EI331 Signals and Systems Lecture 26 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 26 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

May 28, 2019

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Contents

• 1. Singularities, Poles and Zeros
• 2. Residue
• 3. Behavior at Infinity
• 4. Evaluation of Definite Integrals Using Residues

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Singularities

Recall a point z0 ∈ C is a singular point or a singularity of a function f if f is not analytic at z0.

• Example. sin z

z

and 1

z have a singularity at z = 0, since they are

not defined at z = 0. If f(z) is singular at z0, but analytic on a deleted open disk 0 < |z − z0| < r for some r > 0, then z0 is called an isolated singularity of f.

• Example. 1

z and e

1 z have an isolated singularity at z = 0.

Example.

1 sin 1

z has a singularity at z = 0. It also has singularities

at the roots of sin 1

z = 0, or zn = 1 nπ, n ∈ Z \ {0}. Since z = 0 is

the limit zn as |n| → ∞. Any deleted open disk centered at 0 contains infinitely many zn’s, so 0 is not an isolated singularity.

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Classification of Isolated Singularities

At an isolated singularity z0, f is analytic on a deleted open disk 0 < |z − z0| < r, so it has a Laurent series expansion f(z) =

∞

• n=−∞

cn(z − z0)n =

∞

• n=0

cn(z − z0)n +

∞

• n=1

c−n(z − z0)−n φ(z) = ∞

n=0 cn(z − z0)n is the regular or analytic part of f

ψ(z) = ∞

n=1 c−n(z − z0)−n is the singular or principal part of f

• If ψ = 0, then z0 is called a removable singularity of f
• If ψ has finitely many terms, then z0 is called a pole of f.

◮ If m ∈ N is the largest integer s.t. c−m = 0 but c−n = 0 for n > m, then z0 is a pole of order m. ◮ A pole of order 1 is called a simple pole.

• If ψ has infinitely many terms, then z0 is called an essential

singularity of f.

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Removable Singularity

At a removable singularity z0, the Laurent series of f reduces to a power series f(z) =

∞

• n=0

cn(z − z0)n, 0 < |z − z0| < r The power series defines an analytic function F(z) on B(z0, r) s.t. F(z) = f(z) when z = z0 If we define f(z0) = lim

z→z0 f(z) = c0, then f = F is analytic at z0,

whence the name “removable”.

• Example. z = 0 is a removable singularity of sin z

z , since

sin z z = 1 z

∞

• n=0

(−1)n (2n + 1)!z2n+1 =

∞

• n=0

(−1)n (2n + 1)!z2n We can define sin 0 = 1.

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Pole

At a pole of order m, the Laurent series of f is f(z) =

∞

• n=−m

cn(z − z0)n = g(z) (z − z0)m, 0 < |z − z0| < r where g(z) =

∞

• n=0

cn−m(z − z0)n is analytic on B(z0, r) and g(z0) = c−m = 0. Note f(z) ≈

c−m (z−z0)m, so lim z→z0 |f(z)| = +∞, also written lim z→z0 f(z) = ∞

Conversely, if f(z) =

g(z) (z−z0)m for some analytic function g with

g(z0) = 0, then z0 is a pole of order m of f.

• Example. f(z) =

z−2 (z2+1)(z−1)3 has a pole of order 3 at z = 1, two

simple poles at z = ±j.

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Essential Singularity

Theorem (Weierstrass). If z0 ∈ C is an essential singularity of f, then given any A ∈ C, there is a sequence zn s.t. zn → z0 and f(zn) → A, as n → ∞.

• NB. As a consequence, lim

z→z0 f(z) does not exist.

• Example. f(z) = e1/z has an essential singularity at z = 0, since

f(z) = 1 +

∞

• n=1

1 n!z−n We can verify Weierestrass’s Theorem: Given A ∈ C,

• if A = 0, then zn =

1 log A+j2nπ → 0 and f(zn) = elog A+j2nπ = A

• if A = 0, then zn = − 1

n → 0 and f(zn) = e−n → 0

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Zeros and Poles

Let f be analytic and not identically zero on a domain D.

