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.