Applied Complex Analysis

Revision MATH34001

Applied Complex Analysis

Mike Simon

26/09/2011

§1 Revision

Definition 1.1The function f(z) of the complex variable z at the point z = a is differentiable iff the limit

lim

(f(z)− f(a)

z − a

)

exists. This limit is called the derivative of f at z = a, and is denoted f ′(a) ordf

dz

∣∣∣∣z=a

Definition 1.2If the derivative f ′(a) exists for all a in domain D, we say f is regular, analytic, or differentiablein D.

Definition 1.3If f ′(z) exists ∀z ∈ C, then, f is said to be entire.Examples: ez, cos z, sin z, polynomials.

Any function defined by a power series (eg. f(z) =∞∑

n=0

anzn, |z| < R), is regular inside the circle

of convergence.

This could instead be∞∑

n=0

an(z − b)n for |z − b| < R

R could be ∞, finite, or zero.

(a) R =∞ ⇒ f(z) defined as a power series converges for all z ie. f is entire.(b) R = 1 → R is the distance from the centre to the nearest singularity.(c) R = 0 ⇒ Useless

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Revision MATH34001

We could expand f(z) = 11−z around a point other than 0. ie. z = −1 (powers of z+1).

f(z) =1

2− (1 + z)=

1

2

1

1− z+12

=1

2

∞∑n=0

(z + 1

2

)n

which has radius of convergence 2, as expected.

If f(z) and g(z) are regular in D, then so is(i) f(z) + g(z)(ii) f(z)g(z)(iii) f(z)/g(z) provided you avoid points where g(z) = 0.

Derivatives are evaluated using the rules as for real functions.

Example

h(z) =ez cos z

(z2 + 9) sin z

ez, cos z, z2 + 9 and sin z are all entire, so h is regular everywhere, except where (z2 + 9) sin zvanishes. ie.

z2 + 0 = 0 ⇒ z = ±3isin z = 0 ⇒ z = nπ

Definition 1.4An isolated point where a function f fails to be regular is called a singular point, or singularity.There are different types of singularity. (see later.)

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The Function ln(z) MATH34001

§2 The Function ln(z)

If w = ln z then z = ew, write z = reiθ, (r = |z| ≥ 0) and w = u+ iv. Then,

reiθ = eu+iv = eueiv

Equality of numbers in polar form ⇒ r = eu, u = ln r and θ = v + 2mπ,m ∈ Z.ie.

v = arg(z)− 2mπ and u = |z|

Example

ln(1 + i√3) = ln

{2(12 + 1

2 i√3)}

= ln(2ei

π3

)= ln 2 + ln

(ei

π3

)= ln 2 + iπ3 + 2nπi, n ∈ Z

ln(−4) = ln(4eiπ

)= ln 4 + ln eiπ

= 2 ln 2 + iπ(1 + 2n), n ∈ ZThe problem is seen if you move z around the origin by a full turn, so we put in a barrier toprevent this happening. We call it a branch cut. It runs from the origin to infinity. What thismeans is that we choose a rande of values for arg(z).Eg. We choose:(a) 0 ≤ arg(z) < 2π(b) −π ≤ arg(z) < π(c) Any other range of length 2π

Usually (b) is referred to as the principle branch.

Compare f = ln z above and below the cut. ie. z = −x+ iε,−x− iεf(−x+ iε) = ln | − x+ iε|+ i arg(−x+ iε) = lnx+ iπf(−x− iε) = ln | − x− iε|+ i arg(−x− iε) = lnx− iπ

Hence, f(z) is discontinuous across the branch cut. (and [ln z]abovebelow = 2iπ)If we use an alternative definition of the branch (eg. 0 ≤ arg z < 2π.) Then we find the samejump across the cut.

More Revision

(1) The function k1(z) =ez − 3

z(z − 1)has singular points where z = 0 and z = 1. However,

the function k2(z) =ez − 1

z(z − 1)has only a singular point at z = 1, z = 0 is not a singularity

in this case, the function looks like 00 there and (as with real functions) use l’Hopitals Rule to

find the value at z = 0.

(2) Every complex number z can be expressed in polar form z = reiθ where r = |z| andθ = arg(z)(+2mπ). Consider r = 1, ie the circle |z| = 1.

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About Branch Cuts

(1) A branch cut runs between the branch points; for ln(z), the branch points are 0 and ∞.(2) For a given function, the branch points are fixed but we can choose how to join them.(3) ln(z) has a branch point at 0, therefore ln(z − a) has a branch point at a.

Putting branch cuts together

Consider f(z) = lnz + 1

z − 1= ln(z + 1) − ln(z − 1). ln(z + 1) has a branch cut joining -1 to ∞.

ln(z − 1) has a branch cut joining +1 to ∞.Unrelated branch cuts are possible but complicated to use. Instead, for example, choose eachcut to go along the axis to the right. This choice could be achieved by 0 ≤ arg(z ± 1) < 2πln(z − 1) = ln r1 + iθ1, ln(z + 1) = ln r2 + iθ2, 0 ≤ θ1, θ2 < 2πie. f(z) = ln r2 − ln r1 + i(θ2 − θ1)We now compare the values of f(X ± iε)

At X + iε we have,

r1 = |z − 1| = X − 1 θ1 = 0

r2 = |z + 1| = X + 1 θ2 = 0

So, f(X + iε) = lnX + 1

X − 1

At X − iε we have,

r1 = |z − 1| = X − 1 θ1 = 2π

r2 = |z + 1| = X + 1 θ2 = 2π

So, f(X − iε) = lnX + 1

X − 1

As the function f(z) is continuous across this part of the real axis, the branch cut is not neededthere, (and we dispose of it.)We are left with a cut between -1 and +1; do we need it?

for X ± iε, r1 = 1−X r2 = 1 +Xfor X + iε θ1 = π θ2 = 0for X − iε θ1 = π θ2 = 2π

so,

f(X + iε) = ln1 +X

1−X− iπ

f(X − iε) = ln1 +X

1−X+ iπ

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different answers, therefore cut must stay.

In general, try to make cuts go insane direction, and then, with luck, the cuts may cancel incertain regions.

NB: The function f(z) = lnz + 1

z − 1has a cut between +1 and -1.

Behaviour as z →∞

With a suitable branch cut definition all the singularities of f(z) lie inside |z| = 2, and thereforef(z) has a laurent expansion (in inverse powers of z.)

f(z) =∞∑

n=0

anz−n (|z| > 2)

where the coefficients an are to be determined.

f(z) = a0 +a1z

+a2z2

+ ...

Now, for z real and positive,

f(x) = lnx+ 1

x− 1

= ln1 + 1

x

1− 1x

= ln(1 + ε)− ln(1− ε)

=

(ε− 1

2ε2 +

1

3ε3 − 1

4ε4 + ...

)−(−ε− 1

2ε2 − 1

3ε3 − 1

4ε4 − ...

)= 2

(ε+

1

3ε3 +

1

5ε5 + ...

)= 2

(1

x+

1

3x3+

1

5x5+ ...

)Thus,

a0 = 0a2n = 0 ∀n

a1 = 2a2n+1 =2

2n+ 1∀n

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The function zα MATH34001

§3 The function zα

We define

zα = eα ln(z)

= eα{ln |z|+i arg(z)+i2nπ}

= eα ln |z|eiα{arg(z)+2nπ}

= |z|αeiα{arg(z)+2nπ}

So, for instance,

(−8) 13 = (8eiπ)

13

= (8eiπ+i2mπ)13

= 2eiπ3 (1+2m)

m = 0 ⇒ 2e13 iπ = 2

(cos

π

3+ i sin

π

3

)= 1 + i

√3

m = 1 ⇒ 2eiπ = −2

m = 2 ⇒ 2e53 iπ = 1− i

√3

These are equally spaced around the circle of radius 2.For α not an integer, zα will have multivaluedness (because ln z has.) We make zα unique bydefining a branch, as before.So, for example,

zα = (reiθ)α = rαeiαθ, −π < θ ≤ π

Again, we can compare value just above/below the cut.

At −X + iε r = |z| = X, θ = π

At −X − iε r = |z| = X, θ = −π

(−X + iε)α = Xαeiαπ

(−X − iε)α = Xαe−iαπ

The jump across the cut is therefore

[zα]abovebelow = Xα{e−iαπ − e−iαπ

}= Xα × 2i sin(απ)

Hence, α = integer, sin(απ) = 0 ∴ jump is 0 ∴ zα continuous across axis, ∴ branch cut is notneeded.However, α = integer, sin(απ) = 0 ∴ zα not continuous across axis, ∴ branch cut is needed.