• A point z0 ∈ D is called a zero of f if f(z0) = 0.
• z0 is called a zero of order m, if the Taylor series of f at z0 is

f(z) =

∞

• n=m

cn(z − z0)n, cm = 0 A zero of order 1 is called a simple zero

• f(z) ≈ cm(z − z0)m, so z0 is an isolated zero.
• z0 is a zero of order m of f, iff f(z) = (z − z0)mφ(z), where φ is

analytic and φ(z0) = 0.

• z0 is a zero of order m of f, iff f (n)(z0) = 0 for n < m, and

f (m)(z0) = 0.

• Theorem. f(z) has a zero of order m at z0 iff

1 f(z) has a pole of

• rder m at z0.

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Zeros and Poles

• Example. f(z) = z(z − 1)3 and g(z) =

1 f(z) = 1 z(z−1)3

• f has a simple zero at z = 0, g has a simple pole at z = 0
• f a zero of order 3 at z = 1, g has a pole of order 3 at z = 1
• Example. f(z) = ez−1

z2

has a singularity at z = 0. f(z) = 1 z2

∞

• n=1

zn n! =

∞

• n=−1

zn (n + 2)! z = 0 is a pole of order 1, not of order 2!

• Example. f(z) = sinh z

z3

has a singularity at z = 0, f(z) = 1 z3

∞

• n=0

z2n+1 (2n + 1)! =

∞

• n=−1

z2n (2n + 3)! z = 0 is a pole of order 2, not of order 3!

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Quotient of Functions

• Theorem. Suppose f has a zero of order m at z0 and g has a

zero of order n at z0. Let h(z) = f(z)/g(z).

• If n > m, h(z) has a pole of order n − m at z0.
• If n < m, h(z) has a zero of order m − n at z0.
• If n = m, h(z) is analytic at z0 and h(z0) = 0.
• Proof. f and g take the following form,

f(z) = (z − z0)mf1(z), g(z) = (z − z0)ng1(z) where f1(z) and g1(z) are analytic and nonzero at z0. Thus h(z) = (z − z0)n−mh1(z) where h1(z) = f1(z)/g1(z) is analytic and nonzero at z0.

• Example. ez − 1 has a simple zero at z = 0, and z2 has a zero of
• rder 2 at z = 0, so ez−1

z2

has a pole of order 1 at z = 0.

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Contents

• 1. Singularities, Poles and Zeros
• 2. Residue
• 3. Behavior at Infinity
• 4. Evaluation of Definite Integrals Using Residues

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Residue

Suppose f has an isolated singularity at z0 ∈ C, and f is analytic

• n 0 < |z − z0| < R with Laurent series expansion,

f(z) =

∞

• n=−∞

cn(z − z0)n Let C be the positively oriented circle |z − z0| = r ∈ (0, R). Term-by-term integration yields

• C

f(z)dz =

∞

• n=−∞

cn

• C

(z − z0)ndz =

∞

• n=−∞

cnj2πδ[n + 1] = j2πc−1 The residue of f at z0 ∈ C is Res(f, z0) = 1 j2π

• C

f(z)dz = c−1

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Residue

• Example. f(z) = sin z

z

has a removable singularity at z = 0,

sin z z

=

∞

• n=0

(−1)n (2n+1)!z2n =

⇒ Res(f, 0) = 0 In general, Res(f, z0) = 0 if z0 ∈ C is a removable singularity.

• Example. e1/z has an essential singularity at 0.

e1/z =

∞

• n=0

1 n!z−n =

⇒ Res(e−z, 0) = 1

• Example. f(z) =

1 z(1−z) has a simple pole at z = 0,

f(z) = 1

z ∞

• n=0

zn =

∞

• n=−1

zn on 0 < |z| < 1 = ⇒ Res(f, 0) = 1

• Warning. f(z) = − 1

z2 1 1−1/z = − ∞

• n=2

z−n on 1 < |z| < ∞ with c−1 = 0, but Res(f, 0) = 0, since the definition or residue requires using Laurent series on 0 < |z − z0| < r

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Residue at Simple Poles

If z0 is a simple pole, then f(z) = g(z) z − z0 , where g(z) is analytic and nonzero at z0. By Cauchy’s Theorem Res(f, z0) = 1 j2π

• C

g(z) z − z0 dz = g(z0) = lim

z→z0(z − z0)f(z)

If f(z) = φ(z)

