13-Cauchy’sresiduetheorem ComplexVariables

Complex Variables

13 - Cauchy’s residue theorem

Jean-Baptiste Campesato

October 30^th, 2020 and November 2^nd, 2020

Theorem 1(Cauchy’s residue theorem).

Let𝑈 ⊂ ℂbe open. Let𝑆 ⊂ 𝑈be finite. Assume that𝑓 ∶ 𝑈 ⧵ 𝑆 → ℂis holomorphic/analytic.

Let𝛾 ∶ [𝑎, 𝑏] → ℂbe a positively oriented piecewise smooth simple closed curve on𝑈 ⧵ 𝑆^⋆ whose inside^† is entirely included in𝑈. Then^‡

∫_𝛾𝑓 (𝑧)d𝑧 = 2𝑖𝜋 ∑

𝑧∈Inside(𝛾)

Res(𝑓 , 𝑧) Corollary 2. Let𝑈 ⊂ ℂbe open and simply-connected. Let𝑆 ⊂ 𝑈be finite.

Assume that𝑓 ∶ 𝑈 ⧵ 𝑆 → ℂis holomorphic/analytic.

Let𝛾 ∶ [𝑎, 𝑏] → ℂbe a positively oriented piecewise smooth simple closed curve on𝑈 ⧵ 𝑆. Then

∫_𝛾𝑓 (𝑧)d𝑧 = 2𝑖𝜋 ∑

𝑧∈Inside(𝛾)

Res(𝑓 , 𝑧) Proof of Cauchy’s residue theorem.

𝛾

𝑧₁

𝑧₂

𝑧₃ 𝑧₄

We may find pairwise disjoints disks𝐷_𝑟

𝑘(𝑧_𝑘) ⊂ 𝑈 where{𝑧₁, … , 𝑧_𝑛}are the points of𝑆enclosed in𝛾.

We apply Green’s theorem to𝑇 =Inside(𝛾) ⧵ (

𝑛

⋃𝑖=1

𝐷_𝑟

𝑘(𝑧_𝑘)

), then

∫_𝛾𝑓 (𝑧)d𝑧 −

𝑛

∑𝑘=1∫_𝛾

𝑘

𝑓 (𝑧)d𝑧 = 𝑖

∬_𝑇(

𝜕𝑓

𝜕𝑥 + 𝑖𝜕𝑓

𝜕𝑦 )= 0 where𝛾_𝑘∶ [0, 1] → ℂis defined by𝛾_𝑘(𝑡) = 𝑧_𝑘+ 𝑟_𝑘𝑒^{2𝑖𝜋𝑡}.

The last equality is due to the Cauchy–Riemann equations.

Then∫_𝛾𝑓 (𝑧)d𝑧 =

𝑛

∑𝑘=1∫_𝛾

𝑘

𝑓 (𝑧)d𝑧 = 2𝑖𝜋

𝑛

∑𝑘=1

Res(𝑓 , 𝑧_𝑘). ■

⋆i.e.𝛾doesn’t pass through any point of𝑆.

†See Jordan’s curve theorem, September 28.

‡The following sum is finite since Res(𝑓 , 𝑧) ≠ 0only for𝑧 ∈ 𝑆.

Corollary 3. Let𝑆 ⊂ ℂbe finite. Assume that𝑓 ∶ ℂ ⧵ 𝑆 → ℂis holomorphic/analytic. Then Res(𝑓 , ∞) +∑

𝑧∈𝑆

Res(𝑓 , 𝑧) = 0 Remark 4. We may rewrite the above conclusion as

∑

𝑧∈̂ℂ

Res(𝑓 , 𝑧) = 0.

Proof. Take𝑟 > 0such that𝑆 ⊂ 𝐷_𝑟(0)and define𝛾 ∶ [0, 1] → ℂby𝛾(𝑡) = 𝑟𝑒^{2𝑖𝜋𝑡}. Then

𝑧∈𝑆∑

Res(𝑓 , 𝑧) = 1

2𝑖𝜋 ∫_𝛾𝑓 (𝑧)d𝑧 by Cauchy’s residue theorem

= −Res(𝑓 , ∞)

■ Example 5.

∫

+∞

−∞

𝑃 (𝑥)

𝑄(𝑥)d𝑥(rational function)

Assume that𝑃 and𝑄are two polynomials and that𝑄has no real root. Set𝑓 (𝑧) = 𝑃 (𝑧) 𝑄(𝑧). We know that

∫

+∞

−∞

𝑃 (𝑥)

𝑄(𝑥)d𝑥is convergent if and only if deg𝑄 ≥deg𝑃 + 2, let’s assume the latter.