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The function zα MATH34001

A word of caution:Consider f(z) = z

12 with −π < arg(z) ≤ π and let z = −1 + i

√3

Then,

z2 = 1− 2i√3− 3

= −2− 2i√3

= 4

(−1

2− 1

2i√3

)= 4e−2iπ

3

(Not 4e4iπ3 because this angle is outside the allowed range.)

So,

(z2)12 =

(4e−2π i

3

) 12

= 2e−iπ3

= 2

(1

2− i√3

2

)= 1− i

√3

= −z

This suggests being careful with (z1z2)12

We define(z1z2)

12 = z

121 z

122

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Contour Integrals and Cauchy’s Theorem MATH34001

§4 Contour Integrals and Cauchy’s Theorem

Let SN =

N−1∑j=0

f(ξj) dzj

Then we define

limN→∞

SN =

B

A γ

f(z) dz

provided max |dzj | → 0 as N →∞and also provided it does not matter how the subdivision is done.

Notes:(1) Same definition as for real integrals(2) Such an integral is a complex number just as a real integral is a real number.(3) If we want to, we can write,

f(z) = u(x, y) + iv(x, y) so thatf dz = (u+ iv)(dx+ idy) = (udx− vdy) + i(vdx+ udy)Thus, the complex integral can be thought of as two real integrals

B

A γ

(udx− vdy) and

A

B γ

(vdx+ udy)

where the path γ is used to express y = y(x)(4) Clearly, from the definition,

B

A γ

f(z) dz = −A

B γ

f(z) dz

(5) Contour Integrals are not in general evaluated by a use of the definition.(6) If required, we can estimate the size of a contour integral using∣∣∣∣∣∣∣

B

A γ

f(z) dz

∣∣∣∣∣∣∣ ≤ max(A,B)alongγ

|f(z)| × length of AB

We often evaluate contour integrals by parameterisation.

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Example 4.1

B

A

z2 dz

Use z = 3eiθ with 0 ≤ θ < π2 , so, dz = 3ieiθdθ

So,

θ=π2ˆ

θ=0

(3eiθ

)23ieiθdθ = 27i

π2ˆ

0

e3iθ dθ

= 27i

[1

3ie3iθ

]π2

0

= 9(e3i

π2 − 1

)= −9(1 + i)

In general,

B

A

z2 dz becomes (with z = z(s))

s2ˆ

s1

f(z(s))dz

dsds

where, provided we have chosen sensibly, out path γ from A to B becomes the interval (s1, s2).

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Example 4.2

The path AB lies on the real axis between 0 and 2. Use z = x as a parameterisation.

ˆ B

A

dz

z + 1=

ˆ 2

0

dx

x+ 1

Example 4.3

The path AB lies in the imaginary axis between 0 and 2i. Use z = iy as a parameterisation.

ˆ B′

A′

dz

z + 1=

ˆ 2

0

idy

iy + 1

=

ˆ 2

0

dy

y − i

=

ˆ 2

0

(y

y2 + 1+

i

y2 + 1

)dy

=

[1

2ln(y2 + 1)

]20

+ i[tan−1 y

]20

=1

2ln 5 + i tan−1 2

Theorem (Cauchy) 4.4

If γ is a simple closed curve in the complex plane, and if the function f(z) is regular inside andon γ, then ˛

f(z) dz = 0

So, for an entire function f(z) ˆγ

f(z) dz = 0

for any closed loop γ.For a function with a cut, we can achieve the same sort of result if we avoid the cut. For example,for f(z) = (z2 − 1)

12 with an appropriate branch definition, we have

˛f(z) dz = 0

Consider f(z) =ln(z)

z2 + 4−π < arg(z) ≤ π

This has a cut along the negative real axis, and poles at ±2i. For γ avoiding the singularitiesand branch cut, the integral is 0.

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Corollary (To Cauchy’s Theorem) 4.5

Let A and B be two points in C, and let γ and γ′ be two different paths joining them.

ˆ B

A γ

f(z) dz −ˆ B

A γ′f(z) dz =

(ˆ B

A γ

+

ˆ A

B γ′

)f(z) dz =

˛γγ′

f(z), dz = 0

provided f(z) is regular inside and on a loop.

Thus, we can move an integral from γ to γ′ without changing its value, provided we don’t crossany singularities on the way. (Obviously we don’t want any singularities on the path we startwith nor where we finish.)

For instance, considerln(

z+1z−1

)z2 + 1

= f(z) has a cut and simple poles at ±i. With appropriate

choice of branches, we have a cut on (-1, 1)Suppose we want to integrate around the contour γ (An irregular path around the singularitiesand poles). It is easier to use the contour γ′, containing all singularities, but as a regular circle.

ˆγ

=

ˆγ′

as there are no singularities between γ and γ′. It is easier to use γ′ since it involves just oneparameterisation. In addition, we can make the circle γ′ as big as we like, and, with a big circle,use asymptopic results if we want.

Cauchy’s Residue Theorem 4.6

For a function which is regular in a vicinity of a point z0, we can write

f(z) =

∞∑n=0

an(z − z0)n

where (d

dz

)n

f(z) = ann! + an+1g(n)(z − z0) + an+2h(n)(z − z0)2 + ...

∴ z = z0 ⇒((

d

dz

)n

f(z)

)z=z0

= ann!

Hence, we can find an. As before the Taylor Series is valid for |z − z0| < R where R is thedistance from z0 to the nearest singularity. If f(z) has an isolated singularity at z0, we can write

f(z) =∞∑

n=N

an(z − z0)n

f(z) regular at z0 ⇒ f(z) =

∞∑n=0

an(z − z0)n (|z − z0| < R)

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f(z) has an isolated singularity at z0 ⇒ f(z) =∞∑

n=−N

an(z − z0)n (0 < |z − z0| < R) (*)

Here radius of convergence R = distance to nearest other singularity.Rarely N =∞ (very difficult eg. sin 1

2 )More usually N is a finite integer, and we say that f(z) has a pole of order N at z = z0N = 1, we usually say this is a simple pole.In this course, we also meet double and triple poles.

eg. tan z =sin z

cos zwhich is regular everywhere except at z = (n+ 1

2 )π, n ∈ Z which are all simple

poles.

Whereas f(z) =ez

z2(z + 1)2has double poles at z = 0,−1 (and no other singularities.)

The terma−1

z − z0is more important than the others, and the (special) coefficient a−1 is called

the residue of f(z) at z = z0.How do we get hold of it in general?Recall for a Taylor Series

n!an =

[(d

dz

)n

f(z)

]z=z0

so, for (*)

(z − z0)Nf(z) =

∞∑n=−N

an(z − z0)n+N

where this right-hand side is a Taylor Series, and a−1 is the coefficient at (z − z0)N−1. Hence

(N − 1)!a−1 =

[(d

dz

)N−1 {(z − z0)

Nf(z)}]

z=z0

Hence the residue (which we somemtimes write as Res(f(z), z0))Most of the poles you meet are simple (N = 1) and so

Res(f(z), z0) = [(z − z0)f(z)]z=z0

= limz→z0

((z − z0)f(z))

eg. f(z) =sin z

z2has a simple pole at z = 0, at which the residue is(

sinz

z

)z=0

= limz→0

sinz

z= 1

If the function is of the formg(z)

z − z0(where g(z0) = 0) then clearly residue = g(z0).

If the function looks likeg(z)

(z − z1)(z − z2) . . . (z − zN ), then finding the residues is like using the

cover-up rule for partial fractions.

12 Paul Russell


If the function is f(z) =g(z)

h(z)where h(z0) = 0, g(z0) = 0, h′(z0) = 0

then

residue = limz→z0

(z − z0)f(z)

= limz→z0

{(z − z0) ·

g(z0) + (z − z0)g′(z0) + ...

h(z0) + (z − z0)h′(z0) + ...

}=

g(z0)

h′(z0)

Example 4.7

f(z) =1

z3 + 1has 3 simple poles where z3 = −1 ie. z = e(2n+1)iπ

3

The residues are (1

(z3 + 1)′

)poles

=

(1

3z2

)poles

=( z

3z3

)poles

=

(−1

3z

)poles

= −1

3e(2n+1)iπ

3

If the pole is not simple, (eg. N = 2) the easiest way to find the residue is often to examine theoriginal series.

Cauchy’s Residue Theorem 4.8

Let C be a simple closed curve taken anticlockwise, and let f(z) be regular inside and on Cexcept for a finite number of poles z1, z2, ..., zm

˛C

f(z) dz = 2πim∑j=1

Res{f(z), zj}

Proof (Stage 1)Consider just one pole inside C.Cε is a simple closed curve just around the singularity. We know

˛C

f(z) dz =

˛Cε

f(z) dz

because there are no singularities between C and Cε. We choose Cε to be circular and ofradius ε. We find the integral by parameterising z = z1 + εeiθ.