ψ(z), where φ and ψ are analytic at z0, φ(z0) = 0, and ψ

has a simple zero at z0, then Res(f, z0) = φ(z0) ψ′(z0) since Res(f, z0) = lim

z→z0(z − z0)f(z) = lim z→z0

φ(z)

ψ(z)−ψ(z0) z−z0

= φ(z0) ψ′(z0)

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Residue at Simple Poles

• Example. f(z) =

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

Res(f, 0) = lim

z→0 zf(z) = 1,

Res(f, 1) = lim

z→1(z − 1)f(z) = −1

Recall the partial fraction expansion of f(z) takes the form f(z) = A z + B z − 1 where A = lim

z→0 zf(z) = Res(f, 0),

B = lim

z→1(z − 1)f(z) = Res(f, 1)

• Example. f(z) =

1 sin z has simple poles at z = kπ, k ∈ Z. Note

f(z) = φ(z)

ψ(z), where φ(z) = 1 is analytic and nonzero at z = kπ, and

ψ(z) = sin z has a simple zero at z = kπ. Thus Res(f, kπ) = lim

z→kπ

1 (sin z)′ = lim

z→kπ

1 cos z = (−1)k

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Residue at Higher Order Poles

If z0 is a pole of order m, then f(z) = g(z) (z − z0)m where g(z) is analytic and nonzero at z0. By Cauchy’s Integral Formula Res(f, z0) = 1 j2π

• C

g(z) (z − z0)mdz = g(m−1)(z0) (m − 1)! For n ≥ m, the product rule for differentiation yields dn−1 dzn−1[(z−z0)nf(z)]

• z=z0

= dn−1 dzn−1[(z−z0)n−mg(z)]

• z=z0

= (n − 1)! (m − 1)!g(m−1)(z0) so Res(f, z0) = lim

z→z0

1 (n − 1)! dn−1 dzn−1[(z − z0)nf(z)], n ≥ m

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Residue at Higher Order Poles

• Example. f(z) =

1 z(z2 + 1)(z − 2)2 has a pole of order 2 at z = 2. Res(f, 2) = lim

z→2

d dz[(z − 2)2f(z)] = lim

z→2

d dz

• 1

z(z2 + 1)

• = lim

z→2

−(3z2 + 1) z2(z2 + 1)2 = − 13 100

• Example. f(z) = ez sinh z

z5 has a pole of order 4 at z = 0. Res(f, 0) = lim

z→0

1 4! d4 dz4[z5f(z)] = lim

z→0

1 4! d4 dz4 e2z − 1 2

• =

24 2 · 4! = 1 3 We can also use Laurent series expansion to find Res(f, 0) f(z) = e2z − 1 2z5 = 1 2z5

∞

• n=1

1 n!(2z)n = · · · + 1 2z5 · 1 4!(2z)4 + . . .

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Residue Theorem

• Theorem. Let γ be a piecewise smooth Jordan curve with

interior D. If f is analytic on D \ {z1, z2, . . . , zn}, and continuous

• n ¯

D \ {z1, z2, . . . , zn}, then

• γ

f(z)dz = j2π

n

• k=1

Res(f, zk)

D γ z1 C1 z2 C2 z3 C3

• Proof. Draw a small circle Ck centered at

zk, k = 1, 2, . . . , n, s.t. the Ck’s and their interiors are in D, and they are disjoint. By Cauchy’s Theorem and the definition

• f residue,
• γ

f(z)dz =

n

• k=1
• Ck

f(z)dz = j2π

n

• k=1

Res(f, zk)

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Example

Evaluate

• C

zez z2 − 1dz, where C is the positively oriented circle |z| = 2.

• Solution. f(z) =

zez z2−1 has two simple poles z = ±1, both in the

interior of C, so

• C

f(z)dz = j2π[Res(f, 1) + Res(f, −1)] For the residues, Res(f, 1) = lim

z→1(z − 1)f(z) = lim z→1

zez z + 1 = e 2, Res(f, −1) = zez (z2 − 1)′

• z=−1 = zez

2z

• z=−1 = e−1

2 Thus

• C

f(z)dz = j2π e 2 + j2πe−1 2 = j2π cosh 1

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Example

Evaluate

• C

z z4 − 1dz, where C is the positively oriented circle |z| = 2.