Then lim

|𝑧|→∞𝑧𝑓 (𝑧) = 0.

ℜ ℑ

−𝑅 𝑅

Define𝛾_𝑅∶ [0, 1] → ℂby𝛾_𝑅(𝑡) = 𝑅𝑒^𝑖𝜋𝑡.

For𝑅 > 0big enough, all the poles of𝑓whose imaginary part is positive are included within the upper-half disk centered at0and of radius𝑅.

Then, by Cauchy’s residue theorem,

∫

𝑅

−𝑅

𝑓 (𝑧)d𝑧 +

∫_𝛾

𝑅

𝑓 (𝑧)d𝑧 = 2𝑖𝜋

𝑧s.t.∑ℑ(𝑧)>0

Res(𝑓 , 𝑧).

But|∫_𝛾

𝑅

𝑓|≤ 𝜋𝑅sup

𝛾_𝑅

|𝑓 | −−−−−→

𝑅→+∞ 0since lim

|𝑧|→∞𝑧𝑓 (𝑧) = 0.

Hence

∫

+∞

−∞

𝑃 (𝑥)

𝑄(𝑥)d𝑥 = 2𝑖𝜋

𝑧s.t.∑ℑ(𝑧)>0

Res( 𝑃 𝑄, 𝑧

). Example 6. Compute

∫

+∞

−∞

d𝑥

(𝑥²+ 𝑎²)(𝑥²+ 𝑏²) où0 < 𝑎 < 𝑏.

The poles of𝑓 (𝑧) = 1

(𝑧²+ 𝑎²)(𝑧²+ 𝑏²) are−𝑖𝑎,𝑖𝑎,−𝑖𝑏and𝑖𝑏which are simple. Hence

∫

+∞

−∞

d𝑥

(𝑥²+ 𝑎²)(𝑥²+ 𝑏²) = 2𝑖𝜋Res(𝑓 , 𝑖𝑎) + 2𝑖𝜋Res(𝑓 , 𝑖𝑏)

= 2𝑖𝜋

2𝑖(𝑏²− 𝑎²)𝑎+ 2𝑖𝜋 2𝑖(𝑎²− 𝑏²)𝑏

= 𝜋

𝑎𝑏(𝑎 + 𝑏)

Example 7.

∫

2𝜋 0

𝑅(cos𝑡,sin𝑡)d𝑡.

Set𝑧 = 𝑒^𝑖𝑡then cos(𝑡) = ¹

2(𝑧 +¹𝑧), sin(𝑡) = ¹

2𝑖(𝑧 −¹𝑧), and ^d𝑧_𝑖𝑧 =d𝑡.

Set𝑓 (𝑧) = ¹

𝑖𝑧𝑅 (¹2(𝑧 + ¹𝑧) ,2𝑖¹ (𝑧 −¹𝑧)). Then

∫

2𝜋 0

𝑅(cos𝑡,sin𝑡)d𝑡 =

∫_𝑆1

𝑓 (𝑧)d𝑧 = 2𝑖𝜋

𝑧∈𝐷∑₁(0)

Res(𝑓 , 𝑧).

Example 8. Compute

∫

2𝜋 0

𝑎

𝑎²+sin²𝑡d𝑡where𝑎 > 0.

Set𝑓 (𝑧) = 1 𝑖𝑧

𝑎 𝑎²− ¹

4(𝑧 −¹𝑧)

2 = − 4𝑖𝑎𝑧

(𝑧²+ 2𝑎𝑧 − 1)(𝑧²− 2𝑎𝑧 − 1).

Note that the singularity at0of the LHS is removable since we may extend𝑓 through0using the RHS, so that Res(𝑓 , 0) = 0.

Then, the only poles of𝑓 within the unit disk are𝑧₁= −𝑎 + √𝑎²+ 1and𝑧₂= 𝑎 − √𝑎²+ 1which are simple.

Hence

∫

2𝜋 0

𝑎

𝑎²+sin²𝑡d𝑡 = 2𝑖𝜋Res(𝑓 , 𝑧₁) + 2𝑖𝜋Res(𝑓 , 𝑧₂) = 2𝜋

√𝑎²+ 1

Remark 9(Jordan’s lemma). The following trick called,Jordan’s lemma, can be very useful.