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Thus, dz = εieiθ dθ = i(z − z1) dθAlso,

f(z) =

∞∑n=−N

an(z − z1)n

=a−N

(z − z1)N+

a1−N

(z − z1)N−1+ ...+

a−1

z − z1+ a0 + a1(z − z1)

Sof(z)

idz =

a−N

(z − z1)N−1+

a1−N

(z − z1)N−2+ ...

Now observe

ˆ 2π

0

(z − z1)n dθ = εn

ˆ 2π

0

einθ dθ

= εn[einθ

in

]2π0

=ε

in

[e2iπn − 1

]= 0

So,

˛f(z)z, dz = i

ˆ 2π

0

a−1 dθ

= 2iπa−1

= 2iπ × Residue

(Stage 2)Let there be m poles inside the contour C. Shrink C down onto the poles (vaccuum packed)- A combination of circles around singularities and ’pipes’ between singularities.

The contributions to

˛C

f(z) dz from the pipes will cancel leaving Stage 1.

Example 4.9

f(z) =ez

z(z − 1)(z − 2)has poles at z = 0, 1, ... and

˛C

f(z) dz = 2πi{Res(f(z), 0) +Res(f(z), 2)}

= 2πi

{1

2+

1

2e2}

= πi(1 + e2)

14 Paul Russell


Cauchy’s Integral Formula 4.10

Let C be a simple closed curve taken anti-clockwise, and let f(z) be regular inside and on C.Then, for any point z = a inside C,

f(a) =1

2πi

˛C

f(z)

z − 1dz

Notes:If a is outside C, the RHS = 0 (No singularities in C)If a is on C, the RHS = 1

2f(a)

Proof: Putf(z)

z − ainto Cauchy’s Residue Theorem.

Similarly,

f(a) =n!

2πi

˛f(z)

(z − a)n+1dz

Liouville’s Theorem 4.11

Let f(z) be entire and bounded. Then f(z) is constant.Bounded ⇒ |f(z)| < M ∀z for some M

Proof

Consider

˛C

f(z)

(z − a)2dz, C is any closed curve surrounding a.

For convenience, we consider C to be a circle, centre a, radius R >> 1.Then

|f ′(a)| =

∣∣∣∣˛C

f(z)

(z − a)2dz

∣∣∣∣≤ 2πR×max

on C

∣∣∣∣ f(z)

(z − a)2

∣∣∣∣=

2π

R×max

on C|f(z)|

=2πM

R

as |f ′(a)| < 2πM

R∀R, |f ′(a)| = 0, f ′(a) = 0.

We see this is true ∀a ∴ f(z) = const.

15 Paul Russell

Real definite integrals using countour integrals MATH34001

§5 Real definite integrals using countour integrals

Integrals of the form

ˆ 2π

0

F (cos θ, sin θ) dθ

We will use z = eiθ. Then,

(i) (0, 2π) becomes unit circle |z| = 1. [NB: Sometimes an integral´ π0

or´ π/20

: In this case,

we can sometimes use symmetry to turn this into´ 2π0

](ii)

cos θ =1

2(eiθ + e−iθ) =

1

2

(z +

1

z

)sin θ =

1

2i(eiθ − e−iθ) =

1

2i

(z − 1

z

)(iii)

z = eiθ ⇒ dz = ieiθdθ = izdθ so, dθ =dz

izHence, whole integral can be re-expressed as a contour integral in terms of z.

Example 5.1

Jn =

ˆ π

0

(1 + 2 cos θ)n cos(nθ) dθ

As integrand is even,

Jn =1

2

ˆ π

−π

(1 + 2 cos θ)n cos(nθ)dθ

=1

2

ˆ π

−π

(1 + 2 cos θ)neinθdθ

(As this has added ini

2

ˆ π

−π

(1 + 2 cos θ)n sin(nθ)dθ

=i

2

ˆ π

−π

even(θ)odd(θ)dθ

=i

2

ˆ π

−π

odd(θ)dθ = 0

)So,

Jn =1

2

˛|z|=1

(1 + z +

1

z

)n

zndz

iz

=1

2i

˛|z|=1

(z2 + z + 1)ndz

z

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The integrand (1 + z + z2)n1

zis regular inside |z| = 1 except for a simple pole at z = 0, with

residue (1 + z + z2)∣∣z=0

= 1Thus,

Jn =1

2i× 2iπ × 1 = 1

Integrals of the form

ˆ ∞

−∞f(x)dx by a D-contour

We integrate

˛D

f(z)dz if possible

NB 1: It is important to avoid having cos(αx) or sin(alphax) in f(x). Instead,

ˆ ∞

−∞f(x)dx =

ˆ ∞

−∞cos(αx)g(x)dx = Re

ˆ ∞

−∞eiαxg(x)dx

and then

˛D

eiαzg(z)dz.

NB 2: Using the¸D

we are introducing an integral around a semi circle, and we hope that thiscontributes zero as R→∞.NB 3: Using

¸D

we have a closed contour, and we hope that we can use the residue theorem toevaluate this.

Example 5.2

K =

ˆ ∞

−∞

dx

x4 + 1, we consider

˛D

dz

z4 + 1Clearly, ˛

=

ˆD1

+

ˆD2

=

ˆ R

−R

dx

x4 + 1+

ˆD2

and, ∣∣∣∣ˆD2

∣∣∣∣ ≤ πR× max|z|=R

∣∣∣∣ 1

z4 + 1

∣∣∣∣ ≤ πR

R4 − 1= O(R−3)→ 0 as R→∞

So, as R→∞ ˛D

dz

z4 + 1→ K as R→∞

Now, for R > 1,

˛dz

z4 + 1= 2πi

∑insideD

Residues

= 2πi

[Res

(1

z4 + 1, eiπ/4

)+Res

(1

z4 + 1, e3iπ/4

)]

17 Paul Russell


Residues are1

z4 + 1=

z

4z4= −1

4z at these poles

giving,

˛D

= 2πi

(−1

4

)(eiπ/4 + e3iπ/4)

= 2πi

(−1

4

)(i√2)

=π√2

2

Thus, K =π√2

2

Using a shift of contour 5.3

Given

ˆ ∞

−∞e−x2

dx =√π, find

ˆ ∞

−∞e−x2

cos(2ax) dx

Consider

˛e−z2

dz around rectangular contour. Clearly,´C1

=´ x−x

e−x2

dx (→√π as x→∞)

Also,

ˆC3

e−z2

dz =

ˆ −x

x

e−(x+ia)2 dx

= −ˆ x

−x

e−x2−2iax+a2

dx

= −ea2

ˆ x

−x

e−x2

e−2iax dx

= −ea2

ˆ x

−x

e−x2

{cos(2ax)− i sin(2ax)} dx

= −ea2

ˆ x

−x

e−x2

cos(2ax) dx

∣∣∣∣ˆC2

∣∣∣∣ ≤ a×maxC2

∣∣∣e−z2∣∣∣ = a× max

0<y<a

∣∣∣e−(X+iy)2∣∣∣

= a× max0<y<a

∣∣∣e−X2−2iXy+y2∣∣∣

= a× max0<y<a

e−X2+y2

= ae−X2+a2

→ 0 as X →∞

Similarly for C4.So, ˆ

e−z2

dz →√π − ea

2

ˆ ∞

−∞e−x2

cos(2ax) dx

18 Paul Russell


as x→∞Now, e−z2

is entire so ˆe−z2

dz = 0

Hence, ˆ ∞

−∞e−x2

cos(2ax) dx =√πe−a2

Later on, we will use this in the form

ˆ ∞

−∞e−αx2

cos(βx) dx =

√π

αe−

β2

4α

Keyhole Contours 5.4

Find ˆ ∞

0

x

x4 + 1dx

Trick: ˛z

z4 + 1ln(z) dz

Split contour into 4 parts (circle with ’keyhole’ around branch.)

