4The calculus of residues

IB Complex Methods

4.4 Jordan’s lemma

So far, we have bounded the semi-circular integral in a rather crude way, with

Z

γ

R

f(z)e

iλz

dz

≤ R sup

|z|=R

|f(z)|.

This works for most cases. However, sometimes we need a more subtle argument.

Lemma

(Jordan’s lemma)

.

Suppose that

f

is an analytic function, except for a

finite number of singularities, and that f(z) → 0 as |z| → ∞.

γ

R

γ

0

R

−R R

×

×

×

Then for any real constant λ > 0, we have

Z

γ

R

f(z)e

iλz

dz → 0

as R → ∞, where γ

R

is the semicircle of radius R in the upper half-plane.

For

λ <

0, the same conclusion holds for the semicircular

γ

0

R

in the lower

half-plane.

Such integrals arise frequently in Fourier transforms, as we shall see in the

next chapter.

How can we prove this result?

The result is easy to show if in fact

f

(

z

) =

o

(

|z|

−1

) as

|z| → ∞

, i.e.

f(z)

|z|

→

0

as |z| → ∞, since |e

iλz

| = e

−λ Im z

≤ 1 on γ

R

. So

Z

γ

R

f(z)e

iλz

dz

≤ πR · o(R

−1

) = o(1) → 0.

But for functions decaying less rapidly than

o

(

|z|

−1

) (e.g. if

f

(

z

) =

1

z

), we need

to prove Jordan’s lemma, which extends the result to any function

f

(

z

) that

tends to zero at infinity.

Proof. The proof relies on the fact that for θ ∈

0,

π

2

, we have

sin θ ≥

2θ

π

.

So we get

Z

γ

R

f(z)e

iλz

dz

=

Z

π

0

f(Re

iθ

)e

iλRe

iθ

iRe

iθ

dθ

≤ R

Z

π

0

|f(Re

iθ

)|

e

iλRe

iθ

dθ

≤ 2R sup

z∈γ

R

|f(z)|

Z

π/2

0

e

−λR sin θ

dθ

≤ 2R sup

z∈γ

R

|f(z)|

Z

π/2

0

e

−2λRθ/π

dθ

=

π

λ

(1 −e

−λR

) sup

z∈γ

R

|f(z)|

→ 0,

as required. Same for γ

0

R

when λ < 0.

Note that for most cases, we don’t actually need it, but can just bound it by

R sup

|z|=R

f(z).

Example. Suppose we want to compute

I =

Z

∞

0

cos αx

1 + x

2

dx,

where α is a positive real constant. We consider the following contour:

γ

0

γ

R

−R R

×

i

and we compute

Re

Z

γ

e

iαz

1 + z

2

dz.

Along

γ

0

, we obtain 2

I

as

R → ∞

. On

γ

R

, we do not actually need Jordan’s

lemma to show that we obtain zero in the limit, but we can still use it to save

ink. So

I =

1

2

Re

2πi res

z=i

e

iαz

1 + z

2

=

1

2

Re

2πi

e

−α

2i

=

1

2

πe

−α

.

Note that taking the real part does not do anything here — the result of the

integral is completely real. This is since the imaginary part of

R

γ

e

iαz

1+z

2

d

z

is

integrating

sin αz

1+z

2

, which is odd and vanishes.

Note that if we had attempted to integrate

R

γ

cos αz

1+z

2

d

z

directly, we would

have found that

R

γ

R

6→ 0. In fact, cos αz is unbounded at ∞.

Example. We want to find

I =

Z

∞

−∞

sin x

x

dx.

This time, we do require Jordan’s lemma. Here we have an extra complication

— while

sin x

x

is well-behaved at the origin, to perform the contour integral, we

need to integrate

e

iz

z

instead, which is singular at the origin.

Instead, we split the integral in half, and write

Z

∞

−∞

= lim

ε→0

R→∞

Z

−ε

−R

sin x

x

dx +

Z

R

ε

sin x

x

dx

!

= Im lim

ε→0

R→∞

Z

−ε

−R

e

iz

x

dx +

Z

R

ε

e

iz

x

dx

!

We now let

C

be the contour from

−R

to

−ε

, then round a semi-circle

C

ε

to

ε

,

then to

R

, then returning via a semi-circle

C

R

of radius

R

. Then

C

misses all

our singularities.

−R

−ε

C

ε

ε

R

C

R

×

Since C misses all singularities, we must have

Z

−ε

−R

e

iz

z

dz +

Z

R

ε

e

iz

z

dz = −

Z

C

ε

e

iz

z

dz −

Z

C

R

e

iz

z

dz.

By Jordan’s lemma, the integral around

C

R

vanishes as

R → ∞

. On

C

ε

, we

substitute z = εe

iθ

and e

iz

= 1 + O(ε) to obtain

Z

C

ε

e

iz

z

dz =

Z

0

π

1 + O(ε)

εe

iθ

iεe

iθ

dθ = −iπ + O(ε).

Hence, in the limit ε → 0 and R → ∞, we get

Z

∞

−∞

sin x

x

dx = Im(iπ) = π.

Similarly, we can compute

Z

∞

−∞

sin

2

x

x

2

dx = π.

Alternatively, we notice that

sin z

z

has a removable singularity at the origin.

Removing the singularity, the integrand is completely analytic. Therefore the

original integral is equivalent to the integral along this path:

We can then write

sin z

=

e

iz

−e

−iz

2i

, and then apply our standard techniques and

Jordan’s lemma.