Let𝛾_𝑅∶ [0, 𝜋] → ℂbe defined by𝛾_𝑅(𝑡) = 𝑅𝑒^𝑖𝑡(i.e. upper half circle centered at0of radius𝑅).

|∫_𝛾

𝑅

𝑔(𝑧)𝑒^𝑖𝑧d𝑧

|=

|∫

𝜋 0

𝑔(𝑅𝑒^𝑖𝑡)𝑒^{𝑖𝑅(cos𝑡+𝑖sin𝑡)}𝑖𝑅𝑒^𝑖𝑡d𝑡

|

≤∫

𝜋

0 |𝑔(𝑅𝑒^𝑖𝑡)𝑒^{𝑖𝑅(cos𝑡+𝑖sin𝑡)}𝑖𝑅𝑒^𝑖𝑡|d𝑡

≤ 𝑅sup

𝛾_𝑅

|𝑔|∫

𝜋 0

𝑒^{−𝑅sin𝑡}d𝑡

= 2𝑅sup

𝛾_𝑅

|𝑔|∫

𝜋/2 0

𝑒^{−𝑅sin𝑡}d𝑡

≤ 2𝑅sup

𝛾_𝑅

|𝑔|∫

𝜋/2 0

𝑒^{−2𝑅𝑡/𝜋}d𝑡 by Jordan’s inequality:∀𝑥 ∈ [0,𝜋 2 ], 2

𝜋𝑥 ≤sin(𝑥) ≤ 𝑥

≤ 𝜋sup

𝛾_𝑅 |𝑔| (1 − 𝑒^−𝑅)

≤ 𝜋sup

𝛾_𝑅

|𝑔|

Example 10.

∫

+∞

−∞

𝑃 (𝑥) 𝑄(𝑥)𝑒^𝑖𝑎𝑥d𝑥.

We want to compute𝐼(𝑎) ≔

∫

+∞

−∞

𝑃 (𝑥)

𝑄(𝑥)𝑒^𝑖𝑎𝑥d𝑥where𝑃 , 𝑄are real polynomials and𝑎 ∈ ℝ.

Assume that𝑄has no real root.

Note that𝐼(−𝑎) = 𝐼(𝑎), so we may restrict our attention to𝑎 > 0.

The integral is convergent if and only if deg𝑄 ≥deg𝑃 + 1(integration by parts), so we assume the latter.

ℜ ℑ

−𝑅 𝑅

Define𝛾_𝑅∶ [0, 1] → ℂby𝛾_𝑅(𝑡) = 𝑅𝑒^𝑖𝜋𝑡.

For𝑅 > 0big enough, all the poles of𝑓 (𝑧) = ^{𝑃 (𝑧)}

𝑄(𝑧)𝑒^𝑖𝑎𝑧whose imaginary part is positive are included within the upper-half disk centered at0and of radius𝑅.

Then∫

𝑅

−𝑅

𝑓 (𝑧)d𝑧 +

∫_𝛾

𝑅

𝑓 (𝑧)d𝑧 = 2𝑖𝜋 ∑

𝑧s.t.ℑ(𝑧)>0

Res(𝑓 , 𝑧) (∗) But, by Jordan’s lemma,

|∫_𝛾

𝑅

𝑃 (𝑧) 𝑄(𝑧)𝑒^𝑖𝑧d𝑧

|≤ 𝜋sup

𝛾_𝑅

|𝑃 /𝑄| −−−−−→

𝑅→+∞ 0 Hence, by taking𝑅 → +∞in(∗), we get

∫

+∞

−∞

𝑃 (𝑥)

𝑄(𝑥)𝑒^𝑖𝑎𝑥d𝑥 = 2𝑖𝜋 ∑

𝑧s.t.ℑ(𝑧)>0

Res( 𝑃 (𝑧) 𝑄(𝑧)𝑒^𝑖𝑎𝑧, 𝑧

)

Example 11. Compute

∫

+∞

−∞

cos(𝑥)

𝑥²+ 𝛼²d𝑥where𝛼 > 0.

Note that

∫

+∞

−∞

cos(𝑥)

𝑥²+ 𝛼²d𝑥 = ℜ (∫

+∞

−∞

𝑒^𝑖𝑥 𝑥²+ 𝛼²d𝑥

). The poles of𝑓 (𝑧) = 𝑒^𝑖𝑧

𝑧²+ 𝛼² are𝑖𝛼and−𝑖𝛼which are simple.

By the previous slide,

∫

+∞

−∞

𝑒^𝑖𝑧

𝑧²+ 𝛼²d𝑧 = 2𝑖𝜋Res(𝑓 , 𝑖𝛼) = 2𝑖𝜋𝑒^−𝛼

2𝑖𝛼 = 𝜋𝑒^−𝛼 𝛼 . Hence

∫

+∞

−∞

cos(𝑥)

𝑥²+ 𝛼²d𝑥 = 𝜋𝑒^−𝛼 𝛼 .