The introduction of ln(z), which means we need a cut, and we are forced to use a ’keyhole’contour. As C is a closed loop, use residue theorem.There are simple poles at z = ei(2m+1)π/4 (m = 0, 1, 2, 3)(The branch of the ln(z) is defined as 0 ≤ arg(z) < 2π which is consistent with the pole positions.)The residues are

z ln(z)

(z4 + 1)′=

z ln(z)

4z3=

z2 ln(z)

4z4= −1

4z2 ln(z) =

1

4(i(2m+ 1)

π

4)e(2m+1)π/2

m = 0 (eiπ/2) gives −1

4i(i

π

4) =

π

16

m = 1 (e3iπ/4) gives −1

4(−i)(3iπ

4) = −3π

16

m = 2 ... ... ... =5π

16

m = 3 ... ... ... =−7π16

So ˛C

z ln(z)

z4 + 1dz = 2iπ(1− 3 + 5− 7)

π

16= −1

2π2i

Provided C1 is large enough, and C3 is small enough. Now,

˛C

z ln(z)

z4 + 1dz =

ˆC1

+

ˆC2

+

ˆC3

+

ˆC4

19 Paul Russell


∣∣∣∣ˆC1

∣∣∣∣ ≤ 2πR×maxC1

z ln(z)

z4 + 1≤ 2πR2

R4 − 1maxC1

| ln(z)|

| ln(z)| = |ln |z|+ i arg(z)|

=√ln |z|2 + arg2(z)

≤√ln2 |z|+ 4π2

≤ |ln |z||+ 2π∣∣∣∣ˆC1

∣∣∣∣ ≤ 2πR2(ln(R) + 2π)

R4 − 1= O

(ln(R)

R2

)→ 0 as R→∞

Also, ∣∣∣∣ˆC3

∣∣∣∣ ≤ 2πεmaxC3

∣∣∣∣z ln(z)z4 + 1

∣∣∣∣≤ 2πε2

1− ε4maxC3

|ln(z)|

≤ 2πε2

1− ε4

(ln

1

ε+ 2π

)= O

(ε2 ln

1

ε

)→ 0 as ε→ 0

On C4, z = xei0 and so, ˆC4

z ln(z)

z4 + 1dz =

ˆ R

ε

x ln(x)

x4 + 1dx

On C2, z = xei2π, so,

ˆC2

z ln(z)

z4 + 1dz = −

ˆ R

ε

x(ln(x) + 2iπ)

x4 + 1dx

Hence,

ˆC2

+

ˆC4

=

ˆ R

ε

x

x4 + 1{ln(x)− (ln(x) + 2iπ)} dx

= −2iπˆ R

ε

x

x4 + 1dx

Thus,

−1

2iπ2 =

ˆC

=

ˆC1

+

ˆC2

+

ˆC3

+

ˆC4

= −2iπˆ ∞

0

x

x4 + 1dx

giving

ˆ ∞

0

x

x4 + 1dx =

π

4

20 Paul Russell


Notes:

(a) Try

˛z

z4 + 1dz around a quarter of a circle.

(b) In this case we introduced ln(z) and the branch cut, but in cases where there is alreadya branch cut, you do not need to add one.

Consider

ˆ 1

−1

(1− x

1 + x

)1/6

dx. The function f(z) =

(z − 1

1 + z

)1/6

has (with suitable branch defi-

nition) a cut linking ±1. We will consider an integral

˛f(z) dz

where the contour C encircles the cut. We know by Cauchy’s Theorem that

˛C1

f(z) dz =

˛C2

f(z) dz =

(ˆγ1

+

ˆγ2

+

ˆγ3

+

ˆγ4

)f(z) dz

(C2 is a dumbell around cut)We have

f(z) =

(r1e

iθ1

r2eiθ2

)1/6

where z − 1 = r1eiθ1 and z + 1 = r2e

iθ2 and we choose 0 ≤ θ1,2 < 2π (to give the desired finitecut.) So, on −1 < x < 1.

f(x+ iε) =

((1− x)eiπ

(1 + x)ei0

)1/6

=

(1− x

1 + x

)1/6

eiπ6

and, f(x− iε) =

(1− x

1 + x

)1/6

e−iπ6

Hence, ˆγ1

f(z) dz =z=x+iε

ˆ 1+ε

1−δ

(1− x

1 + x

)1/6

eiπ6 dx

and, ˆγ3

f(z) dz =z=x−iε

ˆ 1−δ

ε−1

(1− x

1 + x

)1/6

e−iπ6 dx

Thus, ˆγ1∪γ3

f(z) dz =(e−iπ

6 − eiπ6

)︸︷︷︸−2i sin π

6 =−i

ˆ 1−δ

−1+ε

(1− x

1 + x

)1/6

dx

21 Paul Russell


Also, ∣∣∣∣ˆγ2

f(z) dz

∣∣∣∣ ≤ 2πε

{max

∣∣∣∣z − 1

z + 1

∣∣∣∣}1/6

≤ 2πε

{max |z − 1|min |z + 1|

}1/6

≤ 2πε

{2 + ε

ε

}1/6

= O(ε5/6

)→ 0 as ε→ 0

Similarly, ∣∣∣∣ˆγ4

∣∣∣∣ ≤ 2πδ

{maxγ4

∣∣∣∣z − 1

z + 1

∣∣∣∣}1/6

≤ 2πδ

{δ

2− δ

}1/6

= O(δ7/6

)→ 0 as δ → 0

So, we let ε→ 0 and δ → 0, so that

˛C

f(z) dz → −iˆ 1

−1

(1− x

1 + x

)1/6

dx

We now find this value in another way

˛C

f(z) dz =

˛CR

f(z) dz

where CR is a large circle |z| = R.For large z

f(z) =

(z − 1

z + 1

)1/6

=

(1− 1/z

1 + 1/z

)1/6

=

(1− 1

z

)1/6(1 +

1

z

)−1/6

=

{1− 1

6z+O(z−2)

}2

= 1− 1

3z+O(z−2)

22 Paul Russell


So write

f(z) = 1− 1

3z+ g(z)

We know g(z) = O(z−2) ie. |g(z)| ≤ M

|z|2for some M > 0

So,

˛CR

f(z) dz =

˛CR

{1− 1

3z+ g(z)

}dz

= 0− 1

3(2πi) +

˛CR

g(z) dz

and, ∣∣∣∣˛CR

g(z) dz

∣∣∣∣ ≤ 2πRmax |g(z)| = 2πRM

R2= O(R−1)→ 0 as R→∞

Thus ˛CR

f(z) dz = −2iπ

3

So, ˆ 1

−1

(1− x

1 + x

)1/6

=2π

3

Another integral which uses a keyhole contour.We will later need

I(α) =

ˆ ∞

0

1

xα(x+ 1)dx

We will consider

f(z) =1

zα(z + 1)

and

˛C

f(z) dz where C is a keyhole contour.

We define the branch for z−alpha by z = reiθ with 0 ≤ θ < 2πNB: The cut is outside C, but there is a simple pole inside C at -1. Residue at pole =(reiθ

)−α∣∣∣z=−1

= exp(−πα)Therefore ˆ

C

f(z) dz = 2πie−iπα

(circles need arguing away.)NB: You discover we need 0 < α < 1 in order to argue away both. As an aside we see anotherargument for this range of α.

I(α) =

ˆ ∞

0

dx

xα(x+ 1)

23 Paul Russell


For small x, 1 + x ≈ 1 so

I ≈ˆ0

dx

xα=

[x1−α

1− α

]0

which can only be evaluated if α < 1. Converseley, when x is large x+ 1 ≈ x

∴ I(α) ≈ˆ ∞ dx

xα+1=

[x−α

−α

]∞which can only be evaluated if α > 0

ˆγ1

dz

zα(z + 1)=

ˆ R

ε

dr

(rei0)α(r + 1)

=

ˆ R

ε

dr

rα(r + 1)

and

ˆγ3

dz

zα(z + 1)=

ˆ ε

R

dr

rα(r + 1)

= −ˆ R

ε

dr

rα(r + 1)e−2iπα

Hence,

˛C

f(z) dz =(1− e−2iπα

)ˆ R

ε

dr

rα(r + 1)+ vanishing terms

=(1− e−2iπα

)ˆ ∞

0

dr

rα(r + 1)

Hence,

ˆ ∞

0

dr

rα(r + 1)=

¸Cf(z) dz

1− e−2iπα

=2iπe−iπα

1− e−2iπα

=2iπ

eiπα − e−ıπα

=π

sin(πα)

Hence,

I(α) =π

sin(πα)

24 Paul Russell

Analytic Continuation MATH34001

§6 Analytic Continuation

Consider∞∑

n=0

zn, we know this exists for |z| < 1 and then this =1

1− z

These two functions∞∑

n=0

zn and1

1− zare equal in |z| < 1 but

1

1− zis defined and regular over

the whole complex plane, except for the point 1. This process of analytic continuation (AC) isabout extending the domain of definition of a function whilst still keeping its regularity.

Example 6.1

Consider

g(z) =

ˆ ∞

0

e−tz cos(t) dt

For z = x+ iy, |e−zt| = e−tx and so

|g(z)| ≤ˆ ∞

0

e−tx cos(t) dt

which clearly exists for all x > 0Thus g(z) should exist for Re(z) > 0Actually

g(z) =z

z2 + 1(Re(z) > 0)

z

z2 + 1is regular on C \ {±i}.

Again, we can viewz

z2 + 1as being AC of g(z) into the whole plane.