• Solution. f(z) =

z z4−1 has four simple poles at z = ±1, ±j, all in

the interior of C. For the residues, Res(f, ±1) = lim

z→±1

z [z4 − 1]′ = lim

z→±1

1 4z2 = 1 4 Res(f, ±j) = lim

z→±j

z [z4 − 1]′ = lim

z→±j

1 4z2 = −1 4 Thus

• C

f(z)dz = j2π

• z∈{±1,±j}

Res(f, z) = j2π 1 4 + 1 4 − 1 4 − 1 4

• = 0
• NB. Can also use Laurent series on 1 < |z| < ∞, f(z) =

∞

• n=0

1 z4n+3.

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Example

Evaluate

• C

ez z(z − 1)2dz, where C is the positively oriented circle |z| = 2.

• Solution. f(z) =

ez z(z−1)2 has a simple pole at z = 0, and a pole of

• rder 2 at z = 1, both in the interior of C, so
• C

f(z)dz = j2π[Res(f, 0) + Res(f, 1)] For the residues, Res(f, 0) = lim

z→0 zf(z) = lim z→0

ez (z − 1)2 = 1 Res(f, 1) = lim

z→1

d dz[(z − 1)2f(z)] = lim

z→1

d dz ez z = lim

z→1

ez(z − 1) z2 = 0 Thus

• C

f(z)dz = j2π

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Example

Evaluate

• C

z2 sin 1 z

• dz, where C is the positively oriented

circle |z| = 1.

• Solution. f(z) = z2 sin

1

z

• has an essential singularity at z = 0,

which lies in the interior of C. Expanding f(z) into Laurent series f(z) = z2

∞

• n=0

(−1)n (2n + 1)!z−(2n+1) = z − 1 6z + . . . so Res(f, 0) = −1 6 and

• C

f(z)dz = j2π Res(f, 0) = −jπ 3

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Example

Evaluate

• C

1 z(z − 2)4dz, where C is the positively oriented circle |z − 2| = 1.

• Solution. f(z) =

1 z(z−2)4 has a simple pole at z = 0 and a pole of

• rder 4 at z = 2. Only the pole z = 2 lies in the interior of C, so
• C

f(z)dz = j2π Res(f, 2) For the residue at z = 2, Res(f, 2) = lim

z→2

1 3! d3 dz3[(z − 2)4f(z)] = lim

z→2(−1)3z−4 = − 1

16 Thus

• C

f(z)dz = −jπ 8

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Contents

• 1. Singularities, Poles and Zeros
• 2. Residue
• 3. Behavior at Infinity
• 4. Evaluation of Definite Integrals Using Residues

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Stereographic Projection and the Point at Infinity

S x y z P C N S Introduce a 3D unit sphere S, called the Riemann sphere The stereographic projection sending z to the intersection of zN and S is a one-to-one correspondence between C and S \ {N}. As |z| → ∞, the corresponding point P on S approaches N. Introduce a point at infinity, denoted ∞, that corresponds to N ¯ C = C ∪ {∞} is called the extended complex plane Arithmetics z ± ∞ = ∞ ± z = ∞ (z = ∞) z · ∞ = ∞ · z = ∞ (z = 0)

z ∞ = 0 (z = ∞) ∞ z = ∞ (z = ∞) z 0 = ∞ (z = 0)

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Stereographic Projection and the Point at Infinity

S C x y

N S

The open disk B(0, 1) ⊂ C ⇐ ⇒ the southern hemisphere of S The region 0 < |z| ≤ ∞ of ¯ C ⇐ ⇒ the northern hemisphere of S An open disk centered at 0 ⇐ ⇒ an open “disk” centered at S An open disk centered at ∞ ⇐ ⇒ an open “disk” centered at N Define the deleted open disk B∗(∞, R) = {z ∈ C : R < |z| < ∞}

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Singularity at ∞

If f(z) is analytic on B∗(∞, R), i.e. on R < |z| < ∞, then ∞ is an isolated singularity of f. Use the mapping ζ = 1

z to define a function

φ(ζ) = f 1 ζ

• = f(z)

The deleted open disk R < |z| < ∞ is mapped to the deleted

• pen disk 0 < |ζ| < 1

R

We call z = ∞ a removable singularity, a pole of order m, or an essential singularity of f, if ζ = 0 is a removable singularity, a pole of order m, or an essential singularity of φ.

