(
0 x < 0
e
−ax
x > 0
,
We’ve already done Fourier transforms in IB Methods, so we will no spend
any more time on it. We move on to the new exciting topic of Laplace transforms.