The process of analytic continuation can be looked as as follows: We start with a complexfunction defined in one way, and regular over some domain D of the complex plane; we then findanother way of expressing the same quantity, which is regular over a different domain, allowingus to widen our usage of the function.

Definition 6.2

Suppose(a) f(z) regular in D(b) g(z) regular in E(c) f(z) = g(z) in D ∩ E = ϕ

Then, g(z) is the AC of f(z) into E

This process can be repeated

25 Paul Russell


Eg. AC from f1 in D1 to f2 in D2, and from f2 in D2 to f3 in D3.Beware:

If the picture is as shown and there is a branch cut, AC would not bring you back to f1(z)

Now some important results about AC:

Theorem 6.3

ACs are unique.

ProofSuppose there are two ACs of f(z) into E.ie. g1(z) and g2(z) exist such that

(a) f(z) regular in D(b) g1(z) and g2(z) both regular in E(c) f(z) = g1(z) and f(z) = g2(z) in D ∩ E = ∅

Consider h(z) = g1(z)− g2(z)Then,

(b) ⇒ h(z) regular in E

(c) ⇒ h(z) = 0 in D ∩ E

26 Paul Russell


Contradiction!h(z) ≡ 0 ∀z ∈ E ⇒ g1(z) ≡ g2(z)

Hence, g(z) is unique �

Actually, we don’t need h ≡ 0 over whole of D ∩ E in order to show h ≡ 0 in E. In fact, weonly need h ≡ 0 on some small arc L inside D ∩E. This means we can do analytic continuationreplacing condition (c) by

(c) f(z) = g(z) on L ⊂ D ∩ E

Example 6.4

Consider f(x) = 1, g(x) = cos2(z) + sin2(z)Both f and g are regular over C. Take D = E = C. (So, D ∩ E = C), and take L = smallsection of real axis.

on L, f = gAC ⇒ f ≡ g over whole of C

The same approach works for lots of real identities.

Example 6.5

f(z) = 1, g(z) = | cos2(z)|+ | sin2(z)|Again, f regular in C and f = g on any part of the real axis. g is not regular and in fact, g ≡ fover any other part of C (ie. Im(z) = 0).

Example 6.6

Consider

f(α) =

ˆ ∞

0

xα

x2 + 1dx, g(α) =

π

2 cos(πα2

)for complex α,

We know (Ex 5) that f(α) = g(α) for α ∈ (−1, 1). We need to find which regions of thecomplex α-plane f(α) and g(α) are regular in g(α) is regular in the whole of C, except for α =odd integer.What is the domain of regularity of f(α)?

xα = exp{α ln(z)} = exp{(αr + iαi) ln(x)}

So, |xα| = exp{αr ln(x)} = xαr

ˆ ∞

0

xα

x2 + 1dx will converge provided

ˆxRe(α)

x2 + 1dx does

ie. −1 < Re(α) < 1So, the overlap region (where f(α) and g(α) are both regular) is the strip −1 < Re(α) < 1, andwe use L ≡ α ∈ (−1, 1)(Clearly L lies in the overlap region.)

27 Paul Russell


Then, AC ⇒ f(α) ≡ g(α) in whole overlap region.So now, ˆ ∞

0

xα

x2 + 1dx ≡ π

2 cos(απ2

) (−1 < Re(x) < 1)

Let α = 2iβ, then ˆ ∞

0

x2iβ

x2 + 1≡ π

2 cos(πiβ)≡ π

2 cosh(πβ)

LHS(x = eu) =

ˆ ∞

−∞

e2iβu

e2u + 1eu du

=

ˆ ∞

−∞

e2iβu

eu + e−udu

=1

2

ˆ ∞

−∞

e2iβu

cosh(u)du

=1

2

ˆ ∞

−∞

cos(2βu)

cosh(u)du (∵ sin(2βu) odd)

Hence, ˆ ∞

−∞

cos(2βu)

cosh(u)du =

π

cosh(πβ)

Shwarz’s Reflection Principle 6.7

Consider a symmetrical domain, as shown. z ∈ D ⇒ z ∈ D.Suppose:=

(a) f(z) regular in D(b) f(z) is real on some section L of the real axis within D.

Then f(z) = f(z) ∀z ∈ DConsider g(z) = f(z), we wish to prove that g ≡ f in D. It can be shown (via C-R equations)that g(z) is regular because f(z) is.Also, on L

g(z) = f(z) = f(z) = f(z)

28 Paul Russell


because z is real on L.Hence, AC ⇒ f ≡ g in D.In particular, f is real on every part of the real axis in D.As most complex functions are ’extensions’ of real functions, this ’real for real z’ is not usuallya major restriction.

Contact Continuation 6.8

Suppose fi regular in Di where D1 and D2 are in contact along line L and fi(z) continuousin Di ∪ L and f1(z) = f2(z) on L.Then, each fi(z) is the AC of the other.

Schwarz’s Reflection Principle - Strong Form 6.9

We can use f(z) = f(z) to AC from Im(z) > 0 to Im(z) < 0Imagine the following:-

(1) f(z) is regular in Im(z) > 0, and continuous in Im(z) ≥ 0, except for a simple pole atz = 3i, with residue 1− i

(2) f(z) is real, Re(z)(3) f(z)→ 1, |z| → ∞, Im(z) > 0

Exclude the pole with a small circle, then use (1), (2) and the reflection principle to AC intolower half plane by

f(z) = f(z)

Also, (3) ⇒f(z)→ 1 = 1 as |z| → ∞ in Im(z) < 0

The AC means f(z) is regular in Im(z) < 0 except for a possible singularity at −3i.We know that, near z = 3i, f(z) = 1−i

z−3i + g(z) where g(z) is regular everywhere in Im(z) > 0

AC ⇒ F (z) =1− i

z − 3i+ g(z)

So, values in Im(z) < 0 are given by

f(z) =1− i

z − 3i+ g(z)

=1 + i

z + 3i+ g(z)

We know g(z) is regular everywhere in Im(z) < 0. So f(z) clearly has a simple pole at −3i withresidue 1 + i.Summary: f(z) regular in C except for simple poles at ±3i with residue (1± 3i) and f(z)→ 1as |z| → ∞Consider

h(z) = f(z)− 1 + i

z + 3i− 1− i

z − 3i

29 Paul Russell


Then h(z) is entire and h(z)→ 1 as |z| → ∞Hence, h(z) ≡ 1 (∵ Liouville’s Theorem.)So,

f(z) = h(z) +1 + i

z + 3i+

1− i

z − 3i

= 1 +2z + 6

z2 + 9

=z2 + 2z + 15

z2 + 9

30 Paul Russell

Functions Defined by Integrals MATH34001

§7 Functions Defined by Integrals

Consider

f(z) =

ˆ ∞

0

e−zt

t+ 1dt

Note: Functions defined as

f(z) =

ˆ b

a

F (z, t) dt

If, for a certain z, F (z, t) has a non integrable singularity, we would not expect f(z) to bedifferentiable for that z.It is true that, for z ∈ D ⊂ C, f(z) will be regular provided F (z, t) is suitably well behaved fora ≤ t ≤ b (or 0 ≤ t <∞) and for all z ∈ D

f(z) =

ˆ ∞

0

e−zt

t+ 1dt

should be regular for Re(z) ≥ 0 or Re(z) > 0 (If Re(z) < 0, | exp{−zt} → ∞ as t → ∞ so wewould not expect the integral to exist.)If z = 0, we have ˆ ∞

0

1

t+ 1dt

which does not exist; so f(z) regular for Re(z) > 0Consider

g(z) =

ˆ ∞

0

e−u

u+ zdu

when z is real and positive, we have

g(z) =

ˆ ∞

0

e−zt

tz + zz dt

=

ˆ ∞

0

e−zt

t+ 1dt

= f(z)

g(z) will be regular except when u+ z vanishes. u ∈ (0,∞) so this can only happen for z real ¡0. Domains of regularity as shown. f(z) in red and g(z) in green.

g(z) is the AC of f(z) into whole plane, except negative real axis.

31 Paul Russell

Gamma Function Γ(z) MATH34001

§8 Gamma Function Γ(z)

We define

Γ(z) =

ˆ ∞

0

e−ttz−1 dt

This is regular for Re(z) > 0

Note:

ˆ ε

0

tz−1 dt =

[tz

z

]ε0

and tz can be evaluated at t = 0 only when Re(z) ≥ 0, but at

z = 0 you cannot evaluate tz

z .