• Example. f1(z) = z−3 has a removable singularity at z = ∞;

f2(z) = z3 has a pole of order 3 at ∞; f3(z) = ez has an essential singularity at z = ∞.

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Residue at ∞

Suppose f(z) is analytic on R < |z| < ∞. Let C be the positively

• riented circle |z| = r > R. The residue of f at ∞ is

Res(f, ∞) = 1 j2π

• C− f(z)dz = − 1

j2π

• C

f(z)dz

C z1 z2 z3

If the Laurent series of f(z) on R < |z| < ∞ is f(z) =

∞

• n=−∞

cnzn Term-by-term integration yields Res(f, ∞) = −c−1

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Residue at ∞

• Theorem. If f is analytic on C \ {z1, z2, . . . , zn}, then

n

• k=1

Res(f, zk) + Res(f, ∞) = 0

C z1 z2 z3

• Proof. Let C be a circle |z| = r > max

1≤k≤n |zk|.

By the Residue Theorem, 1 j2π

• C

f(z)dz =

n

• k=1

Res(f, zk) Note f is analytic on r < |z| < ∞. By definition, Res(f, ∞) = − 1 j2π

• C

f(z)dz

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Residue at ∞

• Theorem. If ∞ is an isolated singularity of f, then

Res(f, ∞) = − Res 1 ζ2 f 1 ζ

• , 0
• Proof. Suppose f is analytic on R < |z| < ∞ with Laurent series

f(z) =

∞

• n=−∞

cnzn Then the Laurent series of φ(ζ) =

1 ζ2 f

• 1

ζ

• n 0 < |ζ| < 1

R is

φ(ζ) = 1 ζ2 f 1 ζ

• =

∞

• n=−∞

cnζ−n−2 Thus Res(f, ∞) = −c−1 = − Res(φ, 0)

• NB. Can also be obtained by a change of variable z = 1

ζ in the

integral formula.

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Example

Evaluate

• C

z z4 − 1dz, where C is the positively oriented circle |z| = 2.

• Solution. f(z) =

z z4−1 has four simple poles at z = ±1, ±j, all in

the interior of C, so

• C

f(z)dz = j2π[Res(f, 1) + Res(f, −1) + Res(f, j) + Res(f, −j)] Since f is analytic in the exterior of C,

• C

f(z)dz = −j2π Res(f, ∞) = j2π Res 1 z2 f 1 z

• , 0
• = j2π Res
• z

1 − z4, 0

• = 0

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Example

Evaluate

• C

dz (z + j)10(z − 1)(z − 3), where C is the positively

• riented circle |z| = 2.
• Solution. f(z) =

1 (z+i)10(z−1)(z−3) has two simple poles at z = 1, 3, a

pole of order 10 at z = −j, so Res(f, 1) + Res(f, 3) + Res(f, −j) + Res(f, ∞) = 0. Since 1 and −j are in the interior of C,

• C

f(z)dz = j2π[Res(f, 1) + Res(f, −j)] = −j2π[Res(f, 3) + Res(f, ∞)] = −j2π

• 1

(z + j)10(z − 1)

• z=3 +

z10 (1 + zj)10(1 − z)(1 − 3z)

• z=0
• = −

jπ (3 + j)10

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Contents

• 1. Singularities, Poles and Zeros
• 2. Residue
• 3. Behavior at Infinity
• 4. Evaluation of Definite Integrals Using Residues

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Recipe

We use the following recipe to compute some real integrals

• f(x)dx by the Residue Theorem.
• 1. Find a complex function F(z) closely related to f(x). Often

F(x) = f(x) or Re F(x) = f(x) for x ∈ R, e.g. F(z) = ejz for f(x) = cos x.

• 2. Pick a contour (Jordan curve) C that includes the interval
• n the real axis in the integral, or part of it if it is infinite.
• 3. Break the contour into pieces and evaluate
• F(z)dz along

the pieces not on the real axis.

• 4. Use the Residue Theorem to evaluate
• C

F(z)dz

• 5. Combine the results in 3 and 4 to deduce the value of
• f(x)dx, which may involve taking limits.