Example 12.

∫

+∞

0

𝑥^𝑝

1 + 𝑥^𝑛d𝑥, 𝑛, 𝑝 ∈ ℕ.

We know that the integral

∫

+∞

0

𝑥^𝑝

1 + 𝑥^𝑛d𝑥is convergent if and only if𝑛 ≥ 𝑝 + 2.

Set𝑓 (𝑧) = 𝑧^𝑝

1 + 𝑧^𝑛 and𝑎 = 𝑒^𝑖𝜋𝑛.

We consider the following sector of the circle centered at0and of radius 𝑅, such that the only pole of𝑓 enclosed in its inside is𝑎.

ℜ ℑ

0 𝑅

𝑎²𝑅 𝑎

Let𝛾 ∶ [0,^2𝜋𝑛 ] → ℂbe defined by𝛾(𝑡) = 𝑅𝑒^𝑖𝑡. By the residue theorem,2𝑖𝜋Res(𝑓 , 𝑎) =

∫_[0,𝑅]𝑓 +

∫_𝛾𝑓 +

∫_[𝑎2𝑅,0]

𝑓.

• Res(𝑓 , 𝑎) = ^𝑎𝑝

𝑛𝑎^𝑛−1 = ^𝑎𝑝+1

𝑛𝑎^𝑛 = −^𝑎𝑝+1

𝑛 .

• ∫

[𝑎²𝑅,0]

𝑓 (𝑧)d𝑧 = −𝑎²

∫

𝑅 0

𝑎^2𝑝𝑡^𝑝

1 + 𝑎^2𝑛𝑡^𝑛d𝑡 = −𝑎^2(𝑝+1)

∫

𝑅 0

𝑡^𝑝 1 + 𝑡^𝑛d𝑡

• |∫_𝛾𝑓 (𝑧)d𝑧

|≤ 2𝜋 𝑛 𝑅sup

𝛾

|𝑓 | −−−−−→

𝑅→+∞ 0since lim

𝑧→∞𝑧𝑓 (𝑧) = 0.

Hence, by taking the limit as𝑅 → +∞, we get−2𝑖𝜋𝑎^𝑝+1 𝑛 =

∫

+∞

0

𝑥^𝑝

1 + 𝑥^𝑛d𝑥 − 𝑎^2(𝑝+1)

∫

+∞

0

𝑥^𝑝 1 + 𝑥^𝑛d𝑥.

Finally

∫

+∞

0

𝑥^𝑝

1 + 𝑥^𝑛d𝑥 = 2𝑖𝜋 𝑛

𝑎^𝑝+1

𝑎^2(𝑝+1)− 1 = 𝜋 𝑛

2𝑖

𝑎^𝑝+1− 𝑎^−(𝑝+1) = 𝜋 𝑛sin^(𝑝+1)𝜋

𝑛

.

Example 13.

∫

+∞

0

sin𝑡 𝑡 d𝑡.

ℜ ℑ

−𝑅 −𝑟 𝑟 𝑅

Define𝛾_𝑅∶ [0, 1] → ℂby𝛾_𝑅(𝑡) = 𝑅𝑒^𝑖𝜋𝑡and𝛾_𝑟∶ [0, 1] → ℂby𝛾_𝑟(𝑡) = 𝑟𝑒^𝑖𝜋𝑡. By Cauchy’s integral theorem

∫_𝛾

𝑅

𝑒^𝑖𝑧 𝑧 d𝑧 −

∫_𝛾

𝑟

𝑒^𝑖𝑧 𝑧 d𝑧 +

∫

−𝑟

−𝑅

𝑒^𝑖𝑡 𝑡 d𝑡 +

∫

𝑅 𝑟

𝑒^𝑖𝑡 𝑡 d𝑡 = 0.

• By Jordan’s lemma:

|∫_𝛾

𝑅

𝑒^𝑖𝑧 𝑧 d𝑧

|≤ 𝜋sup

𝛾_𝑅

|1/𝑧| −−−−−→

𝑅→+∞ 0

• Since0is a simple pole of𝑓 (𝑧) = ^𝑒𝑖𝑧

𝑧 , we have that𝑓 (𝑧) =Res(𝑓 , 0)𝑧⁻¹+ 𝑔(𝑧)where𝑔is holomorphic.