Properties: 8.1

(a) Γ(z) is real for real z, and hence, Γ(z) = Γ(z) by Schwartz Reflection Principle.(b)

Γ(1) =

ˆ ∞

0

e−t dt = 1

Γ(2) =

ˆ ∞

0

te−t dt = 1

(c)

Γ(n+ 1) =

ˆ ∞

0

tne−t dt

=[tn(−e−t

)]∞0−ˆ ∞

0

(−e−t

) (ntn−1

)dt

= n

ˆ ∞

0

tn−1e−t dt

= nΓ(n)

so, Γ(2) = Γ(1) = 1, Γ(3) = 2Γ(2) = 2, Γ(n+ 1) = n!We view Γ(z) as the extension of the factorial function to the complex plane.(d) Γ(z + 1) = zΓ(z) provided Re(z) > 0. This allows us to AC to Re(z) < 0(e) It is hard to find Γ(z) for z not a positive integer or 1

2 (positive integer)

Γ

(1

2

)=

ˆ ∞

0

e−tt−1/2 dt

=

ˆ ∞

0

e−u2

u−1(2udu) (t = u2)

= 2

ˆ ∞

0

e−u2

du

=

ˆ ∞

−∞e−u2

du

=√π

32 Paul Russell


ie.

Γ

(1

2

)=

√π

Γ

(3

2

)=

1

2Γ

(1

2

)=

1

2

√π

Γ

(5

2

)=

3

2Γ

(3

2

)=

3

4

√π

Γ

(n+

1

2

)=

(n− 1

2

)Γ

(n− 1

2

)=

(2n− 1

2

)(2n− 3

2

). . .

(3

2

)(1

2

)√π

=

(2n

2n

)(2n− 1

2

)(2n− 2

2n− 2

). . .

(4

4

)(3

2

)(2

2

)(1

2

)√π

=(2n)!

√π

22n(n)(n− 1) . . . (2)(1)

=(2n)!

√π

22nn!

(f) We could write Γ(z) as (z − 1)!. (We usually don’t)(g) The Laplace Transform of a function f(t) is

F (z) =

ˆ ∞

0

f(t) dt

and for f(t) = tα, we have

F (z) =

ˆ ∞

0

tαe−zt dt

ˆ ∞

0

(uz

)α= e−u

(du

z

)

=

ˆ ∞

0

uαe−u du

zα+1

=Γ(α+ 1)

zα+1

AC of Γ(z) into Re(z) < 0 8.2

We have,

Γ(z) =Γ(z + 1)

z(Re(z) > 0)

33 Paul Russell


Γ(z + 1) is regular in Re(z + 1) > 0 Re(z) > −1

∴ RHS is regular in Re(z) > −1 except for simple pole (of residue Γ(1) = 1) at z = 0Similarly,

Γ(z + 1) =Γ(z + 2)

z + 1

So Γ(z) =Γ(z + 2)

z(z + 1)

RHS is regular in Re(z) > −2 except for simple poles at z = 0,−1

Carrying on

Γ(z) =Γ(z + n+ 1)

z(z + 1) . . . (z + n)

which is regular in Re(z) > −n− 1 except for simple poles at 0,−1, ...In this way, Γ(z) can be extended to be regular in the whole complex plane, except for n =0,−1,−2, ... where there are simple poles with residue

limz→−n

{(z + n)Γ(z)} =Γ(1)

(−n)(1− n)(2− n) . . . (−2)(−1)

=1

(−1)nn(n− 1) . . . (2)(1)

=(−1)n

n!

The Reflection Formula 8.3

Γ(z) is regular in C except for z = 0,−1, ...Γ(z + 1) is regular in C except for 1− z = 0,−1, ... ie. z = 1, 2, 3, ...Γ(z)Γ(1−z) is regular in C except at each integer sin(πz) is zero at each integer. So the function

g(z) = sin(πz)Γ(z)Γ(1− z)

has no singularities.If we know what g(z) does as z →∞ we could try to use Liouville’s Theorem.If g(z) was a constant, its value would be g(1/2) = sin(π/2)(Γ(1/2))2 = π

34 Paul Russell


Consider

Γ(z)Γ(1− z) =

(ˆ ∞

0

e−αxz−1 dx

)(ˆ ∞

0

e−yy−z dy

)=

ˆ ∞

0

e−xxz−1

ˆ ∞

0

e−yy−z dy dx

(y = vx) =

ˆ ∞

0

e−xxz−1

ˆ ∞

0

e−vxv−z dv dx

=

ˆ ∞

0

(ˆ ∞

0

e−vxv−z dv

)dx

=

ˆ ∞

0

v−z

(ˆ ∞

0

e−x(1+v) dx

)dv

=

ˆ ∞

0

dv

v2(1 + v)

=π

sin(πz)(0 < z < 1)

Then, AC ⇒ Γ(z)Γ(1− z) ≡ πsin(πz) ∀z ∈ R \ Z

So, for example

Γ(−1/2) =π

sin(πz)Γ(1− z)

=π

sin(−π/2)Γ(3/2)

=π

(−1)(12

√π)

= −2√π

and

Γ

(−3

2

)=

π

sin(πz)Γ(1− z)

∣∣∣∣z=−3/2

=π

(+1)(34

√π)

=

(4

3

)√π

35 Paul Russell


Example 8.4

(a)

Γ

(n+

1

2

)=

(2n)!√π

22nn!, so

Γ

(1

2− n

)=

π

sin{π(n+ 1

2

)}Γ(n+ 1

2

)=

(−1)n√π22nn!

(2n)!

(b)

Γ

(1

2+ iy

)Γ

(1

2− iy

)=

π

sin{π(12 + iy

)}=

π

cos(πiy)

=π

cosh(πy)

Also,

Γ

(1

2− iy

)= Γ

(1

2+ iy

)= Γ

(1

2+ iy

)by Schwartz.Hence,

LHS =

∣∣∣∣Γ(1

2+ iy

)∣∣∣∣2 ie.

∣∣∣∣Γ(1

2+ iy

)∣∣∣∣ =

√π

cosh(πy)

36 Paul Russell

Integral Transforms MATH34001

§9 Integral Transforms

Definition 9.1

Suppose f(x) is defined on [a, b] (or any interval.) Then,

F (y) =

ˆ b

a

f(x)k(x, y) dx

is said to be the integral transform of the function f(x), with respect to the kernel k(x, y) overthe interval (a, b).

Every choice of k(x, y) and (a, b) defines an integral transform. The important ones are the onesfor which

(a) an inversion formula exists.(b) other useful properties exist.

In this course, we focus on Laplace and Fourier transforms.

Example 9.2

(i) The Laplace Transform is,

F (y) =

ˆ ∞

0

f(x)e−xy dx

Hence, the Laplace transform of f(x) = xn is

F (y) =

ˆ ∞

0

xne−xy dx

=

ˆ ∞

0

(u

y

)n

e−u dy

y

= y−n−1

ˆ ∞

0

une−u du

=Γ(n+ 1)

yn+1

(ii) The Fourier Transform is,

F (y) =

ˆ ∞

−∞f(x)e−ixy dx

37 Paul Russell


so that the Fourier Transform of e−λx2

, (λ > 0) is

F (y) =

ˆ ∞

−∞e−λx2−ixy dx

=

ˆ ∞

−∞e−λx2

cos(xy) dx

=

√π

λexp

(− y2

4λ

)Left: f(x), Right: F (x), green λ = 0.5, black λ = 1, red λ = 10.

Function and transform obey opposite trends.See handout for ’justification’ of form of Fourier Transform and its inverse.

Restriction of f(x) 9.3

Cearly we want f(x) such that F (y) exists. The conditionˆ ∞

−∞|f(x)| dx ≤ ∞

ie. f(x) is absolutely integrable is sufficient to make F (y) exist. However, this excludes all theordinary functions we are interested in. (eg. xn, eαx, cos(αx) etc.) Instead, we think of y asbeing a complex quantity, and then we can get around this restriction.An example with

´|f(x)| dx <∞

f(x) =1

1 + x2

(⇒ˆ|f(x)| dx = π <∞

)and then

F (y) =

ˆ ∞

−∞

e−ixy

1 + x2dx

we evaluate this using a D-shaped contour. (Paying attention to (a) the size of e−ixy and(b) the direction of the contour for the residue theorem. We find

F (y) = πe−|y|

38 Paul Russell


even because f(x) is)According to our inversion formula, we should have

f(x) =1

2π

ˆ ∞

−∞F (y)e+ixy dy

=1

2

ˆ ∞

−∞e−|y|+ixy dy

=1

2

ˆ ∞

−∞e−|y| cos(xy) dy

=

ˆ ∞

0

e−|y| cos(xy) dy

=

ˆ ∞

0

e−y cos(xy) dy

= Re

(ˆ ∞

0

e−y+ixy dy

)= Re

[e−(1−ix)y

−(1− ix)