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2π R(cos θ, sin θ)dθ

R(cos θ, sin θ) is a rational function of cos θ and sin θ. Let z = ejθ. Then dz = jejθdθ = jzdθ cos θ = 1 2(ejθ + e−jθ) = z2 + 1 2z , sin θ = 1 2j(ejθ − e−jθ) = z2 − 1 2jz Thus 2π R(cos θ, sin θ)dθ =

• |z|=1

R z2 + 1 2z , z2 − 1 2jz dz jz =

• |z|=1

f(z)dz f(z) is a rational function of z. By the Residue Theorem, I = j2π

n

• k=1

Res(f, zk) where z1, . . . , zk are the singularity of f in the interior of |z| = 1.

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Example

Evaluate I = 2π cos(2θ)dθ 1 − 2p cos θ + p2dθ, where 0 < p < 1.

• Solution. Let z = ejθ and C the unit circle |z| = 1 with positive
• rientation. Note

cos(kθ) = 1 2(ejkθ + e−jkθ) = zk + z−k 2 , dz = jejθdθ = jzdθ Thus I =

• C

(z2 + z−2)/2 1 − 2pz+z−1

2

+ p2 · dz jz =

• C

1 + z4 2jz2(1 − pz)(z − p)dz =

• C

f(z)dz Since f(z) has two poles z = 0 and z = p in the interior of C, I = j2π[Res(f, 0) + Res(f, p)] = 2πp2 1 − p2

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∞

−∞ R(x)dx

R(x) = N(x)

D(x) is a rational function of x, where N, D are polynomials

with deg D ≥ deg N + 2, and R has no singularity on the real axis. x y

−r r Cr z1 z2 z3

• 1. Pick a r large enough s.t. the upper

half disk centered at 0 contains all the singularities z1, . . . , zK, of R(z) in the upper half plane (we don’t care about those in the lower half plane) 2. r

−r

R(x)dx +

• Cr

R(z)dz = j2π

K

• k=1

Res(R, zk)

• 3. As r → ∞,
• Cr

R(z)dz → 0 by the condition deg D ≥ deg N + 2 4. ∞

−∞

R(x)dx = j2π

K

• k=1

Res(R, zk)

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Example

Evaluate I = ∞

−∞

x2dx (x2 + a2)(x2 + b2), where a, b > 0

• Solution. R(z) =

z2 (z2 + a2)(z2 + b2) has four simple poles at z = ±ja, ±jb. Two of them, ja, jb, are in the upper half plane. The residues are Res(R, ja) = lim

z→ja(z − ja)R(z) =

a 2j(a2 − b2) Res(R, jb) = lim

z→jb(z − jab)R(z) =

−b 2j(a2 − b2) Thus I = j2π[Res(R, ja) + Res(R, jb)] = π a + b

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∞

−∞ R(x)ejaxdx

R(x) = N(x)

D(x) is a rational function of x, where deg D ≥ deg N + 1,

and R has no singularity on the real axis. x y

−r r Cr z1 z2 z3

• 1. Pick a r large enough s.t. the upper

half disk centered at 0 contains all the singularities z1, . . . , zK, of R(z) in the upper half plane (we don’t care about those in the lower half plane) 2. r

−r

R(x)ejaxdx +

• Cr

R(z)ejazdz = j2π

K

• k=1

Res[R(z)ejaz, zk]

• 3. As r → ∞,
• Cr

R(z)ejazdz → 0 by deg D ≥ deg N + 1 4. ∞

−∞

R(x)ejaxdx = j2π

K

• k=1

Res[R(z)ejaz, zk]

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Example

Evaluate I = ∞ x sin x x2 + a2dx, where a > 0

• Solution. R(z) =

z z2 + a2 has two simple poles at z = ±ja. The pole ja is in the upper half plane, Res[R(z)ejz, ja] = lim

z→ja(z − ja)R(z)ejz = e−a

2 Thus ∞

−∞

xejx x2 + a2dx = j2π Res[R(z)ejz, ja] = jπe−a By symmetry, I = 1 2 ∞

−∞

x sin x x2 + a2dx = 1 2Im ∞

−∞

xejx x2 + a2dx = π 2e−a