Then∫_𝛾

𝑟

𝑓 (𝑧)d𝑧 =

∫_𝛾

𝑟

Res(𝑓 , 0)𝑧⁻¹d𝑧 +

∫_𝛾

𝑟

𝑔(𝑧)d𝑧but

∫_𝛾

𝑟

Res(𝑓 , 0)𝑧⁻¹d𝑧 =Res(𝑓 , 0)𝑖𝜋and

|∫_𝛾

𝑟

𝑔(𝑧)d𝑧

|≤ 𝜋𝑟sup

𝛾_𝑟

|𝑔| −−−→

𝑟→0 0.

Hence

∫_𝛾

𝑟

𝑓 (𝑧)d𝑧 −−−→

𝑟→0 Res(𝑓 , 0)𝑖𝜋 = 𝑖𝜋.

∫

𝑅 𝑟

sin𝑡 𝑡 d𝑡 = 1

2𝑖 ∫

𝑅 𝑟

𝑒^𝑖𝑡− 𝑒^−𝑖𝑡 𝑡 d𝑡 = 1

2𝑖 ∫

𝑅 𝑟

𝑒^𝑖𝑡 𝑡 d𝑡−1

2𝑖 ∫

𝑅 𝑟

𝑒^−𝑖𝑡 𝑡 d𝑡 = 1

2𝑖 ∫

𝑅 𝑟

𝑒^𝑖𝑡 𝑡 d𝑡+1

2𝑖 ∫

−𝑟

−𝑅

𝑒^𝑖𝑡 𝑡 d𝑡 = 1

2𝑖 (∫_𝛾𝑟 𝑒^𝑖𝑧

𝑧 d𝑧 −

∫_𝛾

𝑅

𝑒^𝑖𝑧 𝑧 d𝑧

) Hence, taking𝑟 → 0and𝑅 → +∞we get that

∫

+∞

0

sin𝑡 𝑡 d𝑡 = 𝜋

2.

Example 14.

∫

+∞

0

(log𝑡)² 1 + 𝑡² d𝑡.

We set𝑓 (𝑧) = ^(log𝑧)2

1+𝑧² but for that we need to fix a branch of the logarithm.

Let’s fix log∶ ℂ ⧵ {𝑖𝑦 ∶ 𝑦 ≤ 0} → ℂdefined by log𝑧 =log|𝑧| + 𝑖Arg(𝑧)where Arg(𝑧) ∈ (−^𝜋2, 3^𝜋

2).

ℜ ℑ

−𝑅 −𝑟 𝑟 𝑅

𝑖

Define𝛾_𝑅∶ [0, 1] → ℂby𝛾_𝑅(𝑡) = 𝑅𝑒^𝑖𝜋𝑡and𝛾_𝑟∶ [0, 1] → ℂby𝛾_𝑟(𝑡) = 𝑟𝑒^𝑖𝜋𝑡. By Cauchy’s residue theorem

∫_𝛾

𝑅

𝑓 (𝑧)d𝑧 −

∫_𝛾

𝑟

𝑓 (𝑧)d𝑧 +

∫

−𝑟

−𝑅

𝑓 (𝑧)d𝑧 +

∫

𝑅 𝑟

𝑓 (𝑧)d𝑧 = 2𝑖𝜋Res(𝑓 , 𝑖) = −𝜋³ 4 .

• Since|log𝑧| ≤ |ln𝑟| + 𝜋on𝛾_𝑟,

∫_𝛾

𝑟

𝑓 (𝑧)d𝑧 ≤ 𝜋𝑟(|ln𝑟| + 𝜋)²

1 + 𝑟² −−−−−−−−→

𝑟→+∞or0 0.

• Since𝑧 = 𝑡𝑒^𝑖𝜋on[−𝑅, −𝑟], we have

∫

−𝑟

−𝑅

𝑓 (𝑧)d𝑧 =

∫

𝑅 𝑟

(ln𝑡 + 𝑖𝜋)² 1 + 𝑡² d𝑡 =

∫

𝑅 𝑟

(ln𝑡)²

1 + 𝑡²d𝑡 + 2𝑖𝜋

∫

𝑅 𝑟

ln𝑡 1 + 𝑡²d𝑡 −

∫

𝑅 𝑟

𝜋² 1 + 𝑡²d𝑡 By taking the limits𝑟 → 0and𝑅 → +∞we get

∫

+∞

0

(ln𝑡)²

1 + 𝑡²d𝑡+2𝑖𝜋

∫

+∞

0

ln𝑡

1 + 𝑡²d𝑡−𝜋³ 2 +

∫

+∞

0

(ln𝑡)²

1 + 𝑡²d𝑡 = −𝜋³ 4 . Considering the real part, we get

∫

+∞

0

(ln𝑡)²

1 + 𝑡²d𝑡 = 𝜋³ 8 .