]∞0

= Re1

1− ix

=1

1 + x2

Fourier Cosine and sine Transforms 9.4

Let f(x) be defined for x > 0, and let

ˆ ∞

0

|f(x)| dx < ∞

Define the ’even extension’ of f(x) by

fe(x) =

{f(x) (x > 0)

f(−x) (x < 0)(Then also,

ˆ ∞

−∞|fe(x)| dx <∞

)

Fe(k) =

ˆ ∞

−∞fe(x)e

−ikx dx =

ˆ ∞

−∞fe(x) cos(kx) dx

= 2

ˆ ∞

0

fe(x) cos(kx) dx

= 2

ˆ ∞

0

f(x) cos(kx) dx

39 Paul Russell


Also, Fe(k) is even and

fe(x) =1

2π

ˆ ∞

−∞Fe(k)e

ikx dk

=1

2π

ˆ ∞

−∞Fe(k) cos(kx) dk

=1

π

ˆ ∞

0

Fe(k) cos(kx) dk

x > 0 ⇒ f(x) = fe(x) =1

π

ˆ ∞

0

Fe(k) cos(kx) dk

Write Fe(k) = 2F (k)⇒

F (k) =

ˆ ∞

0

f(x) cos(kx) dx (FCT)

f(x) =2

π

ˆ ∞

0

F (k) cos(kx) dk (Inverse FCT)

Similarly, if we create the odd extension fo(x) we derive the fourier sine transform

F (k) =

ˆ ∞

0

f(x) sin(kx) dx (FST)

f(x) =2

π

ˆ ∞

0

F (k) sin(kx) dk (Inverse FST)

We will see how to choose which of these transforms to use. Be careful: the three FTs aredifferent. eg. Consider f(x) = e−x, (x > 0)

FCT is Fc(k) =

ˆ ∞

0

e−x cos(kx) dx =1

1 + k2

FST is Fs(k) =

ˆ ∞

0

e−x sin(kx) dx =k

1 + k2

and if you suppose that f(x) = 0, x < 0 then the FT isˆ ∞

−∞f(x)e−ikx dx =

1

1 + ik

Analytic behaviour of f(x) 9.5

In general F (k) may not be well defined in the complex k-plane. Eg. for f(x) = 11+x2 , F (k)

does not exist if Im(k) = 0,

F (k) =

ˆ ∞

−∞

1

1 + x2e−ikx dx

=

ˆ ∞

−∞

1

1 + x2e−ix(k1+ik2) dx

=

ˆ ∞

−∞

1

1 + x2ek2x−ik1x dx

40 Paul Russell


If k2 > 0, ek2x →∞ as x→∞If k2 < 0, ek2x →∞ as x→ −∞

Suppose instead that f(x) =

{O (e−ax) x→∞O(ebx)

x→ −∞for some a, b,, then F (k) converges (and is therefore regular) provided −a < Im(k) < b. Thisstrip has thickness a+ b, so exists if a+ b > 0. We therefore come up with a strip of regularityof F (k). In this case, we can invert the transform by

f(x) =1

2π

ˆ ∞+ic

−∞+ic

F (k)e+ikx dk

where −a < c < b.

Example 9.6

Heaviside function

H(x) =

{0 x < 0

1 x > 0

We cannot evaluate

ˆ ∞

−∞H(x)eikx dx for real k.

For this function we can use a = 0 and b =∞. So the FT is regular in this case for 0 < Im(k) <∞.In this strip we have

ˆ ∞

−infty

H(x)eikx dx =

ˆ ∞

0

eikx dx

=

ˆ ∞

0

ei(k1+ik2)x dx

=

ˆ ∞

0

e(−k2+ik1)x dx

=1

k2 − ik1

=1

ik2 + k1

=i

k, (k2 = Im(k) > 0)

This clearly will not work on the real axis as the Fourier Transform has a pole there.According to our understanding of the Fourier Inversion, we should have

H(x) =1

2π

ˆ ∞+ic

−∞+ic

i

keikx dk (any c > 0)

For x > 0 we close the contour below giving¸= −2πi× Res = −2πi(i) = 2π. Argue away the

arc with Jordan’s Lemma ∴ H(x) = 1(x > 0)

41 Paul Russell


For x < 0 we close the contour above ∴¸

= 0 (since integral is regular everywhere inside.)Arguing away is as before ∴ H(x) = 0 (x < 0)

Laplace Transform 9.7

Let

f(x) =

{O (eax) (x→∞)

0 x < 0

Then F (k) =´∞−∞f(x)e+ikx dx is regular for a < Im(k) <∞ with inversion formula

f(x) =1

2π

ˆ ∞+ic

−∞+ic

F (k)e−ikx dk

We now replace k by pi

∴ F (p) = F (k) =

ˆ ∞

0

f(x)e−px dx

and

f(x) =1

2π

ˆ c−i∞

c+i∞F (p)epx idp

F (p) is now regular fr Re(p) > 0, and we turn contour round, so we write

f(x) =1

2πi

ˆ c+i∞

c−i∞F (p)epx dp

Inverse Laplace Transform → Bromwich’s Contour.

Example 9.8

f(x) =

{0 x < 0

sin(x) x > 0

so we can use a = 0 in our exponential bound, therefore

F (p) =

ˆ ∞

0

e−px sin(x) dx

should exist for Re(p) > 0.Actually,

F (p) = Im

ˆ ∞

0

e−pxeix dx = Im1

p− i=

1

p2 + 1

This has simple poles at ±i, and no other singularities. (∴ is regular for Re(p) > 0). Check:

ˆ c+i∞

c−i∞

1

p2 + 1epx dx

42 Paul Russell


for any c > 0.For x > 0, we cannot close the contour to the right, therefore, close to the left.

˛= 2πi

∑±i

Residues = 2πi

{eix

2i+

e−ix

−2i

}= π

{eix − e−ix

}= 2iπ sin(x)

giving inverse = sin(x), (x > 0)For x < 0 we close to the right, therefore

¸= 0

Arguing away for both x < 0 and x > 0 is by our simple bound, therefore inverse = 0 (x < 0)

Transforms and Derivatives 9.9

Consider

F (k) =

ˆ ∞

−∞f(x)eikx dx

Then,

dF

dk=

d

dk

ˆ ∞

−∞f(x)eikx dx

=

ˆ ∞

−∞ixf(x)eikx dx

= FT of (ixf(x))

Differentiation of FT = multiplication of original function. Conversely,

FT of f ′(x) =

ˆ ∞

−∞f ′(x)eikx dx

=[f(x)eikx

]∞−∞ − ik

ˆ ∞

−∞f(x)eikx dx

We assume that f(x)→ 0 as x→ ±∞ we see that

FT of f ′ = ikF (k)

If we do this again, we see that

FT of (f ′′) = (−ik)2F (k) = −k2F (k)

provided f, f ′ → 0 as x→ ±∞.

What about Laplace Transforms?

F (p) =

ˆ ∞

0

f(x)e−px dx ⇒ dF

dp= −ˆ ∞

0

xf(x)e−px dx = LT of (−xf(x))

43 Paul Russell


Also, LT of f ′(x) isˆ ∞

0

f ′(x)e−px dx =[f(x)e−px

]∞0

+ p

ˆ ∞

0

e−pxf(x) dx

= −f(0) + pF (p)

and LT of f ′′(x) is

−f ′(0) + p(LT of f ′) = −f ′(0)− pf(0) + p2F (p)

An ODE solved by Laplace Transform 9.10

Considerdy

dt− y = t−1/2 (t > 0)(1)

subject to y(0) = 0 (2)Let

Y (p) =

ˆ ∞

0

y(t)e−pt dt (= LT of y(t))

LT (1) ⇒

−y(0)︸︷︷︸=O(z)

+pY (p)

=

ˆ ∞

0

t−1/2e−pt dt

=

ˆ ∞

0

(u

p

)−1/2

e−u du

p

=1

p1/2

ˆ ∞

0

u−1/2e−u du

=Γ(12

)p1/2

=

√π

pℜ(p) > 0

Thus,

Y (p) =1

p− 1

√π

p

NB: We want Y (p) to be regular in Re(p) > a for some a so, we choose the branch cut of p−1/2

to lie along the ive real axis (and we also see a > 1)Thus we use −π < arg(p) ≤ π and

y(t) =1

2πi

ˆ c+i∞

c−i∞

1

p− 1

√π

pept dp

and we need to do a keyhole contour to close the integral˛... dp = 2πi

(√πet)

44 Paul Russell


We argue away arcs conventionally.On the section above the branch cut, (γ3)

ˆγ3

=

ˆ ε

R

1

−r − 1

√π

reiπe−rt(−dr)

= −1

i

ˆ R

ε

e−rt

r + 1

√π

rdr

= i

ˆ R

ε

e−rt

r + 1

√π

rdr

Below the branch cut (γ5) arg(p) = −π, so write p = re−iπ to give

ˆγ5

=

ˆ R

ε

e−rt

−r − 1

√π

re−iπ(−dr)

= i

ˆ R

ε

e−rt

r + 1

√π

rdr

Now let R→∞ and ε→ 0, this gives

2πi(√πet) =

˛=

ˆγ1

+2i

ˆ ∞

0

e−rt

r + 1

√π

rdr

= 2πiy(t) + 2i

ˆ ∞

0

e−rt

r + 1

√π

rdr

Thus,

y(t) =√πet − 1√

π

ˆ ∞

0

e−rt dr

(r + 1)√r

=√πet − 2√

π

ˆ ∞

0

e−ts2

1 + s2ds

We cannot find this analytically - it may not matter.

45 Paul Russell

Solving PDEs using transforms MATH34001

§10 Solving PDEs using transforms

We study the heat equation for u(x, t)

∂u

∂t= α

∂2u

∂x2(10.1)

0 < x <∞ or −∞ < x <∞ and 0 < t <∞, together with some critical state.

u(x, 0) = f(x) (known)

We first assume we have an infinite problem. (ie. −∞ < x <∞) and write

U(k, t) =

ˆ ∞

−∞u(x, t)eikx dx

together with U(k, t) =

ˆ ∞

−∞f(x)eikx dx

= F (k) (known)

We have FT of∂u

∂t= α

∂2u

∂x2⇒ ∂U

∂t= α(−k2U)

provided∂u

∂x→ 0 as x→ ±∞

Thus, U(k, t) = Ae−αk2t for some A, and A = u(k, 0) = F (k) ie.

U(k, t) = F (k)e−αk2t

f(x) is known (in principle)⇒ F (k) is known⇒ U(k, t) is known⇒ u(x, t) is known (inverse FT)

Consider

f(x) =

{0 x < 0

u0e−ax x > 0

⇒ = u0

ˆ ∞

0

e−ax+ikx dx

=u0

a− ik

=iu0

k + ia

Showing U has a simple pole at k = −ia. Then

u(x, t) =iu0

2π

ˆ ∞

−∞

e−αk2t−ikx

k + iadk

46 Paul Russell


But, ∣∣∣e−αk2t∣∣∣→∞ as Im(k)→ ±∞

so we cannot close the contour above or below, we can use another approach in the limit a→ 0+.Then, f(x)→ u0H(x) and F (k)→ iu0k and the pole has moved up onto the real axis, wherewe want to itnegrate.Use the contour

#which has a semi-circle of radius epsilon around the −∞ pole.

Look at

u(x, t) =iu0

2π

$ ∞

−∞e−αk2t−ikx dk

k

then,

∂u

∂x=

u0

2π

$ ∞

−∞e−αk2t−ikx dk

=

(ˆ ∞

−∞

)← no longer a pole to hop over!

=u0

2π

ˆ ∞

−∞e−αk2t cos(kx) dk

=u0

2π

√π

αtexp

(−x2

4αt

)(text earlier in course)

This gives

u(x, t) =

ˆ ∞

−∞

u0

2π

√π

αtexp

(−s2

4αt

)ds

write s = v√4αt to get

u(x, t) =u0√π

ˆ x/(4αt)

−∞e−v2

dv

=1

2u0

(1 + erf

(x√4αt

))where

erf(x) =2√π

ˆ x

0

e−v2

dv

(1) erf is an odd function and erf(0) = 0(2) erf(∞) = 1

We can also solve for u(x, t) analytically for

f(x) = e−x2/c2 or f(x) = δ(x)

This delta-function is such that δ(x) = 0 for x = 0 with

ˆ b

−a

δ(x)dx = 1

The δ(x) has the substitution propertyˆ ∞

−∞δ(x− c)f(x) dx = f(c)

47 Paul Russell


Using Fourier Transforms for a semi-infinite heat-conduction problem 10.1

Because we only have (10.1) over 0 < x <∞ we cannot use the complex Fourier Transform.

ˆ ∞

0

∂2u

∂x2eikx dx =

[∂u

∂xeikx

]∞0

− ik

ˆ ∞

0

∂u

∂xeikx dx

=

[(∂u

∂x− iku

)eikx

]∞0

− k2ˆ ∞

0

ueikx dx

Again, assume u and ∂u∂x → 0 as x→∞. Then

ˆ ∞

0

∂2u

∂x2eikx dx =

(−∂u

∂x+ iku

)x=0

− k2ˆ ∞

0

ueikx dx

Re ⇒ˆ ∞

0

∂2u

∂x2cos(kx) dx =

(−∂u

∂x

)x=0

− k2ˆ ∞

0

u cos(kx) dx (10.9)

Im ⇒ˆ ∞

0

∂2u

∂x2sin(kx) dx = ku|x=0 − k2

ˆ ∞

0

u sin(kx) dx (10.10)

This tells us that if we know about u(0, t) we use a Fourier Sine Transform; if however outinformation is about ∂u

∂x , then a Fourier Cosine Transform is appropriate.

Example 10.2

We solve (10.1) and (10.2) together with

u(0, t) = g(t) (10.13)

We use an FST and write

U(k, t) =

ˆ ∞

0

u(x, t) sin(kx) dx

Then (10.1) ⇒

∂U

∂t= α

{kg(t)− k2U

}or

∂U

∂t+ αk2U = αkg(t)

with (10.2) ⇒ U(k, 0) = F (k)These two provide a unique solution (if g and f are known)For example, if f ≡ 0 and g ≡ constant ≡ u0

Ie.U(k, t) =

u0

k

(1− e−αk2t

)sin(kx) dk

48 Paul Russell


Again, we can neither close the contour above nor below (∵ e−αk2t); instead

∂u

∂x=

2u0

π

ˆ ∞

0

(1− e−αk2t

)cos(ku) dk

=2u0

πlimε→0

ˆ ∞

0

(e−kε − e−αk2t

)cos(kx) dk

=2u0

πlimε→0

(Re

1

ε+ ik− 1

2

√π

αtexp(−x2/(4αt))

)dk

=u0

π

√π

αtexp(−x2/(4αt))

Hence,

u(x, t) =u0

π

√π

αt

ˆ ∞

x

exp(−s2/(4αt)) ds

=u0

π

√π

αt

ˆ ∞

x/√4αt

e−v2√4αt dv

=2u0√π

ˆ ∞

x/√4αt

e−v2

dv

and2√π

ˆ ∞

x

e−v2

dv = 1− erf(x) = erfc(x)

We now tackle the same problem using a Laplace Transform with respect to time.Write

U(x, p) =

ˆ ∞

0

u(x, t)e−pt dt

Then,

(10.1)⇒ α∂2U

∂x2=

ˆ ∞

0

α∂2u

∂x2e−pt dt

=

ˆ ∞

0

∂u

∂te−pt dt

= pU − u(x, 0)︸︷︷︸f(x)

(10.17)

together with an end condition u(0, t) = g(t) so that

U(0, p) = G(p) (10.18)

(10.17) and (10.18) represent a second order ODEwith only one condition. We need anothercondition!

49 Paul Russell


Boundedness Condition 10.3

Suppose we know that|u(x, t)| ≤ Mebt, ∀x, t > 0 (10.19)

Then,

|U | =

∣∣∣∣ˆ ∞

0

ue−pt dt

∣∣∣∣≤

ˆ ∞

0

|u|e−pt dt

≤ˆ ∞

0

Mebt−pt dt

=M

p− bRe(p) > b (10.20)

We see that x is not part of (10.19) or (10.20); in particular, (10.20) applies as x → ∞, and sowe use

|U | is bounded as x→∞ (10.21)

We will solve (10.17) with (10.18) and (10.21) under the assumption that f(x) = 0Thus,

∂2U

∂x2=

p

αU

So,

U(x, p) = U1(p) exp

{x

√p

α

}+ U2(p) exp

{−x√

p

α

}To invert the Laplace Transform, we need to think of p as complex and thus we must define abranch for

√p. We use −π < arg(p) ≤ π (The branch cut avoids the Bromwich integral.)

So, −π2 < arg(

√p) ≤ π

2and so, Re(

√p) ≥ 0

This means exp{x√

pα

}→∞ as α→∞, but boundedness does not allow this. Hence, U1(p) = 0,

and then, U2(p) = G(p)So,

U(x, p) = G(p) exp

{−x√

p

α

}and hence,

u(x, t) =1

2πi

ˆ c+i∞

c−i∞G(p) exp

{−x√

p

α

}ept dp

We need to choose c so that this contour is to the right of all singularities.

Continued on handout.

End of Course! (YAY)

50 Paul Russell

Documents

Applied Complex Analysis